EUCLID’S ELEMENTS OF GEOMETRY

EUCLID’S ELEMENTS OF GEOMETRY
The Greek text of J.L. Heiberg (1883–1885)
from Euclidis Elementa, edidit et Latine interpretatus est I.L. Heiberg, in aedibus
B.G. Teubneri, 1883–1885
edited, and provided with a modern English translation, by
Richard Fitzpatrick

First edition - 2007
Revised and corrected - 2008
ISBN 978-0-6151-7984-1

Contents
Introduction 4
Book 1 5
Book 2 49
Book 3 69
Book 4 109
Book 5 129
Book 6 155
Book 7 193
Book 8 227
Book 9 253
Book 10 281
Book 11 423
Book 12 471
Book 13 505
Greek-English Lexicon 539

Introduction
Euclid’s Elements is by far the most famous mathematical work of classical antiquity, and also has the distinction
of being the world’s oldest continuously used mathematical textbook. Little is known about the author, beyond
the fact that he lived in Alexandria around 300 BCE. The main subjects of the work are geometry, proportion, and
number theory.
Most of the theorems appearing in the Elements were not discovered by Euclid himself, but were the work of
earlier Greek mathematicians such as Pythagoras (and his school), Hippocrates of Chios, Theaetetus of Athens, and
Eudoxus of Cnidos. However, Euclid is generally credited with arranging these theorems in a logical manner, so as to
demonstrate (admittedly, not always with the rigour demanded by modern mathematics) that they necessarily follow
from five simple axioms. Euclid is also credited with devising a number of particularly ingenious proofs of previously
discovered theorems: e.g., Theorem 48 in Book 1.
The geometrical constructions employed in the Elements are restricted to those which can be achieved using a
straight-rule and a compass. Furthermore, empirical proofs by means of measurement are strictly forbidden: i.e.,
any comparison of two magnitudes is restricted to saying that the magnitudes are either equal, or that one is greater
than the other.
The Elements consists of thirteen books. Book 1 outlines the fundamental propositions of plane geometry, including
the three cases in which triangles are congruent, various theorems involving parallel lines, the theorem regarding
the sum of the angles in a triangle, and the Pythagorean theorem. Book 2 is commonly said to deal with “geometric
algebra”, since most of the theorems contained within it have simple algebraic interpretations. Book 3 investigates
circles and their properties, and includes theorems on tangents and inscribed angles. Book 4 is concerned with regular
polygons inscribed in, and circumscribed around, circles. Book 5 develops the arithmetic theory of proportion.
Book 6 applies the theory of proportion to plane geometry, and contains theorems on similar figures. Book 7 deals
with elementary number theory: e.g., prime numbers, greatest common denominators, etc. Book 8 is concerned with
geometric series. Book 9 contains various applications of results in the previous two books, and includes theorems
on the infinitude of prime numbers, as well as the sum of a geometric series. Book 10 attempts to classify incommensurable
(i.e., irrational) magnitudes using the so-called “method of exhaustion”, an ancient precursor to integration.
Book 11 deals with the fundamental propositions of three-dimensional geometry. Book 12 calculates the relative
volumes of cones, pyramids, cylinders, and spheres using the method of exhaustion. Finally, Book 13 investigates the
five so-called Platonic solids.
This edition of Euclid’s Elements presents the definitive Greek text—i.e., that edited by J.L. Heiberg (1883–
1885)—accompanied by a modern English translation, as well as a Greek-English lexicon. Neither the spurious
books 14 and 15, nor the extensive scholia which have been added to the Elements over the centuries, are included.
The aim of the translation is to make the mathematical argument as clear and unambiguous as possible, whilst still
adhering closely to the meaning of the original Greek. Text within square parenthesis (in both Greek and English)
indicates material identified by Heiberg as being later interpolations to the original text (some particularly obvious or
unhelpful interpolations have been omitted altogether). Text within round parenthesis (in English) indicates material
which is implied, but not actually present, in the Greek text.
My thanks to Mariusz Wodzicki (Berkeley) for typesetting advice, and to Sam Watson & Jonathan Fenno (U.
Mississippi), and Gregory Wong (UCSD) for pointing out a number of errors in Book 1.
4

ELEMENTS BOOK 1
Fundamentals of Plane Geometry Involving
Straight-Lines
5

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ÎÇÖÓ.
αʹ.Σημεῖόνἐστιν,οὗμέροςοὐθέν.
Definitions
1. A point is that of which there is no part.
βʹ.Γραμμὴδὲμῆκοςἀπλατές.
2. And a line is a length without breadth.
γʹ.Γραμμῆςδὲπέρατασημεῖα.
3. And the extremities of a line are points.
δʹ.Εὐθεῖαγραμμήἐστιν,ἥτιςἐξἴσουτοῖςἐφ᾿ἑαυτῆς 4. A straight-line is (any) one which lies evenly with
σημείοιςκεῖται.
points on itself.
εʹ.᾿Επιφάνειαδέἐστιν,ὃμῆκοςκαὶπλάτοςμόνονἔχει. 5. And a surface is that which has length and breadth
ϛʹ.᾿Επιφανείαςδὲπέραταγραμμαί.
only.
ζʹ. ᾿Επίπεδος ἐπιφάνειά ἐστιν, ἥτις ἐξ ἴσου ταῖς ἐφ᾿ 6. And the extremities of a surface are lines.
ἑαυτῆςεὐθείαιςκεῖται.
7. A plane surface is (any) one which lies evenly with
ηʹ.᾿Επίπεδοςδὲγωνίαἐστὶνἡἐνἐπιπέδῳδύογραμμῶν the straight-lines on itself.
ἁπτομένων ἀλλήλων καὶ μὴ ἐπ᾿ εὐθείας κειμένων πρὸς 8. And a plane angle is the inclination of the lines to
ἀλλήλαςτῶνγραμμῶνκλίσις.
one another, when two lines in a plane meet one another,
θʹ.῞Οτανδὲαἱπεριέχουσαιτὴνγωνίανγραμμαὶεὐθεῖαι and are not lying in a straight-line.
ὦσιν,εὐθύγραμμοςκαλεῖταιἡγωνία.
9. And when the lines containing the angle are
ιʹ.῞Οταν δὲ εὐθεῖα ἐπ᾿ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς straight then the angle is called rectilinear.
γωνίαςἴσαςἀλλήλαιςποιῇ,ὀρθὴἑκατέρατῶνἴσωνγωνιῶν 10. And when a straight-line stood upon (another)
ἐστι, καὶ ἡ ἐφεστηκυῖα εὐθεῖα κάθετος καλεῖται, ἐφ᾿ ἣν straight-line makes adjacent angles (which are) equal to
ἐφέστηκεν.
one another, each of the equal angles is a right-angle, and
ιαʹ.Ἀμβλεῖαγωνίαἐστὶνἡμείζωνὀρθῆς.
the former straight-line is called a perpendicular to that
ιβʹ.᾿Οξεῖαδὲἡἐλάσσωνὀρθῆς.
upon which it stands.
ιγʹ.῞Οροςἐστίν,ὅτινόςἐστιπέρας.
11. An obtuse angle is one greater than a right-angle.
ιδʹ.Σχῆμάἐστιτὸὑπότινοςἤτινωνὅρωνπεριεχόμενον. 12. And an acute angle (is) one less than a right-angle.
ιεʹ. Κύκλος ἐστὶ σχῆμα ἐπίπεδον ὑπὸ μιᾶς γραμμῆς 13. A boundary is that which is the extremity of someπεριεχόμενον
[ἣ καλεῖται περιφέρεια], πρὸς ἣν ἀφ᾿ ἑνὸς thing.
σημείου τῶν ἐντὸς τοῦ σχήματος κειμένων πᾶσαι αἱ 14. A figure is that which is contained by some boundπροσπίπτουσαιεὐθεῖαι[πρὸςτὴντοῦκύκλουπεριφέρειαν]
ary or boundaries.
ἴσαιἀλλήλαιςεἰσίν.
15. A circle is a plane figure contained by a single line
ιϛʹ.Κέντρονδὲτοῦκύκλουτὸσημεῖονκαλεῖται. [which is called a circumference], (such that) all of the
ιζʹ.Διάμετροςδὲτοῦκύκλουἐστὶνεὐθεῖάτιςδιὰτοῦ straight-lines radiating towards [the circumference] from
κέντρου ἠγμένη καὶ περατουμένη ἐφ᾿ ἑκάτερα τὰ μέρη one point amongst those lying inside the figure are equal
ὑπὸτῆςτοῦκύκλουπεριφερείας,ἥτιςκαὶδίχατέμνειτὸν to one another.
κύκλον.
16. And the point is called the center of the circle.
ιηʹ.῾Ημικύκλιονδέἐστιτὸπεριεχόμενονσχῆμαὑπότε 17. And a diameter of the circle is any straight-line,
τῆς διαμέτρουκαὶτῆς ἀπολαμβανομένηςὑπ᾿αὐτῆςπερι- being drawn through the center, and terminated in each
φερείας. κέντρονδὲτοῦἡμικυκλίουτὸαὐτό,ὃκαὶτοῦ direction by the circumference of the circle. (And) any
κύκλουἐστίν. such (straight-line) also cuts the circle in half. †
ιθʹ. Σχήματα εὐθύγραμμά ἐστι τὰ ὑπὸ εὐθειῶν πε- 18. And a semi-circle is the figure contained by the
ριεχόμενα,τρίπλευραμὲντὰὑπὸτριῶν,τετράπλευραδὲτὰ diameter and the circumference cuts off by it. And the
ὑπὸτεσσάρων,πολύπλευραδὲτὰὑπὸπλειόνωνἢτεσσάρων center of the semi-circle is the same (point) as (the center
εὐθειῶνπεριεχόμενα.
of) the circle.
κʹ. Τῶν δὲ τριπλεύρων σχημάτων ἰσόπλευρον μὲν 19. Rectilinear figures are those (figures) contained
τρίγωνόνἐστιτὸτὰςτρεῖςἴσαςἔχονπλευράς,ἰσοσκελὲς by straight-lines: trilateral figures being those contained
δὲτὸτὰςδύομόναςἴσαςἔχονπλευράς,σκαληνὸνδὲτὸ by three straight-lines, quadrilateral by four, and multiτὰςτρεῖςἀνίσουςἔχονπλευράς.
lateral by more than four.
καʹ῎Ετιδὲτῶντριπλεύρωνσχημάτωνὀρθογώνιονμὲν 20. And of the trilateral figures: an equilateral trianτρίγωνόνἐστιτὸἔχονὀρθὴνγωνίαν,ἀμβλυγώνιονδὲτὸ
gle is that having three equal sides, an isosceles (triangle)
ἔχονἀμβλεῖανγωνίαν,ὀξυγώνιονδὲτὸτὰςτρεῖςὀξείας that having only two equal sides, and a scalene (triangle)
ἔχονγωνίας.
that having three unequal sides.
6

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
κβʹ.Τὼνδὲτετραπλεύρωνσχημάτωντετράγωνονμέν 21. And further of the trilateral figures: a right-angled
ἐστιν, ὃἰσόπλευρόντέἐστικαὶὀρθογώνιον,ἑτερόμηκες triangle is that having a right-angle, an obtuse-angled
δέ,ὃὀρθογώνιονμέν,οὐκἰσόπλευρονδέ,ῥόμβοςδέ,ὃ (triangle) that having an obtuse angle, and an acute-
ἰσόπλευρονμέν,οὐκὀρθογώνιονδέ,ῥομβοειδὲςδὲτὸτὰς angled (triangle) that having three acute angles.
ἀπεναντίονπλευράςτεκαὶγωνίαςἴσαςἀλλήλαιςἔχον,ὃ 22. And of the quadrilateral figures: a square is that
οὔτεἰσόπλευρόνἐστινοὔτεὀρθογώνιον·τὰδὲπαρὰταῦτα which is right-angled and equilateral, a rectangle that
τετράπλευρατραπέζιακαλείσθω.
which is right-angled but not equilateral, a rhombus that
κγʹ. Παράλληλοί εἰσιν εὐθεῖαι, αἵτινες ἐν τῷ αὐτῷ which is equilateral but not right-angled, and a rhomboid
ἐπιπέδῳοὖσαικαὶἐκβαλλόμεναιεἰςἄπειρονἐφ᾿ἑκάτερα that having opposite sides and angles equal to one anτὰμέρηἐπὶμηδέτερασυμπίπτουσινἀλλήλαις.
other which is neither right-angled nor equilateral. And
let quadrilateral figures besides these be called trapezia.
ØÑغ
23. Parallel lines are straight-lines which, being in the
same plane, and being produced to infinity in each direction,
meet with one another in neither (of these directions).
† This should really be counted as a postulate, rather than as part of a definition.
Postulates
αʹ. ᾿Ηιτήσθω ἀπὸ παντὸς σημείου ἐπὶ πᾶν σημεῖον 1. Let it have been postulated † to draw a straight-line
εὐθεῖανγραμμὴνἀγαγεῖν.
from any point to any point.
βʹ. Καὶ πεπερασμένην εὐθεῖαν κατὰ τὸ συνεχὲς ἐπ᾿ 2. And to produce a finite straight-line continuously
εὐθείαςἐκβαλεῖν.
in a straight-line.
γʹ.Καὶπαντὶκέντρῳκαὶδιαστήματικύκλονγράφεσθαι. 3. And to draw a circle with any center and radius.
δʹ.Καὶπάσαςτὰςὀρθὰςγωνίαςἴσαςἀλλήλαιςεἶναι. 4. And that all right-angles are equal to one another.
εʹ.Καὶἐὰνεἰςδύοεὐθείαςεὐθεῖαἐμπίπτουσατὰςἐντὸς 5. And that if a straight-line falling across two (other)
καὶἐπὶτὰαὐτὰμέρηγωνίαςδύοὀρθῶνἐλάσσοναςποιῇ, straight-lines makes internal angles on the same side
ἐκβαλλομέναςτὰςδύοεὐθείαςἐπ᾿ἄπειρονσυμπίπτειν,ἐφ᾿ (of itself whose sum is) less than two right-angles, then
ἃμέρηεἰσὶναἱτῶνδύοὀρθῶνἐλάσσονες.
the two (other) straight-lines, being produced to infinity,
meet on that side (of the original straight-line) that the
(sum of the internal angles) is less than two right-angles
(and do not meet on the other side). ‡
† The Greek present perfect tense indicates a past action with present significance. Hence, the 3rd-person present perfect imperativeÀØ×Û
ÃÓÒÒÒÓº
could be translated as “let it be postulated”, in the sense “let it stand as postulated”, but not “let the postulate be now brought forward”. The
literal translation “let it have been postulated” sounds awkward in English, but more accurately captures the meaning of the Greek.
‡ This postulate effectively specifies that we are dealing with the geometry of flat, rather than curved, space.
Common Notions
αʹ.Τὰτῷαὐτῷἴσακαὶἀλλήλοιςἐστὶνἴσα.
1. Things equal to the same thing are also equal to
βʹ.Καὶἐὰνἴσοιςἴσαπροστεθῇ,τὰὅλαἐστὶνἴσα. one another.
γʹ.Καὶἐὰνἀπὸἴσωνἴσαἀφαιρεθῇ,τὰκαταλειπόμενά 2. And if equal things are added to equal things then
ἐστινἴσα.
the wholes are equal.
δʹ.Καὶτὰἐφαρμόζονταἐπ᾿ἀλλήλαἴσαἀλλήλοιςἐστίν. 3. And if equal things are subtracted from equal things
εʹ.Καὶτὸὅλοντοῦμέρουςμεῖζόν[ἐστιν]. then the remainders are equal. †
4. And things coinciding with one another are equal
to one another.
5. And the whole [is] greater than the part.
† As an obvious extension of C.N.s 2 & 3—if equal things are added or subtracted from the two sides of an inequality then the inequality remains
7

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
an inequality of the same type.. Proposition 1
᾿Επὶ τῆς δοθείσης εὐθείας πεπερασμένης τρίγωνον To construct an equilateral triangle on a given finite
ἰσόπλευρονσυστήσασθαι.
straight-line.
Γ
C
∆
Α
Β Ε D A B
E
῎ΕστωἡδοθεῖσαεὐθεῖαπεπερασμένηἡΑΒ.
Let AB be the given finite straight-line.
Δεῖ δὴ ἐπὶ τῆς ΑΒ εὐθείας τρίγωνον ἰσόπλευρον So it is required to construct an equilateral triangle on
συστήσασθαι.
the straight-line AB.
Κέντρῳ μὲν τῷ Α διαστήματι δὲ τῷ ΑΒ κύκλος Let the circle BCD with center A and radius AB have
γεγράφθωὁΒΓΔ,καὶπάλινκέντρῳμὲντῷΒδιαστήματιδὲ been drawn [Post. 3], and again let the circle ACE with
τῷΒΑκύκλοςγεγράφθωὁΑΓΕ,καὶἀπὸτοῦΓσημείου, center B and radius BA have been drawn [Post. 3]. And
καθ᾿ὃτέμνουσινἀλλήλουςοἱκύκλοι,ἐπίτὰΑ,Βσημεῖα let the straight-lines CA and CB have been joined from
ἐπεζεύχθωσανεὐθεῖαιαἱΓΑ,ΓΒ.
the point C, where the circles cut one another, † to the
ΚαὶἐπεὶτὸΑσημεῖονκέντρονἐστὶτοῦΓΔΒκύκλου, points A and B (respectively) [Post. 1].
ἴσηἐστὶνἡΑΓτῇΑΒ·πάλιν,ἐπεὶτὸΒσημεῖονκέντρον And since the point A is the center of the circle CDB,
ἐστὶτοῦΓΑΕκύκλου,ἴσηἐστὶνἡΒΓτῇΒΑ.ἐδείχθηδὲ AC is equal to AB [Def. 1.15]. Again, since the point
καὶἡΓΑτῇΑΒἴση·ἑκατέραἄρατῶνΓΑ,ΓΒτῇΑΒἐστιν B is the center of the circle CAE, BC is equal to BA
ἴση.τὰδὲτῷαὐτῷἴσακαὶἀλλήλοιςἐστὶνἴσα·καὶἡΓΑἄρα [Def. 1.15]. But CA was also shown (to be) equal to AB.
τῇΓΒἐστινἴση·αἱτρεῖςἄρααἱΓΑ,ΑΒ,ΒΓἴσαιἀλλήλαις Thus, CA and CB are each equal to AB. But things equal
εἰσίν. to the same thing are also equal to one another [C.N. 1].
᾿ΙσόπλευρονἄραἐστὶτὸΑΒΓτρίγωνον.καὶσυνέσταται Thus, CA is also equal to CB. Thus, the three (straight-
ἐπὶτῆςδοθείσηςεὐθείαςπεπερασμένηςτῆςΑΒ.ὅπερἔδει lines) CA, AB, and BC are equal to one another.
ποιῆσαι.
Thus, the triangle ABC is equilateral, and has been
constructed on the given finite straight-line AB. (Which
is) the very thing it was required to do.
† The assumption that the circles do indeed cut one another should be counted as an additional postulate. There is also an implicit assumption
that two straight-lines cannot share a common segment.
. Proposition 2 †
Πρὸςτῷδοθέντισημείῳτῇδοθείσῃεὐθείᾳἴσηνεὐθεῖαν To place a straight-line equal to a given straight-line
θέσθαι.
at a given point (as an extremity).
῎ΕστωτὸμὲνδοθὲνσημεῖοντὸΑ,ἡδὲδοθεῖσαεὐθεῖα Let A be the given point, and BC the given straight-
ἡΒΓ·δεῖδὴπρὸςτῷΑσημείῳτῇδοθείσῃεὐθείᾳτῇΒΓ line. So it is required to place a straight-line at point A
ἴσηνεὐθεῖανθέσθαι.
equal to the given straight-line BC.
᾿ΕπεζεύχθωγὰρἀπὸτοῦΑσημείουἐπίτὸΒσημεῖον For let the straight-line AB have been joined from
εὐθεῖαἡΑΒ,καὶσυνεστάτωἐπ᾿αὐτῆςτρίγωνονἰσόπλευρον point A to point B [Post. 1], and let the equilateral trianτὸΔΑΒ,καὶἐκβεβλήσθωσανἐπ᾿εὐθείαςταῖςΔΑ,ΔΒ
gle DAB have been been constructed upon it [Prop. 1.1].
8

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
εὐθεῖαιαἱΑΕ,ΒΖ,καὶκέντρῳμὲντῷΒδιαστήματιδὲτῷ And let the straight-lines AE and BF have been pro-
ΒΓκύκλοςγεγράφθωὁΓΗΘ,καὶπάλινκέντρῳτῷΔκαὶ duced in a straight-line with DA and DB (respectively)
διαστήματιτῷΔΗκύκλοςγεγράφθωὁΗΚΛ.
[Post. 2]. And let the circle CGH with center B and radius
BC have been drawn [Post. 3], and again let the circle
GKL with center D and radius DG have been drawn
[Post. 3].
Γ
C
Κ
Θ
∆
Β
K
H
D
B
Α
Η
A
G
Ζ
F
Λ
L
Ε
E
᾿ΕπεὶοὖντὸΒσημεῖονκέντρονἐστὶτοῦΓΗΘ,ἴσηἐστὶν Therefore, since the point B is the center of (the cir-
ἡΒΓτῇΒΗ.πάλιν,ἐπεὶτὸΔσημεῖονκέντρονἐστὶτοῦ cle) CGH, BC is equal to BG [Def. 1.15]. Again, since
ΗΚΛκύκλου,ἴσηἐστὶνἡΔΛτῇΔΗ,ὧνἡΔΑτῇΔΒἴση the point D is the center of the circle GKL, DL is equal
ἐστίν. λοιπὴἄραἡΑΛλοιπῇτῇΒΗἐστινἴση. ἐδείχθηδὲ to DG [Def. 1.15]. And within these, DA is equal to DB.
καὶἡΒΓτῇΒΗἴση·ἑκατέραἄρατῶνΑΛ,ΒΓτῇΒΗἐστιν Thus, the remainder AL is equal to the remainder BG
ἴση. τὰδὲτῷαὐτῷἴσακαὶἀλλήλοιςἐστὶνἴσα·καὶἡΑΛ [C.N. 3]. But BC was also shown (to be) equal to BG.
ἄρατῇΒΓἐστινἴση.
Thus, AL and BC are each equal to BG. But things equal
ΠρὸςἄρατῷδοθέντισημείῳτῷΑτῇδοθείσῃεὐθείᾳ to the same thing are also equal to one another [C.N. 1].
τῇΒΓἴσηεὐθεῖακεῖταιἡΑΛ·ὅπερἔδειποιῆσαι. Thus, AL is also equal to BC.
Thus, the straight-line AL, equal to the given straightline
BC, has been placed at the given point A. (Which
is) the very thing it was required to do.
† This proposition admits of a number of different cases, depending on the relative positions of the point A and the line BC. In such situations,
Euclid invariably only considers one particular case—usually, the most difficult—and leaves the remaining cases as exercises for the reader.
. Proposition 3
Δύο δοθεισῶν εὐθειῶν ἀνίσων ἀπὸ τῆς μείζονος τῇ For two given unequal straight-lines, to cut off from
ἐλάσσονιἴσηνεὐθεῖανἀφελεῖν.
the greater a straight-line equal to the lesser.
῎ΕστωσαναἱδοθεῖσαιδύοεὐθεῖαιἄνισοιαἱΑΒ,Γ,ὧν Let AB and C be the two given unequal straight-lines,
μείζωνἔστωἡΑΒ·δεῖδὴἀπὸτῆςμείζονοςτῆςΑΒτῇ of which let the greater be AB. So it is required to cut off
ἐλάσσονιτῇΓἴσηνεὐθεῖανἀφελεῖν.
a straight-line equal to the lesser C from the greater AB.
ΚείσθωπρὸςτῷΑσημείῳτῇΓεὐθείᾳἴσηἡΑΔ·καὶ Let the line AD, equal to the straight-line C, have
κέντρῳμὲντῷΑδιαστήματιδὲτῷΑΔκύκλοςγεγράφθω been placed at point A [Prop. 1.2]. And let the circle
ὁΔΕΖ.
DEF have been drawn with center A and radius AD
ΚαὶἐπεὶτὸΑσημεῖονκέντρονἐστὶτοῦΔΕΖκύκλου, [Post. 3].
9

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ἴσηἐστὶνἡΑΕτῇΑΔ·ἀλλὰκαὶἡΓτῇΑΔἐστινἴση. And since point A is the center of circle DEF , AE
ἑκατέραἄρατῶνΑΕ,ΓτῇΑΔἐστινἴση·ὥστεκαὶἡΑΕ is equal to AD [Def. 1.15]. But, C is also equal to AD.
τῇΓἐστινἴση.
Thus, AE and C are each equal to AD. So AE is also
equal to C [C.N. 1].
Γ
C
∆
D
Α
Ε
Β
A
E
B
Ζ
ΔύοἄραδοθεισῶνεὐθειῶνἀνίσωντῶνΑΒ,Γἀπὸτῆς Thus, for two given unequal straight-lines, AB and C,
μείζονοςτῆςΑΒτῇἐλάσσονιτῇΓἴσηἀφῄρηταιἡΑΕ·ὅπερ the (straight-line) AE, equal to the lesser C, has been cut
ἔδειποιῆσαι.
off from the greater AB. (Which is) the very thing it was
required to do.
. Proposition 4
᾿Εὰνδύοτρίγωνατὰςδύοπλευρὰς[ταῖς]δυσὶπλευραῖς If two triangles have two sides equal to two sides, re-
ἴσαςἔχῃἑκατέρανἑκατέρᾳκαὶτὴνγωνίαντῇγωνίᾳἴσην spectively, and have the angle(s) enclosed by the equal
ἔχῃ τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην, καὶ τὴν straight-lines equal, then they will also have the base
βάσιντῂβάσειἴσηνἕξει,καὶτὸτρίγωνοντῷτριγώνῳἴσον equal to the base, and the triangle will be equal to the tri-
ἔσται,καὶαἱλοιπαὶγωνίαιταῖςλοιπαῖςγωνίαιςἴσαιἔσονται angle, and the remaining angles subtended by the equal
ἑκατέραἑκατέρᾳ,ὑφ᾿ἃςαἱἴσαιπλευραὶὑποτείνουσιν. sides will be equal to the corresponding remaining angles.
Α
∆
A
D
F
Β
Γ Ε Ζ B
C E
F
῎Εστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο πλευρὰς Let ABC and DEF be two triangles having the two
τὰςΑΒ,ΑΓταῖςδυσὶπλευραῖςταῖςΔΕ,ΔΖἴσαςἔχοντα sides AB and AC equal to the two sides DE and DF , re-
ἑκατέρανἑκατέρᾳτὴνμὲνΑΒτῇΔΕτὴνδὲΑΓτῇΔΖ spectively. (That is) AB to DE, and AC to DF . And (let)
καὶγωνίαντὴνὑπὸΒΑΓγωνίᾳτῇὑπὸΕΔΖἴσην. λέγω, the angle BAC (be) equal to the angle EDF . I say that
ὅτικαὶβάσιςἡΒΓβάσειτῇΕΖἴσηἐστίν, καὶτὸΑΒΓ the base BC is also equal to the base EF , and triangle
τρίγωνοντῷΔΕΖτριγώνῳἴσονἔσται,καὶαἱλοιπαὶγωνίαι ABC will be equal to triangle DEF , and the remaining
ταῖςλοιπαῖςγωνίαιςἴσαιἔσονταιἑκατέραἑκατέρᾳ,ὑφ᾿ἃς angles subtended by the equal sides will be equal to the
αἱἴσαιπλευραὶὑποτείνουσιν,ἡμὲνὑπὸΑΒΓτῇὑπὸΔΕΖ, corresponding remaining angles. (That is) ABC to DEF ,
ἡδὲὑπὸΑΓΒτῇὑπὸΔΖΕ.
and ACB to DFE.
᾿ΕφαρμοζομένουγὰρτοῦΑΒΓτριγώνουἐπὶτὸΔΕΖ For if triangle ABC is applied to triangle DEF , † the
τρίγωνονκαὶτιθεμένουτοῦμὲνΑσημείουἐπὶτὸΔσημεῖον point A being placed on the point D, and the straight-line
10

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
τῆςδὲΑΒεὐθείαςἐπὶτὴνΔΕ,ἐφαρμόσεικαὶτὸΒσημεῖον AB on DE, then the point B will also coincide with E,
ἐπὶτὸΕδιὰτὸἴσηνεἶναιτὴνΑΒτῇΔΕ·ἐφαρμοσάσηςδὴ on account of AB being equal to DE. So (because of)
τῆςΑΒἐπὶτὴνΔΕἐφαρμόσεικαὶἡΑΓεὐθεῖαἐπὶτὴνΔΖ AB coinciding with DE, the straight-line AC will also
διὰτὸἴσηνεἶναιτὴνὑπὸΒΑΓγωνίαντῇὑπὸΕΔΖ·ὥστεκαὶ coincide with DF , on account of the angle BAC being
τὸΓσημεῖονἐπὶτὸΖσημεῖονἐφαρμόσειδιὰτὸἴσηνπάλιν equal to EDF . So the point C will also coincide with the
εἶναιτὴνΑΓτῇΔΖ.ἀλλὰμὴνκαὶτὸΒἐπὶτὸΕἐφηρμόκει· point F , again on account of AC being equal to DF . But,
ὥστεβάσιςἡΒΓἐπὶβάσιντὴνΕΖἐφαρμόσει. εἰγὰρτοῦ point B certainly also coincided with point E, so that the
μὲνΒἐπὶτὸΕἐφαρμόσαντοςτοῦδὲΓἐπὶτὸΖἡΒΓβάσις base BC will coincide with the base EF . For if B coin-
ἐπὶτὴνΕΖοὐκἐφαρμόσει,δύοεὐθεῖαιχωρίονπεριέξουσιν· cides with E, and C with F , and the base BC does not
ὅπερἐστὶνἀδύνατον. ἐφαρμόσειἄραἡΒΓβάσιςἐπὶτὴν coincide with EF , then two straight-lines will encompass
ΕΖκαὶἴσηαὐτῇἔσται·ὥστεκαὶὅλοντὸΑΒΓτρίγωνον an area. The very thing is impossible [Post. 1]. ‡ Thus,
ἐπὶὅλοντὸΔΕΖτρίγωνονἐφαρμόσεικαὶἴσοναὐτῷἔσται, the base BC will coincide with EF , and will be equal to
καὶαἱλοιπαὶγωνίαιἐπὶτὰςλοιπὰςγωνίαςἐφαρμόσουσικαὶ it [C.N. 4]. So the whole triangle ABC will coincide with
ἴσαιαὐταῖςἔσονται,ἡμὲνὑπὸΑΒΓτῇὑπὸΔΕΖἡδὲὑπὸ the whole triangle DEF , and will be equal to it [C.N. 4].
ΑΓΒτῇὑπὸΔΖΕ.
And the remaining angles will coincide with the remain-
᾿Εὰν ἄρα δύο τρίγωνα τὰς δύο πλευρὰς [ταῖς] δύο ing angles, and will be equal to them [C.N. 4]. (That is)
πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ καὶ τὴν γωνίαν τῇ ABC to DEF , and ACB to DFE [C.N. 4].
γωνίᾳἴσηνἔχῃτὴνὑπὸτῶνἴσωνεὐθειῶνπεριεχομένην, Thus, if two triangles have two sides equal to two
καὶ τὴν βάσιν τῂ βάσει ἴσην ἕξει, καὶ τὸ τρίγωνον τῷ sides, respectively, and have the angle(s) enclosed by the
τριγώνῳ ἴσον ἔσται, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς equal straight-line equal, then they will also have the base
γωνίαιςἴσαιἔσονταιἑκατέραἑκατέρᾳ,ὑφ᾿ἃςαἱἴσαιπλευραὶ equal to the base, and the triangle will be equal to the tri-
ὑποτείνουσιν·ὅπερἔδειδεῖξαι.
angle, and the remaining angles subtended by the equal
sides will be equal to the corresponding remaining angles.
(Which is) the very thing it was required to show.
† The application of one figure to another should be counted as an additional postulate.
‡ Since Post. 1 implicitly assumes that the straight-line joining two given points is unique.
. Proposition 5
Τῶνἰσοσκελῶντριγώνωναἱτρὸςτῇβάσειγωνίαιἴσαι For isosceles triangles, the angles at the base are equal
ἀλλήλαιςεἰσίν,καὶπροσεκβληθεισῶντῶνἴσωνεὐθειῶναἱ to one another, and if the equal sides are produced then
ὑπὸτὴνβάσινγωνίαιἴσαιἀλλήλαιςἔσονται.
the angles under the base will be equal to one another.
Α
A
Β
Γ
B
C
Ζ
Η
F
G
∆ Ε
D
E
῎Εστω τρίγωνον ἰσοσκελὲς τὸ ΑΒΓ ἴσην ἔχον τὴν Let ABC be an isosceles triangle having the side AB
ΑΒπλευρὰντῇΑΓπλευρᾷ,καὶπροσεκβεβλήσθωσανἐπ᾿ equal to the side AC, and let the straight-lines BD and
εὐθείαςταῖςΑΒ,ΑΓεὐθεῖαιαἱΒΔ,ΓΕ·λέγω,ὅτιἡμὲν CE have been produced in a straight-line with AB and
ὑπὸΑΒΓγωνίατῇὑπὸΑΓΒἴσηἐστίν,ἡδὲὑπὸΓΒΔτῇ AC (respectively) [Post. 2]. I say that the angle ABC is
ὑπὸΒΓΕ.
equal to ACB, and (angle) CBD to BCE.
Εἰλήφθω γὰρ ἐπὶ τῆς ΒΔ τυχὸν σημεῖον τὸ Ζ, καὶ For let the point F have been taken at random on BD,
ἀφῃρήσθωἀπὸτῆςμείζονοςτῆςΑΕτῇἐλάσσονιτῇΑΖ and let AG have been cut off from the greater AE, equal
11

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ἴσηἡΑΗ,καὶἐπεζεύχθωσαναἱΖΓ,ΗΒεὐθεῖαι. to the lesser AF [Prop. 1.3]. Also, let the straight-lines
᾿ΕπεὶοὖνἴσηἐστὶνἡμὲνΑΖτῇΑΗἡδὲΑΒτῇΑΓ, FC and GB have been joined [Post. 1].
δύοδὴαἱΖΑ,ΑΓδυσὶταῖςΗΑ,ΑΒἴσαιεἰσὶνἑκατέρα In fact, since AF is equal to AG, and AB to AC,
ἑκατέρᾳ·καὶγωνίανκοινὴνπεριέχουσιτὴνὑπὸΖΑΗ·βάσις the two (straight-lines) FA, AC are equal to the two
ἄραἡΖΓβάσειτῇΗΒἴσηἐστίν,καὶτὸΑΖΓτρίγωνοντῷ (straight-lines) GA, AB, respectively. They also encom-
ΑΗΒτριγώνῳἴσονἔσται,καὶαἱλοιπαὶγωνίαιταῖςλοιπαῖς pass a common angle, FAG. Thus, the base FC is equal
γωνίαιςἴσαιἔσονταιἑκατέραἑκατέρᾳ,ὑφ᾿ἃςαἱἴσαιπλευραὶ to the base GB, and the triangle AFC will be equal to the
ὑποτείνουσιν,ἡμὲνὑπὸΑΓΖτῇὑπὸΑΒΗ,ἡδὲὑπὸΑΖΓ triangle AGB, and the remaining angles subtendend by
τῇὑπὸΑΗΒ.καὶἐπεὶὅληἡΑΖὅλῃτῇΑΗἐστινἴση,ὧν the equal sides will be equal to the corresponding remain-
ἡΑΒτῇΑΓἐστινἴση,λοιπὴἄραἡΒΖλοιπῇτῇΓΗἐστιν ing angles [Prop. 1.4]. (That is) ACF to ABG, and AFC
ἴση.ἐδείχθηδὲκαὶἡΖΓτῇΗΒἴση·δύοδὴαἱΒΖ,ΖΓδυσὶ to AGB. And since the whole of AF is equal to the whole
ταῖςΓΗ,ΗΒἴσαιεἰσὶνἑκατέραἑκατέρᾳ·καὶγωνίαἡὑπὸ of AG, within which AB is equal to AC, the remainder
ΒΖΓγωνίᾳτῃὑπὸΓΗΒἴση,καὶβάσιςαὐτῶνκοινὴἡΒΓ· BF is thus equal to the remainder CG [C.N. 3]. But FC
καὶτὸΒΖΓἄρατρίγωνοντῷΓΗΒτριγώνῳἴσονἔσται,καὶ was also shown (to be) equal to GB. So the two (straightαἱλοιπαὶγωνίαιταῖςλοιπαῖςγωνίαιςἴσαιἔσονταιἑκατέρα
lines) BF , FC are equal to the two (straight-lines) CG,
ἑκατέρᾳ,ὑφ᾿ἃςαἱἴσαιπλευραὶὑποτείνουσιν·ἴσηἄραἐστὶν GB, respectively, and the angle BFC (is) equal to the
ἡμὲνὑπὸΖΒΓτῇὑπὸΗΓΒἡδὲὑπὸΒΓΖτῇὑπὸΓΒΗ. angle CGB, and the base BC is common to them. Thus,
ἐπεὶοὖνὅληἡὑπὸΑΒΗγωνίαὅλῃτῇὑπὸΑΓΖγωνίᾳ the triangle BFC will be equal to the triangle CGB, and
ἐδείχθηἴση,ὧνἡὑπὸΓΒΗτῇὑπὸΒΓΖἴση,λοιπὴἄραἡ the remaining angles subtended by the equal sides will be
ὑπὸΑΒΓλοιπῇτῇὑπὸΑΓΒἐστινἴση·καίεἰσιπρὸςτῇ equal to the corresponding remaining angles [Prop. 1.4].
βάσειτοῦΑΒΓτριγώνου. ἐδείχθηδὲκαὶἡὑπὸΖΒΓτῇ Thus, FBC is equal to GCB, and BCF to CBG. There-
ὑπὸΗΓΒἴση·καίεἰσινὑπὸτὴνβάσιν.
fore, since the whole angle ABG was shown (to be) equal
Τῶνἄραἰσοσκελῶντριγώνωναἱτρὸςτῇβάσειγωνίαι to the whole angle ACF , within which CBG is equal to
ἴσαιἀλλήλαιςεἰσίν,καὶπροσεκβληθεισῶντῶνἴσωνεὐθειῶν BCF , the remainder ABC is thus equal to the remainder
αἱὑπὸτὴνβάσινγωνίαιἴσαιἀλλήλαιςἔσονται·ὅπερἔδει ACB [C.N. 3]. And they are at the base of triangle ABC.
δεῖξαι.
And FBC was also shown (to be) equal to GCB. And
they are under the base.
Thus, for isosceles triangles, the angles at the base are
equal to one another, and if the equal sides are produced
then the angles under the base will be equal to one another.
(Which is) the very thing it was required to show.
¦. Proposition 6
᾿Εὰντριγώνουαἱδύογωνίαιἴσαιἀλλήλαιςὦσιν, καὶ If a triangle has two angles equal to one another then
αἱὑπὸτὰςἴσαςγωνίαςὑποτείνουσαιπλευραὶἴσαιἀλλήλαις the sides subtending the equal angles will also be equal
ἔσονται.
to one another.
Α
A
∆
D
Β
Γ
῎ΕστωτρίγωνοντὸΑΒΓἴσηνἔχοντὴνὑπὸΑΒΓγωνίαν Let ABC be a triangle having the angle ABC equal
τῇὑπὸΑΓΒγωνίᾳ·λέγω,ὅτικαὶπλευρὰἡΑΒπλευρᾷτῇ to the angle ACB. I say that side AB is also equal to side
ΑΓἐστινἴση.
AC.
B
C
12

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ΕἰγὰρἄνισόςἐστινἡΑΒτῇΑΓ,ἡἑτέρααὐτῶνμείζων For if AB is unequal to AC then one of them is
ἐστίν.ἔστωμείζωνἡΑΒ,καὶἀφῃρήσθωἀπὸτῆςμείζονος greater. Let AB be greater. And let DB, equal to
τῆςΑΒτῇἐλάττονιτῇΑΓἴσηἡΔΒ,καὶἐπεζεύχθωἡΔΓ. the lesser AC, have been cut off from the greater AB
᾿ΕπεὶοὖνἴσηἐστὶνἡΔΒτῇΑΓκοινὴδὲἡΒΓ,δύοδὴ [Prop. 1.3]. And let DC have been joined [Post. 1].
αἱΔΒ,ΒΓδύοταῖςΑΓ,ΓΒἴσαιεἰσὶνἑκατέραἑκατέρᾳ,καὶ Therefore, since DB is equal to AC, and BC (is) comγωνίαἡὑπὸΔΒΓγωνίᾳτῇὑπὸΑΓΒἐστινἴση·βάσιςἄραἡ
mon, the two sides DB, BC are equal to the two sides
ΔΓβάσειτῇΑΒἴσηἐστίν,καὶτὸΔΒΓτρίγωνοντῷΑΓΒ AC, CB, respectively, and the angle DBC is equal to the
τριγώνῳἴσονἔσται,τὸἔλασσοντῷμείζονι·ὅπερἄτοπον· angle ACB. Thus, the base DC is equal to the base AB,
οὐκἄραἄνισόςἐστινἡΑΒτῇΑΓ·ἴσηἄρα.
and the triangle DBC will be equal to the triangle ACB
᾿Εὰνἄρατριγώνουαἱδὑογωνίαιἴσαιἀλλήλαιςὦσιν,καὶ [Prop. 1.4], the lesser to the greater. The very notion (is)
αἱὑπὸτὰςἴσαςγωνίαςὑποτείνουσαιπλευραὶἴσαιἀλλήλαις absurd [C.N. 5]. Thus, AB is not unequal to AC. Thus,
ἔσονται·ὅπερἔδειδεῖξαι. (it is) equal. †
Thus, if a triangle has two angles equal to one another
then the sides subtending the equal angles will also be
equal to one another. (Which is) the very thing it was
required to show.
† Here, use is made of the previously unmentioned common notion that if two quantities are not unequal then they must be equal. Later on, use
is made of the closely related common notion that if two quantities are not greater than or less than one another, respectively, then they must be
equal to one another.
Þ.
Proposition 7
᾿Επὶτῆςαὐτῆςεὐθείαςδύοταῖςαὐταῖςεὐθείαιςἄλλαι On the same straight-line, two other straight-lines
δύοεὐθεῖαιἴσαιἑκατέραἑκατέρᾳοὐσυσταθήσονταιπρὸς equal, respectively, to two (given) straight-lines (which
ἄλλῳκαὶἄλλῳσημείῳἐπὶτὰαὐτὰμέρητὰαὐτὰπέρατα meet) cannot be constructed (meeting) at a different
ἔχουσαιταῖςἐξἀρχῆςεὐθείαις.
point on the same side (of the straight-line), but having
the same ends as the given straight-lines.
Γ
C
∆
D
Α
Β
Εἰγὰρδυνατόν,ἐπὶτῆςαὐτῆςεὐθείαςτῆςΑΒδύοταῖς For, if possible, let the two straight-lines AC, CB,
αὐταῖςεὐθείαιςταῖςΑΓ,ΓΒἄλλαιδύοεὐθεῖαιαἱΑΔ,ΔΒ equal to two other straight-lines AD, DB, respectively,
ἴσαιἑκατέραἑκατέρᾳσυνεστάτωσανπρὸςἄλλῳκαὶἄλλῳ have been constructed on the same straight-line AB,
σημείῳτῷτεΓκαὶΔἐπὶτὰαὐτὰμέρητὰαὐτὰπέρατα meeting at different points, C and D, on the same side
ἔχουσαι,ὥστεἴσηνεἶναιτὴνμὲνΓΑτῇΔΑτὸαὐτὸπέρας (of AB), and having the same ends (on AB). So CA is
ἔχουσαναὐτῇτὸΑ,τὴνδὲΓΒτῇΔΒτὸαὐτὸπέραςἔχου- equal to DA, having the same end A as it, and CB is
σαναὐτῇτὸΒ,καὶἐπεζεύχθωἡΓΔ.
equal to DB, having the same end B as it. And let CD
᾿ΕπεὶοὖνἴσηἐστὶνἡΑΓτῇΑΔ,ἴσηἐστὶκαὶγωνίαἡ have been joined [Post. 1].
ὑπὸΑΓΔτῇὑπὸΑΔΓ·μείζωνἄραἡὑπὸΑΔΓτῆςὑπὸ Therefore, since AC is equal to AD, the angle ACD
ΔΓΒ·πολλῷἄραἡὑπὸΓΔΒμείζωνἐστίτῆςὑπὸΔΓΒ. is also equal to angle ADC [Prop. 1.5]. Thus, ADC (is)
πάλινἐπεὶἴσηἐστὶνἡΓΒτῇΔΒ,ἴσηἐστὶκαὶγωνίαἡ greater than DCB [C.N. 5]. Thus, CDB is much greater
ὑπὸΓΔΒγωνίᾳτῇὑπὸΔΓΒ.ἐδείχθηδὲαὐτῆςκαὶπολλῷ than DCB [C.N. 5]. Again, since CB is equal to DB, the
μείζων·ὅπερἐστὶνἀδύνατον.
angle CDB is also equal to angle DCB [Prop. 1.5]. But
Οὐκἄραἐπὶτῆςαὐτῆςεὐθείαςδύοταῖςαὐταῖςεὐθείαις it was shown that the former (angle) is also much greater
A
B
13

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ἄλλαιδύοεὐθεῖαιἴσαιἑκατέραἑκατέρᾳσυσταθήσονταιπρὸς (than the latter). The very thing is impossible.
ἄλλῳκαὶἄλλῳσημείῳἐπὶτὰαὐτὰμέρητὰαὐτὰπέρατα Thus, on the same straight-line, two other straight-
ἔχουσαιταῖςἐξἀρχῆςεὐθείαις·ὅπερἔδειδεῖξαι. lines equal, respectively, to two (given) straight-lines
(which meet) cannot be constructed (meeting) at a different
point on the same side (of the straight-line), but
having the same ends as the given straight-lines. (Which
is) the very thing it was required to show.
. Proposition 8
᾿Εὰνδύοτρίγωνατὰςδύοπλευρὰς[ταῖς]δύοπλευραῖς If two triangles have two sides equal to two sides, re-
ἴσαςἔχῃἑκατέρανἑκατέρᾳ,ἔχῃδὲκαὶτὴνβάσιντῇβάσει spectively, and also have the base equal to the base, then
ἴσην,καὶτὴνγωνίαντῇγωνίᾳἴσηνἕξειτὴνὑπὸτῶνἴσων they will also have equal the angles encompassed by the
εὐθειῶνπεριεχομένην.
equal straight-lines.
Α
A
D
G
∆ Η
Ζ
Γ
C
F
Ε
E
Β
B
῎Εστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο πλευρὰς Let ABC and DEF be two triangles having the two
τὰςΑΒ,ΑΓταῖςδύοπλευραῖςταῖςΔΕ,ΔΖἴσαςἔχοντα sides AB and AC equal to the two sides DE and DF ,
ἑκατέρανἑκατέρᾳ,τὴνμὲνΑΒτῇΔΕτὴνδὲΑΓτῇΔΖ· respectively. (That is) AB to DE, and AC to DF . Let
ἐχέτωδὲκαὶβάσιντὴνΒΓβάσειτῇΕΖἴσην·λέγω,ὅτικαὶ them also have the base BC equal to the base EF . I say
γωνίαἡὑπὸΒΑΓγωνίᾳτῇὑπὸΕΔΖἐστινἴση. that the angle BAC is also equal to the angle EDF .
᾿ΕφαρμοζομένουγὰρτοῦΑΒΓτριγώνουἐπὶτὸΔΕΖ For if triangle ABC is applied to triangle DEF , the
τρίγωνονκαὶτιθεμένουτοῦμὲνΒσημείουἐπὶτὸΕσημεῖον point B being placed on point E, and the straight-line
τῆςδὲΒΓεὐθείαςἐπὶτὴνΕΖἐφαρμόσεικαὶτὸΓσημεῖον BC on EF , then point C will also coincide with F , on
ἐπὶτὸΖδιὰτὸἴσηνεἶναιτὴνΒΓτῇΕΖ·ἐφαρμοσάσηςδὴ account of BC being equal to EF . So (because of) BC
τῆςΒΓἐπὶτὴνΕΖἐφαρμόσουσικαὶαἱΒΑ,ΓΑἐπὶτὰςΕΔ, coinciding with EF , (the sides) BA and CA will also co-
ΔΖ.εἰγὰρβάσιςμὲνἡΒΓἐπὶβάσιντὴνΕΖἐφαρμόσει,αἱ incide with ED and DF (respectively). For if base BC
δὲΒΑ,ΑΓπλευραὶἐπὶτὰςΕΔ,ΔΖοὐκἐφαρμόσουσιν coincides with base EF , but the sides AB and AC do not
ἀλλὰπαραλλάξουσινὡςαἱΕΗ,ΗΖ,συσταθήσονταιἐπὶτῆς coincide with ED and DF (respectively), but miss like
αὐτῆςεὐθείαςδύοταῖςαὐταῖςεὐθείαιςἄλλαιδύοεὐθεῖαι EG and GF (in the above figure), then we will have con-
ἴσαιἑκατέραἑκατέρᾳπρὸςἄλλῳκαὶἄλλῳσημείῳἐπὶτὰ structed upon the same straight-line, two other straightαὐτὰμέρητὰαὐτὰπέραταἔχουσαι.
οὐσυνίστανταιδέ· lines equal, respectively, to two (given) straight-lines,
οὐκἄραἐφαρμοζομένηςτῆςΒΓβάσεωςἐπὶτὴνΕΖβάσιν and (meeting) at a different point on the same side (of
οὐκἐφαρμόσουσικαὶαἱΒΑ,ΑΓπλευραὶἐπὶτὰςΕΔ,ΔΖ. the straight-line), but having the same ends. But (such
ἐφαρμόσουσινἄρα·ὥστεκαὶγωνίαἡὑπὸΒΑΓἐπὶγωνίαν straight-lines) cannot be constructed [Prop. 1.7]. Thus,
τὴνὑπὸΕΔΖἐφαρμόσεικαὶἴσηαὐτῇἔσται.
the base BC being applied to the base EF , the sides BA
᾿Εὰν ἄρα δύο τρίγωνα τὰς δύο πλευρὰς [ταῖς] δύο and AC cannot not coincide with ED and DF (respecπλευραῖςἴσαςἔχῃἑκατέρανἑκατέρᾳκαὶτὴνβάσιντῇβάσει
tively). Thus, they will coincide. So the angle BAC will
ἴσηνἔχῃ,καὶτὴνγωνίαντῇγωνίᾳἴσηνἕξειτὴνὑπὸτῶν also coincide with angle EDF , and will be equal to it
ἴσωνεὐθειῶνπεριεχομένην·ὅπερἔδειδεῖξαι. [C.N. 4].
Thus, if two triangles have two sides equal to two
side, respectively, and have the base equal to the base,
14

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
. Proposition
Τὴνδοθεῖσανγωνίανεὐθύγραμμονδίχατεμεῖν.
Α
then they will also have equal the angles encompassed
by the equal straight-lines. (Which is) the very thing it
was required to show.
9
To cut a given rectilinear angle in half.
A
∆
Ε
D
E
Β
Ζ
Γ
῎ΕστωἡδοθεῖσαγωνίαεὐθύγραμμοςἡὑπὸΒΑΓ.δεῖ Let BAC be the given rectilinear angle. So it is reδὴαὐτὴνδίχατεμεῖν.
quired to cut it in half.
ΕἰλήφθωἐπὶτῆςΑΒτυχὸνσημεῖοντὸΔ,καὶἀφῃρήσθω Let the point D have been taken at random on AB,
ἀπὸτῆςΑΓτῇΑΔἴσηἡΑΕ,καὶἐπεζεύχθωἡΔΕ,καὶ and let AE, equal to AD, have been cut off from AC
συνεστάτωἐπὶτῆςΔΕτρίγωνονἰσόπλευροντὸΔΕΖ,καὶ [Prop. 1.3], and let DE have been joined. And let the
ἐπεζεύχθωἡΑΖ·λέγω,ὅτιἡὑπὸΒΑΓγωνίαδίχατέτμηται equilateral triangle DEF have been constructed upon
ὑπὸτῆςΑΖεὐθείας.
DE [Prop. 1.1], and let AF have been joined. I say that
᾿ΕπεὶγὰρἴσηἐστὶνἡΑΔτῇΑΕ,κοινὴδὲἡΑΖ,δύοδὴ the angle BAC has been cut in half by the straight-line
αἱΔΑ,ΑΖδυσὶταῖςΕΑ,ΑΖἴσαιεἰσὶνἑκατέραἑκατέρᾳ. AF .
καὶβάσιςἡΔΖβάσειτῇΕΖἴσηἐστίν·γωνίαἄραἡὑπὸ For since AD is equal to AE, and AF is common,
ΔΑΖγωνίᾳτῇὑπὸΕΑΖἴσηἐστίν.
the two (straight-lines) DA, AF are equal to the two
῾ΗἄραδοθεῖσαγωνίαεὐθύγραμμοςἡὑπὸΒΑΓδίχα (straight-lines) EA, AF , respectively. And the base DF
τέτμηταιὑπὸτῆςΑΖεὐθείας·ὅπερἔδειποιῆσαι. is equal to the base EF . Thus, angle DAF is equal to
angle EAF [Prop. 1.8].
Thus, the given rectilinear angle BAC has been cut in
half by the straight-line AF . (Which is) the very thing it
was required to do.
. Proposition 10
Τὴνδοθεῖσανεὐθεῖανπεπερασμένηνδίχατεμεῖν. To cut a given finite straight-line in half.
῎ΕστωἡδοθεῖσαεὐθεῖαπεπερασμένηἡΑΒ·δεῖδὴτὴν Let AB be the given finite straight-line. So it is re-
ΑΒεὐθεῖανπεπερασμένηνδίχατεμεῖν.
quired to cut the finite straight-line AB in half.
Συνεστάτωἐπ᾿αὐτῆςτρίγωνονἰσόπλευροντὸΑΒΓ,καὶ Let the equilateral triangle ABC have been conτετμήσθωἡὑπὸΑΓΒγωνίαδίχατῇΓΔεὐθείᾳ·λέγω,ὅτι
structed upon (AB) [Prop. 1.1], and let the angle ACB
ἡΑΒεὐθεῖαδίχατέτμηταικατὰτὸΔσημεῖον. have been cut in half by the straight-line CD [Prop. 1.9].
᾿ΕπεὶγὰρἴσηἐστὶνἡΑΓτῇΓΒ,κοινὴδὲἡΓΔ,δύοδὴ I say that the straight-line AB has been cut in half at
αἱΑΓ,ΓΔδύοταῖςΒΓ,ΓΔἴσαιεἰσὶνἑκατέραἑκατέρᾳ·καὶ point D.
γωνίαἡὑπὸΑΓΔγωνίᾳτῇὑπὸΒΓΔἴσηἐστίν·βάσιςἄρα For since AC is equal to CB, and CD (is) common,
B
F
C
15

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ἡΑΔβάσειτῇΒΔἴσηἐστίν.
Γ
the two (straight-lines) AC, CD are equal to the two
(straight-lines) BC, CD, respectively. And the angle
ACD is equal to the angle BCD. Thus, the base AD
is equal to the base BD [Prop. 1.4].
C
Α
∆
Β
῾ΗἄραδοθεῖσαεὐθεῖαπεπερασμένηἡΑΒδίχατέτμηται Thus, the given finite straight-line AB has been cut
κατὰτὸΔ·ὅπερἔδειποιῆσαι.
in half at (point) D. (Which is) the very thing it was
required to do.
. Proposition 11
Τῇδοθείσῃεὐθείᾳἀπὸτοῦπρὸςαὐτῇδοθέντοςσημείου To draw a straight-line at right-angles to a given
πρὸςὀρθὰςγωνίαςεὐθεῖανγραμμὴνἀγαγεῖν.
straight-line from a given point on it.
Ζ
A
D
F
B
Α
∆ Γ Ε
Β
῎ΕστωἡμὲνδοθεῖσαεὐθεῖαἡΑΒτὸδὲδοθὲνσημεῖον Let AB be the given straight-line, and C the given
ἐπ᾿αὐτῆςτὸΓ·δεῖδὴἀπὸτοῦΓσημείουτῇΑΒεὐθείᾳ point on it. So it is required to draw a straight-line from
πρὸςὀρθὰςγωνίαςεὐθεῖανγραμμὴνἀγαγεῖν.
the point C at right-angles to the straight-line AB.
ΕἰλήφθωἐπὶτῆςΑΓτυχὸνσημεῖοντὸΔ,καὶκείσθω Let the point D be have been taken at random on AC,
τῇ ΓΔ ἴση ἡ ΓΕ, καὶ συνεστάτω ἐπὶ τῆς ΔΕ τρίγωνον and let CE be made equal to CD [Prop. 1.3], and let the
ἰσόπλευροντὸΖΔΕ,καὶἐπεζεύχθωἡΖΓ·λέγω,ὅτιτῇ equilateral triangle FDE have been constructed on DE
δοθείσῃεὐθείᾳτῇΑΒἀπὸτοῦπρὸςαὐτῇδοθέντοςσημείου [Prop. 1.1], and let FC have been joined. I say that the
τοῦΓπρὸςὀρθὰςγωνίαςεὐθεῖαγραμμὴἦκταιἡΖΓ. straight-line FC has been drawn at right-angles to the
᾿ΕπεὶγὰρἴσηἐστὶνἡΔΓτῇΓΕ,κοινὴδὲἡΓΖ,δύο given straight-line AB from the given point C on it.
δὴαἱΔΓ,ΓΖδυσὶταῖςΕΓ,ΓΖἴσαιεἰσὶνἑκατέραἑκατέρᾳ· For since DC is equal to CE, and CF is common,
καὶβάσιςἡΔΖβάσειτῇΖΕἴσηἐστίν·γωνίαἄραἡὑπὸ the two (straight-lines) DC, CF are equal to the two
ΔΓΖγωνίᾳτῇὑπὸΕΓΖἴσηἐστίν·καίεἰσινἐφεξῆς. ὅταν (straight-lines), EC, CF , respectively. And the base DF
δὲ εὐθεῖα ἐπ᾿ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας is equal to the base FE. Thus, the angle DCF is equal
ἀλλήλαιςποιῇ,ὀρθὴἑκατέρατῶνἴσωνγωνιῶνἐστιν·ὀρθὴ to the angle ECF [Prop. 1.8], and they are adjacent.
ἄραἐστὶνἑκατέρατῶνὑπὸΔΓΖ,ΖΓΕ.
But when a straight-line stood on a(nother) straight-line
A
D
C
E
B
16

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
Τῇ ἄρα δοθείσῃ εὐθείᾳ τῇ ΑΒ ἀπὸ τοῦ πρὸς αὐτῇ makes the adjacent angles equal to one another, each of
δοθέντοςσημείουτοῦΓπρὸςὀρθὰςγωνίαςεὐθεῖαγραμμὴ the equal angles is a right-angle [Def. 1.10]. Thus, each
ἦκταιἡΓΖ·ὅπερἔδειποιῆσαι.
of the (angles) DCF and FCE is a right-angle.
Thus, the straight-line CF has been drawn at rightangles
to the given straight-line AB from the given point
C on it. (Which is) the very thing it was required to do.
. Proposition 12
᾿Επὶτὴνδοθεῖσανεὐθεῖανἄπειρονἀπὸτοῦδοθέντος To draw a straight-line perpendicular to a given infiσημείου,ὃμήἐστινἐπ᾿αὐτῆς,κάθετονεὐθεῖανγραμμὴν
nite straight-line from a given point which is not on it.
ἀγαγεῖν.
Ζ
F
Γ
C
Α
Η
Θ
∆
Ε
Β
῎ΕστωἡμὲνδοθεῖσαεὐθεῖαἄπειροςἡΑΒτὸδὲδοθὲν Let AB be the given infinite straight-line and C the
σημεῖον,ὃμήἐστινἐπ᾿αὐτῆς,τὸΓ·δεῖδὴἐπὶτὴνδοθεῖσαν given point, which is not on (AB). So it is required to
εὐθεῖανἄπειροντὴνΑΒἀπὸτοῦδοθέντοςσημείουτοῦΓ, draw a straight-line perpendicular to the given infinite
ὃμήἐστινἐπ᾿αὐτῆς,κάθετονεὐθεῖανγραμμὴνἀγαγεῖν. straight-line AB from the given point C, which is not on
ΕἰλήφθωγὰρἐπὶτὰἕτεραμέρητῆςΑΒεὐθείαςτυχὸν (AB).
σημεῖοντὸΔ,καὶκέντρῳμὲντῷΓδιαστήματιδὲτῷΓΔ For let point D have been taken at random on the
κύκλοςγεγράφθωὁΕΖΗ,καὶτετμήσθωἡΕΗεὐθεῖαδίχα other side (to C) of the straight-line AB, and let the
κατὰ τὸ Θ, καὶ ἐπεζεύχθωσαν αἱ ΓΗ, ΓΘ, ΓΕ εὐθεῖαι· circle EFG have been drawn with center C and radius
λέγω,ὅτιἐπὶτὴνδοθεῖσανεὐθεῖανἄπειροντὴνΑΒἀπὸ CD [Post. 3], and let the straight-line EG have been cut
τοῦδοθέντοςσημείουτοῦΓ,ὃμήἐστινἐπ᾿αὐτῆς,κάθετος in half at (point) H [Prop. 1.10], and let the straight-
ἦκταιἡΓΘ.
lines CG, CH, and CE have been joined. I say that the
᾿ΕπεὶγὰρἴσηἐστὶνἡΗΘτῇΘΕ,κοινὴδὲἡΘΓ,δύο (straight-line) CH has been drawn perpendicular to the
δὴαἱΗΘ,ΘΓδύοταῖςΕΘ,ΘΓἴσαιεἱσὶνἑκατέραἑκατέρᾳ· given infinite straight-line AB from the given point C,
καὶβάσιςἡΓΗβάσειτῇΓΕἐστινἴση·γωνίαἄραἡὑπὸ which is not on (AB).
ΓΘΗγωνίᾳτῇὑπὸΕΘΓἐστινἴση.καίεἰσινἐφεξῆς.ὅταν For since GH is equal to HE, and HC (is) common,
δὲ εὐθεῖα ἐπ᾿ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας the two (straight-lines) GH, HC are equal to the two
ἀλλήλαιςποιῇ,ὀρθὴἑκατέρατῶνἴσωνγωνιῶνἐστιν,καὶ (straight-lines) EH, HC, respectively, and the base CG
ἡἐφεστηκυῖαεὐθεῖακάθετοςκαλεῖταιἐφ᾿ἣνἐφέστηκεν. is equal to the base CE. Thus, the angle CHG is equal
᾿ΕπὶτὴνδοθεῖσανἄραεὐθεῖανἄπειροντὴνΑΒἀπὸτοῦ to the angle EHC [Prop. 1.8], and they are adjacent.
δοθέντοςσημείουτοῦΓ,ὃμήἐστινἐπ᾿αὐτῆς,κάθετος But when a straight-line stood on a(nother) straight-line
ἦκταιἡΓΘ·ὅπερἔδειποιῆσαι.
makes the adjacent angles equal to one another, each of
the equal angles is a right-angle, and the former straightline
is called a perpendicular to that upon which it stands
[Def. 1.10].
Thus, the (straight-line) CH has been drawn perpendicular
to the given infinite straight-line AB from the
A
G
H
D
E
B
17

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
given point C, which is not on (AB). (Which is) the very
thing it was required to do.
. Proposition 13
᾿Εὰνεὐθεῖαἐπ᾿εὐθεῖανσταθεῖσαγωνίαςποιῇ,ἤτοιδύο If a straight-line stood on a(nother) straight-line
ὀρθὰςἢδυσὶνὀρθαῖςἴσαςποιήσει.
makes angles, it will certainly either make two rightangles,
or (angles whose sum is) equal to two rightangles.
Ε
Α
E
A
∆ Β Γ
D B C
Εὐθεῖα γάρ τις ἡ ΑΒ ἐπ᾿ εὐθεῖαν τὴν ΓΔ σταθεῖσα For let some straight-line AB stood on the straightγωνίαςποιείτωτὰςὑπὸΓΒΑ,ΑΒΔ·λὲγω,ὅτιαἱὑπὸΓΒΑ,
line CD make the angles CBA and ABD. I say that
ΑΒΔγωνίαιἤτοιδύοὀρθαίεἰσινἢδυσὶνὀρθαῖςἴσαι. the angles CBA and ABD are certainly either two right-
ΕἰμὲνοὖνἴσηἐστὶνἡὑπὸΓΒΑτῇὑπὸΑΒΔ,δύοὀρθαί angles, or (have a sum) equal to two right-angles.
εἰσιν.εἰδὲοὔ,ἤχθωἀπὸτοῦΒσημείουτῇΓΔ[εὐθείᾳ]πρὸς In fact, if CBA is equal to ABD then they are two
ὀρθὰςἡΒΕ·αἱἄραὑπὸΓΒΕ,ΕΒΔδύοὀρθαίεἰσιν·καὶ right-angles [Def. 1.10]. But, if not, let BE have been
ἐπεὶἡὑπὸΓΒΕδυσὶταῖςὑπὸΓΒΑ,ΑΒΕἴσηἐστίν,κοινὴ drawn from the point B at right-angles to [the straightπροσκείσθωἡὑπὸΕΒΔ·αἱἄραὑπὸΓΒΕ,ΕΒΔτρισὶταῖς
line] CD [Prop. 1.11]. Thus, CBE and EBD are two
ὑπὸΓΒΑ,ΑΒΕ,ΕΒΔἴσαιεἰσίν. πάλιν,ἐπεὶἡὑπὸΔΒΑ right-angles. And since CBE is equal to the two (anδυσὶταῖςὑπὸΔΒΕ,ΕΒΑἴσηἐστίν,κοινὴπροσκείσθωἡ
gles) CBA and ABE, let EBD have been added to both.
ὑπὸΑΒΓ·αἱἄραὑπὸΔΒΑ,ΑΒΓτρισὶταῖςὑπὸΔΒΕ,ΕΒΑ, Thus, the (sum of the angles) CBE and EBD is equal to
ΑΒΓἴσαιεἰσίν.ἐδείχθησανδὲκαὶαἱὑπὸΓΒΕ,ΕΒΔτρισὶ the (sum of the) three (angles) CBA, ABE, and EBD
ταῖςαὐταῖςἴσαι·τὰδὲτῷαὐτῷἴσακαὶἀλλήλοιςἐστὶνἴσα· [C.N. 2]. Again, since DBA is equal to the two (anκαὶαἱὑπὸΓΒΕ,ΕΒΔἄραταῖςὑπὸΔΒΑ,ΑΒΓἴσαιεἰσίν·
gles) DBE and EBA, let ABC have been added to both.
ἀλλὰαἱὑπὸΓΒΕ,ΕΒΔδύοὀρθαίεἰσιν·καὶαἱὑπὸΔΒΑ, Thus, the (sum of the angles) DBA and ABC is equal to
ΑΒΓἄραδυσὶνὀρθαῖςἴσαιεἰσίν.
the (sum of the) three (angles) DBE, EBA, and ABC
᾿Εὰνἄραεὐθεῖαἐπ᾿εὐθεῖανσταθεῖσαγωνίαςποιῇ,ἤτοι [C.N. 2]. But (the sum of) CBE and EBD was also
δύοὀρθὰςἢδυσὶνὀρθαῖςἴσαςποιήσει·ὅπερἔδειδεῖξαι. shown (to be) equal to the (sum of the) same three (angles).
And things equal to the same thing are also equal
to one another [C.N. 1]. Therefore, (the sum of) CBE
and EBD is also equal to (the sum of) DBA and ABC.
But, (the sum of) CBE and EBD is two right-angles.
Thus, (the sum of) ABD and ABC is also equal to two
right-angles.
Thus, if a straight-line stood on a(nother) straightline
makes angles, it will certainly either make two rightangles,
or (angles whose sum is) equal to two rightangles.
(Which is) the very thing it was required to show.
18

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
. Proposition 14
᾿Εὰν πρός τινι εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ δύο If two straight-lines, not lying on the same side, make
εὐθεῖαιμὴἐπὶτὰαὐτὰμέρηκείμεναιτὰςἐφεξῆςγωνίας adjacent angles (whose sum is) equal to two right-angles
δυσὶνὀρθαῖςἴσαςποιῶσιν,ἐπ᾿εὐθείαςἔσονταιἀλλήλαιςαἱ with some straight-line, at a point on it, then the two
εὐθεῖαι.
straight-lines will be straight-on (with respect) to one another.
Α
Ε
A
E
Γ Β ∆
C
B
D
ΠρὸςγάρτινιεὐθείᾳτῇΑΒκαὶτῷπρὸςαὐτῇσημείῳ For let two straight-lines BC and BD, not lying on the
τῷΒδύοεὐθεῖαιαἱΒΓ,ΒΔμὴἐπὶτὰαὐτὰμέρηκείμεναι same side, make adjacent angles ABC and ABD (whose
τὰςἐφεξῆςγωνίαςτὰςὑπὸΑΒΓ,ΑΒΔδύοὀρθαῖςἴσας sum is) equal to two right-angles with some straight-line
ποιείτωσαν·λέγω,ὅτιἐπ᾿εὐθείαςἐστὶτῇΓΒἡΒΔ. AB, at the point B on it. I say that BD is straight-on with
ΕἰγὰρμήἐστιτῇΒΓἐπ᾿εὐθείαςἡΒΔ,ἔστωτῇΓΒ respect to CB.
ἐπ᾿εὐθείαςἡΒΕ.
For if BD is not straight-on to BC then let BE be
᾿ΕπεὶοὖνεὐθεῖαἡΑΒἐπ᾿εὐθεῖαντὴνΓΒΕἐφέστηκεν, straight-on to CB.
αἱἄραὑπὸΑΒΓ,ΑΒΕγωνίαιδύοὀρθαῖςἴσαιεἰσίν·εἰσὶδὲ Therefore, since the straight-line AB stands on the
καὶαἱὑπὸΑΒΓ,ΑΒΔδύοὀρθαῖςἴσαι·αἱἄραὑπὸΓΒΑ, straight-line CBE, the (sum of the) angles ABC and
ΑΒΕταῖςὑπὸΓΒΑ,ΑΒΔἴσαιεἰσίν. κοινὴἀφῃρήσθωἡ ABE is thus equal to two right-angles [Prop. 1.13]. But
ὑπὸΓΒΑ·λοιπὴἄραἡὑπὸΑΒΕλοιπῇτῇὑπὸΑΒΔἐστιν (the sum of) ABC and ABD is also equal to two right-
ἴση,ἡἐλάσσωντῇμείζονι·ὅπερἐστὶνἀδύνατον. οὐκἄρα angles. Thus, (the sum of angles) CBA and ABE is equal
ἐπ᾿εὐθείαςἐστὶνἡΒΕτῇΓΒ.ὁμοίωςδὴδείξομεν,ὅτιοὐδὲ to (the sum of angles) CBA and ABD [C.N. 1]. Let (an-
ἄλλητιςπλὴντῆςΒΔ·ἐπ᾿εὐθείαςἄραἐστὶνἡΓΒτῇΒΔ. gle) CBA have been subtracted from both. Thus, the re-
᾿Εὰν ἄρα πρός τινι εὐθείᾳ καὶ τῷ πρὸς αὐτῇσημείῳ mainder ABE is equal to the remainder ABD [C.N. 3],
δύοεὐθεῖαιμὴἐπὶαὐτὰμέρηκείμεναιτὰςἐφεξῆςγωνίας the lesser to the greater. The very thing is impossible.
δυσὶνὀρθαῖςἴσαςποιῶσιν,ἐπ᾿εὐθείαςἔσονταιἀλλήλαιςαἱ Thus, BE is not straight-on with respect to CB. Simiεὐθεῖαι·ὅπερἔδειδεῖξαι.
larly, we can show that neither (is) any other (straightline)
than BD. Thus, CB is straight-on with respect to
BD.
Thus, if two straight-lines, not lying on the same side,
make adjacent angles (whose sum is) equal to two rightangles
with some straight-line, at a point on it, then the
two straight-lines will be straight-on (with respect) to
one another. (Which is) the very thing it was required
to show.
. Proposition 15
᾿Εὰνδύοεὐθεῖαιτέμνωσινἀλλήλας,τὰςκατὰκορυφὴν If two straight-lines cut one another then they make
γωνίαςἴσαςἀλλήλαιςποιοῦσιν.
the vertically opposite angles equal to one another.
19

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ΔύογὰρεὐθεῖαιαἱΑΒ,ΓΔτεμνέτωσανἀλλήλαςκατὰ For let the two straight-lines AB and CD cut one anτὸΕσημεῖον·λέγω,ὅτιἴσηἐστὶνἡμὲνὑπὸΑΕΓγωνίατῇ
other at the point E. I say that angle AEC is equal to
ὑπὸΔΕΒ,ἡδὲὑπὸΓΕΒτῇὑπὸΑΕΔ.
(angle) DEB, and (angle) CEB to (angle) AED.
Α
A
∆
Ε
Γ
D
E
C
Β
᾿ΕπεὶγὰρεὐθεῖαἡΑΕἐπ᾿εὐθεῖαντὴνΓΔἐφέστηκε For since the straight-line AE stands on the straightγωνίαςποιοῦσατὰςὑπὸΓΕΑ,ΑΕΔ,αἱἄραὑπὸΓΕΑ,ΑΕΔ
line CD, making the angles CEA and AED, the (sum
γωνίαιδυσὶνὀρθαῖςἴσαιεἰσίν.πάλιν,ἐπεὶεὐθεῖαἡΔΕἐπ᾿ of the) angles CEA and AED is thus equal to two rightεὐθεῖαντὴνΑΒἐφέστηκεγωνίαςποιοῦσατὰςὑπὸΑΕΔ,
angles [Prop. 1.13]. Again, since the straight-line DE
ΔΕΒ,αἱἄραὑπὸΑΕΔ,ΔΕΒγωνίαιδυσὶνὀρθαῖςἴσαιεἰσίν. stands on the straight-line AB, making the angles AED
ἐδείχθησανδὲκαὶαἱὑπὸΓΕΑ,ΑΕΔδυσὶνὀρθαῖςἴσαι·αἱ and DEB, the (sum of the) angles AED and DEB is
ἄραὑπὸΓΕΑ,ΑΕΔταῖςὑπὸΑΕΔ,ΔΕΒἴσαιεἰσίν.κοινὴ thus equal to two right-angles [Prop. 1.13]. But (the sum
ἀφῃρήσθωἡὑπὸΑΕΔ·λοιπὴἄραἡὑπὸΓΕΑλοιπῇτῇὑπὸ of) CEA and AED was also shown (to be) equal to two
ΒΕΔἴσηἐστίν·ὁμοίωςδὴδειχθήσεται,ὅτικαὶαἱὑπὸΓΕΒ, right-angles. Thus, (the sum of) CEA and AED is equal
ΔΕΑἴσαιεἰσίν.
to (the sum of) AED and DEB [C.N. 1]. Let AED have
᾿Εὰνἄραδύοεὐθεῖαιτέμνωσινἀλλήλας,τὰςκατὰκο- been subtracted from both. Thus, the remainder CEA is
ρυφὴνγωνίαςἴσαςἀλλήλαιςποιοῦσιν·ὅπερἔδειδεῖξαι. equal to the remainder BED [C.N. 3]. Similarly, it can
be shown that CEB and DEA are also equal.
Thus, if two straight-lines cut one another then they
make the vertically opposite angles equal to one another.
(Which is) the very thing it was required to show.
¦. Proposition 16
Παντὸςτριγώνουμιᾶςτῶνπλευρῶνπροσεκβληθείσης For any triangle, when one of the sides is produced,
ἡἐκτὸςγωνίαἑκατέραςτῶνἐντὸςκαὶἀπεναντίονγωνιῶν the external angle is greater than each of the internal and
μείζωνἐστίν.
opposite angles.
῎ΕστωτρίγωνοντὸΑΒΓ,καὶπροσεκβεβλήσθωαὐτοῦ Let ABC be a triangle, and let one of its sides BC
μίαπλευρὰἡΒΓἐπὶτὸΔ·λὲγω,ὅτιἡἐκτὸςγωνίαἡὑπὸ have been produced to D. I say that the external angle
ΑΓΔμείζωνἐστὶνἑκατέραςτῶνἐντὸςκαὶἀπεναντίοντῶν ACD is greater than each of the internal and opposite
ὑπὸΓΒΑ,ΒΑΓγωνιῶν.
angles, CBA and BAC.
ΤετμήσθωἡΑΓδίχακατὰτὸΕ,καὶἐπιζευχθεῖσαἡΒΕ Let the (straight-line) AC have been cut in half at
ἐκβεβλήσθωἐπ᾿εὐθείαςἐπὶτὸΖ,καὶκείσθωτῇΒΕἴσηἡ (point) E [Prop. 1.10]. And BE being joined, let it have
ΕΖ,καὶἐπεζεύχθωἡΖΓ,καὶδιήχθωἡΑΓἐπὶτὸΗ. been produced in a straight-line to (point) F . † And let
᾿ΕπεὶοὖνἴσηἐστὶνἡμὲνΑΕτῇΕΓ,ἡδὲΒΕτῇΕΖ,δύο EF be made equal to BE [Prop. 1.3], and let FC have
δὴαἱΑΕ,ΕΒδυσὶταῖςΓΕ,ΕΖἴσαιεἰσὶνἑκατέραἑκατέρᾳ· been joined, and let AC have been drawn through to
καὶγωνίαἡὑπὸΑΕΒγωνίᾳτῇὑπὸΖΕΓἴσηἐστίν·κατὰ (point) G.
κορυφὴνγάρ·βάσιςἄραἡΑΒβάσειτῇΖΓἴσηἐστίν,καὶτὸ Therefore, since AE is equal to EC, and BE to EF ,
ΑΒΕτρίγωνοντῷΖΕΓτριγώνῳἐστὶνἴσον,καὶαἱλοιπαὶ the two (straight-lines) AE, EB are equal to the two
B
20

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
γωνίαιταῖςλοιπαῖςγωνίαιςἴσαιεἰσὶνἑκατέραἑκατέρᾳ,ὑφ᾿ (straight-lines) CE, EF , respectively. Also, angle AEB
ἃςαἱἴσαιπλευραὶὑποτείνουσιν·ἴσηἄραἐστὶνἡὑπὸΒΑΕ is equal to angle FEC, for (they are) vertically opposite
τῇὑπὸΕΓΖ.μείζωνδέἐστινἡὑπὸΕΓΔτῆςὑπὸΕΓΖ· [Prop. 1.15]. Thus, the base AB is equal to the base FC,
μείζωνἄραἡὑπὸΑΓΔτῆςὑπὸΒΑΕ.῾ΟμοίωςδὴτῆςΒΓ and the triangle ABE is equal to the triangle FEC, and
τετμημένηςδίχαδειχθήσεταικαὶἡὑπὸΒΓΗ,τουτέστινἡ the remaining angles subtended by the equal sides are
ὑπὸΑΓΔ,μείζωνκαὶτῆςὑπὸΑΒΓ. equal to the corresponding remaining angles [Prop. 1.4].
Thus, BAE is equal to ECF . But ECD is greater than
ECF . Thus, ACD is greater than BAE. Similarly, by
having cut BC in half, it can be shown (that) BCG—that
is to say, ACD—(is) also greater than ABC.
Α Ζ
A
F
Ε
E
Β
Γ
∆
B
C
D
Η
G
Παντὸς ἄρα τριγώνου μιᾶς τῶν πλευρῶν προσεκ- Thus, for any triangle, when one of the sides is proβληθείσης
ἡ ἐκτὸς γωνία ἑκατέρας τῶν ἐντὸς καὶ ἀπε- duced, the external angle is greater than each of the inναντίονγωνιῶνμείζωνἐστίν·ὅπερἔδειδεῖξαι.
ternal and opposite angles. (Which is) the very thing it
was required to show.
† The implicit assumption that the point F lies in the interior of the angle ABC should be counted as an additional postulate.
Þ. Proposition 17
Παντὸςτριγώνουαἱδύογωνίαιδύοὀρθῶνἐλάσσονές For any triangle, (the sum of) two angles taken toεἰσιπάντῇμεταλαμβανόμεναι.
gether in any (possible way) is less than two right-angles.
Α
A
Β
Γ
∆
῎ΕστωτρίγωνοντὸΑΒΓ·λέγω,ὅτιτοῦΑΒΓτριγώνου Let ABC be a triangle. I say that (the sum of) two
αἱδύογωνίαιδύοὀρθῶνἐλάττονέςεἰσιπάντῃμεταλαμ- angles of triangle ABC taken together in any (possible
βανόμεναι.
way) is less than two right-angles.
B
C
D
21

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
᾿ΕκβεβλήσθωγὰρἡΒΓἐπὶτὸΔ. For let BC have been produced to D.
ΚαὶἐπεὶτριγώνουτοῦΑΒΓἐκτόςἐστιγωνίαἡὑπὸ And since the angle ACD is external to triangle ABC,
ΑΓΔ,μείζωνἐστὶτῆςἐντὸςκαὶἀπεναντίοντῆςὑπὸΑΒΓ. it is greater than the internal and opposite angle ABC
κοινὴπροσκείσθωἡὑπὸΑΓΒ·αἱἄραὑπὸΑΓΔ,ΑΓΒτῶν [Prop. 1.16]. Let ACB have been added to both. Thus,
ὑπὸΑΒΓ,ΒΓΑμείζονέςεἰσιν. ἀλλ᾿αἱὑπὸΑΓΔ,ΑΓΒ the (sum of the angles) ACD and ACB is greater than
δύοὀρθαῖςἴσαιεἰσίν· αἱἄραὑπὸΑΒΓ,ΒΓΑδύοὀρθῶν the (sum of the angles) ABC and BCA. But, (the sum of)
ἐλάσσονέςεἰσιν.ὁμοίωςδὴδείξομεν,ὅτικαὶαἱὑπὸΒΑΓ, ACD and ACB is equal to two right-angles [Prop. 1.13].
ΑΓΒδύοὀρθῶνἐλάσσονέςεἰσικαὶἔτιαἱὑπὸΓΑΒ,ΑΒΓ. Thus, (the sum of) ABC and BCA is less than two right-
Παντὸςἄρατριγώνουαἱδύογωνίαιδύοὀρθῶνἐλάσς- angles. Similarly, we can show that (the sum of) BAC
ονέςεἰσιπάντῇμεταλαμβανόμεναι·ὅπερἔδειδεῖξαι. and ACB is also less than two right-angles, and further
(that the sum of) CAB and ABC (is less than two rightangles).
Thus, for any triangle, (the sum of) two angles taken
together in any (possible way) is less than two rightangles.
(Which is) the very thing it was required to show.
. Proposition 18
Παντὸςτριγώνουἡμείζωνπλευρὰτὴνμείζοναγωνίαν In any triangle, the greater side subtends the greater
ὑποτείνει.
angle.
Α
A
∆
D
Β
Γ
῎Εστω γὰρ τρίγωνον τὸ ΑΒΓ μείζονα ἔχον τὴν ΑΓ For let ABC be a triangle having side AC greater than
πλευρὰντῆςΑΒ·λέγω,ὅτικαὶγωνίαἡὑπὸΑΒΓμείζων AB. I say that angle ABC is also greater than BCA.
ἐστὶτῆςὑπὸΒΓΑ·
For since AC is greater than AB, let AD be made
᾿ΕπεὶγὰρμείζωνἐστὶνἡΑΓτῆςΑΒ,κείσθωτῇΑΒἴση equal to AB [Prop. 1.3], and let BD have been joined.
ἡΑΔ,καὶἐπεζεύχθωἡΒΔ.
And since angle ADB is external to triangle BCD, it
ΚαὶἐπεὶτριγώνουτοῦΒΓΔἐκτόςἐστιγωνίαἡὑπὸ is greater than the internal and opposite (angle) DCB
ΑΔΒ,μείζωνἐστὶτῆςἐντὸςκαὶἀπεναντίοντῆςὑπὸΔΓΒ· [Prop. 1.16]. But ADB (is) equal to ABD, since side
ἴσηδὲἡὑπὸΑΔΒτῇὑπὸΑΒΔ,ἐπεὶκαὶπλευρὰἡΑΒτῇ AB is also equal to side AD [Prop. 1.5]. Thus, ABD is
ΑΔἐστινἴση·μείζωνἄρακαὶἡὑπὸΑΒΔτῆςὑπὸΑΓΒ· also greater than ACB. Thus, ABC is much greater than
πολλῷἄραἡὑπὸΑΒΓμείζωνἐστὶτῆςὑπὸΑΓΒ. ACB.
Παντὸς ἄρα τριγώνου ἡ μείζων πλευρὰ τὴν μείζονα Thus, in any triangle, the greater side subtends the
γωνίανὑποτείνει·ὅπερἔδειδεῖξαι.
greater angle. (Which is) the very thing it was required
to show.
. Proposition 19
Παντὸς τριγώνου ὑπὸ τὴν μείζονα γωνίαν ἡ μείζων In any triangle, the greater angle is subtended by the
πλευρὰὑποτείνει.
greater side.
῎ΕστωτρίγωνοντὸΑΒΓμείζοναἔχοντὴνὑπὸΑΒΓ Let ABC be a triangle having the angle ABC greater
γωνίαντῆςὑπὸΒΓΑ·λέγω,ὅτικαὶπλευρὰἡΑΓπλευρᾶς than BCA. I say that side AC is also greater than side
τῆςΑΒμείζωνἐστίν.
AB.
B
C
22

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
Εἰγὰρμή,ἤτοιἴσηἐστὶνἡΑΓτῇΑΒἢἐλάσσων·ἴση For if not, AC is certainly either equal to, or less than,
μὲνοὖνοὐκἔστινἡΑΓτῇΑΒ·ἴσηγὰρἂνἦνκαὶγωνίαἡ AB. In fact, AC is not equal to AB. For then angle ABC
ὑπὸΑΒΓτῇὑπὸΑΓΒ·οὐκἔστιδέ·οὐκἄραἴσηἐστὶνἡΑΓ would also have been equal to ACB [Prop. 1.5]. But it
τῇΑΒ.οὐδὲμὴνἐλάσσωνἐστὶνἡΑΓτῆςΑΒ·ἐλάσσων is not. Thus, AC is not equal to AB. Neither, indeed, is
γὰρἂνἦνκαὶγωνίαἡὑπὸΑΒΓτῆςὑπὸΑΓΒ·οὐκἔστι AC less than AB. For then angle ABC would also have
δέ·οὐκἄραἐλάσσωνἐστὶνἡΑΓτῆςΑΒ.ἐδείχθηδέ,ὅτι been less than ACB [Prop. 1.18]. But it is not. Thus, AC
οὐδὲἴσηἐστίν.μείζωνἄραἐστὶνἡΑΓτῆςΑΒ.
is not less than AB. But it was shown that (AC) is not
equal (to AB) either. Thus, AC is greater than AB.
Α
A
Β
B
Γ
Παντὸςἄρατριγώνουὑπὸτὴνμείζοναγωνίανἡμείζων Thus, in any triangle, the greater angle is subtended
πλευρὰὑποτείνει·ὅπερἔδειδεῖξαι.
by the greater side. (Which is) the very thing it was required
to show.
. Proposition 20
Παντὸςτριγώνουαἱδύοπλευραὶτῆςλοιπῆςμείζονές In any triangle, (the sum of) two sides taken toεἰσιπάντῃμεταλαμβανόμεναι.
gether in any (possible way) is greater than the remaining
(side).
∆
D
C
Α
A
Β
Γ
῎Εστω γὰρ τρίγωνον τὸ ΑΒΓ· λέγω, ὅτι τοῦ ΑΒΓ For let ABC be a triangle. I say that in triangle ABC
τριγώνουαἱδύοπλευραὶτῆςλοιπῆςμείζονέςεἰσιπάντῃ (the sum of) two sides taken together in any (possible
μεταλαμβανόμεναι,αἱμὲνΒΑ,ΑΓτῆςΒΓ,αἱδὲΑΒ,ΒΓ way) is greater than the remaining (side). (So), (the sum
τῆςΑΓ,αἱδὲΒΓ,ΓΑτῆςΑΒ.
of) BA and AC (is greater) than BC, (the sum of) AB
B
C
23

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ΔιήχθωγὰρἡΒΑἐπὶτὸΔσημεῖον,καὶκείσθωτῇΓΑ and BC than AC, and (the sum of) BC and CA than
ἴσηἡΑΔ,καὶἐπεζεύχθωἡΔΓ.
AB.
᾿ΕπεὶοὖνἴσηἐστὶνἡΔΑτῇΑΓ,ἴσηἐστὶκαὶγωνία For let BA have been drawn through to point D, and
ἡὑπὸΑΔΓτῇὑπὸΑΓΔ·μείζωνἄραἡὑπὸΒΓΔτῆςὑπὸ let AD be made equal to CA [Prop. 1.3], and let DC
ΑΔΓ·καὶἐπεὶτρίγωνόνἐστιτὸΔΓΒμείζοναἔχοντὴνὑπὸ have been joined.
ΒΓΔγωνίαντῆςὑπὸΒΔΓ,ὑπὸδὲτὴνμείζοναγωνίανἡ Therefore, since DA is equal to AC, the angle ADC
μείζωνπλευρὰὑποτείνει,ἡΔΒἄρατῆςΒΓἐστιμείζων.ἴση is also equal to ACD [Prop. 1.5]. Thus, BCD is greater
δὲἡΔΑτῇΑΓ·μείζονεςἄρααἱΒΑ,ΑΓτῆςΒΓ·ὁμοίως than ADC. And since DCB is a triangle having the angle
δὴδείξομεν,ὅτικαὶαἱμὲνΑΒ,ΒΓτῆςΓΑμείζονέςεἰσιν, BCD greater than BDC, and the greater angle subtends
αἱδὲΒΓ,ΓΑτῆςΑΒ.
the greater side [Prop. 1.19], DB is thus greater than
Παντὸς ἄρα τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς BC. But DA is equal to AC. Thus, (the sum of) BA and
μείζονέςεἰσιπάντῃμεταλαμβανόμεναι·ὅπερἔδειδεῖξαι. AC is greater than BC. Similarly, we can show that (the
sum of) AB and BC is also greater than CA, and (the
sum of) BC and CA than AB.
Thus, in any triangle, (the sum of) two sides taken together
in any (possible way) is greater than the remaining
(side). (Which is) the very thing it was required to show.
. Proposition 21
᾿Εὰντριγώνουἐπὶμιᾶςτῶνπλευρῶνἀπὸτῶνπεράτων If two internal straight-lines are constructed on one
δύοεὐθεῖαιἐντὸςσυσταθῶσιν,αἱσυσταθεῖσαιτῶνλοιπῶν of the sides of a triangle, from its ends, the constructed
τοῦτριγώνουδύοπλευρῶνἐλάττονεςμὲνἔσονται,μείζονα (straight-lines) will be less than the two remaining sides
δὲγωνίανπεριέξουσιν.
of the triangle, but will encompass a greater angle.
Α
∆
Ε
A
D
E
Β
Γ
ΤριγώνουγὰρτοῦΑΒΓἐπὶμιᾶςτῶνπλευρῶντῆςΒΓ For let the two internal straight-lines BD and DC
ἀπὸτῶνπεράτωντῶνΒ,Γδύοεὐθεῖαιἐντὸςσυνεστάτωσαν have been constructed on one of the sides BC of the triαἱΒΔ,ΔΓ·λέγω,ὅτιαἱΒΔ,ΔΓτῶνλοιπῶντοῦτριγώνου
angle ABC, from its ends B and C (respectively). I say
δύοπλευρῶντῶνΒΑ,ΑΓἐλάσσονεςμένεἰσιν,μείζοναδὲ that BD and DC are less than the (sum of the) two reγωνίανπεριέχουσιτὴνὑπὸΒΔΓτῆςὑπὸΒΑΓ.
maining sides of the triangle BA and AC, but encompass
ΔιήχθωγὰρἡΒΔἐπὶτὸΕ.καὶἐπεὶπαντὸςτριγώνου an angle BDC greater than BAC.
αἱδύοπλευραὶτῆςλοιπῆςμείζονέςεἰσιν, τοῦΑΒΕ ἄρα For let BD have been drawn through to E. And since
τριγώνου αἱ δύο πλευραὶ αἱ ΑΒ, ΑΕ τῆς ΒΕ μείζονές in any triangle (the sum of any) two sides is greater than
εἰσιν· κοινὴπροσκείσθωἡΕΓ·αἱἄραΒΑ,ΑΓτῶνΒΕ, the remaining (side) [Prop. 1.20], in triangle ABE the
ΕΓμείζονέςεἰσιν. πάλιν,ἐπεὶτοῦΓΕΔτριγώνουαἱδύο (sum of the) two sides AB and AE is thus greater than
πλευραὶαἱΓΕ,ΕΔτῆςΓΔμείζονέςεἰσιν,κοινὴπροσκείσθω BE. Let EC have been added to both. Thus, (the sum
ἡΔΒ·αἱΓΕ,ΕΒἄρατῶνΓΔ,ΔΒμείζονέςεἰσιν. ἀλλὰ of) BA and AC is greater than (the sum of) BE and EC.
τῶνΒΕ,ΕΓμείζονεςἐδείχθησαναἱΒΑ,ΑΓ·πολλῷἄρααἱ Again, since in triangle CED the (sum of the) two sides
ΒΑ,ΑΓτῶνΒΔ,ΔΓμείζονέςεἰσιν.
CE and ED is greater than CD, let DB have been added
Πάλιν,ἐπεὶπαντὸςτριγώνουἡἐκτὸςγωνίατῆςἐντὸς to both. Thus, (the sum of) CE and EB is greater than
καὶ ἀπεναντίον μείζων ἐστίν, τοῦ ΓΔΕ ἄρα τριγώνου ἡ (the sum of) CD and DB. But, (the sum of) BA and
ἐκτὸςγωνίαἡὑπὸΒΔΓμείζωνἐστὶτῆςὑπὸΓΕΔ.διὰ AC was shown (to be) greater than (the sum of) BE and
ταὐτὰτοίνυνκαὶτοῦΑΒΕτριγώνουἡἐκτὸςγωνίαἡὑπὸ EC. Thus, (the sum of) BA and AC is much greater than
B
C
24

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ΓΕΒμείζωνἐστὶτῆςὑπὸΒΑΓ.ἀλλὰτῆςὑπὸΓΕΒμείζων (the sum of) BD and DC.
ἐδείχθηἡὑπὸΒΔΓ·πολλῷἄραἡὑπὸΒΔΓμείζωνἐστὶ Again, since in any triangle the external angle is
τῆςὑπὸΒΑΓ.
greater than the internal and opposite (angles) [Prop.
᾿Εὰν ἄρα τριγώνου ἐπὶ μιᾶς τῶν πλευρῶν ἀπὸ τῶν 1.16], in triangle CDE the external angle BDC is thus
περάτωνδύοεὐθεῖαιἐντὸςσυσταθῶσιν,αἱσυσταθεῖσαιτῶν greater than CED. Accordingly, for the same (reason),
λοιπῶν τοῦ τριγώνου δύο πλευρῶν ἐλάττονες μέν εἰσιν, the external angle CEB of the triangle ABE is also
μείζοναδὲγωνίανπεριέχουσιν·ὅπερἔδειδεῖξαι. greater than BAC. But, BDC was shown (to be) greater
than CEB. Thus, BDC is much greater than BAC.
Thus, if two internal straight-lines are constructed on
one of the sides of a triangle, from its ends, the constructed
(straight-lines) are less than the two remaining
sides of the triangle, but encompass a greater angle.
(Which is) the very thing it was required to show.
. Proposition 22
᾿Εκ τριῶν εὐθειῶν, αἵ εἰσιν ἴσαι τρισὶ ταῖς δοθείσαις To construct a triangle from three straight-lines which
[εὐθείαις],τρίγωνονσυστήσασθαι·δεῖδὲτὰςδύοτῆςλοιπῆς are equal to three given [straight-lines]. It is necessary
μείζοναςεἶναιπάντῃμεταλαμβανομένας[διὰτὸκαὶπαντὸς for (the sum of) two (of the straight-lines) taken together
τριγώνουτὰςδύοπλευρὰςτῆςλοιπῆςμείζοναςεἶναιπάντῃ in any (possible way) to be greater than the remaining
μεταλαμβανομένας].
(one), [on account of the (fact that) in any triangle (the
sum of) two sides taken together in any (possible way) is
greater than the remaining (one) [Prop. 1.20] ].
Α
A
Β
B
Γ
C
Κ
K
∆
Ζ
Η
Θ
Ε
D
F
G
H
E
Λ
L
῎ΕστωσαναἱδοθεῖσαιτρεῖςεὐθεῖαιαἱΑ,Β,Γ,ὧναἱ Let A, B, and C be the three given straight-lines, of
δύοτῆςλοιπῆςμείζονεςἔστωσανπάντῃμεταλαμβανόμεναι, which let (the sum of) two taken together in any (possible
αἱμὲνΑ,ΒτῆςΓ,αἱδὲΑ,ΓτῆςΒ,καὶἔτιαἱΒ,ΓτῆςΑ· way) be greater than the remaining (one). (Thus), (the
δεῖδὴἐκτῶνἴσωνταῖςΑ,Β,Γτρίγωνονσυστήσασθαι. sum of) A and B (is greater) than C, (the sum of) A and
᾿ΕκκείσθωτιςεὐθεῖαἡΔΕπεπερασμένημὲνκατὰτὸ C than B, and also (the sum of) B and C than A. So
ΔἄπειροςδὲκατὰτὸΕ,καὶκείσθωτῇμὲνΑἴσηἡΔΖ, it is required to construct a triangle from (straight-lines)
τῇδὲΒἴσηἡΖΗ,τῇδὲΓἴσηἡΗΘ·καὶκέντρῳμὲντῷ equal to A, B, and C.
Ζ,διαστήματιδὲτῷΖΔκύκλοςγεγράφθωὁΔΚΛ·πάλιν Let some straight-line DE be set out, terminated at D,
κέντρῳμὲντῷΗ,διαστήματιδὲτῷΗΘκύκλοςγεγράφθω and infinite in the direction of E. And let DF made equal
ὁΚΛΘ,καὶἐπεζεύχθωσαναἱΚΖ,ΚΗ·λέγω,ὅτιἐκτριῶν to A, and FG equal to B, and GH equal to C [Prop. 1.3].
εὐθειῶντῶνἴσωνταῖςΑ,Β,Γτρίγωνονσυνέσταταιτὸ And let the circle DKL have been drawn with center F
ΚΖΗ.
and radius FD. Again, let the circle KLH have been
᾿ΕπεὶγὰρτὸΖσημεῖονκέντρονἐστὶτοῦΔΚΛκύκλου, drawn with center G and radius GH. And let KF and
ἴσηἐστὶνἡΖΔτῇΖΚ·ἀλλὰἡΖΔτῇΑἐστινἴση. καὶἡ KG have been joined. I say that the triangle KFG has
25

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ΚΖἄρατῇΑἐστινἴση. πάλιν,ἐπεὶτὸΗσημεῖονκέντρον been constructed from three straight-lines equal to A, B,
ἐστὶτοῦΛΚΘκύκλου,ἴσηἐστὶνἡΗΘτῇΗΚ·ἀλλὰἡΗΘ and C.
τῇΓἐστινἴση·καὶἡΚΗἄρατῇΓἐστινἴση.ἐστὶδὲκαὶἡ For since point F is the center of the circle DKL, FD
ΖΗτῇΒἴση·αἱτρεῖςἄραεὐθεῖαιαἱΚΖ,ΖΗ,ΗΚτρισὶταῖς is equal to FK. But, FD is equal to A. Thus, KF is also
Α,Β,Γἴσαιεἰσίν.
equal to A. Again, since point G is the center of the circle
᾿Εκ τριῶν ἄρα εὐθειῶν τῶν ΚΖ, ΖΗ, ΗΚ, αἵ εἰσιν LKH, GH is equal to GK. But, GH is equal to C. Thus,
ἴσαιτρισὶταῖςδοθείσαιςεὐθείαιςταῖςΑ,Β,Γ,τρίγωνον KG is also equal to C. And FG is also equal to B. Thus,
συνέσταταιτὸΚΖΗ·ὅπερἔδειποιῆσαι. the three straight-lines KF , FG, and GK are equal to A,
B, and C (respectively).
Thus, the triangle KFG has been constructed from
the three straight-lines KF , FG, and GK, which are
equal to the three given straight-lines A, B, and C (respectively).
(Which is) the very thing it was required to
do.
. Proposition 23
Πρὸς τῇ δοθείσῃ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ To construct a rectilinear angle equal to a given rectiτῇδοθείσῃγωνίᾳεὐθυγράμμῳἴσηνγωνίανεὐθύγραμμον
linear angle at a (given) point on a given straight-line.
συστήσασθαι.
∆
D
Γ
C
Ε
E
Ζ
F
Α
Η Β
A
G B
῎ΕστωἡμὲνδοθεῖσαεὐθεῖαἡΑΒ,τὸδὲπρὸςαὐτῇ Let AB be the given straight-line, A the (given) point
σημεῖοντὸΑ,ἡδὲδοθεῖσαγωνίαεὐθύγραμμοςἡὑπὸΔΓΕ· on it, and DCE the given rectilinear angle. So it is reδεῖδὴπρὸςτῇδοθείσῃεὐθείᾳτῇΑΒκαὶτῷπρὸςαὐτῇ
quired to construct a rectilinear angle equal to the given
σημείῳτῷΑτῇδοθείσῃγωνίᾳεὐθυγράμμῳτῇὑπὸΔΓΕ rectilinear angle DCE at the (given) point A on the given
ἴσηνγωνίανεὐθύγραμμονσυστήσασθαι.
straight-line AB.
Εἰλήφθωἐφ᾿ἑκατέραςτῶνΓΔ,ΓΕτυχόντασημεῖατὰ Let the points D and E have been taken at random
Δ,Ε,καὶἐπεζεύχθωἡΔΕ·καὶἐκτριῶνεὐθειῶν,αἵεἰσιν on each of the (straight-lines) CD and CE (respectively),
ἴσαιτρισὶταῖςΓΔ,ΔΕ,ΓΕ,τρίγωνονσυνεστάτωτὸΑΖΗ, and let DE have been joined. And let the triangle AFG
ὥστεἴσηνεἶναιτὴνμὲνΓΔτῇΑΖ,τὴνδὲΓΕτῇΑΗ,καὶ have been constructed from three straight-lines which are
ἔτιτὴνΔΕτῇΖΗ. equal to CD, DE, and CE, such that CD is equal to AF ,
᾿ΕπεὶοὖνδύοαἱΔΓ,ΓΕδύοταῖςΖΑ,ΑΗἴσαιεἰσὶν CE to AG, and further DE to FG [Prop. 1.22].
ἑκατέραἑκατέρᾳ,καὶβάσιςἡΔΕβάσειτῇΖΗἴση,γωνία Therefore, since the two (straight-lines) DC, CE are
ἄραἡὑπὸΔΓΕγωνίᾳτῇὑπὸΖΑΗἐστινἴση.
equal to the two (straight-lines) FA, AG, respectively,
ΠρὸςἄρατῇδοθείσῃεὐθείᾳτῇΑΒκαὶτῷπρὸςαὐτῇ and the base DE is equal to the base FG, the angle DCE
σημείῳτῷΑτῇδοθείσῃγωνίᾳεὐθυγράμμῳτῇὑπὸΔΓΕ is thus equal to the angle FAG [Prop. 1.8].
ἴσηγωνίαεὐθύγραμμοςσυνέσταταιἡὑπὸΖΑΗ·ὅπερἔδει Thus, the rectilinear angle FAG, equal to the given
ποιῆσαι.
rectilinear angle DCE, has been constructed at the
(given) point A on the given straight-line AB. (Which
26

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
is) the very thing it was required to do.
. Proposition 24
᾿Εὰνδύοτρίγωνατὰςδύοπλευρὰς[ταῖς]δύοπλευραῖς If two triangles have two sides equal to two sides, re-
ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ, τὴν δὲ γωνίαν τῆς γωνίας spectively, but (one) has the angle encompassed by the
μείζοναἔχῃτὴνὑπὸτῶνἴσωνεὐθειῶνπεριεχομένην,καὶ equal straight-lines greater than the (corresponding) anτὴνβάσιντῆςβάσεωςμείζοναἕξει.
gle (in the other), then (the former triangle) will also
have a base greater than the base (of the latter).
Α ∆
A D
Γ
Β
Η
Ζ
Ε
῎Εστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο πλευρὰς Let ABC and DEF be two triangles having the two
τὰςΑΒ,ΑΓταῖςδύοπλευραῖςταῖςΔΕ,ΔΖἴσαςἔχοντα sides AB and AC equal to the two sides DE and DF ,
ἑκατέρανἑκατέρᾳ,τὴνμὲνΑΒτῇΔΕτὴνδὲΑΓτῇΔΖ,ἡ respectively. (That is), AB (equal) to DE, and AC to
δὲπρὸςτῷΑγωνίατῆςπρὸςτῷΔγωνίαςμείζωνἔστω· DF . Let them also have the angle at A greater than the
λέγω,ὅτικαὶβάσιςἡΒΓβάσεωςτῆςΕΖμείζωνἐστίν. angle at D. I say that the base BC is also greater than
᾿Επεὶ γὰρ μείζων ἡ ὑπὸ ΒΑΓ γωνία τῆς ὑπὸ ΕΔΖ the base EF .
γωνίας,συνεστάτωπρὸςτῇΔΕεὐθείᾳκαὶτῷπρὸςαὐτῇ For since angle BAC is greater than angle EDF ,
σημείῳτῷΔτῇὑπὸΒΑΓγωνίᾳἴσηἡὑπὸΕΔΗ,καὶκείσθω let (angle) EDG, equal to angle BAC, have been
ὁποτέρᾳτῶνΑΓ,ΔΖἴσηἡΔΗ,καὶἐπεζεύχθωσαναἱΕΗ, constructed at the point D on the straight-line DE
ΖΗ. [Prop. 1.23]. And let DG be made equal to either of
᾿ΕπεὶοὖνἴσηἐστὶνἡμὲνΑΒτῇΔΕ,ἡδὲΑΓτῇΔΗ, AC or DF [Prop. 1.3], and let EG and FG have been
δύοδὴαἱΒΑ,ΑΓδυσὶταῖςΕΔ,ΔΗἴσαιεἰσὶνἑκατέρα joined.
ἑκατέρᾳ·καὶγωνίαἡὑπὸΒΑΓγωνίᾳτῇὑπὸΕΔΗἴση· Therefore, since AB is equal to DE and AC to DG,
βάσιςἄραἡΒΓβάσειτῇΕΗἐστινἴση. πάλιν,ἐπεὶἴση the two (straight-lines) BA, AC are equal to the two
ἐστὶνἡΔΖτῇΔΗ,ἴσηἐστὶκαὶἡὑπὸΔΗΖγωνίατῇὑπὸ (straight-lines) ED, DG, respectively. Also the angle
ΔΖΗ·μείζωνἄραἡὑπὸΔΖΗτῆςὑπὸΕΗΖ·πολλῷἄρα BAC is equal to the angle EDG. Thus, the base BC
μείζωνἐστὶνἡὑπὸΕΖΗτῆςὑπὸΕΗΖ.καὶἐπεὶτρίγωνόν is equal to the base EG [Prop. 1.4]. Again, since DF
ἐστιτὸΕΖΗμείζοναἔχοντὴνὑπὸΕΖΗγωνίαντῆςὑπὸ is equal to DG, angle DGF is also equal to angle DFG
ΕΗΖ,ὑπὸδὲτὴνμείζοναγωνίανἡμείζωνπλευρὰὑποτείνει, [Prop. 1.5]. Thus, DFG (is) greater than EGF . Thus,
μείζωνἄρακαὶπλευρὰἡΕΗτῆςΕΖ.ἴσηδὲἡΕΗτῇΒΓ· EFG is much greater than EGF . And since triangle
μείζωνἄρακαὶἡΒΓτῆςΕΖ.
EFG has angle EFG greater than EGF , and the greater
᾿Εὰνἄραδύοτρίγωνατὰςδύοπλευρὰςδυσὶπλευραῖς angle is subtended by the greater side [Prop. 1.19], side
ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ, τὴν δὲ γωνίαν τῆς γωνίας EG (is) thus also greater than EF . But EG (is) equal to
μείζοναἔχῃτὴνὑπὸτῶνἴσωνεὐθειῶνπεριεχομένην,καὶ BC. Thus, BC (is) also greater than EF .
τὴνβάσιντῆςβάσεωςμείζοναἕξει·ὅπερἔδειδεῖξαι. Thus, if two triangles have two sides equal to two
sides, respectively, but (one) has the angle encompassed
by the equal straight-lines greater than the (corresponding)
angle (in the other), then (the former triangle) will
also have a base greater than the base (of the latter).
C
B
G
F
E
27

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
(Which is) the very thing it was required to show.
. Proposition 25
᾿Εὰνδύοτρίγωνατὰςδύοπλευρὰςδυσὶπλευραῖςἴσας If two triangles have two sides equal to two sides,
ἔχῃἑκατέρανἑκατέρᾳ,τὴνδὲβάσιντῆςβάσεωςμείζονα respectively, but (one) has a base greater than the base
ἔχῃ,καὶτὴνγωνίαντῆςγωνίαςμείζοναἕξειτὴνὑπὸτῶν (of the other), then (the former triangle) will also have
ἴσωνεὐθειῶνπεριεχομένην.
the angle encompassed by the equal straight-lines greater
than the (corresponding) angle (in the latter).
Α
A
Γ
C
Β
∆
B
D
Ε
Ζ
῎Εστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο πλευρὰς Let ABC and DEF be two triangles having the two
τὰςΑΒ,ΑΓταῖςδύοπλευραῖςταῖςΔΕ,ΔΖἴσαςἔχοντα sides AB and AC equal to the two sides DE and DF ,
ἑκατέρανἑκατέρᾳ,τὴνμὲνΑΒτῇΔΕ,τὴνδὲΑΓτῇΔΖ· respectively (That is), AB (equal) to DE, and AC to DF .
βάσιςδὲἡΒΓβάσεωςτῆςΕΖμείζωνἔστω·λέγω,ὅτικαὶ And let the base BC be greater than the base EF . I say
γωνίαἡὑπὸΒΑΓγωνίαςτῆςὑπὸΕΔΖμείζωνἐστίν. that angle BAC is also greater than EDF .
Εἰγὰρμή,ἤτοιἴσηἐστὶναὐτῇἢἐλάσσων·ἴσημὲνοὖν For if not, (BAC) is certainly either equal to, or less
οὐκἔστινἡὑπὸΒΑΓτῇὑπὸΕΔΖ·ἴσηγὰρἂνἦνκαὶβάσις than, (EDF ). In fact, BAC is not equal to EDF . For
ἡΒΓβάσειτῇΕΖ·οὐκἔστιδέ. οὐκἄραἴσηἐστὶγωνίαἡ then the base BC would also have been equal to the base
ὑπὸΒΑΓτῇὑπὸΕΔΖ·οὐδὲμὴνἐλάσσωνἐστὶνἡὑπὸΒΑΓ EF [Prop. 1.4]. But it is not. Thus, angle BAC is not
τῆςὑπὸΕΔΖ·ἐλάσσωνγὰρἂνἦνκαὶβάσιςἡΒΓβάσεως equal to EDF . Neither, indeed, is BAC less than EDF .
τῆςΕΖ·οὐκἔστιδέ·οὐκἄραἐλάσσωνἐστὶνἡὑπὸΒΑΓ For then the base BC would also have been less than the
γωνίατῆςὑπὸΕΔΖ.ἐδείχθηδέ,ὅτιοὐδὲἴση·μείζωνἄρα base EF [Prop. 1.24]. But it is not. Thus, angle BAC is
ἐστὶνἡὑπὸΒΑΓτῆςὑπὸΕΔΖ.
not less than EDF . But it was shown that (BAC is) not
᾿Εὰνἄραδύοτρίγωνατὰςδύοπλευρὰςδυσὶπλευραῖς equal (to EDF ) either. Thus, BAC is greater than EDF .
ἴσας ἔχῃ ἑκατέραν ἑκάτερᾳ, τὴν δὲ βασίν τῆς βάσεως Thus, if two triangles have two sides equal to two
μείζοναἔχῃ,καὶτὴνγωνίαντῆςγωνίαςμείζοναἕξειτὴν sides, respectively, but (one) has a base greater than the
ὑπὸτῶνἴσωνεὐθειῶνπεριεχομένην·ὅπερἔδειδεῖξαι. base (of the other), then (the former triangle) will also
have the angle encompassed by the equal straight-lines
greater than the (corresponding) angle (in the latter).
(Which is) the very thing it was required to show.
¦. Proposition 26
᾿Εὰνδύοτρίγωνατὰςδύογωνίαςδυσὶγωνίαιςἴσαςἔχῃ If two triangles have two angles equal to two angles,
ἑκατέρανἑκατέρᾳκαὶμίανπλευρὰνμιᾷπλευρᾷἴσηνἤτοιτὴν respectively, and one side equal to one side—in fact, eiπρὸςταῖςἴσαιςγωνίαιςἢτὴνὑποτείνουσανὑπὸμίαντῶν
ther that by the equal angles, or that subtending one of
ἴσωνγωνιῶν,καὶτὰςλοιπὰςπλευρὰςταῖςλοιπαῖςπλευραῖς the equal angles—then (the triangles) will also have the
ἴσαςἕξει[ἑκατέρανἑκατέρᾳ]καὶτὴνλοιπὴνγωνίαντῇλοιπῇ remaining sides equal to the [corresponding] remaining
γωνίᾳ.
sides, and the remaining angle (equal) to the remaining
῎ΕστωδύοτρίγωνατὰΑΒΓ,ΔΕΖτὰςδύογωνίαςτὰς angle.
E
F
28

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ὑπὸ ΑΒΓ, ΒΓΑ δυσὶ ταῖς ὑπὸ ΔΕΖ, ΕΖΔ ἴσας ἔχοντα Let ABC and DEF be two triangles having the two
ἑκατέρανἑκατέρᾳ, τὴν μὲν ὑπὸΑΒΓ τῇ ὑπὸΔΕΖ, τὴν angles ABC and BCA equal to the two (angles) DEF
δὲὑπὸΒΓΑτῇὑπὸΕΖΔ·ἐχέτωδὲκαὶμίανπλευρὰνμιᾷ and EFD, respectively. (That is) ABC (equal) to DEF ,
πλευρᾷ ἴσην, πρότερον τὴν πρὸς ταῖς ἴσαις γωνίαις τὴν and BCA to EFD. And let them also have one side equal
ΒΓτῇΕΖ·λέγω,ὅτικαὶτὰςλοιπὰςπλευρὰςταῖςλοιπαῖς to one side. First of all, the (side) by the equal angles.
πλευραῖςἴσαςἕξειἑκατέρανἑκατέρᾳ,τὴνμὲνΑΒτῇΔΕ (That is) BC (equal) to EF . I say that they will have
τὴνδὲΑΓτῇΔΖ,καὶτὴνλοιπὴνγωνίαντῇλοιπῇγωνίᾳ, the remaining sides equal to the corresponding remainτὴνὑπὸΒΑΓτῇὑπὸΕΔΖ.
ing sides. (That is) AB (equal) to DE, and AC to DF .
And (they will have) the remaining angle (equal) to the
remaining angle. (That is) BAC (equal) to EDF .
∆
Β
Η
Α
Θ Γ Ε
Ζ
D
A
G
E
B H C
ΕἰγὰρἄνισόςἐστινἡΑΒτῇΔΕ,μίααὐτῶνμείζων For if AB is unequal to DE then one of them is
ἐστίν.ἔστωμείζωνἡΑΒ,καὶκείσθωτῇΔΕἴσηἡΒΗ,καὶ greater. Let AB be greater, and let BG be made equal
ἐπεζεύχθωἡΗΓ.
to DE [Prop. 1.3], and let GC have been joined.
᾿ΕπεὶοὖνἴσηἐστὶνἡμὲνΒΗτῇΔΕ,ἡδὲΒΓτῇΕΖ,δύο Therefore, since BG is equal to DE, and BC to EF ,
δὴαἱΒΗ,ΒΓδυσὶταῖςΔΕ,ΕΖἴσαιεἰσὶνἑκατέραἑκατέρᾳ· the two (straight-lines) GB, BC † are equal to the two
καὶγωνίαἡὑπὸΗΒΓγωνίᾳτῇὑπὸΔΕΖἴσηἐστίν·βάσις (straight-lines) DE, EF , respectively. And angle GBC is
ἄραἡΗΓβάσειτῇΔΖἴσηἐστίν,καὶτὸΗΒΓτρίγωνοντῷ equal to angle DEF . Thus, the base GC is equal to the
ΔΕΖτριγώνῳἴσονἐστίν,καὶαἱλοιπαὶγωνίαιταῖςλοιπαῖς base DF , and triangle GBC is equal to triangle DEF ,
γωνίαιςἴσαιἔσονται,ὑφ᾿ἃςαἱἴσαιπλευραὶὑποτείνουσιν· and the remaining angles subtended by the equal sides
ἴσηἄραἡὑπὸΗΓΒγωνίατῇὑπὸΔΖΕ.ἀλλὰἡὑπὸΔΖΕ will be equal to the (corresponding) remaining angles
τῇὑπὸΒΓΑὑπόκειταιἴση·καὶἡὑπὸΒΓΗἄρατῇὑπὸΒΓΑ [Prop. 1.4]. Thus, GCB (is equal) to DFE. But, DFE
ἴσηἐστίν,ἡἐλάσσωντῇμείζονι·ὅπερἀδύνατον. οὐκἄρα was assumed (to be) equal to BCA. Thus, BCG is also
ἄνισόςἐστινἡΑΒτῇΔΕ.ἴσηἄρα.ἔστιδὲκαὶἡΒΓτῇΕΖ equal to BCA, the lesser to the greater. The very thing
ἴση·δύοδὴαἱΑΒ,ΒΓδυσὶταῖςΔΕ,ΕΖἴσαιεἰσὶνἑκατέρα (is) impossible. Thus, AB is not unequal to DE. Thus,
ἑκατέρᾳ·καὶγωνίαἡὑπὸΑΒΓγωνίᾳτῇὑπὸΔΕΖἐστιν (it is) equal. And BC is also equal to EF . So the two
ἴση·βάσιςἄραἡΑΓβάσειτῇΔΖἴσηἐστίν,καὶλοιπὴγωνία (straight-lines) AB, BC are equal to the two (straight-
ἡὑπὸΒΑΓτῇλοιπῇγωνίᾳτῇὑπὸΕΔΖἴσηἐστίν. lines) DE, EF , respectively. And angle ABC is equal to
Ἀλλὰδὴπάλινἔστωσαναἱὑπὸτὰςἴσαςγωνίαςπλευραὶ angle DEF . Thus, the base AC is equal to the base DF ,
ὑποτείνουσαιἴσαι,ὡςἡΑΒτῇΔΕ·λέγωπάλιν,ὅτικαὶαἱ and the remaining angle BAC is equal to the remaining
λοιπαὶπλευραὶταῖςλοιπαῖςπλευραῖςἴσαιἔσονται,ἡμὲνΑΓ angle EDF [Prop. 1.4].
τῇΔΖ,ἡδὲΒΓτῇΕΖκαὶἔτιἡλοιπὴγωνίαἡὑπὸΒΑΓ But, again, let the sides subtending the equal angles
τῇλοιπῇγωνίᾳτῇὑπὸΕΔΖἴσηἐστίν.
be equal: for instance, (let) AB (be equal) to DE. Again,
ΕἰγὰρἄνισόςἐστινἡΒΓτῇΕΖ,μίααὐτῶνμείζωνἐστίν. I say that the remaining sides will be equal to the remain-
ἔστωμείζων,εἰδυνατόν,ἡΒΓ,καὶκείσθωτῇΕΖἴσηἡΒΘ, ing sides. (That is) AC (equal) to DF , and BC to EF .
καὶἐπεζεύχθωἡΑΘ.καὶἐπὲιἴσηἐστὶνἡμὲνΒΘτῇΕΖ Furthermore, the remaining angle BAC is equal to the
ἡδὲΑΒτῇΔΕ,δύοδὴαἱΑΒ,ΒΘδυσὶταῖςΔΕ,ΕΖἴσαι remaining angle EDF .
εἰσὶνἑκατέραἑκαρέρᾳ·καὶγωνίαςἴσαςπεριέχουσιν·βάσις For if BC is unequal to EF then one of them is
ἄραἡΑΘβάσειτῇΔΖἴσηἐστίν,καὶτὸΑΒΘτρίγωνοντῷ greater. If possible, let BC be greater. And let BH be
ΔΕΖτριγώνῳἴσονἐστίν,καὶαἱλοιπαὶγωνίαιταῖςλοιπαῖς made equal to EF [Prop. 1.3], and let AH have been
γωνίαιςἴσαιἔσονται,ὑφ᾿ἃςαἱἴσαςπλευραὶὑποτείνουσιν· joined. And since BH is equal to EF , and AB to DE,
ἴσηἄραἐστὶνἡὑπὸΒΘΑγωνίατῇὑπὸΕΖΔ.ἀλλὰἡὑπὸ the two (straight-lines) AB, BH are equal to the two
F
29

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ΕΖΔτῇὑπὸΒΓΑἐστινἴση·τριγώνουδὴτοῦΑΘΓἡἐκτὸς (straight-lines) DE, EF , respectively. And the angles
γωνίαἡὑπὸΒΘΑἴσηἐστὶτῇἐντὸςκαὶἀπεναντίοντῇὑπὸ they encompass (are also equal). Thus, the base AH is
ΒΓΑ·ὅπερἀδύνατον.οὐκἄραἄνισόςἐστινἡΒΓτῇΕΖ·ἴση equal to the base DF , and the triangle ABH is equal to
ἄρα. ἐστὶδὲκαὶἡΑΒτῇΔΕἴση. δύοδὴαἱΑΒ,ΒΓδύο the triangle DEF , and the remaining angles subtended
ταῖςΔΕ,ΕΖἴσαιεἰσὶνἑκατέραἑκατέρᾳ·καὶγωνίαςἴσας by the equal sides will be equal to the (corresponding)
περιέχουσι·βάσιςἄραἡΑΓβάσειτῇΔΖἴσηἐστίν,καὶτὸ remaining angles [Prop. 1.4]. Thus, angle BHA is equal
ΑΒΓτρίγωνοντῷΔΕΖτριγώνῳἴσονκαὶλοιπὴγωνίαἡ to EFD. But, EFD is equal to BCA. So, in triangle
ὑπὸΒΑΓτῇλοιπῂγωνίᾳτῇὑπὸΕΔΖἴση.
AHC, the external angle BHA is equal to the internal
᾿Εὰνἄραδύοτρίγωνατὰςδύογωνίαςδυσὶγωνίαιςἴσας and opposite angle BCA. The very thing (is) impossi-
ἔχῃἑκατέρανἑκατέρᾳκαὶμίανπλευρὰνμιᾷπλευρᾷἴσην ble [Prop. 1.16]. Thus, BC is not unequal to EF . Thus,
ἤτοιτὴνπρὸςταῖςἴσαιςγωνίαις,ἢτὴνὑποτείνουσανὑπὸ (it is) equal. And AB is also equal to DE. So the two
μίαντῶνἴσωνγωνιῶν,καὶτὰςλοιπὰςπλευρὰςταῖςλοιπαῖς (straight-lines) AB, BC are equal to the two (straightπλευραῖςἴσαςἕξεικαὶτὴνλοιπὴνγωνίαντῇλοιπῇγωνίᾳ·
lines) DE, EF , respectively. And they encompass equal
ὅπερἔδειδεῖξαι.
angles. Thus, the base AC is equal to the base DF , and
triangle ABC (is) equal to triangle DEF , and the remaining
angle BAC (is) equal to the remaining angle
EDF [Prop. 1.4].
Thus, if two triangles have two angles equal to two
angles, respectively, and one side equal to one side—in
fact, either that by the equal angles, or that subtending
one of the equal angles—then (the triangles) will also
have the remaining sides equal to the (corresponding) remaining
sides, and the remaining angle (equal) to the remaining
angle. (Which is) the very thing it was required
to show.
† The Greek text has “BG, BC”, which is obviously a mistake.
Þ. Proposition 27
᾿Εὰν εἰςδύοεὐθείαςεὐθεῖαἐμπίπτουσατὰςἐναλλὰξ If a straight-line falling across two straight-lines
γωνίαςἴσαςἀλλήλαιςποιῇ,παράλληλοιἔσονταιἀλλήλαιςαἱ makes the alternate angles equal to one another then
εὐθεῖαι.
the (two) straight-lines will be parallel to one another.
Α Ε Β
A
E
B
Η
G
Γ
Ζ
∆
C
F
D
ΕἰςγὰρδύοεὐθείαςτὰςΑΒ,ΓΔεὐθεῖαἐμπίπτουσαἡ For let the straight-line EF , falling across the two
ΕΖτὰςἐναλλὰξγωνίαςτὰςὑπὸΑΕΖ,ΕΖΔἴσαςἀλλήλαις straight-lines AB and CD, make the alternate angles
ποιείτω·λέγω,ὅτιπαράλληλόςἐστινἡΑΒτῇΓΔ. AEF and EFD equal to one another. I say that AB and
Εἰγὰρμή,ἐκβαλλόμεναιαἱΑΒ,ΓΔσυμπεσοῦνταιἤτοι CD are parallel.
ἐπὶτὰΒ,ΔμέρηἢἐπὶτὰΑ,Γ.ἐκβεβλήσθωσανκαὶσυμ- For if not, being produced, AB and CD will certainly
πιπτέτωσανἐπὶτὰΒ,ΔμέρηκατὰτὸΗ.τριγώνουδὴτοῦ meet together: either in the direction of B and D, or (in
ΗΕΖἡἐκτὸςγωνίαἡὑπὸΑΕΖἴσηἐστὶτῇἐντὸςκαὶἀπε- the direction) of A and C [Def. 1.23]. Let them have
ναντίοντῇὑπὸΕΖΗ·ὅπερἐστὶνἀδύνατον·οὐκἄρααἱΑΒ, been produced, and let them meet together in the di-
ΔΓἐκβαλλόμεναισυμπεσοῦνταιἐπὶτὰΒ,Δμέρη.ὁμοίως rection of B and D at (point) G. So, for the triangle
30

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
δὴδειχθήσεται,ὅτιοὐδὲἐπὶτὰΑ,Γ·αἱδὲἐπὶμηδέτερατὰ GEF , the external angle AEF is equal to the interior
μέρησυμπίπτουσαιπαράλληλοίεἰσιν·παράλληλοςἄραἐστὶν and opposite (angle) EFG. The very thing is impossible
ἡΑΒτῇΓΔ.
[Prop. 1.16]. Thus, being produced, AB and CD will not
᾿Εὰνἄραεἰςδύοεὐθείαςεὐθεῖαἐμπίπτουσατὰςἐναλλὰξ meet together in the direction of B and D. Similarly, it
γωνίαςἴσαςἀλλήλαιςποιῇ,παράλληλοιἔσονταιαἱεὐθεῖαι· can be shown that neither (will they meet together) in
ὅπερἔδειδεῖξαι.
(the direction of) A and C. But (straight-lines) meeting
in neither direction are parallel [Def. 1.23]. Thus, AB
and CD are parallel.
Thus, if a straight-line falling across two straight-lines
makes the alternate angles equal to one another then
the (two) straight-lines will be parallel (to one another).
(Which is) the very thing it was required to show.
. Proposition 28
᾿Εὰν εἰς δύο εὐθείας εὐθεῖα ἐμπίπτουσα τὴν ἐκτὸς If a straight-line falling across two straight-lines
γωνίαντῇἐντὸςκαὶἀπεναντίονκαὶἐπὶτὰαὐτὰμέρηἴσην makes the external angle equal to the internal and oppoποιῇἢτὰςἐντὸςκαὶἐπὶτὰαὐτὰμέρηδυσὶνὀρθαῖςἴσας,
site angle on the same side, or (makes) the (sum of the)
παράλληλοιἔσονταιἀλλήλαιςαἱεὐθεῖαι.
internal (angles) on the same side equal to two rightangles,
then the (two) straight-lines will be parallel to
one another.
Ε
E
Α
Η
Β
A
G
B
Γ
Θ
∆
C
H
D
Ζ
ΕἰςγὰρδύοεὐθείαςτὰςΑΒ,ΓΔεὐθεῖαἐμπίπτουσαἡ For let EF , falling across the two straight-lines AB
ΕΖτὴνἐκτὸςγωνίαντὴνὑπὸΕΗΒτῇἐντὸςκαὶἀπεναντίον and CD, make the external angle EGB equal to the inγωνίᾳτῇὑπὸΗΘΔἴσηνποιείτωἢτὰςἐντὸςκαὶἐπὶτὰ
ternal and opposite angle GHD, or the (sum of the) inαὐτὰμέρητὰςὑπὸΒΗΘ,ΗΘΔδυσὶνὀρθαῖςἴσας·λέγω,
ternal (angles) on the same side, BGH and GHD, equal
ὅτιπαράλληλόςἐστινἡΑΒτῇΓΔ.
to two right-angles. I say that AB is parallel to CD.
᾿ΕπεὶγὰρἴσηἐστὶνἡὑπὸΕΗΒτῇὑπὸΗΘΔ,ἀλλὰἡὑπὸ For since (in the first case) EGB is equal to GHD, but
ΕΗΒτῇὑπὸΑΗΘἐστινἴση,καὶἡὑπὸΑΗΘἄρατῇὑπὸ EGB is equal to AGH [Prop. 1.15], AGH is thus also
ΗΘΔἐστινἴση·καίεἰσινἐναλλάξ·παράλληλοςἄραἐστὶνἡ equal to GHD. And they are alternate (angles). Thus,
ΑΒτῇΓΔ. AB is parallel to CD [Prop. 1.27].
Πάλιν,ἐπεὶαἱὑπὸΒΗΘ,ΗΘΔδύοὀρθαῖςἴσαιεἰσίν, Again, since (in the second case, the sum of) BGH
εἰσὶδὲκαὶαἱὑπὸΑΗΘ,ΒΗΘδυσὶνὀρθαῖςἴσαι,αἱἄρα and GHD is equal to two right-angles, and (the sum
ὑπὸΑΗΘ,ΒΗΘταῖςὑπὸΒΗΘ,ΗΘΔἴσαιεἰσίν· κοινὴ of) AGH and BGH is also equal to two right-angles
ἀφῃρήσθωἡὑπὸΒΗΘ·λοιπὴἄραἡὑπὸΑΗΘλοιπῇτῇ [Prop. 1.13], (the sum of) AGH and BGH is thus equal
ὑπὸΗΘΔἐστινἴση· καίεἰσινἐναλλάξ· παράλληλοςἄρα to (the sum of) BGH and GHD. Let BGH have been
ἐστὶνἡΑΒτῇΓΔ.
subtracted from both. Thus, the remainder AGH is equal
᾿Εὰνἄραεἰςδύοεὐθείαςεὐθεῖαἐμπίπτουσατὴνἐκτὸς to the remainder GHD. And they are alternate (angles).
γωνίαντῇἐντὸςκαὶἀπεναντίονκαὶἐπὶτὰαὐτὰμέρηἴσην Thus, AB is parallel to CD [Prop. 1.27].
F
31

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ποιῇἢτὰςἐντὸςκαὶἐπὶτὰαὐτὰμέρηδυσὶνὀρθαῖςἴσας, Thus, if a straight-line falling across two straight-lines
παράλληλοιἔσονταιαἱεὐθεῖαι·ὅπερἔδειδεῖξαι.
makes the external angle equal to the internal and opposite
angle on the same side, or (makes) the (sum of the)
internal (angles) on the same side equal to two rightangles,
then the (two) straight-lines will be parallel (to
one another). (Which is) the very thing it was required
to show.
. Proposition 29
῾Ηεἰςτὰςπαραλλήλουςεὐθείαςεὐθεῖαἐμπίπτουσατάς A straight-line falling across parallel straight-lines
τεἐναλλὰξγωνίαςἴσαςἀλλήλαιςποιεῖκαὶτὴνἐκτὸςτῇ makes the alternate angles equal to one another, the ex-
ἐντὸςκαὶἀπεναντίονἴσηνκαὶτὰςἐντὸςκαὶἐπὶτὰαὐτὰ ternal (angle) equal to the internal and opposite (angle),
μέρηδυσὶνὀρθαῖςἴσας.
and the (sum of the) internal (angles) on the same side
equal to two right-angles.
Ε
E
Α
Η
Β
A
G
B
Γ
Θ
∆
C
H
D
Ζ
Εἰς γὰρ παραλλήλους εὐθείας τὰς ΑΒ, ΓΔ εὐθεῖα For let the straight-line EF fall across the parallel
ἐμπιπτέτωἡΕΖ·λέγω, ὅτιτὰςἐναλλὰξγωνίαςτὰςὑπὸ straight-lines AB and CD. I say that it makes the alter-
ΑΗΘ,ΗΘΔἴσαςποιεῖκαὶτὴνἐκτὸςγωνίαντὴνὑπὸΕΗΒ nate angles, AGH and GHD, equal, the external angle
τῇἐντὸςκαὶἀπεναντίοντῇὑπὸΗΘΔἴσηνκαὶτὰςἐντὸς EGB equal to the internal and opposite (angle) GHD,
καὶἐπὶτὰαὐτὰμέρητὰςὑπὸΒΗΘ,ΗΘΔδυσὶνὀρθαῖς and the (sum of the) internal (angles) on the same side,
ἴσας.
BGH and GHD, equal to two right-angles.
ΕἰγὰρἄνισόςἐστινἡὑπὸΑΗΘτῇὑπὸΗΘΔ,μίααὐτῶν For if AGH is unequal to GHD then one of them is
μείζωνἐστίν.ἔστωμείζωνἡὑπὸΑΗΘ·κοινὴπροσκείσθω greater. Let AGH be greater. Let BGH have been added
ἡὑπὸΒΗΘ·αἱἄραὑπὸΑΗΘ,ΒΗΘτῶνὑπὸΒΗΘ,ΗΘΔ to both. Thus, (the sum of) AGH and BGH is greater
μείζονέςεἰσιν. ἀλλὰαἱὑπὸΑΗΘ,ΒΗΘδυσὶνὀρθαῖςἴσαι than (the sum of) BGH and GHD. But, (the sum of)
εἰσίν. [καὶ]αἱἄραὑπὸΒΗΘ,ΗΘΔδύοὀρθῶνἐλάσσονές AGH and BGH is equal to two right-angles [Prop 1.13].
εἰσιν. αἱ δὲ ἀπ᾿ ἐλασσόνων ἢ δύο ὀρθῶν ἐκβαλλόμεναι Thus, (the sum of) BGH and GHD is [also] less than
εἰς ἄπειρον συμπίπτουσιν· αἱ ἄρα ΑΒ, ΓΔ ἐκβαλλόμεναι two right-angles. But (straight-lines) being produced to
εἰςἄπειρονσυμπεσοῦνται· οὐσυμπίπτουσιδὲδιὰτὸπα- infinity from (internal angles whose sum is) less than two
ραλλήλουςαὐτὰςὑποκεῖσθαι·οὐκἄραἄνισόςἐστινἡὑπὸ right-angles meet together [Post. 5]. Thus, AB and CD,
ΑΗΘτῇὑπὸΗΘΔ·ἴσηἄρα.ἀλλὰἡὑπὸΑΗΘτῇὑπὸΕΗΒ being produced to infinity, will meet together. But they do
ἐστινἴση·καὶἡὑπὸΕΗΒἄρατῇὑπὸΗΘΔἐστινἴση·κοινὴ not meet, on account of them (initially) being assumed
προσκείσθωἡὑπὸΒΗΘ·αἱἄραὑπὸΕΗΒ,ΒΗΘταῖςὑπὸ parallel (to one another) [Def. 1.23]. Thus, AGH is not
ΒΗΘ,ΗΘΔἴσαιεἰσίν.ἀλλὰαἱὑπὸΕΗΒ,ΒΗΘδύοὀρθαῖς unequal to GHD. Thus, (it is) equal. But, AGH is equal
ἴσαιεἰσίν·καὶαἱὑπὸΒΗΘ,ΗΘΔἄραδύοὀρθαῖςἴσαιεἰσίν. to EGB [Prop. 1.15]. And EGB is thus also equal to
῾Ηἄραεἰςτὰςπαραλλήλουςεὐθείαςεὐθεῖαἐμπίπτουσα GHD. Let BGH be added to both. Thus, (the sum of)
τάςτεἐναλλὰξγωνίαςἴσαςἀλλήλαιςποιεῖκαὶτὴνἐκτὸς EGB and BGH is equal to (the sum of) BGH and GHD.
τῇἐντὸςκαὶἀπεναντίονἴσηνκαὶτὰςἐντὸςκαὶἐπὶτὰαὐτὰ But, (the sum of) EGB and BGH is equal to two right-
F
32

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
μέρηδυσὶνὀρθαῖςἴσας·ὅπερἔδειδεῖξαι.
angles [Prop. 1.13]. Thus, (the sum of) BGH and GHD
is also equal to two right-angles.
Thus, a straight-line falling across parallel straightlines
makes the alternate angles equal to one another, the
external (angle) equal to the internal and opposite (angle),
and the (sum of the) internal (angles) on the same
side equal to two right-angles. (Which is) the very thing
it was required to show.
Ð. Proposition 30
Αἱτῇαὐτῇεὐθείᾳπαράλληλοικαὶἀλλήλαιςεἰσὶπαράλλη- (Straight-lines) parallel to the same straight-line are
λοι.
also parallel to one another.
Α
Η
Β
A
G
B
Ε
Θ
Ζ
E
H
F
Γ
Κ
∆
C
K
D
῎ΕστωἑκατέρατῶνΑΒ,ΓΔτῇΕΖπαράλληλος·λέγω, Let each of the (straight-lines) AB and CD be parallel
ὅτικαὶἡΑΒτῇΓΔἐστιπαράλληλος.
to EF . I say that AB is also parallel to CD.
᾿ΕμπιπτέτωγὰρεἰςαὐτὰςεὐθεῖαἡΗΚ.
For let the straight-line GK fall across (AB, CD, and
ΚαὶἐπεὶεἰςπαραλλήλουςεὐθείαςτὰςΑΒ,ΕΖεὐθεῖα EF ).
ἐμπέπτωκεν ἡ ΗΚ, ἴση ἄρα ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ. And since the straight-line GK has fallen across the
πάλιν, ἐπεὶεἰςπαραλλήλουςεὐθείαςτὰςΕΖ, ΓΔεὐθεῖα parallel straight-lines AB and EF , (angle) AGK (is) thus
ἐμπέπτωκενἡΗΚ,ἴσηἐστὶνἡὑπὸΗΘΖτῇὑπὸΗΚΔ. equal to GHF [Prop. 1.29]. Again, since the straight-line
ἐδείχθηδὲκαὶἡὑπὸΑΗΚτῇὑπὸΗΘΖἴση.καὶἡὑπὸΑΗΚ GK has fallen across the parallel straight-lines EF and
ἄρατῇὑπὸΗΚΔἐστινἴση·καίεἰσινἐναλλάξ.παράλληλος CD, (angle) GHF is equal to GKD [Prop. 1.29]. But
ἄραἐστὶνἡΑΒτῇΓΔ.
AGK was also shown (to be) equal to GHF . Thus, AGK
[Αἱἄρατῇαὐτῇεὐθείᾳπαράλληλοικαὶἀλλήλαιςεἰσὶ is also equal to GKD. And they are alternate (angles).
παράλληλοι·]ὅπερἔδειδεῖξαι. Thus, AB is parallel to CD [Prop. 1.27].
[Thus, (straight-lines) parallel to the same straightline
are also parallel to one another.] (Which is) the very
thing it was required to show.
Ð. Proposition 31
Διὰτοῦδοθέντοςσημείουτῇδοθείσῃεὐθείᾳπαράλληλον To draw a straight-line parallel to a given straight-line,
εὐθεῖανγραμμὴνἀγαγεῖν.
through a given point.
῎ΕστωτὸμὲνδοθὲνσημεῖοντὸΑ,ἡδὲδοθεῖσαεὐθεῖα Let A be the given point, and BC the given straight-
ἡΒΓ·δεῖδὴδιὰτοῦΑσημείουτῇΒΓεὐθείᾳπαράλληλον line. So it is required to draw a straight-line parallel to
εὐθεῖανγραμμὴνἀγαγεῖν. the straight-line BC, through the point A.
ΕἰλήφθωἐπὶτῆςΒΓτυχὸνσημεῖοντὸΔ,καὶἐπεζεύχθω Let the point D have been taken a random on BC, and
ἡΑΔ·καὶσυνεστάτωπρὸςτῇΔΑεὐθείᾳκαὶτῷπρὸςαὐτῇ let AD have been joined. And let (angle) DAE, equal to
σημείῳ τῷ Α τῇ ὑπὸ ΑΔΓ γωνίᾳ ἴση ἡ ὑπὸ ΔΑΕ· καὶ angle ADC, have been constructed on the straight-line
33

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ἐκβεβλήσθωἐπ᾿εὐθείαςτῇΕΑεὐθεῖαἡΑΖ.
Ε
Α
Ζ
DA at the point A on it [Prop. 1.23]. And let the straightline
AF have been produced in a straight-line with EA.
E
A
F
Β
∆
Γ
ΚαὶἐπεὶεἰςδύοεὐθείαςτὰςΒΓ,ΕΖεὐθεῖαἐμπίπτουσα And since the straight-line AD, (in) falling across the
ἡ ΑΔ τὰς ἐναλλὰξ γωνίας τὰς ὑπὸ ΕΑΔ, ΑΔΓ ἴσας two straight-lines BC and EF , has made the alternate
ἀλλήλαιςπεποίηκεν,παράλληλοςἄραἐστὶνἡΕΑΖτῇΒΓ. angles EAD and ADC equal to one another, EAF is thus
ΔιὰτοῦδοθέντοςἄρασημείουτοῦΑτῇδοθείσῃεὐθείᾳ parallel to BC [Prop. 1.27].
τῇΒΓπαράλληλοςεὐθεῖαγραμμὴἦκταιἡΕΑΖ·ὅπερἔδει Thus, the straight-line EAF has been drawn parallel
ποιῆσαι. to the given straight-line BC, through the given point A.
(Which is) the very thing it was required to do.
Ð. Proposition 32
Παντὸςτριγώνουμιᾶςτῶνπλευρῶνπροσεκβληθείσης In any triangle, (if) one of the sides (is) produced
ἡἐκτὸςγωνίαδυσὶταῖςἐντὸςκαὶἀπεναντίονἴσηἐστίν,καὶ (then) the external angle is equal to the (sum of the) two
αἱἐντὸςτοῦτριγώνουτρεῖςγωνίαιδυσὶνὀρθαῖςἴσαιεἰσίν. internal and opposite (angles), and the (sum of the) three
internal angles of the triangle is equal to two right-angles.
Α
Ε
B
A
D
E
C
Β
Γ
∆
῎ΕστωτρίγωνοντὸΑΒΓ,καὶπροσεκβεβλήσθωαὐτοῦ Let ABC be a triangle, and let one of its sides BC
μίαπλευρὰἡΒΓἐπὶτὸΔ·λέγω,ὅτιἡἐκτὸςγωνίαἡὑπὸ have been produced to D. I say that the external angle
ΑΓΔἴσηἐστὶδυσὶταῖςἐντὸςκαὶἀπεναντίονταῖςὑπὸΓΑΒ, ACD is equal to the (sum of the) two internal and oppo-
ΑΒΓ,καὶαἱἐντὸςτοῦτριγώνουτρεῖςγωνίαιαἱὑπὸΑΒΓ, site angles CAB and ABC, and the (sum of the) three
ΒΓΑ,ΓΑΒδυσὶνὀρθαῖςἴσαιεἰσίν.
internal angles of the triangle—ABC, BCA, and CAB—
῎ΗχθωγὰρδιὰτοῦΓσημείουτῇΑΒεὐθείᾳπαράλληλος is equal to two right-angles.
ἡΓΕ.
For let CE have been drawn through point C parallel
ΚαὶἐπεὶπαράλληλόςἐστινἡΑΒτῇΓΕ,καὶεἰςαὐτὰς to the straight-line AB [Prop. 1.31].
ἐμπέπτωκενἡΑΓ,αἱἐναλλὰξγωνίαιαἱὑπὸΒΑΓ,ΑΓΕἴσαι And since AB is parallel to CE, and AC has fallen
ἀλλήλαιςεἰσίν.πάλιν,ἐπεὶπαράλληλόςἐστινἡΑΒτῇΓΕ, across them, the alternate angles BAC and ACE are
καὶεἰςαὐτὰςἐμπέπτωκενεὐθεῖαἡΒΔ,ἡἐκτὸςγωνίαἡ equal to one another [Prop. 1.29]. Again, since AB is
ὑπὸΕΓΔἴσηἐστὶτῇἐντὸςκαὶἀπεναντίοντῇὑπὸΑΒΓ. parallel to CE, and the straight-line BD has fallen across
ἐδείχθηδὲκαὶἡὑπὸΑΓΕτῇὑπὸΒΑΓἴση·ὅληἄραἡὑπὸ them, the external angle ECD is equal to the internal
ΑΓΔγωνίαἴσηἐστὶδυσὶταῖςἐντὸςκαὶἀπεναντίονταῖςὑπὸ and opposite (angle) ABC [Prop. 1.29]. But ACE was
ΒΑΓ,ΑΒΓ.
also shown (to be) equal to BAC. Thus, the whole an-
B
C
D
34

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ΚοινὴπροσκείσθωἡὑπὸΑΓΒ·αἱἄραὑπὸΑΓΔ,ΑΓΒ gle ACD is equal to the (sum of the) two internal and
τρισὶταῖςὑπὸΑΒΓ,ΒΓΑ,ΓΑΒἴσαιεἰσίν.ἀλλ᾿αἱὑπὸΑΓΔ, opposite (angles) BAC and ABC.
ΑΓΒδυσὶνὀρθαῖςἴσαιεἰσίν·καὶαἱὑπὸΑΓΒ,ΓΒΑ,ΓΑΒ Let ACB have been added to both. Thus, (the sum
ἄραδυσὶνὀρθαῖςἴσαιεἰσίν.
of) ACD and ACB is equal to the (sum of the) three
Παντὸς ἄρα τριγώνου μιᾶς τῶν πλευρῶν προσεκ- (angles) ABC, BCA, and CAB. But, (the sum of) ACD
βληθείσηςἡἐκτὸςγωνίαδυσὶταῖςἐντὸςκαὶἀπεναντίον and ACB is equal to two right-angles [Prop. 1.13]. Thus,
ἴσηἐστίν, καὶαἱἐντὸςτοῦτριγώνουτρεῖςγωνίαιδυσὶν (the sum of) ACB, CBA, and CAB is also equal to two
ὀρθαῖςἴσαιεἰσίν·ὅπερἔδειδεῖξαι.
right-angles.
Thus, in any triangle, (if) one of the sides (is) produced
(then) the external angle is equal to the (sum of
the) two internal and opposite (angles), and the (sum of
the) three internal angles of the triangle is equal to two
right-angles. (Which is) the very thing it was required to
show.
Ð. Proposition 33
Αἱτὰςἴσαςτεκαὶπαραλλήλουςἐπὶτὰαὐτὰμέρηἐπι- Straight-lines joining equal and parallel (straightζευγνύουσαιεὐθεῖαικαὶαὐταὶἴσαιτεκαὶπαράλληλοίεἰσιν.
lines) on the same sides are themselves also equal and
parallel.
Β
Α
B
A
∆ Γ D C
῎ΕστωσανἴσαιτεκαὶπαράλληλοιαἱΑΒ,ΓΔ,καὶἐπι- Let AB and CD be equal and parallel (straight-lines),
ζευγνύτωσαναὐτὰςἐπὶτὰαὐτὰμέρηεὐθεῖαιαἱΑΓ,ΒΔ· and let the straight-lines AC and BD join them on the
λέγω,ὅτικαὶαἱΑΓ,ΒΔἴσαιτεκαὶπαράλληλοίεἰσιν. same sides. I say that AC and BD are also equal and
᾿ΕπεζεύχθωἡΒΓ.καὶἐπεὶπαράλληλόςἐστινἡΑΒτῇ parallel.
ΓΔ,καὶεἰςαὐτὰςἐμπέπτωκενἡΒΓ,αἱἐναλλὰξγωνίαιαἱ Let BC have been joined. And since AB is paral-
ὑπὸΑΒΓ,ΒΓΔἴσαιἀλλήλαιςεἰσίν.καὶἐπεὶἴσηἐστὶνἡΑΒ lel to CD, and BC has fallen across them, the alterτῇΓΔκοινὴδὲἡΒΓ,δύοδὴαἱΑΒ,ΒΓδύοταῖςΒΓ,ΓΔἴσαι
nate angles ABC and BCD are equal to one another
εἰσίν·καὶγωνίαἡὑπὸΑΒΓγωνίᾳτῇὑπὸΒΓΔἴση·βάσις [Prop. 1.29]. And since AB is equal to CD, and BC
ἄραἡΑΓβάσειτῇΒΔἐστινἴση,καὶτὸΑΒΓτρίγωνοντῷ is common, the two (straight-lines) AB, BC are equal
ΒΓΔτριγώνῳἴσονἐστίν,καὶαἱλοιπαὶγωνίαιταῖςλοιπαῖς to the two (straight-lines) DC, CB. † And the angle ABC
γωνίαιςἴσαιἔσονταιἑκατέραἑκατέρᾳ,ὑφ᾿ἃςαἱἴσαιπλευραὶ is equal to the angle BCD. Thus, the base AC is equal
ὑποτείνουσιν·ἴσηἄραἡὑπὸΑΓΒγωνίατῇὑπὸΓΒΔ.καὶ to the base BD, and triangle ABC is equal to triangle
ἐπεὶεἰςδύοεὐθείαςτὰςΑΓ,ΒΔεὐθεῖαἐμπίπτουσαἡΒΓ DCB ‡ , and the remaining angles will be equal to the
τὰςἐναλλὰξγωνίαςἴσαςἀλλήλαιςπεποίηκεν,παράλληλος corresponding remaining angles subtended by the equal
ἄραἐστὶνἡΑΓτῇΒΔ.ἐδείχθηδὲαὐτῇκαὶἴση. sides [Prop. 1.4]. Thus, angle ACB is equal to CBD.
Αἱἄρατὰςἴσαςτεκαὶπαραλλήλουςἐπὶτὰαὐτὰμέρη Also, since the straight-line BC, (in) falling across the
ἐπιζευγνύουσαιεὐθεῖαικαὶαὐταὶἴσαιτεκαὶπαράλληλοί two straight-lines AC and BD, has made the alternate
εἰσιν·ὅπερἔδειδεῖξαι.
angles (ACB and CBD) equal to one another, AC is thus
parallel to BD [Prop. 1.27]. And (AC) was also shown
(to be) equal to (BD).
Thus, straight-lines joining equal and parallel (straight-
35

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
lines) on the same sides are themselves also equal and
parallel. (Which is) the very thing it was required to
show.
† The Greek text has “BC, CD”, which is obviously a mistake.
‡ The Greek text has “DCB”, which is obviously a mistake.
Ð. Proposition 34
Τῶνπαραλληλογράμμωνχωρίωναἱἀπεναντίονπλευραί In parallelogrammic figures the opposite sides and angles
τεκαὶγωνίαιἴσαιἀλλήλαιςεἰσίν,καὶἡδιάμετροςαὐτὰδίχα are equal to one another, and a diagonal cuts them in half.
τέμνει.
Α
Β
A
B
Γ
∆
῎ΕστωπαραλληλόγραμμονχωρίοντὸΑΓΔΒ,διάμετρος Let ACDB be a parallelogrammic figure, and BC its
δὲαὐτοῦἡΒΓ·λέγω,ὅτιτοῦΑΓΔΒπαραλληλογράμμουαἱ diagonal. I say that for parallelogram ACDB, the oppo-
ἀπεναντίονπλευραίτεκαὶγωνίαιἴσαιἀλλήλαιςεἰσίν,καὶἡ site sides and angles are equal to one another, and the
ΒΓδιάμετροςαὐτὸδίχατέμνει.
diagonal BC cuts it in half.
᾿ΕπεὶγὰρπαράλληλόςἐστινἡΑΒτῇΓΔ,καὶεἰςαὐτὰς For since AB is parallel to CD, and the straight-line
ἐμπέπτωκενεὐθεῖαἡΒΓ,αἱἐναλλὰξγωνίαιαἱὑπὸΑΒΓ, BC has fallen across them, the alternate angles ABC and
ΒΓΔἴσαιἀλλήλαιςεἰσίν.πάλινἐπεὶπαράλληλόςἐστινἡΑΓ BCD are equal to one another [Prop. 1.29]. Again, since
τῇΒΔ,καὶεἰςαὐτὰςἐμπέπτωκενἡΒΓ,αἱἐναλλὰξγωνίαι AC is parallel to BD, and BC has fallen across them,
αἱὑπὸΑΓΒ,ΓΒΔἴσαιἀλλήλαιςεἰσίν.δύοδὴτρίγωνάἐστι the alternate angles ACB and CBD are equal to one
τὰΑΒΓ,ΒΓΔτὰςδύογωνίαςτὰςὑπὸΑΒΓ,ΒΓΑδυσὶ another [Prop. 1.29]. So ABC and BCD are two triταῖςὑπὸΒΓΔ,ΓΒΔἴσαςἔχονταἑκατέρανἑκατέρᾳκαὶμίαν
angles having the two angles ABC and BCA equal to
πλευρὰνμιᾷπλευρᾷἴσηντὴνπρὸςταῖςἴσαιςγωνίαιςκοινὴν the two (angles) BCD and CBD, respectively, and one
αὐτῶντὴνΒΓ·καὶτὰςλοιπὰςἄραπλευρὰςταῖςλοιπαῖς side equal to one side—the (one) by the equal angles and
ἴσαςἕξειἑκατέρανἑκατέρᾳκαὶτὴνλοιπὴνγωνίαντῇλοιπῇ common to them, (namely) BC. Thus, they will also
γωνίᾳ·ἴσηἄραἡμὲνΑΒπλευρὰτῇΓΔ,ἡδὲΑΓτῇΒΔ, have the remaining sides equal to the corresponding reκαὶἔτιἴσηἐστὶνἡὑπὸΒΑΓγωνίατῇὑπὸΓΔΒ.καὶἐπεὶ
maining (sides), and the remaining angle (equal) to the
ἴσηἐστὶνἡμὲνὑπὸΑΒΓγωνίατῇὑπὸΒΓΔ,ἡδὲὑπὸΓΒΔ remaining angle [Prop. 1.26]. Thus, side AB is equal to
τῇὑπὸΑΓΒ,ὅληἄραἡὑπὸΑΒΔὅλῃτῇὑπὸΑΓΔἐστιν CD, and AC to BD. Furthermore, angle BAC is equal
ἴση.ἐδείχθηδὲκαὶἡὑπὸΒΑΓτῇὑπὸΓΔΒἴση. to CDB. And since angle ABC is equal to BCD, and
Τῶν ἄρα παραλληλογράμμων χωρίων αἱ ἀπεναντίον CBD to ACB, the whole (angle) ABD is thus equal to
πλευραίτεκαὶγωνίαιἴσαιἀλλήλαιςεἰσίν.
the whole (angle) ACD. And BAC was also shown (to
Λέγωδή,ὅτικαὶἡδιάμετροςαὐτὰδίχατέμνει.ἐπεὶγὰρ be) equal to CDB.
ἴσηἐστὶνἡΑΒτῇΓΔ,κοινὴδὲἡΒΓ,δύοδὴαἱΑΒ,ΒΓ Thus, in parallelogrammic figures the opposite sides
δυσὶταῖςΓΔ,ΒΓἴσαιεἰσὶνἑκατέραἑκατέρᾳ·καὶγωνίαἡ and angles are equal to one another.
ὑπὸΑΒΓγωνίᾳτῇὑπὸΒΓΔἴση. καὶβάσιςἄραἡΑΓτῇ And, I also say that a diagonal cuts them in half. For
ΔΒἴση.καὶτὸΑΒΓ[ἄρα]τρίγωνοντῷΒΓΔτριγώνῳἴσον since AB is equal to CD, and BC (is) common, the two
ἐστίν.
(straight-lines) AB, BC are equal to the two (straight-
῾Η ἄρα ΒΓ διάμετρος δίχα τέμνει τὸ ΑΒΓΔ παραλ- lines) DC, CB † , respectively. And angle ABC is equal to
ληλόγραμμον·ὅπερἔδειδεῖξαι.
angle BCD. Thus, the base AC (is) also equal to DB,
C
D
36

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
and triangle ABC is equal to triangle BCD [Prop. 1.4].
Thus, the diagonal BC cuts the parallelogram ACDB ‡
in half. (Which is) the very thing it was required to show.
† The Greek text has “CD, BC”, which is obviously a mistake.
‡ The Greek text has “ABCD”, which is obviously a mistake.
Ð. Proposition 35
Τὰπαραλληλόγραμματὰἐπὶτῆςαὐτῆςβάσεωςὄντακαὶ Parallelograms which are on the same base and be-
ἐνταῖςαὐταῖςπαραλλήλοιςἴσαἀλλήλοιςἐστίν.
tween the same parallels are equal † to one another.
Α ∆ Ε Ζ A D
E F
Η
G
Β Γ
B C
῎Εστω παραλληλόγραμμα τὰ ΑΒΓΔ, ΕΒΓΖ ἐπὶ τῆς Let ABCD and EBCF be parallelograms on the same
αὐτῆςβάσεωςτῆςΒΓκαὶἐνταῖςαὐταῖςπαραλλήλοιςταῖς base BC, and between the same parallels AF and BC. I
ΑΖ,ΒΓ·λέγω,ὅτιἴσονἐστὶτὸΑΒΓΔτῷΕΒΓΖπαραλλη- say that ABCD is equal to parallelogram EBCF .
λογράμμῳ.
For since ABCD is a parallelogram, AD is equal to
᾿ΕπεὶγὰρπαραλληλόγραμμόνἐστιτὸΑΒΓΔ,ἴσηἐστὶν BC [Prop. 1.34]. So, for the same (reasons), EF is also
ἡΑΔτῇΒΓ.διὰτὰαὐτὰδὴκαὶἡΕΖτῇΒΓἐστινἴση· equal to BC. So AD is also equal to EF . And DE is
ὥστεκαὶἡΑΔτῇΕΖἐστινἴση·καὶκοινὴἡΔΕ·ὅληἄρα common. Thus, the whole (straight-line) AE is equal to
ἡΑΕὅλῃτῇΔΖἐστινἴση. ἔστιδὲκαὶἡΑΒτῇΔΓἴση· the whole (straight-line) DF . And AB is also equal to
δύοδὴαἱΕΑ,ΑΒδύοταῖςΖΔ,ΔΓἴσαιεἰσὶνἑκατέρα DC. So the two (straight-lines) EA, AB are equal to
ἑκατέρᾳ·καὶγωνίαἡὑπὸΖΔΓγωνίᾳτῇὑπὸΕΑΒἐστιν the two (straight-lines) FD, DC, respectively. And angle
ἴσηἡἐκτὸςτῇἐντός·βάσιςἄραἡΕΒβάσειτῇΖΓἴσηἐστίν, FDC is equal to angle EAB, the external to the interκαὶτὸΕΑΒτρίγωνοντῷΔΖΓτριγώνῳἴσονἔσται·κοινὸν
nal [Prop. 1.29]. Thus, the base EB is equal to the base
ἀφῃρήσθωτὸΔΗΕ·λοιπὸνἄρατὸΑΒΗΔτραπέζιονλοιπῷ FC, and triangle EAB will be equal to triangle DFC
τῷΕΗΓΖτραπεζίῳἐστὶνἴσον·κοινὸνπροσκείσθωτὸΗΒΓ [Prop. 1.4]. Let DGE have been taken away from both.
τρίγωνον·ὅλονἄρατὸΑΒΓΔπαραλληλόγραμμονὅλῳτῷ Thus, the remaining trapezium ABGD is equal to the re-
ΕΒΓΖπαραλληλογράμμῳἴσονἐστίν.
maining trapezium EGCF . Let triangle GBC have been
Τὰἄραπαραλληλόγραμματὰἐπὶτῆςαὐτῆςβάσεωςὄντα added to both. Thus, the whole parallelogram ABCD is
καὶἐνταῖςαὐταῖςπαραλλήλοιςἴσαἀλλήλοιςἐστίν·ὅπερἔδει equal to the whole parallelogram EBCF .
δεῖξαι.
Thus, parallelograms which are on the same base and
between the same parallels are equal to one another.
(Which is) the very thing it was required to show.
† Here, for the first time, “equal” means “equal in area”, rather than “congruent”.
Ц. Proposition 36
Τὰπαραλληλόγραμματὰἐπὶἴσωνβάσεωνὄντακαὶἐν Parallelograms which are on equal bases and between
ταῖςαὐταῖςπαραλλήλοιςἴσαἀλλήλοιςἐστίν.
the same parallels are equal to one another.
῎Εστω παραλληλόγραμματὰ ΑΒΓΔ, ΕΖΗΘ ἐπὶ ἴσων Let ABCD and EFGH be parallelograms which are
βάσεωνὄντατῶνΒΓ,ΖΗκαὶἐνταῖςαὐταῖςπαραλλήλοις on the equal bases BC and FG, and (are) between the
ταῖς ΑΘ, ΒΗ· λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓΔ παραλ- same parallels AH and BG. I say that the parallelogram
37

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ληλόγραμμοντῷΕΖΗΘ.
Α ∆
Ε
Θ
ABCD is equal to EFGH.
A D E H
Β Γ Ζ Η B C
F G
᾿ΕπεζεύχθωσανγὰραἱΒΕ,ΓΘ.καὶἐπεὶἴσηἐστὶνἡ For let BE and CH have been joined. And since BC is
ΒΓτῇΖΗ,ἀλλὰἡΖΗτῇΕΘἐστινἴση,καὶἡΒΓἄρατῇ equal to FG, but FG is equal to EH [Prop. 1.34], BC is
ΕΘἐστινἴση. εἰσὶδὲκαὶπαράλληλοι.καὶἐπιζευγνύουσιν thus equal to EH. And they are also parallel, and EB and
αὐτὰςαἱΕΒ,ΘΓ·αἱδὲτὰςἴσαςτεκαὶπαραλλήλουςἐπὶ HC join them. But (straight-lines) joining equal and parτὰαὐτὰμέρηἐπιζευγνύουσαιἴσαιτεκαὶπαράλληλοίεἰσι
allel (straight-lines) on the same sides are (themselves)
[καὶαἱΕΒ,ΘΓἄραἴσαιτέεἰσικαὶπαράλληλοι]. παραλ- equal and parallel [Prop. 1.33] [thus, EB and HC are
ληλόγραμμονἄραἐστὶτὸΕΒΓΘ.καίἐστινἴσοντῷΑΒΓΔ· also equal and parallel]. Thus, EBCH is a parallelogram
βάσιντεγὰραὐτῷτὴναὐτὴνἔχειτὴνΒΓ,καὶἐνταῖςαὐταῖς [Prop. 1.34], and is equal to ABCD. For it has the same
παραλλήλοιςἐστὶναὐτῷταῖςΒΓ,ΑΘ.δὶατὰαὐτὰδὴκαὶτὸ base, BC, as (ABCD), and is between the same paral-
ΕΖΗΘτῷαὐτῷτῷΕΒΓΘἐστινἴσον·ὥστεκαὶτὸΑΒΓΔ lels, BC and AH, as (ABCD) [Prop. 1.35]. So, for the
παραλληλόγραμμοντῷΕΖΗΘἐστινἴσον.
same (reasons), EFGH is also equal to the same (par-
Τὰἄραπαραλληλόγραμματὰἐπὶἴσωνβάσεωνὄντακαὶ allelogram) EBCH [Prop. 1.34]. So that the parallelo-
ἐνταῖςαὐταῖςπαραλλήλοιςἴσαἀλλήλοιςἐστίν· ὅπερἔδει gram ABCD is also equal to EFGH.
δεῖξαι.
Thus, parallelograms which are on equal bases and
between the same parallels are equal to one another.
(Which is) the very thing it was required to show.
ÐÞ. Proposition 37
Τὰτρίγωνατὰἐπὶτῆςαὐτῆςβάσεωςὄντακαὶἐνταῖς Triangles which are on the same base and between
αὐταῖςπαραλλήλοιςἴσαἀλλήλοιςἐστίν.
the same parallels are equal to one another.
Α ∆
A D
Ε
Ζ E
F
Β Γ
B
C
῎ΕστωτρίγωνατὰΑΒΓ,ΔΒΓἐπὶτῆςαὐτῆςβάσεωςτῆς Let ABC and DBC be triangles on the same base BC,
ΒΓκαὶἐνταῖςαὐταῖςπαραλλήλοιςταῖςΑΔ,ΒΓ·λέγω,ὅτι and between the same parallels AD and BC. I say that
ἴσονἐστὶτὸΑΒΓτρίγωνοντῷΔΒΓτριγώνῳ.
triangle ABC is equal to triangle DBC.
᾿ΕκβεβλήσθωἡΑΔἐφ᾿ἑκάτερατὰμέρηἐπὶτὰΕ,Ζ,καὶ Let AD have been produced in both directions to E
διὰμὲντοῦΒτῇΓΑπαράλληλοςἤχθωἡΒΕ,δὶαδὲτοῦΓτῇ and F , and let the (straight-line) BE have been drawn
ΒΔπαράλληλοςἤχθωἡΓΖ.παραλληλόγραμμονἄραἐστὶν through B parallel to CA [Prop. 1.31], and let the
ἑκάτεροντῶνΕΒΓΑ,ΔΒΓΖ·καίεἰσινἴσα·ἐπίτεγὰρτῆς (straight-line) CF have been drawn through C parallel
αὐτῆςβάσεώςεἰσιτῆςΒΓκαὶἐνταῖςαὐταῖςπαραλλήλοις to BD [Prop. 1.31]. Thus, EBCA and DBCF are both
ταῖςΒΓ,ΕΖ·καίἐστιτοῦμὲνΕΒΓΑπαραλληλογράμμου parallelograms, and are equal. For they are on the same
ἥμισυτὸΑΒΓτρίγωνον·ἡγὰρΑΒδιάμετροςαὐτὸδίχα base BC, and between the same parallels BC and EF
τέμνει· τοῦδὲΔΒΓΖπαραλληλογράμμουἥμισυτὸΔΒΓ [Prop. 1.35]. And the triangle ABC is half of the paralτρίγωνον·
ἡγὰρΔΓδιάμετροςαὐτὸδίχατέμνει. [τὰδὲ lelogram EBCA. For the diagonal AB cuts the latter in
38

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
τῶνἴσωνἡμίσηἴσαἀλλήλοιςἐστίν].ἴσονἄραἐστὶτὸΑΒΓ half [Prop. 1.34]. And the triangle DBC (is) half of the
τρίγωνοντῷΔΒΓτριγώνῳ.
parallelogram DBCF . For the diagonal DC cuts the lat-
Τὰἄρατρίγωνατὰἐπὶτῆςαὐτῆςβάσεωςὄντακαὶἐνταῖς ter in half [Prop. 1.34]. [And the halves of equal things
αὐταῖςπαραλλήλοιςἴσαἀλλήλοιςἐστίν·ὅπερἔδειδεῖξαι. are equal to one another.] † Thus, triangle ABC is equal
to triangle DBC.
Thus, triangles which are on the same base and
between the same parallels are equal to one another.
(Which is) the very thing it was required to show.
† This is an additional common notion.
Ð. Proposition 38
Τὰτρίγωνατὰἐπὶἴσωνβάσεωνὄντακαὶἐνταῖςαὐταῖς Triangles which are on equal bases and between the
παραλλήλοιςἴσαἀλλήλοιςἐστίν.
same parallels are equal to one another.
Η Α ∆ Θ
G A D H
Β Γ Ε Ζ B C
E F
῎ΕστωτρίγωνατὰΑΒΓ,ΔΕΖἐπὶἴσωνβάσεωντῶνΒΓ, Let ABC and DEF be triangles on the equal bases
ΕΖκαὶἐνταῖςαὐταῖςπαραλλήλοιςταῖςΒΖ,ΑΔ·λέγω,ὅτι BC and EF , and between the same parallels BF and
ἴσονἐστὶτὸΑΒΓτρίγωνοντῷΔΕΖτριγώνῳ. AD. I say that triangle ABC is equal to triangle DEF .
᾿ΕκβεβλήσθωγὰρἡΑΔἐφ᾿ἑκάτερατὰμέρηἐπὶτὰΗ, For let AD have been produced in both directions
Θ,καὶδιὰμὲντοῦΒτῇΓΑπαράλληλοςἤχθωἡΒΗ,δὶαδὲ to G and H, and let the (straight-line) BG have been
τοῦΖτῇΔΕπαράλληλοςἤχθωἡΖΘ.παραλληλόγραμμον drawn through B parallel to CA [Prop. 1.31], and let the
ἄραἐστὶνἑκάτεροντῶνΗΒΓΑ,ΔΕΖΘ·καὶἴσοντὸΗΒΓΑ (straight-line) FH have been drawn through F parallel
τῷΔΕΖΘ·ἐπίτεγὰρἴσωνβάσεώνεἰσιτῶνΒΓ,ΕΖκαὶ to DE [Prop. 1.31]. Thus, GBCA and DEFH are each
ἐνταῖςαὐταῖςπαραλλήλοιςταῖςΒΖ,ΗΘ·καίἐστιτοῦμὲν parallelograms. And GBCA is equal to DEFH. For they
ΗΒΓΑπαραλληλογράμμουἥμισυτὸΑΒΓτρίγωνον.ἡγὰρ are on the equal bases BC and EF , and between the
ΑΒδιάμετροςαὐτὸδίχατέμνει·τοῦδὲΔΕΖΘπαραλλη- same parallels BF and GH [Prop. 1.36]. And triangle
λογράμμουἥμισυτὸΖΕΔτρίγωνον·ἡγὰρΔΖδίαμετρος ABC is half of the parallelogram GBCA. For the diagoαὐτὸδίχατέμνει[τὰδὲτῶνἴσωνἡμίσηἴσαἀλλήλοιςἐστίν].
nal AB cuts the latter in half [Prop. 1.34]. And triangle
ἴσονἄραἐστὶτὸΑΒΓτρίγωνοντῷΔΕΖτριγώνῳ. FED (is) half of parallelogram DEFH. For the diagonal
Τὰἄρατρίγωνατὰἐπὶἴσωνβάσεωνὄντακαὶἐνταῖς DF cuts the latter in half. [And the halves of equal things
αὐταῖςπαραλλήλοιςἴσαἀλλήλοιςἐστίν·ὅπερἔδειδεῖξαι. are equal to one another.] Thus, triangle ABC is equal
to triangle DEF .
Thus, triangles which are on equal bases and between
the same parallels are equal to one another. (Which is)
the very thing it was required to show.
Ð. Proposition 39
Τὰἴσατρίγωνατὰἐπὶτῆςαὐτῆςβάσεωςὄντακαὶἐπὶ Equal triangles which are on the same base, and on
τὰαὐτὰμέρηκαὶἐνταῖςαὐταῖςπαραλλήλοιςἐστίν. the same side, are also between the same parallels.
῎ΕστωἴσατρίγωνατὰΑΒΓ,ΔΒΓἐπὶτῆςαὐτῆςβάσεως Let ABC and DBC be equal triangles which are on
ὄντακαὶἐπὶτὰαὐτὰμέρητῆςΒΓ·λέγω,ὅτικαὶἐνταῖς the same base BC, and on the same side (of it). I say that
39

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
αὐταῖςπαραλλήλοιςἐστίν.
Α
Ε
∆
they are also between the same parallels.
A
E
D
Β
Γ
᾿ΕπεζεύχθωγὰρἡΑΔ·λέγω,ὅτιπαράλληλόςἐστινἡ For let AD have been joined. I say that AD and BC
ΑΔτῇΒΓ.
are parallel.
Εἰ γὰρ μή, ἤχθω διὰ τοῦ Α σημείου τῇ ΒΓ εὐθείᾳ For, if not, let AE have been drawn through point A
παράλληλοςἡΑΕ,καὶἐπεζεύχθωἡΕΓ.ἴσονἄραἐστὶτὸ parallel to the straight-line BC [Prop. 1.31], and let EC
ΑΒΓ τρίγωνοντῷ ΕΒΓτριγώνῳ· ἐπίτεγὰρτῆςαὐτῆς have been joined. Thus, triangle ABC is equal to triangle
βάσεώςἐστιναὐτῷτῆςΒΓκαὶἐνταῖςαὐταῖςπαραλλήλοις. EBC. For it is on the same base as it, BC, and between
ἀλλὰτὸΑΒΓτῷΔΒΓἐστινἴσον·καὶτὸΔΒΓἄρατῷΕΒΓ the same parallels [Prop. 1.37]. But ABC is equal to
ἴσονἐστὶτὸμεῖζοντῷἐλάσσονι·ὅπερἐστὶνἀδύνατον·οὐκ DBC. Thus, DBC is also equal to EBC, the greater to
ἄραπαράλληλόςἐστινἡΑΕτῇΒΓ.ὁμοίωςδὴδείξομεν, the lesser. The very thing is impossible. Thus, AE is not
ὅτιοὐδ᾿ἄλλητιςπλὴντῆςΑΔ· ἡΑΔ ἄρατῇΒΓἐστι parallel to BC. Similarly, we can show that neither (is)
παράλληλος.
any other (straight-line) than AD. Thus, AD is parallel
Τὰἄραἴσατρίγωνατὰἐπὶτῆςαὐτῆςβάσεωςὄντακαὶ to BC.
ἐπὶτὰαὐτὰμέρηκαὶἐνταῖςαὐταῖςπαραλλήλοιςἐστίν·ὅπερ Thus, equal triangles which are on the same base, and
ἔδειδεῖξαι.
on the same side, are also between the same parallels.
(Which is) the very thing it was required to show.
Ñ. Proposition 40 †
Τὰἴσατρίγωνατὰἐπὶἴσωνβάσεωνὄντακαὶἐπὶτὰαὐτὰ Equal triangles which are on equal bases, and on the
μέρηκαὶἐνταῖςαὐταῖςπαραλλήλοιςἐστίν.
same side, are also between the same parallels.
Α
∆
B
A
D
C
Ζ
F
Β
Γ
Ε
῎ΕστωἴσατρίγωνατὰΑΒΓ,ΓΔΕἐπὶἴσωνβάσεωντῶν Let ABC and CDE be equal triangles on the equal
ΒΓ,ΓΕκαὶἐπὶτὰαὐτὰμέρη.λέγω,ὅτικαὶἐνταῖςαὐταῖς bases BC and CE (respectively), and on the same side
παραλλήλοιςἐστίν.
(of BE). I say that they are also between the same par-
᾿ΕπεζεύχθωγὰρἡΑΔ·λέγω,ὅτιπαράλληλόςἐστινἡ allels.
ΑΔτῇΒΕ.
For let AD have been joined. I say that AD is parallel
Εἰγὰρμή,ἤχθωδιὰτοῦΑτῇΒΕπαράλληλοςἡΑΖ, to BE.
καὶ ἐπεζεύχθω ἡ ΖΕ. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον For if not, let AF have been drawn through A parallel
τῷΖΓΕτριγώνῳ· ἐπίτεγὰρἴσωνβάσεώνεἰσιτῶνΒΓ, to BE [Prop. 1.31], and let FE have been joined. Thus,
ΓΕκαὶἐνταῖςαὐταῖςπαραλλήλοιςταῖςΒΕ,ΑΖ.ἀλλὰτὸ triangle ABC is equal to triangle FCE. For they are on
ΑΒΓτρίγωνονἴσονἐστὶτῷΔΓΕ[τρίγωνῳ]·καὶτὸΔΓΕ equal bases, BC and CE, and between the same paral-
ἄρα[τρίγωνον]ἴσονἐστὶτῷΖΓΕτριγώνῳτὸμεῖζοντῷ lels, BE and AF [Prop. 1.38]. But, triangle ABC is equal
B
C
E
40

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ἐλάσσονι·ὅπερἐστὶνἀδύνατον·οὐκἄραπαράλληλοςἡΑΖ to [triangle] DCE. Thus, [triangle] DCE is also equal to
τῇΒΕ.ὁμοίωςδὴδείξομεν,ὅτιοὐδ᾿ἄλλητιςπλὴντῆςΑΔ· triangle FCE, the greater to the lesser. The very thing is
ἡΑΔἄρατῇΒΕἐστιπαράλληλος.
impossible. Thus, AF is not parallel to BE. Similarly, we
Τὰἄραἴσατρίγωνατὰἐπὶἴσωνβάσεωνὄντακαὶἐπὶτὰ can show that neither (is) any other (straight-line) than
αὐτὰμέρηκαὶἐνταῖςαὐταῖςπαραλλήλοιςἐστίν·ὅπερἔδει AD. Thus, AD is parallel to BE.
δεῖξαι.
Thus, equal triangles which are on equal bases, and
on the same side, are also between the same parallels.
(Which is) the very thing it was required to show.
† This whole proposition is regarded by Heiberg as a relatively early interpolation to the original text.
Ñ. Proposition 41
᾿Εὰν παραλληλόγραμμον τριγώνῳ βάσιν τε ἔχῃ τὴν If a parallelogram has the same base as a triangle, and
αὐτὴν καὶ ἐν ταῖς αὐταῖς παραλλήλοιςᾖ, διπλάσιόν ἐστί is between the same parallels, then the parallelogram is
τὸπαραλληλόγραμμοντοῦτριγώνου.
double (the area) of the triangle.
Α ∆ Ε A D E
Β
Γ
Παραλληλόγραμμονγὰρ τὸ ΑΒΓΔ τριγώνῳ τῷ ΕΒΓ For let parallelogram ABCD have the same base BC
βάσιντεἐχέτωτὴναὐτὴντὴνΒΓκαὶἐνταῖςαὐταῖςπα- as triangle EBC, and let it be between the same parallels,
ραλλήλοιςἔστωταῖςΒΓ,ΑΕ·λέγω,ὅτιδιπλάσιόνἐστιτὸ BC and AE. I say that parallelogram ABCD is double
ΑΒΓΔπαραλληλόγραμμοντοῦΒΕΓτριγώνου.
(the area) of triangle BEC.
᾿ΕπεζεύχθωγὰρἡΑΓ.ἴσονδήἐστιτὸΑΒΓτρίγωνον For let AC have been joined. So triangle ABC is equal
τῷ ΕΒΓ τριγώνῳ· ἐπί τε γὰρ τῆς αὐτῆς βάσεώς ἐστιν to triangle EBC. For it is on the same base, BC, as
αὐτῷτῆςΒΓκαὶἐνταῖςαὐταῖςπαραλλήλοιςταῖςΒΓ,ΑΕ. (EBC), and between the same parallels, BC and AE
ἀλλὰτὸΑΒΓΔπαραλληλόγραμμονδιπλάσιόνἐστιτοῦΑΒΓ [Prop. 1.37]. But, parallelogram ABCD is double (the
τριγώνου· ἡ γὰρ ΑΓ διάμετρος αὐτὸ δίχα τέμνει· ὥστε area) of triangle ABC. For the diagonal AC cuts the forτὸΑΒΓΔπαραλληλόγραμμονκαὶτοῦΕΒΓτριγώνουἐστὶ
mer in half [Prop. 1.34]. So parallelogram ABCD is also
διπλάσιον.
double (the area) of triangle EBC.
᾿Εὰνἄραπαραλληλόγραμμοντριγώνῳβάσιντεἔχῃτὴν Thus, if a parallelogram has the same base as a trianαὐτὴνκαὶἐνταῖςαὐταῖςπαραλλήλοιςᾖ,διπλάσιόνἐστίτὸ
gle, and is between the same parallels, then the parallelπαραλληλόγραμμοντοῦτριγώνου·ὅπερἔδειδεῖξαι.
ogram is double (the area) of the triangle. (Which is) the
very thing it was required to show.
Ñ. Proposition 42
Τῷ δοθέντιτριγώνῳ ἴσονπαραλληλόγραμμονσυστή- To construct a parallelogram equal to a given triangle
σασθαιἐντῇδοθείσῃγωνίᾳεὐθυγράμμῳ.
in a given rectilinear angle.
῎ΕστωτὸμὲνδοθὲντρίγωνοντὸΑΒΓ,ἡδὲδοθεῖσα Let ABC be the given triangle, and D the given rectiγωνίαεὐθύγραμμοςἡΔ·δεῖδὴτῷΑΒΓτριγώνῳἴσονπα-
linear angle. So it is required to construct a parallelogram
ραλληλόγραμμονσυστήσασθαιἐντῇΔγωνίᾳεὐθυγράμμῳ. equal to triangle ABC in the rectilinear angle D.
B
C
41

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
∆
Α
Ζ
Η
D
A
F
G
Β Ε Γ
B E C
ΤετμήσθωἡΒΓδίχακατὰτὸΕ,καὶἐπεζεύχθωἡΑΕ, Let BC have been cut in half at E [Prop. 1.10], and
καὶσυνεστάτωπρὸςτῇΕΓεὐθείᾳκαὶτῷπρὸςαὐτῇσημείῳ let AE have been joined. And let (angle) CEF , equal to
τῷΕτῇΔγωνίᾳἴσηἡὑπὸΓΕΖ,καὶδιὰμὲντοῦΑτῇΕΓ angle D, have been constructed at the point E on the
παράλληλοςἤχθωἡΑΗ,διὰδὲτοῦΓτῇΕΖπαράλληλος straight-line EC [Prop. 1.23]. And let AG have been
ἤχθωἡΓΗ·παραλληλόγραμμονἄραἐστὶτὸΖΕΓΗ.καὶἐπεὶ drawn through A parallel to EC [Prop. 1.31], and let CG
ἴσηἐστὶνἡΒΕτῇΕΓ,ἴσονἐστὶκαὶτὸΑΒΕτρίγωνοντῷ have been drawn through C parallel to EF [Prop. 1.31].
ΑΕΓτριγώνῳ·ἐπίτεγὰρἴσωνβάσεώνεἰσιτῶνΒΕ,ΕΓκαὶ Thus, FECG is a parallelogram. And since BE is equal
ἐνταῖςαὐταῖςπαραλλήλοιςταῖςΒΓ,ΑΗ·διπλάσιονἄραἐστὶ to EC, triangle ABE is also equal to triangle AEC. For
τὸΑΒΓτρίγωνοντοῦΑΕΓτριγώνου.ἔστιδὲκαὶτὸΖΕΓΗ they are on the equal bases, BE and EC, and between
παραλληλόγραμμονδιπλάσιοντοῦΑΕΓτριγώνου·βάσιντε the same parallels, BC and AG [Prop. 1.38]. Thus, triγὰρ
αὐτῷτὴναὐτὴνἔχεικαὶἐν ταῖςαὐταῖςἐστιν αὐτῷ angle ABC is double (the area) of triangle AEC. And
παραλλήλοις·ἴσονἄραἐστὶτὸΖΕΓΗπαραλληλόγραμμον parallelogram FECG is also double (the area) of triangle
τῷΑΒΓτριγώνῳ. καὶἔχειτὴνὑπὸΓΕΖγωνίανἴσηντῇ AEC. For it has the same base as (AEC), and is between
δοθείσῃτῇΔ.
the same parallels as (AEC) [Prop. 1.41]. Thus, paral-
ΤῷἄραδοθέντιτριγώνῳτῷΑΒΓἴσονπαραλληλόγραμ- lelogram FECG is equal to triangle ABC. (FECG) also
μονσυνέσταταιτὸΖΕΓΗἐνγωνίᾳτῇὑπὸΓΕΖ,ἥτιςἐστὶν has the angle CEF equal to the given (angle) D.
ἴσητῇΔ·ὅπερἔδειποιῆσαι.
Thus, parallelogram FECG, equal to the given triangle
ABC, has been constructed in the angle CEF , which
is equal to D. (Which is) the very thing it was required
to do.
Ñ. Proposition 43
Παντὸςπαραλληλογράμμουτῶνπερὶτὴνδιάμετρονπα- For any parallelogram, the complements of the paralραλληλογράμμωντὰπαραπληρώματαἴσαἀλλήλοιςἐστίν.
lelograms about the diagonal are equal to one another.
῎Εστω παραλληλόγραμμον τὸ ΑΒΓΔ, διάμετρος δὲ Let ABCD be a parallelogram, and AC its diagonal.
αὐτοῦἡΑΓ,περὶδὲτὴνΑΓπαραλληλόγραμμαμὲνἔστω And let EH and FG be the parallelograms about AC, and
τὰΕΘ,ΖΗ,τὰδὲλεγόμεναπαραπληρώματατὰΒΚ,ΚΔ· BK and KD the so-called complements (about AC). I
λέγω, ὅτιἴσονἐστὶτὸΒΚπαραπλήρωματῷ ΚΔπαρα- say that the complement BK is equal to the complement
πληρώματι.
KD.
᾿ΕπεὶγὰρπαραλληλόγραμμόνἐστιτὸΑΒΓΔ,διάμετρος For since ABCD is a parallelogram, and AC its diagoδὲ
αὐτοῦ ἡ ΑΓ, ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΓΔ nal, triangle ABC is equal to triangle ACD [Prop. 1.34].
τριγώνῳ. πάλιν, ἐπεὶ παραλληλόγραμμόν ἐστι τὸ ΕΘ, Again, since EH is a parallelogram, and AK is its diagoδιάμετροςδὲαὐτοῦἐστινἡΑΚ,ἴσονἐστὶτὸΑΕΚτρίγωνον
nal, triangle AEK is equal to triangle AHK [Prop. 1.34].
τῷΑΘΚτριγώνῳ. διὰτὰαὐτὰδὴκαὶτὸΚΖΓτρίγωνον So, for the same (reasons), triangle KFC is also equal to
τῷΚΗΓἐστινἴσον. ἐπεὶοὖντὸμὲνΑΕΚτρίγωνοντῷ (triangle) KGC. Therefore, since triangle AEK is equal
ΑΘΚτριγώνῳἐστὶνἴσον,τὸδὲΚΖΓτῷΚΗΓ,τὸΑΕΚ to triangle AHK, and KFC to KGC, triangle AEK plus
τρίγωνονμετὰτοῦΚΗΓἴσονἐστὶτῷΑΘΚτριγώνῳμετὰ KGC is equal to triangle AHK plus KFC. And the
τοῦΚΖΓ·ἔστιδὲκαὶὅλοντὸΑΒΓτρίγωνονὅλῳτῷΑΔΓ whole triangle ABC is also equal to the whole (triangle)
ἴσον·λοιπὸνἄρατὸΒΚπαραπλήρωμαλοιπῷτῷΚΔπαρα- ADC. Thus, the remaining complement BK is equal to
42

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
πληρώματίἐστινἴσον.
Α Θ ∆
the remaining complement KD.
A
H
D
Ε
Κ
Ζ
E
K
F
Β
Η
Γ
Παντὸςἄραπαραλληλογράμμουχωρίουτῶνπερὶτὴν Thus, for any parallelogramic figure, the compleδιάμετρονπαραλληλογράμμωντὰπαραπληρώματαἴσαἀλλή-
ments of the parallelograms about the diagonal are equal
λοιςἐστίν·ὅπερἔδειδεῖξαι.
to one another. (Which is) the very thing it was required
to show.
Ñ. Proposition 44
Παρὰτὴνδοθεῖσανεὐθεῖαντῷδοθέντιτριγώνῳἴσονπα- To apply a parallelogram equal to a given triangle to
ραλληλόγραμμονπαραβαλεῖνἐντῇδοθείσῃγωνίᾳεὐθυγράμ- a given straight-line in a given rectilinear angle.
μῳ.
B
G
C
Γ
∆
C
D
Ζ
Ε
Κ
F E K
Η
Β
Μ
G
B
M
Θ Α Λ
H A L
῎ΕστωἡμὲνδοθεῖσαεὐθεῖαἡΑΒ,τὸδὲδοθὲντρίγωνον Let AB be the given straight-line, C the given trianτὸΓ,ἡδὲδοθεῖσαγωνίαεὐθύγραμμοςἡΔ·δεῖδὴπαρὰ
gle, and D the given rectilinear angle. So it is required to
τὴνδοθεῖσανεὐθεῖαντὴνΑΒτῷδοθέντιτριγώνῳτῷΓ apply a parallelogram equal to the given triangle C to the
ἴσονπαραλληλόγραμμονπαραβαλεῖνἐνἴσῃτῇΔγωνίᾳ. given straight-line AB in an angle equal to (angle) D.
ΣυνεστάτωτῷΓτριγώνῳἴσονπαραλληλόγραμμοντὸ Let the parallelogram BEFG, equal to the triangle C,
ΒΕΖΗἐνγωνίᾳτῇὑπὸΕΒΗ,ἥἐστινἴσητῇΔ·καὶκείσθω have been constructed in the angle EBG, which is equal
ὥστεἐπ᾿εὐθείαςεἶναιτὴνΒΕτῇΑΒ,καὶδιήχθωἡΖΗ to D [Prop. 1.42]. And let it have been placed so that
ἐπὶτὸΘ,καὶδιὰτοῦΑὁποτέρᾳτῶνΒΗ,ΕΖπαράλληλος BE is straight-on to AB. † And let FG have been drawn
ἤχθωἡΑΘ,καὶἐπεζεύχθωἡΘΒ.καὶἐπεὶεἰςπαραλλήλους through to H, and let AH have been drawn through A
τὰςΑΘ,ΕΖεὐθεῖαἐνέπεσενἡΘΖ,αἱἄραὑπὸΑΘΖ,ΘΖΕ parallel to either of BG or EF [Prop. 1.31], and let HB
γωνίαι δυσὶν ὀρθαῖς εἰσιν ἴσαι. αἱ ἄρα ὑπὸ ΒΘΗ, ΗΖΕ have been joined. And since the straight-line HF falls
δύοὀρθῶνἐλάσσονέςεἰσιν· αἱδὲἀπὸἐλασσόνωνἢδύο across the parallels AH and EF , the (sum of the) an-
ὀρθῶνεἰςἄπειρονἐκβαλλόμεναισυμπίπτουσιν·αἱΘΒ,ΖΕ gles AHF and HFE is thus equal to two right-angles
43

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ἄραἐκβαλλόμεναισυμπεσοῦνται.ἐκβεβλήσθωσανκαὶσυμ- [Prop. 1.29]. Thus, (the sum of) BHG and GFE is less
πιπτέτωσανκατὰτὸ Κ, καὶδιὰτοῦΚσημείουὁποτέρᾳ than two right-angles. And (straight-lines) produced to
τῶνΕΑ,ΖΘπαράλληλοςἤχθωἡΚΛ,καὶἐκβεβλήσθωσαν infinity from (internal angles whose sum is) less than two
αἱΘΑ,ΗΒἐπὶτὰΛ,Μσημεῖα. παραλληλόγραμμονἄρα right-angles meet together [Post. 5]. Thus, being pro-
ἐστὶτὸΘΛΚΖ,διάμετροςδὲαὐτοῦἡΘΚ,περὶδὲτὴνΘΚ duced, HB and FE will meet together. Let them have
παραλληλόγραμμαμὲντὰΑΗ,ΜΕ,τὰδὲλεγόμεναπαρα- been produced, and let them meet together at K. And let
πληρώματατὰΛΒ,ΒΖ·ἴσονἄραἐστὶτὸΛΒτῷΒΖ.ἀλλὰ KL have been drawn through point K parallel to either
τὸΒΖτῷΓτριγώνῳἐστὶνἴσον·καὶτὸΛΒἄρατῷΓἐστιν of EA or FH [Prop. 1.31]. And let HA and GB have
ἴσον. καὶἐπεὶἴσηἐστὶνἡὑπὸΗΒΕγωνίατῇὑπὸΑΒΜ, been produced to points L and M (respectively). Thus,
ἀλλὰἡὑπὸΗΒΕτῇΔἐστινἴση,καὶἡὑπὸΑΒΜἄρατῇΔ HLKF is a parallelogram, and HK its diagonal. And
γωνίᾳἐστὶνἴση.
AG and ME (are) parallelograms, and LB and BF the
Παρὰ τὴν δοθεῖσανἄραεὐθεῖαντὴν ΑΒ τῷ δοθέντι so-called complements, about HK. Thus, LB is equal to
τριγώνῳτῷΓἴσονπαραλληλόγραμμονπαραβέβληταιτὸΛΒ BF [Prop. 1.43]. But, BF is equal to triangle C. Thus,
ἐνγωνίᾳτῇὑπὸΑΒΜ,ἥἐστινἴσητῇΔ·ὅπερἔδειποιῆσαι. LB is also equal to C. Also, since angle GBE is equal to
ABM [Prop. 1.15], but GBE is equal to D, ABM is thus
also equal to angle D.
Thus, the parallelogram LB, equal to the given triangle
C, has been applied to the given straight-line AB in
the angle ABM, which is equal to D. (Which is) the very
thing it was required to do.
† This can be achieved using Props. 1.3, 1.23, and 1.31.
Ñ. Proposition 45
Τῷδοθέντιεὐθυγράμμῳἴσονπαραλληλόγραμμονσυστ- To construct a parallelogram equal to a given rectilinήσασθαιἐντῇδοθείσῃγωνίᾳεὐθυγράμμῳ.
ear figure in a given rectilinear angle.
῎Εστω τὸ μὲν δοθὲν εὐθύγραμμον τὸ ΑΒΓΔ, ἡ δὲ Let ABCD be the given rectilinear figure, † and E the
δοθεῖσαγωνίαεὐθύγραμμοςἡΕ·δεῖδὴτῷΑΒΓΔεὐθυ- given rectilinear angle. So it is required to construct a
γράμμῳἴσονπαραλληλόγραμμονσυστήσασθαιἐντῇδοθείσῃ parallelogram equal to the rectilinear figure ABCD in
γωνίᾳτῇΕ. the given angle E.
᾿ΕπεζεύχθωἡΔΒ,καὶσυνεστάτωτῷΑΒΔτριγώνῳ Let DB have been joined, and let the parallelogram
ἴσονπαραλληλόγραμμοντὸΖΘἐντῇὑπὸΘΚΖγωνίᾳ,ἥ FH, equal to the triangle ABD, have been constructed
ἐστινἴσητῇΕ·καὶπαραβεβλήσθωπαρὰτὴνΗΘεὐθεῖαντῷ in the angle HKF , which is equal to E [Prop. 1.42]. And
ΔΒΓτριγώνῳἴσονπαραλληλόγραμμοντὸΗΜἐντῇὑπὸ let the parallelogram GM, equal to the triangle DBC,
ΗΘΜγωνίᾳ,ἥἐστινἴσητῇΕ.καὶἐπεὶἡΕγωνίαἑκατέρᾳ have been applied to the straight-line GH in the angle
τῶνὑπὸΘΚΖ,ΗΘΜἐστινἴση,καὶἡὑπὸΘΚΖἄρατῇὑπὸ GHM, which is equal to E [Prop. 1.44]. And since angle
ΗΘΜἐστινἴση. κοινὴπροσκείσθωἡὑπὸΚΘΗ·αἱἄρα E is equal to each of (angles) HKF and GHM, (an-
ὑπὸΖΚΘ,ΚΘΗταῖςὑπὸΚΘΗ,ΗΘΜἴσαιεἰσίν. ἀλλ᾿αἱ gle) HKF is thus also equal to GHM. Let KHG have
ὑπὸΖΚΘ,ΚΘΗδυσὶνὀρθαῖςἴσαιεἰσίν·καὶαἱὑπὸΚΘΗ, been added to both. Thus, (the sum of) FKH and KHG
ΗΘΜἄραδύοὀρθαῖςἴσαιεἰσίν.πρὸςδήτινιεὐθεῖᾳτῇΗΘ is equal to (the sum of) KHG and GHM. But, (the
καὶτῷπρὸςαὐτῇσημείῳτῷΘδύοεὐθεῖαιαἱΚΘ,ΘΜμὴ sum of) FKH and KHG is equal to two right-angles
ἐπὶτὰαὐτὰμέρηκείμεναιτὰςἐφεξῆςγωνίαςδύοὀρθαῖς [Prop. 1.29]. Thus, (the sum of) KHG and GHM is
ἴσαςποιοῦσιν· ἐπ᾿εὐθείαςἄραἐστὶν ἡ ΚΘτῇ ΘΜ·καὶ also equal to two right-angles. So two straight-lines, KH
ἐπεὶεἰςπαραλλήλουςτὰςΚΜ,ΖΗεὐθεῖαἐνέπεσενἡΘΗ, and HM, not lying on the same side, make adjacent anαἱἐναλλὰξγωνίαιαἱὑπὸΜΘΗ,ΘΗΖἴσαιἀλλήλαιςεἰσίν.
gles with some straight-line GH, at the point H on it,
κοινὴπροσκείσθωἡὑπὸΘΗΛ·αἱἄραὑπὸΜΘΗ,ΘΗΛταῖς (whose sum is) equal to two right-angles. Thus, KH is
ὑπὸΘΗΖ,ΘΗΛἴσαιεἰσιν. ἀλλ᾿αἱὑπὸΜΘΗ,ΘΗΛδύο straight-on to HM [Prop. 1.14]. And since the straight-
ὀρθαῖςἴσαιεἰσίν·καὶαἱὑπὸΘΗΖ,ΘΗΛἄραδύοὀρθαῖς line HG falls across the parallels KM and FG, the al-
ἴσαιεἰσίν·ἐπ᾿εὐθείαςἄραἐστὶνἡΖΗτῇΗΛ.καὶἐπεὶἡ ternate angles MHG and HGF are equal to one another
ΖΚτῇΘΗἴσητεκαὶπαράλληλόςἐστιν,ἀλλὰκαὶἡΘΗτῇ [Prop. 1.29]. Let HGL have been added to both. Thus,
ΜΛ,καὶἡΚΖἄρατῇΜΛἴσητεκαὶπαράλληλόςἐστιν·καὶ (the sum of) MHG and HGL is equal to (the sum of)
44

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ἐπιζευγνύουσιναὐτὰςεὐθεῖαιαἱΚΜ,ΖΛ·καὶαἱΚΜ,ΖΛ HGF and HGL. But, (the sum of) MHG and HGL is
ἄραἴσαιτεκαὶπαράλληλοίεἰσιν· παραλληλόγραμμονἄρα equal to two right-angles [Prop. 1.29]. Thus, (the sum of)
ἐστὶτὸΚΖΛΜ.καὶἐπεὶἴσονἐστὶτὸμὲνΑΒΔτρίγωνον HGF and HGL is also equal to two right-angles. Thus,
τῷΖΘπαραλληλογράμμῳ,τὸδὲΔΒΓτῷΗΜ,ὅλονἄρα FG is straight-on to GL [Prop. 1.14]. And since FK is
τὸΑΒΓΔεὐθύγραμμονὅλῳτῷΚΖΛΜπαραλληλογράμμῳ equal and parallel to HG [Prop. 1.34], but also HG to
ἐστὶνἴσον.
ML [Prop. 1.34], KF is thus also equal and parallel to
ML [Prop. 1.30]. And the straight-lines KM and FL
join them. Thus, KM and FL are equal and parallel as
well [Prop. 1.33]. Thus, KFLM is a parallelogram. And
since triangle ABD is equal to parallelogram FH, and
DBC to GM, the whole rectilinear figure ABCD is thus
equal to the whole parallelogram KFLM.
∆
D
Γ
C
Α
A
Ζ
Β
Η
Λ
Ε
F
G
B
L
E
Κ Θ Μ
K H M
Τῷ ἄραδοθέντιεὐθυγράμμῳτῷ ΑΒΓΔἴσον παραλ- Thus, the parallelogram KFLM, equal to the given
ληλόγραμμονσυνέσταταιτὸΚΖΛΜἐνγωνίᾳτῇὑπὸΖΚΜ, rectilinear figure ABCD, has been constructed in the an-
ἥἐστινἴσητῇδοθείσῃτῇΕ·ὅπερἔδειποιῆσαι.
gle FKM, which is equal to the given (angle) E. (Which
is) the very thing it was required to do.
† The proof is only given for a four-sided figure. However, the extension to many-sided figures is trivial.
Ѧ. Proposition 46
Ἀπὸτῆςδοθείσηςεὐθείαςτετράγωνονἀναγράψαι. To describe a square on a given straight-line.
῎Εστω ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ· δεῖ δὴ ἀπὸ τῆς ΑΒ Let AB be the given straight-line. So it is required to
εὐθείαςτετράγωνονἀναγράψαι.
describe a square on the straight-line AB.
῎ΗχθωτῇΑΒεὐθείᾳἀπὸτοῦπρὸςαὐτῇσημείουτοῦ Let AC have been drawn at right-angles to the
ΑπρὸςὀρθὰςἡΑΓ,καὶκείσθωτῇΑΒἴσηἡΑΔ·καὶδιὰ straight-line AB from the point A on it [Prop. 1.11],
μὲντοῦΔσημείουτῇΑΒπαράλληλοςἤχθωἡΔΕ,διὰ and let AD have been made equal to AB [Prop. 1.3].
δὲτοῦΒσημείουτῇΑΔπαράλληλοςἤχθωἡΒΕ.παραλ- And let DE have been drawn through point D parallel to
ληλόγραμμονἄραἐστὶτὸΑΔΕΒ·ἴσηἄραἐστὶνἡμὲνΑΒ AB [Prop. 1.31], and let BE have been drawn through
τῇΔΕ,ἡδὲΑΔτῇΒΕ.ἀλλὰἡΑΒτῇΑΔἐστινἴση· point B parallel to AD [Prop. 1.31]. Thus, ADEB is a
αἱτέσσαρεςἄρααἱΒΑ,ΑΔ,ΔΕ,ΕΒἴσαιἀλλήλαιςεἰσίν· parallelogram. Therefore, AB is equal to DE, and AD to
ἰσόπλευρονἄραἐστὶτὸΑΔΕΒπαραλληλόγραμμον. λέγω BE [Prop. 1.34]. But, AB is equal to AD. Thus, the four
δή,ὅτικαὶὀρθογώνιον.ἐπεὶγὰρεἰςπαραλλήλουςτὰςΑΒ, (sides) BA, AD, DE, and EB are equal to one another.
ΔΕεὐθεῖαἐνέπεσενἡΑΔ,αἱἄραὑπὸΒΑΔ,ΑΔΕγωνίαι Thus, the parallelogram ADEB is equilateral. So I say
δύοὀρθαῖςἴσαιεἰσίν. ὀρθὴδὲἡὑπὸΒΑΔ·ὀρθὴἄρακαὶ that (it is) also right-angled. For since the straight-line
45

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
ἡὑπὸΑΔΕ. τῶνδὲπαραλληλογράμμωνχωρίων αἱἀπε- AD falls across the parallels AB and DE, the (sum of
ναντίονπλευραίτεκαὶγωνίαιἴσαιἀλλήλαιςεἰσίν·ὀρθὴἄρα the) angles BAD and ADE is equal to two right-angles
καὶἑκατέρατῶνἀπεναντίοντῶνὑπὸΑΒΕ,ΒΕΔγωνιῶν· [Prop. 1.29]. But BAD (is a) right-angle. Thus, ADE
ὀρθογώνιονἄραἐστὶτὸΑΔΕΒ.ἐδείχθηδὲκαὶἰσόπλευρον. (is) also a right-angle. And for parallelogrammic figures,
the opposite sides and angles are equal to one another
[Prop. 1.34]. Thus, each of the opposite angles ABE
and BED (are) also right-angles. Thus, ADEB is rightangled.
And it was also shown (to be) equilateral.
Γ
C
∆
Ε
D
E
Α
Β
Τετράγωνονἄραἐστίν·καίἐστινἀπὸτῆςΑΒεὐθείας Thus, (ADEB) is a square [Def. 1.22]. And it is de-
ἀναγεγραμμένον·ὅπερἔδειποιῆσαι.
scribed on the straight-line AB. (Which is) the very thing
it was required to do.
ÑÞ. Proposition 47
᾿Εντοῖςὀρθογωνίοιςτριγώνοιςτὸἀπὸτῆςτὴνὀρθὴν In right-angled triangles, the square on the side subγωνίανὑποτεινούσηςπλευρᾶςτετράγωνονἴσονἐστὶτοῖς
tending the right-angle is equal to the (sum of the)
ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν πλευρῶν τε- squares on the sides containing the right-angle.
τραγώνοις.
Let ABC be a right-angled triangle having the angle
῎ΕστωτρίγωνονὀρθογώνιοντὸΑΒΓὀρθὴνἔχοντὴν BACa right-angle. I say that the square on BC is equal
ὑπὸΒΑΓγωνίαν· λέγω,ὅτιτὸἀπὸτῆςΒΓτετράγωνον to the (sum of the) squares on BA and AC.
ἴσονἐστὶτοῖςἀπὸτῶνΒΑ,ΑΓτετραγώνοις.
For let the square BDEC have been described on
Ἀναγεγράφθω γὰρ ἀπὸ μὲν τῆς ΒΓ τετράγωνον τὸ BC, and (the squares) GB and HC on AB and AC
ΒΔΕΓ, ἀπὸ δὲ τῶν ΒΑ, ΑΓ τὰ ΗΒ, ΘΓ, καὶ διὰ τοῦ (respectively) [Prop. 1.46]. And let AL have been
Α ὁποτέρᾳ τῶν ΒΔ, ΓΕ παράλληλος ἤχθω ἡ ΑΛ· καὶ drawn through point A parallel to either of BD or CE
ἐπεζεύχθωσαν αἱ ΑΔ, ΖΓ. καὶ ἐπεὶ ὀρθή ἐστιν ἑκατέρα [Prop. 1.31]. And let AD and FC have been joined. And
τῶνὑπὸΒΑΓ,ΒΑΗγωνιῶν,πρὸςδήτινιεὐθείᾳτῇΒΑ since angles BAC and BAG are each right-angles, then
καὶτῷπρὸςαὐτῇσημείῳτῷΑδύοεὐθεῖαιαἱΑΓ,ΑΗμὴ two straight-lines AC and AG, not lying on the same
ἐπὶτὰαὐτὰμέρηκείμεναιτὰςἐφεξῆςγωνίαςδυσὶνὀρθαῖς side, make the adjacent angles with some straight-line
ἴσαςποιοῦσιν·ἐπ᾿εὐθείαςἄραἐστὶνἡΓΑτῇΑΗ.διὰτὰ BA, at the point A on it, (whose sum is) equal to two
αὐτὰδὴκαὶἡΒΑτῇΑΘἐστινἐπ᾿εὐθείας. καὶἐπεὶἴση right-angles. Thus, CA is straight-on to AG [Prop. 1.14].
ἐστὶνἡὑπὸΔΒΓγωνίατῇὑπὸΖΒΑ·ὀρθὴγὰρἑκατέρα· So, for the same (reasons), BA is also straight-on to AH.
κοινὴπροσκείσθωἡὑπὸΑΒΓ·ὅληἄραἡὑπὸΔΒΑὅλῃτῇ And since angle DBC is equal to FBA, for (they are)
ὑπὸΖΒΓἐστινἴση.καὶἐπεὶἴσηἐστὶνἡμὲνΔΒτῇΒΓ,ἡ both right-angles, let ABC have been added to both.
δὲΖΒτῇΒΑ,δύοδὴαἱΔΒ,ΒΑδύοταῖςΖΒ,ΒΓἴσαιεἰσὶν Thus, the whole (angle) DBA is equal to the whole (an-
ἑκατέραἑκατέρᾳ·καὶγωνίαἡὑπὸΔΒΑγωνίᾳτῇὑπὸΖΒΓ gle) FBC. And since DB is equal to BC, and FB to
ἴση·βάσιςἄραἡΑΔβάσειτῇΖΓ[ἐστιν]ἴση,καὶτὸΑΒΔ BA, the two (straight-lines) DB, BA are equal to the
A
B
46

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
τρίγωνοντῷΖΒΓτριγώνῳἐστὶνἴσον·καί[ἐστι]τοῦμὲν two (straight-lines) CB, BF , † respectively. And angle
ΑΒΔτριγώνουδιπλάσιοντὸΒΛπαραλληλόγραμμον·βάσιν DBA (is) equal to angle FBC. Thus, the base AD [is]
τεγὰρτὴναὐτὴνἔχουσιτὴνΒΔκαὶἐνταῖςαὐταῖςεἰσι equal to the base FC, and the triangle ABD is equal to
παραλλήλοιςταῖςΒΔ,ΑΛ·τοῦδὲΖΒΓτριγώνουδιπλάσιον the triangle FBC [Prop. 1.4]. And parallelogram BL
τὸΗΒτετράγωνον·βάσιντεγὰρπάλιντὴναὐτὴνἔχουσι [is] double (the area) of triangle ABD. For they have
τὴνΖΒκαὶἐνταῖςαὐταῖςεἰσιπαραλλήλοιςταῖςΖΒ,ΗΓ. the same base, BD, and are between the same parallels,
[τὰδὲτῶνἴσωνδιπλάσιαἴσαἀλλήλοιςἐστίν·]ἴσονἄραἐστὶ BD and AL [Prop. 1.41]. And square GB is double (the
καὶτὸΒΛπαραλληλόγραμμοντῷΗΒτετραγώνῳ. ὁμοίως area) of triangle FBC. For again they have the same
δὴἐπιζευγνυμένωντῶν ΑΕ, ΒΚδειχθήσεταικαὶτὸΓΛ base, FB, and are between the same parallels, FB and
παραλληλόγραμμονἴσοντῷΘΓτετραγώνῳ·ὅλονἄρατὸ GC [Prop. 1.41]. [And the doubles of equal things are
ΒΔΕΓτετράγωνονδυσὶτοῖςΗΒ,ΘΓτετραγώνοιςἴσον equal to one another.] ‡ Thus, the parallelogram BL is
ἐστίν.καίἐστιτὸμὲνΒΔΕΓτετράγωνονἀπὸτῆςΒΓἀνα- also equal to the square GB. So, similarly, AE and BK
γραφέν,τὰδὲΗΒ,ΘΓἀπὸτῶνΒΑ,ΑΓ.τὸἄραἀπὸτῆς being joined, the parallelogram CL can be shown (to
ΒΓπλευρᾶςτετράγωνονἴσονἐστὶτοῖςἀπὸτῶνΒΑ,ΑΓ be) equal to the square HC.
πλευρῶντετραγώνοις.
Θ
Thus, the whole square
BDEC is equal to the (sum of the) two squares GB and
HC. And the square BDEC is described on BC, and
the (squares) GB and HC on BA and AC (respectively).
Thus, the square on the side BC is equal to the (sum of
the) squares on the sides BA and AC.
H
Κ
K
Η
Α
G
A
Ζ
F
Β
Γ
B
C
∆ Λ Ε
D L E
᾿Εν ἄρα τοῖς ὀρθογωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν Thus, in right-angled triangles, the square on the
ὀρθὴνγωνίανὑποτεινούσηςπλευρᾶςτετράγωνονἴσονἐστὶ side subtending the right-angle is equal to the (sum of
τοῖςἀπὸτῶντὴνὀρθὴν[γωνίαν]περιεχουσῶνπλευρῶντε- the) squares on the sides surrounding the right-[angle].
τραγώνοις·ὅπερἔδειδεῖξαι.
(Which is) the very thing it was required to show.
† The Greek text has “FB, BC”, which is obviously a mistake.
‡ This is an additional common notion.
47

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 1
Ñ. Proposition 48
᾿Εὰντριγώνουτὸἀπὸμιᾶςτῶνπλευρῶντετράγωνον If the square on one of the sides of a triangle is equal
ἴσονᾖτοῖςἀπὸτῶνλοιπῶντοῦτριγώνουδύοπλευρῶν to the (sum of the) squares on the two remaining sides of
τετραγώνοις, ἡ περιεχομένη γωνία ὑπὸ τῶν λοιπῶν τοῦ the triangle then the angle contained by the two remainτριγώνουδύοπλευρῶνὀρθήἐστιν.
ing sides of the triangle is a right-angle.
Γ
C
∆ Α Β
D A B
ΤριγώνουγὰρτοῦΑΒΓτὸἀπὸμιᾶςτῆςΒΓπλευρᾶς For let the square on one of the sides, BC, of triangle
τετράγωνονἴσονἔστωτοῖςἀπὸτῶνΒΑ,ΑΓπλευρῶντε- ABC be equal to the (sum of the) squares on the sides
τραγώνοις·λέγω,ὅτιὀρθήἐστινἡὑπὸΒΑΓγωνία. BA and AC. I say that angle BAC is a right-angle.
῎ΗχθωγὰρἀπὸτοῦΑσημείουτῇΑΓεὐθείᾳπρὸςὀρθὰς For let AD have been drawn from point A at right-
ἡΑΔκαὶκείσθωτῇΒΑἴσηἡΑΔ,καὶἐπεζεύχθωἡΔΓ. angles to the straight-line AC [Prop. 1.11], and let AD
ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ, ἴσον ἐστὶ καὶ τὸ ἀπὸ τῆς have been made equal to BA [Prop. 1.3], and let DC
ΔΑτετράγωνοντῷἀπὸτῆςΑΒτετραγώνῳ. κοινὸνπρο- have been joined. Since DA is equal to AB, the square
σκείσθωτὸἀπὸτῆςΑΓτετράγωνον·τὰἄραἀπὸτῶνΔΑ, on DA is thus also equal to the square on AB. † Let the
ΑΓτετράγωναἴσαἐστὶτοῖςἀπὸτῶνΒΑ,ΑΓτετραγώνοις. square on AC have been added to both. Thus, the (sum
ἀλλὰτοῖςμὲνἀπὸτῶνΔΑ,ΑΓἴσονἐστὶτὸἀπὸτῆςΔΓ· of the) squares on DA and AC is equal to the (sum of
ὀρθὴγάρἐστινἡὑπὸΔΑΓγωνία·τοῖςδὲἀπὸτῶνΒΑ, the) squares on BA and AC. But, the (square) on DC is
ΑΓἴσονἐστὶτὸἀπὸτῆςΒΓ·ὑπόκειταιγάρ·τὸἄραἀπὸ equal to the (sum of the squares) on DA and AC. For anτῆςΔΓτετράγωνονἴσονἐστὶτῷἀπὸτῆςΒΓτετραγώνῳ·
gle DAC is a right-angle [Prop. 1.47]. But, the (square)
ὥστεκαὶπλευρὰἡΔΓτῇΒΓἐστινἴση·καὶἐπεὶἴσηἐστὶν on BC is equal to (sum of the squares) on BA and AC.
ἡΔΑτῇΑΒ,κοινὴδὲἡΑΓ,δύοδὴαἱΔΑ,ΑΓδύοταῖς For (that) was assumed. Thus, the square on DC is equal
ΒΑ,ΑΓἴσαιεἰσίν·καὶβάσιςἡΔΓβάσειτῇΒΓἴση·γωνία to the square on BC. So side DC is also equal to (side)
ἄραἡὑπὸΔΑΓγωνίᾳτῇὑπὸΒΑΓ[ἐστιν]ἴση.ὀρθὴδὲἡ BC. And since DA is equal to AB, and AC (is) com-
ὑπὸΔΑΓ·ὀρθὴἄρακαὶἡὑπὸΒΑΓ.
mon, the two (straight-lines) DA, AC are equal to the
᾿Εὰνἀρὰτριγώνουτὸἀπὸμιᾶςτῶνπλευρῶντετράγωνον two (straight-lines) BA, AC. And the base DC is equal
ἴσονᾖτοῖςἀπὸτῶνλοιπῶντοῦτριγώνουδύοπλευρῶν to the base BC. Thus, angle DAC [is] equal to angle
τετραγώνοις, ἡ περιεχομένη γωνία ὑπὸ τῶν λοιπῶν τοῦ BAC [Prop. 1.8]. But DAC is a right-angle. Thus, BAC
τριγώνουδύοπλευρῶνὀρθήἐστιν·ὅπερἔδειδεῖξαι. is also a right-angle.
Thus, if the square on one of the sides of a triangle is
equal to the (sum of the) squares on the remaining two
sides of the triangle then the angle contained by the remaining
two sides of the triangle is a right-angle. (Which
is) the very thing it was required to show.
† Here, use is made of the additional common notion that the squares of equal things are themselves equal. Later on, the inverse notion is used.
48

ELEMENTS BOOK 2
Fundamentals of Geometric Algebra
49

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 2
Definitions
ÎÇÖÓ.
αʹ. Πᾶν παραλληλόγραμμον ὀρθογώνιον περιέχεσθαι 1. Any rectangular parallelogram is said to be conλέγεται
ὑπὸ δύο τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν tained by the two straight-lines containing the rightεὐθειῶν.
angle.
βʹ.Παντὸςδὲπαραλληλογράμμουχωρίουτῶνπερὶτὴν 2. And in any parallelogrammic figure, let any one
διάμετροναὐτοῦπαραλληλογράμμωνἓνὁποιονοῦνσὺντοῖς whatsoever of the parallelograms about its diagonal,
δυσὶπαραπληρώμασιγνώμωνκαλείσθω.
(taken) with its two complements, be called a gnomon.
. Proposition 1 †
᾿Εὰνὦσιδύοεὐθεῖαι,τμηθῇδὲἡἑτέρααὐτῶνεἰςὁσα- If there are two straight-lines, and one of them is cut
δηποτοῦντμήματα,τὸπεριεχόμενονὀρθογώνιονὑπὸτῶν into any number of pieces whatsoever, then the rectangle
δύοεὐθειῶνἴσονἐστὶτοῖςὑπότετῆςἀτμήτουκαὶἑκάστου contained by the two straight-lines is equal to the (sum
τῶντμημάτωνπεριεχομένοιςὀρθογωνίοις.
Β
Α
∆
Ε
Γ
of the) rectangles contained by the uncut (straight-line),
and every one of the pieces (of the cut straight-line).
A
B D E C
Η
Ζ
Κ
Λ
Θ
῎ΕστωσανδύοεὐθεῖαιαἱΑ,ΒΓ,καὶτετμήσθωἡΒΓ, Let A and BC be the two straight-lines, and let BC
ὡςἔτυχεν,κατὰτὰΔ,Εσημεῖα·λέγω,ὅτιτὸὑπὸτῶνΑ, be cut, at random, at points D and E. I say that the rect-
ΒΓπεριεχομένονὀρθογώνιονἴσονἐστὶτῷτεὑπὸτῶνΑ, angle contained by A and BC is equal to the rectangle(s)
ΒΔπεριεχομένῳὀρθογωνίῳκαὶτῷὑπὸτῶνΑ,ΔΕκαὶἔτι contained by A and BD, by A and DE, and, finally, by A
τῷὑπὸτῶνΑ,ΕΓ.
and EC.
῎ΗχθωγὰρἀπὸτοῦΒτῇΒΓπρὸςὀρθὰςἡΒΖ,καὶ For let BF have been drawn from point B, at rightκείσθωτῇΑἴσηἡΒΗ,καὶδιὰμὲντοῦΗτῇΒΓπαράλληλος
angles to BC [Prop. 1.11], and let BG be made equal
ἤχθωἡΗΘ,διὰδὲτῶνΔ,Ε,ΓτῇΒΗπαράλληλοιἤχθωσαν to A [Prop. 1.3], and let GH have been drawn through
αἱΔΚ,ΕΛ,ΓΘ.
(point) G, parallel to BC [Prop. 1.31], and let DK, EL,
῎ΙσονδήἐστιτὸΒΘτοῖςΒΚ,ΔΛ,ΕΘ.καίἐστιτὸ and CH have been drawn through (points) D, E, and C
μὲνΒΘτὸὑπὸτῶνΑ,ΒΓ·περιέχεταιμὲνγὰρὑπὸτῶν (respectively), parallel to BG [Prop. 1.31].
ΗΒ,ΒΓ,ἴσηδὲἡΒΗτῇΑ·τὸδὲΒΚτὸὑπὸτῶνΑ,ΒΔ· So the (rectangle) BH is equal to the (rectangles)
περιέχεταιμὲνγὰρὑπὸτῶνΗΒ,ΒΔ,ἴσηδὲἡΒΗτῇΑ.τὸ BK, DL, and EH. And BH is the (rectangle contained)
δὲΔΛτὸὑπὸτῶνΑ,ΔΕ·ἴσηγὰρἡΔΚ,τουτέστινἡΒΗ, by A and BC. For it is contained by GB and BC, and BG
τῇΑ.καὶἔτιὁμοίωςτὸΕΘτὸὑπὸτῶνΑ,ΕΓ·τὸἄραὑπὸ (is) equal to A. And BK (is) the (rectangle contained) by
τῶνΑ,ΒΓἴσονἐστὶτῷτεὑπὸΑ,ΒΔκαὶτῷὑπὸΑ,ΔΕ A and BD. For it is contained by GB and BD, and BG
καὶἔτιτῷὑπὸΑ,ΕΓ.
(is) equal to A. And DL (is) the (rectangle contained) by
᾿Εὰνἄραὦσιδύοεὐθεῖαι,τμηθῇδὲἡἑτέρααὐτῶνεἰς A and DE. For DK, that is to say BG [Prop. 1.34], (is)
ὁσαδηποτοῦντμήματα, τὸπεριεχόμενονὀρθογώνιονὑπὸ equal to A. Similarly, EH (is) also the (rectangle conτῶνδύοεὐθειῶνἴσονἐστὶτοῖςὑπότετῆςἀτμήτουκαὶ
tained) by A and EC. Thus, the (rectangle contained)
ἑκάστουτῶντμημάτωνπεριεχομένοιςὀρθογωνίοις· ὅπερ by A and BC is equal to the (rectangles contained) by A
G
F
K
L
H
50

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 2
ἔδειδεῖξαι.
and BD, by A and DE, and, finally, by A and EC.
Thus, if there are two straight-lines, and one of them
is cut into any number of pieces whatsoever, then the
rectangle contained by the two straight-lines is equal
to the (sum of the) rectangles contained by the uncut
(straight-line), and every one of the pieces (of the cut
straight-line). (Which is) the very thing it was required
to show.
† This proposition is a geometric version of the algebraic identity: a(b + c + d + · · · ) = a b + a c + a d + · · · .
. Proposition 2 †
᾿Εὰνεὐθεῖαγραμμὴτμηθῇ,ὡςἔτυχεν,τὸὑπὸτῆςὅλης If a straight-line is cut at random then the (sum of
καὶἑκατέρουτῶντμημάτωνπεριεχόμενονὀρθογώνιονἴσον the) rectangle(s) contained by the whole (straight-line),
ἐστὶτῷἀπὸτῆςὅληςτετραγώνῳ.
and each of the pieces (of the straight-line), is equal to
the square on the whole.
Α
Γ
Β
A
C
B
∆
Ζ
Ε
D F E
ΕὐθεῖαγὰρἡΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰτὸΓ For let the straight-line AB have been cut, at random,
σημεῖον· λέγω, ὅτι τὸ ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον at point C. I say that the rectangle contained by AB and
ὀρθογώνιον μετὰ τοῦ ὑπὸ ΒΑ, ΑΓ περιεχομένου ὀρθο- BC, plus the rectangle contained by BA and AC, is equal
γωνίουἴσονἐστὶτῷἀπὸτῆςΑΒτετραγώνῳ.
to the square on AB.
ἈναγεγράφθωγὰρἀπὸτῆςΑΒτετράγωνοντὸΑΔΕΒ, For let the square ADEB have been described on AB
καὶἤχθωδιὰτοῦΓὁποτέρᾳτῶνΑΔ,ΒΕπαράλληλοςἡ [Prop. 1.46], and let CF have been drawn through C,
ΓΖ. parallel to either of AD or BE [Prop. 1.31].
῎ΙσονδήἐστὶτὸΑΕτοῖςΑΖ,ΓΕ.καίἐστιτὸμὲνΑΕ So the (square) AE is equal to the (rectangles) AF
τὸἀπὸτῆςΑΒτετράγωνον,τὸδὲΑΖτὸὑπὸτῶνΒΑ, and CE. And AE is the square on AB. And AF (is) the
ΑΓπεριεχόμενονὀρθογώνιον·περιέχεταιμὲνγὰρὑπὸτῶν rectangle contained by the (straight-lines) BA and AC.
ΔΑ,ΑΓ,ἴσηδὲἡΑΔτῇΑΒ·τὸδὲΓΕτὸὑπὸτῶνΑΒ, For it is contained by DA and AC, and AD (is) equal to
ΒΓ·ἴσηγὰρἡΒΕτῇΑΒ.τὸἄραὑπὸτῶνΒΑ,ΑΓμετὰ AB. And CE (is) the (rectangle contained) by AB and
τοῦὑπὸτῶνΑΒ,ΒΓἴσονἐστὶτῷἀπὸτῆςΑΒτετραγώνῳ. BC. For BE (is) equal to AB. Thus, the (rectangle con-
᾿Εὰνἄραεὐθεῖαγραμμὴτμηθῇ,ὡςἔτυχεν,τὸὑπὸτῆς tained) by BA and AC, plus the (rectangle contained) by
ὅληςκαὶἑκατέρουτῶντμημάτωνπεριεχόμενονὀρθογώνιον AB and BC, is equal to the square on AB.
ἴσονἐστὶτῷἀπὸτῆςὅληςτετραγώνῳ·ὅπερἔδειδεῖξαι. Thus, if a straight-line is cut at random then the (sum
of the) rectangle(s) contained by the whole (straightline),
and each of the pieces (of the straight-line), is equal
to the square on the whole. (Which is) the very thing it
was required to show.
51

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 2
† This proposition is a geometric version of the algebraic identity: a b + a c = a 2 if a = b + c.
. Proposition 3 †
᾿Εὰνεὐθεῖαγραμμὴτμηθῇ,ὡςἔτυχεν,τὸὑπὸτῆςὅλης If a straight-line is cut at random then the rectangle
καὶἑνὸςτῶντμημάτωνπεριεχόμενονὀρθογώνιονἴσονἐστὶ contained by the whole (straight-line), and one of the
τῷτεὑπὸτῶντμημάτωνπεριεχομένῳὀρθογωνίῳκαὶτῷ pieces (of the straight-line), is equal to the rectangle con-
ἀπὸτοῦπροειρημένουτμήματοςτετραγώνῳ.
tained by (both of) the pieces, and the square on the
aforementioned piece.
Α
Γ
Β
A
C
B
Ζ
∆
Ε
F D E
ΕὐθεῖαγὰρἡΑΒτετμήσθω, ὡςἔτυχεν, κατὰτὸΓ· For let the straight-line AB have been cut, at random,
λέγω,ὅτιτὸὑπὸτῶνΑΒ,ΒΓπεριεχόμενονὀρθογώνιον at (point) C. I say that the rectangle contained by AB
ἴσονἐστὶτῷτεὑπὸτῶνΑΓ,ΓΒπεριεχομένῳὀρθογωνίῳ and BC is equal to the rectangle contained by AC and
μετὰτοῦἀπὸτῆςΒΓτετραγώνου.
CB, plus the square on BC.
ἈναγεγράφθωγὰρἀπὸτῆςΓΒτετράγωνοντὸΓΔΕΒ, For let the square CDEB have been described on CB
καὶδιήχθωἡΕΔἐπὶτὸΖ,καὶδιὰτοῦΑὁποτέρᾳτῶνΓΔ, [Prop. 1.46], and let ED have been drawn through to
ΒΕπαράλληλοςἤχθωἡΑΖ.ἴσονδήἐστιτὸΑΕτοῖςΑΔ, F , and let AF have been drawn through A, parallel to
ΓΕ·καίἐστιτὸμὲνΑΕτὸὑπὸτῶνΑΒ,ΒΓπεριεχόμενον either of CD or BE [Prop. 1.31]. So the (rectangle) AE
ὀρθογώνιον·περιέχεταιμὲνγὰρὑπὸτῶνΑΒ,ΒΕ,ἴσηδὲἡ is equal to the (rectangle) AD and the (square) CE. And
ΒΕτῇΒΓ·τὸδὲΑΔτὸὑπὸτῶνΑΓ,ΓΒ·ἴσηγὰρἡΔΓ AE is the rectangle contained by AB and BC. For it is
τῇΓΒ·τὸδὲΔΒτὸἀπὸτῆςΓΒτετράγωνον·τὸἄραὑπὸ contained by AB and BE, and BE (is) equal to BC. And
τῶνΑΒ,ΒΓπεριεχόμενονὀρθογώνιονἴσονἐστὶτῷὑπὸ AD (is) the (rectangle contained) by AC and CB. For
τῶνΑΓ,ΓΒπεριεχομένῳὀρθογωνίῳμετὰτοῦἀπὸτῆςΒΓ DC (is) equal to CB. And DB (is) the square on CB.
τετραγώνου.
Thus, the rectangle contained by AB and BC is equal to
᾿Εὰνἄραεὐθεῖαγραμμὴτμηθῇ,ὡςἔτυχεν,τὸὑπὸτῆς the rectangle contained by AC and CB, plus the square
ὅληςκαὶἑνὸςτῶντμημάτωνπεριεχόμενονὀρθογώνιονἴσον on BC.
ἐστὶτῷτεὑπὸτῶντμημάτωνπεριεχομένῳὀρθογωνίῳκαὶ Thus, if a straight-line is cut at random then the rectτῷἀπὸτοῦπροειρημένουτμήματοςτετραγώνῳ·ὅπερἔδει
angle contained by the whole (straight-line), and one of
δεῖξαι.
the pieces (of the straight-line), is equal to the rectangle
contained by (both of) the pieces, and the square on the
aforementioned piece. (Which is) the very thing it was
required to show.
† This proposition is a geometric version of the algebraic identity: (a + b) a = a b + a 2 .
. Proposition 4 †
᾿Εὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς If a straight-line is cut at random then the square
ὅληςτετράγωνονἴσονἐστὶτοῖςτεἀπὸτῶντμημάτωντε- on the whole (straight-line) is equal to the (sum of the)
τραγώνοιςκαὶτῷδὶςὑπὸτῶντμημάτωνπεριεχομένῳὀρθο- squares on the pieces (of the straight-line), and twice the
52

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 2
γωνίῳ.
Α
Γ
Β
rectangle contained by the pieces.
A C B
Θ
Η
Κ
H
G
K
∆
Ζ
Ε
D F E
ΕὐθεῖαγὰργραμμὴἡΑΒτετμήσθω,ὡςἔτυχεν,κατὰ For let the straight-line AB have been cut, at random,
τὸΓ.λέγω,ὅτιτὸἀπὸτῆςΑΒτετράγωνονἴσονἐστὶτοῖς at (point) C. I say that the square on AB is equal to
τεἀπὸτῶνΑΓ,ΓΒτετραγώνοιςκαὶτῷδὶςὑπὸτῶνΑΓ, the (sum of the) squares on AC and CB, and twice the
ΓΒπεριεχομένῳὀρθογωνίῳ.
rectangle contained by AC and CB.
ἈναγεγράφθωγὰρἀπὸτῆςΑΒτετράγωνοντὸΑΔΕΒ, For let the square ADEB have been described on AB
καὶἐπεζεύχθωἡΒΔ,καὶδιὰμὲντοῦΓὁποτέρᾳτῶνΑΔ, [Prop. 1.46], and let BD have been joined, and let CF
ΕΒπαράλληλοςἤχθωἡΓΖ,διὰδὲτοῦΗὁποτέρᾳτῶνΑΒ, have been drawn through C, parallel to either of AD or
ΔΕπαράλληλοςἤχθωἡΘΚ.καὶἐπεὶπαράλληλόςἐστινἡ EB [Prop. 1.31], and let HK have been drawn through
ΓΖτῇΑΔ,καὶεἰςαὐτὰςἐμπέπτωκενἡΒΔ,ἡἐκτὸςγωνία G, parallel to either of AB or DE [Prop. 1.31]. And since
ἡὑπὸΓΗΒἴσηἐστὶτῇἐντὸςκαὶἀπεναντίοντῇὑπὸΑΔΒ. CF is parallel to AD, and BD has fallen across them, the
ἀλλ᾿ἡὑπὸΑΔΒτῇὑπὸΑΒΔἐστινἴση,ἐπεὶκαὶπλευρὰἡ external angle CGB is equal to the internal and opposite
ΒΑτῇΑΔἐστινἴση·καὶἡὑπὸΓΗΒἄραγωνιάτῇὑπὸΗΒΓ (angle) ADB [Prop. 1.29]. But, ADB is equal to ABD,
ἐστινἴση·ὥστεκαὶπλευρὰἡΒΓπλευρᾷτῇΓΗἐστινἴση· since the side BA is also equal to AD [Prop. 1.5]. Thus,
ἀλλ᾿ἡμὲνΓΒτῇΗΚἐστινἴση.ἡδὲΓΗτῇΚΒ·καὶἡΗΚ angle CGB is also equal to GBC. So the side BC is
ἄρατῇΚΒἐστινἴση·ἰσόπλευρονἄραἐστὶτὸΓΗΚΒ.λέγω equal to the side CG [Prop. 1.6]. But, CB is equal to
δή,ὅτικαὶὀρθογώνιον. ἐπεὶγὰρπαράλληλόςἐστινἡΓΗ GK, and CG to KB [Prop. 1.34]. Thus, GK is also equal
τῇΒΚ[καὶεἰςαὐτὰςἐμπέπτωκενεὐθεῖαἡΓΒ],αἱἄραὑπὸ to KB. Thus, CGKB is equilateral. So I say that (it is)
ΚΒΓ,ΗΓΒγωνίαιδύοὀρθαῖςεἰσινἴσαι. ὀρθὴδὲἡὑπὸ also right-angled. For since CG is parallel to BK [and the
ΚΒΓ·ὀρθὴἄρακαὶἡὑπὸΒΓΗ·ὥστεκαὶαἱἀπεναντίον straight-line CB has fallen across them], the angles KBC
αἱὑπὸΓΗΚ,ΗΚΒὀρθαίεἰσιν. ὀρθογώνιονἄραἐστὶτὸ and GCB are thus equal to two right-angles [Prop. 1.29].
ΓΗΚΒ·ἐδείχθηδὲκαὶἰσόπλευρον·τετράγωνονἄραἐστίν· But KBC (is) a right-angle. Thus, BCG (is) also a rightκαίἐστινἀπὸτῆςΓΒ.διὰτὰαὐτὰδὴκαὶτὸΘΖτετράγωνόν
angle. So the opposite (angles) CGK and GKB are also
ἐστιν·καίἐστινἀπὸτῆςΘΗ,τουτέστιν[ἀπὸ]τῆςΑΓ·τὰ right-angles [Prop. 1.34]. Thus, CGKB is right-angled.
ἄραΘΖ,ΚΓτετράγωναἀπὸτῶνΑΓ,ΓΒεἰσιν. καὶἐπεὶ And it was also shown (to be) equilateral. Thus, it is a
ἴσονἐστὶτὸΑΗτῷΗΕ,καίἐστιτὸΑΗτὸὑπὸτῶνΑΓ, square. And it is on CB. So, for the same (reasons),
ΓΒ·ἴσηγὰρἡΗΓτῇΓΒ·καὶτὸΗΕἄραἴσονἐστὶτῷ HF is also a square. And it is on HG, that is to say [on]
ὑπὸΑΓ,ΓΒ·τὰἄραΑΗ, ΗΕ ἴσαἐστὶτῷ δὶςὑπὸτῶν AC [Prop. 1.34]. Thus, the squares HF and KC are
ΑΓ,ΓΒ.ἔστιδὲκαὶτὰΘΖ,ΓΚτετράγωναἀπὸτῶνΑΓ, on AC and CB (respectively). And the (rectangle) AG
ΓΒ·τὰἄρατέσσαρατὰΘΖ,ΓΚ,ΑΗ,ΗΕἴσαἐστὶτοῖςτε is equal to the (rectangle) GE [Prop. 1.43]. And AG is
ἀπὸτῶνΑΓ,ΓΒτετραγώνοιςκαὶτῷδὶςὑπὸτῶνΑΓ,ΓΒ the (rectangle contained) by AC and CB. For GC (is)
περιεχομένῳὀρθογωνίῳ. ἀλλὰτὰΘΖ,ΓΚ,ΑΗ,ΗΕὅλον equal to CB. Thus, GE is also equal to the (rectangle
ἐστὶτὸΑΔΕΒ,ὅἐστινἀπὸτῆςΑΒτετράγωνον·τὸἄρα contained) by AC and CB. Thus, the (rectangles) AG
ἀπὸτῆςΑΒτετράγωνονἴσονἐστὶτοῖςτεἀπὸτῶνΑΓ, and GE are equal to twice the (rectangle contained) by
ΓΒτετραγώνοιςκαὶτῷδὶςὑπὸτῶνΑΓ,ΓΒπεριεχομένῳ AC and CB. And HF and CK are the squares on AC
ὀρθογωνίῳ.
and CB (respectively). Thus, the four (figures) HF , CK,
᾿Εὰνἄραεὐθεῖαγραμμὴτμηθῇ,ὡςἔτυχεν,τὸἀπὸτῆς AG, and GE are equal to the (sum of the) squares on
53

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 2
ὅληςτετράγωνονἴσονἐστὶτοῖςτεἀπὸτῶντμημάτωντε- AC and BC, and twice the rectangle contained by AC
τραγώνοιςκαὶτῷδὶςὑπὸτῶντμημάτωνπεριεχομένῳὀρθο- and CB. But, the (figures) HF , CK, AG, and GE are
γωνίῳ·ὅπερἔδειδεῖξαι.
(equivalent to) the whole of ADEB, which is the square
on AB. Thus, the square on AB is equal to the (sum
of the) squares on AC and CB, and twice the rectangle
contained by AC and CB.
Thus, if a straight-line is cut at random then the
square on the whole (straight-line) is equal to the (sum
of the) squares on the pieces (of the straight-line), and
twice the rectangle contained by the pieces. (Which is)
the very thing it was required to show.
† This proposition is a geometric version of the algebraic identity: (a + b) 2 = a 2 + b 2 + 2 a b.
. Proposition 5 ‡
᾿Εὰνεὐθεῖαγραμμὴτμηθῇεἰςἴσακαὶἄνισα,τὸὑπὸτῶν If a straight-line is cut into equal and unequal (pieces)
ἀνίσωντῆςὅληςτμημάτωνπεριεχόμενονὀρθογώνιονμετὰ then the rectangle contained by the unequal pieces of the
τοῦἀπὸτῆςμεταξὺτῶντομῶντετραγώνουἴσονἐστὶτῷ whole (straight-line), plus the square on the (difference)
ἀπὸτῆςἡμισείαςτετραγώνῳ.
between the (equal and unequal) pieces, is equal to the
square on half (of the straight-line).
Α
Γ
∆
Β
A
C
D
B
Κ
Λ
Θ
Μ
Ξ
Ε
Η
Ν
Μ
Ζ
K
L
E
O
H
N
P
ΕὐθεῖαγάρτιςἡΑΒτετμήσθωεἰςμὲνἴσακατὰτὸ For let any straight-line AB have been cut—equally at
Γ,εἰςδὲἄνισακατὰτὸΔ·λέγω,ὅτιτὸὑπὸτῶνΑΔ,ΔΒ C, and unequally at D. I say that the rectangle contained
περιεχόμενονὀρθογώνιονμετὰτοῦἀπὸτῆςΓΔτετραγώνου by AD and DB, plus the square on CD, is equal to the
ἴσονἐστὶτῷἀπὸτῆςΓΒτετραγώνῳ.
square on CB.
ἈναγεγράφθωγὰρἀπὸτῆςΓΒτετράγωνοντὸΓΕΖΒ, For let the square CEFB have been described on CB
καὶἐπεζεύχθωἡΒΕ,καὶδιὰμὲντοῦΔὁποτέρᾳτῶνΓΕ, [Prop. 1.46], and let BE have been joined, and let DG
ΒΖπαράλληλοςἤχθωἡΔΗ,διὰδὲτοῦΘὁποτέρᾳτῶν have been drawn through D, parallel to either of CE or
ΑΒ,ΕΖπαράλληλοςπάλινἤχθωἡΚΜ,καὶπάλινδιὰτοῦΑ BF [Prop. 1.31], and again let KM have been drawn
ὁποτέρᾳτῶνΓΛ,ΒΜπαράλληλοςἤχθωἡΑΚ.καὶἐπεὶἴσον through H, parallel to either of AB or EF [Prop. 1.31],
ἐστὶτὸΓΘπαραπλήρωματῷΘΖπαραπληρώματι,κοινὸν and again let AK have been drawn through A, parallel to
προσκείσθωτὸΔΜ·ὅλονἄρατὸΓΜὅλῳτῷΔΖἴσον either of CL or BM [Prop. 1.31]. And since the comple-
ἐστίν. ἀλλὰτὸΓΜτῷΑΛἴσονἐστίν,ἐπεὶκαὶἡΑΓτῇ ment CH is equal to the complement HF [Prop. 1.43],
ΓΒἐστινἴση·καὶτὸΑΛἄρατῷΔΖἴσονἐστίν. κοινὸν let the (square) DM have been added to both. Thus,
προσκείσθωτὸΓΘ·ὅλονἄρατὸΑΘτῷΜΝΞ † γνώμονι the whole (rectangle) CM is equal to the whole (rect-
ἴσονἐστίν. ἀλλὰτὸΑΘτὸὑπὸτῶνΑΔ,ΔΒἐστιν·ἴση angle) DF . But, (rectangle) CM is equal to (rectangle)
γὰρἡΔΘτῇΔΒ·καὶὁΜΝΞἄραγνώμωνἴσοςἐστὶτῷ AL, since AC is also equal to CB [Prop. 1.36]. Thus,
ὑπὸΑΔ,ΔΒ.κοινὸνπροσκείσθωτὸΛΗ,ὅἐστινἴσοντῷ (rectangle) AL is also equal to (rectangle) DF . Let (rect-
ἀπὸτῆςΓΔ·ὁἄραΜΝΞγνώμωνκαὶτὸΛΗἴσαἐστὶτῷ angle) CH have been added to both. Thus, the whole
ὑπὸτῶνΑΔ,ΔΒπεριεχομένῳὀρθογωνίῳκαὶτῷἀπὸτῆς (rectangle) AH is equal to the gnomon NOP . But, AH
G
M
F
54

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 2
ΓΔτετραγώνῳ.ἀλλὰὁΜΝΞγνώμωνκαὶτὸΛΗὅλονἐστὶ is the (rectangle contained) by AD and DB. For DH
τὸΓΕΖΒτετράγωνον,ὅἐστινἀπὸτῆςΓΒ·τὸἄραὑπὸτῶν (is) equal to DB. Thus, the gnomon NOP is also equal
ΑΔ,ΔΒπεριεχόμενονὀρθογώνιονμετὰτοῦἀπὸτῆςΓΔ to the (rectangle contained) by AD and DB. Let LG,
τετραγώνουἴσονἐστὶτῷἀπὸτῆςΓΒτετραγώνῳ. which is equal to the (square) on CD, have been added to
᾿Εὰνἄραεὐθεῖαγραμμὴτμηθῇεἰςἴσακαὶἄνισα,τὸὑπὸ both. Thus, the gnomon NOP and the (square) LG are
τῶνἀνίσωντῆςὅληςτμημάτωνπεριεχόμενονὀρθογώνιον equal to the rectangle contained by AD and DB, and the
μετὰτοῦἀπὸτῆςμεταξὺτῶντομῶντετραγώνουἴσονἐστὶ square on CD. But, the gnomon NOP and the (square)
τῷἀπὸτῆςἡμισείαςτετραγώνῳ.ὅπερἔδειδεῖξαι. LG is (equivalent to) the whole square CEFB, which is
on CB. Thus, the rectangle contained by AD and DB,
plus the square on CD, is equal to the square on CB.
Thus, if a straight-line is cut into equal and unequal
(pieces) then the rectangle contained by the unequal
pieces of the whole (straight-line), plus the square on the
(difference) between the (equal and unequal) pieces, is
equal to the square on half (of the straight-line). (Which
is) the very thing it was required to show.
† Note the (presumably mistaken) double use of the label M in the Greek text.
‡ This proposition is a geometric version of the algebraic identity: a b + [(a + b)/2 − b] 2 = [(a + b)/2] 2 .
¦. Proposition 6 †
᾿Εὰνεὐθεῖαγραμμὴτμηθῇδίχα,προστεθῇδέτιςαὐτῇ If a straight-line is cut in half, and any straight-line
εὐθεῖαἐπ᾿εὐθείας,τὸὑπὸτῆςὅληςσὺντῇπροσκειμένῃκαὶ added to it straight-on, then the rectangle contained by
τῆςπροσκειμένηςπεριεχόμενονὀρθόγώνιονμετὰτοῦἀπὸ the whole (straight-line) with the (straight-line) having
τῆςἡμισείαςτετραγώνουἴσονἐστὶτῷἀπὸτῆςσυγκειμένης being added, and the (straight-line) having being added,
ἔκτετῆςἡμισείαςκαὶτῆςπροσκειμένηςτετραγώνῳ. plus the square on half (of the original straight-line), is
equal to the square on the sum of half (of the original
straight-line) and the (straight-line) having been added.
Α
Κ
Λ
Γ
Β
Θ
Ν
Ο
Ξ
∆
Μ
A
K
L
C B D
N
H
P
O
M
Ε
Η
Ζ
E G F
ΕὐθεῖαγάρτιςἡΑΒτετμήσθωδίχακατὰτὸΓσημεῖον, For let any straight-line AB have been cut in half at
προσκείσθωδέτιςαὐτῇεὐθεῖαἐπ᾿εὐθείαςἡΒΔ·λέγω, point C, and let any straight-line BD have been added to
ὅτιτὸὑπὸτῶνΑΔ,ΔΒπεριεχόμενονὀρθογώνιονμετὰ it straight-on. I say that the rectangle contained by AD
τοῦἀπὸτῆςΓΒτετραγώνουἴσονἐστὶτῷἀπὸτῆςΓΔτε- and DB, plus the square on CB, is equal to the square
τραγώνῳ.
on CD.
ἈναγεγράφθωγὰρἀπὸτῆςΓΔτετράγωνοντὸΓΕΖΔ, For let the square CEFD have been described on
καὶἐπεζεύχθωἡΔΕ,καὶδιὰμὲντοῦΒσημείουὁποτέρᾳ CD [Prop. 1.46], and let DE have been joined, and
τῶνΕΓ,ΔΖπαράλληλοςἤχθωἡΒΗ,διὰδὲτοῦΘσημείου let BG have been drawn through point B, parallel to
ὁποτέρᾳτῶνΑΒ,ΕΖπαράλληλοςἤχθωἡΚΜ,καὶἔτιδιὰ either of EC or DF [Prop. 1.31], and let KM have
τοῦΑὁποτέρᾳτῶνΓΛ,ΔΜπαράλληλοςἤχθωἡΑΚ. been drawn through point H, parallel to either of AB
᾿ΕπεὶοὖνἴσηἐστὶνἡΑΓτῇΓΒ,ἴσονἐστὶκαὶτὸΑΛ or EF [Prop. 1.31], and finally let AK have been drawn
55

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 2
τῷΓΘ.ἀλλὰτὸΓΘτῷΘΖἴσονἐστίν.καὶτὸΑΛἄρατῷ through A, parallel to either of CL or DM [Prop. 1.31].
ΘΖἐστινἴσον. κοινὸνπροσκείσθωτὸΓΜ·ὅλονἄρατὸ Therefore, since AC is equal to CB, (rectangle) AL is
ΑΜτῷΝΞΟγνώμονίἐστινἴσον.ἀλλὰτὸΑΜἐστιτὸὑπὸ also equal to (rectangle) CH [Prop. 1.36]. But, (rectanτῶνΑΔ,ΔΒ·ἴσηγάρἐστινἡΔΜτῇΔΒ·καὶὁΝΞΟἄρα
gle) CH is equal to (rectangle) HF [Prop. 1.43]. Thus,
γνώμωνἴσοςἐστὶτῷὑπὸτῶνΑΔ,ΔΒ[περιεχομένῳὀρθο- (rectangle) AL is also equal to (rectangle) HF . Let (rectγωνίῳ].
κοινὸνπροσκείσθωτὸΛΗ,ὅἐστινἴσοντῷἀπὸ angle) CM have been added to both. Thus, the whole
τῆςΒΓτετραγώνῳ·τὸἄραὑπὸτῶνΑΔ,ΔΒπεριεχόμενον (rectangle) AM is equal to the gnomon NOP . But, AM
ὀρθογώνιονμετὰτοῦἀπὸτῆςΓΒτετραγώνουἴσονἐστὶ is the (rectangle contained) by AD and DB. For DM is
τῷΝΞΟγνώμονικαὶτῷΛΗ.ἀλλὰὁΝΞΟγνώμωνκαὶ equal to DB. Thus, gnomon NOP is also equal to the
τὸΛΗὅλονἐστὶτὸΓΕΖΔτετράγωνον,ὅἐστινἀπὸτῆς [rectangle contained] by AD and DB. Let LG, which
ΓΔ·τὸἄραὑπὸτῶνΑΔ,ΔΒπεριεχόμενονὀρθογώνιον is equal to the square on BC, have been added to both.
μετὰτοῦἀπὸτῆςΓΒτετραγώνουἴσονἐστὶτῷἀπὸτῆςΓΔ Thus, the rectangle contained by AD and DB, plus the
τετραγώνῳ.
square on CB, is equal to the gnomon NOP and the
᾿Εὰνἄραεὐθεῖαγραμμὴτμηθῇδίχα,προστεθῇδέτις (square) LG. But the gnomon NOP and the (square)
αὐτῇ εὐθεῖα ἐπ᾿ εὐθείας, τὸ ὑπὸ τῆς ὅλης σὺν τῇ προ- LG is (equivalent to) the whole square CEFD, which is
σκειμένῃκαὶτῆςπροσκειμένηςπεριεχόμενονὀρθόγώνιον on CD. Thus, the rectangle contained by AD and DB,
μετὰτοῦἀπὸτῆςἡμισείαςτετραγώνουἴσονἐστὶτῷἀπὸ plus the square on CB, is equal to the square on CD.
τῆςσυγκειμένηςἔκτετῆςἡμισείαςκαὶτῆςπροσκειμένης Thus, if a straight-line is cut in half, and any straightτετραγώνῳ·ὅπερἔδειδεῖξαι.
line added to it straight-on, then the rectangle contained
by the whole (straight-line) with the (straight-line) having
being added, and the (straight-line) having being
added, plus the square on half (of the original straightline),
is equal to the square on the sum of half (of the
original straight-line) and the (straight-line) having been
added. (Which is) the very thing it was required to show.
† This proposition is a geometric version of the algebraic identity: (2 a + b) b + a 2 = (a + b) 2 .
Þ. Proposition 7 †
᾿Εὰνεὐθεῖαγραμμὴτμηθῇ,ὡςἔτυχεν,τὸἀπὸτῆςὅλης If a straight-line is cut at random then the sum of
καὶτὸἀφ᾿ἑνὸςτῶντμημάτωντὰσυναμφότερατετράγωνα the squares on the whole (straight-line), and one of the
ἴσαἐστὶτῷτεδὶςὑπὸτῆςὅληςκαὶτοῦεἰρημένουτμήματος pieces (of the straight-line), is equal to twice the rectanπεριεχομένῳὀρθογωνίῳκαὶτῷἀπὸτοῦλοιποῦτμήματος
gle contained by the whole, and the said piece, and the
τετραγώνῳ.
square on the remaining piece.
Α
Γ
Β
A C B
Λ
L
Θ
Κ
Μ
Η
Ζ
H
K
M
G
F
∆
Ν
Ε
D N E
ΕὐθεῖαγάρτιςἡΑΒτετμήσθω,ὡςἔτυχεν,κατὰτὸΓ For let any straight-line AB have been cut, at random,
σημεῖον·λέγω,ὅτιτὰἀπὸτῶνΑΒ,ΒΓτετράγωναἴσαἐστὶ at point C. I say that the (sum of the) squares on AB and
τῷτεδὶςὑπὸτῶνΑΒ,ΒΓπεριεχομένῳὀρθογωνίῳκαὶτῷ BC is equal to twice the rectangle contained by AB and
56

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 2
ἀπὸτῆςΓΑτετραγώνῳ.
BC, and the square on CA.
ἈναγεγράφθωγὰρἀπὸτῆςΑΒτετράγωνοντὸΑΔΕΒ· For let the square ADEB have been described on AB
καὶκαταγεγράφθωτὸσχῆμα.
[Prop. 1.46], and let the (rest of) the figure have been
᾿ΕπεὶοὖνἴσονἐστὶτὸΑΗτῷΗΕ,κοινὸνπροσκείσθω drawn.
τὸΓΖ·ὅλονἄρατὸΑΖὅλῳτῷΓΕἴσονἐστίν· τὰἄρα Therefore, since (rectangle) AG is equal to (rectan-
ΑΖ,ΓΕδιπλάσιάἐστιτοῦΑΖ.ἀλλὰτὰΑΖ,ΓΕὁΚΛΜ gle) GE [Prop. 1.43], let the (square) CF have been
ἐστιγνώμωνκαὶτὸΓΖτετράγωνον·ὁΚΛΜἄραγνώμων added to both. Thus, the whole (rectangle) AF is equal
καὶτὸΓΖδιπλάσιάἐστιτοῦΑΖ.ἔστιδὲτοῦΑΖδιπλάσιον to the whole (rectangle) CE. Thus, (rectangle) AF plus
καὶτὸδὶςὑπὸτῶνΑΒ,ΒΓ·ἴσηγὰρἡΒΖτῇΒΓ·ὁἄρα (rectangle) CE is double (rectangle) AF . But, (rectan-
ΚΛΜγνώμωνκαὶτὸΓΖτετράγωνονἴσονἐστὶτῷδὶςὑπὸ gle) AF plus (rectangle) CE is the gnomon KLM, and
τῶνΑΒ,ΒΓ.κοινὸνπροσκείσθωτὸΔΗ,ὅἐστινἀπὸτῆς the square CF . Thus, the gnomon KLM, and the square
ΑΓ τετράγωνον· ὁ ἄρα ΚΛΜ γνώμων καὶ τὰ ΒΗ, ΗΔ CF , is double the (rectangle) AF . But double the (rectτετράγωναἴσαἐστὶτῷτεδὶςὑπὸτῶνΑΒ,ΒΓπεριεχομένῳ
angle) AF is also twice the (rectangle contained) by AB
ὀρθογωνίῳκαὶτῷἀπὸτῆςΑΓτετραγώνῳ. ἀλλὰὁΚΛΜ and BC. For BF (is) equal to BC. Thus, the gnomon
γνώμωνκαὶτὰΒΗ,ΗΔτετράγωναὅλονἐστὶτὸΑΔΕΒκαὶ KLM, and the square CF , are equal to twice the (rectτὸΓΖ,ἅἐστινἀπὸτῶνΑΒ,ΒΓτετράγωνα·τὰἄραἀπὸτῶν
angle contained) by AB and BC. Let DG, which is
ΑΒ,ΒΓτετράγωναἴσαἐστὶτῷ[τε]δὶςὑπὸτῶνΑΒ,ΒΓ the square on AC, have been added to both. Thus, the
περιεχομένῳὀρθογωνίῳμετὰτοῦἀπὸτῆςΑΓτετραγώνου. gnomon KLM, and the squares BG and GD, are equal
᾿Εὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ to twice the rectangle contained by AB and BC, and the
τῆςὅληςκαὶτὸἀφ᾿ἑνὸςτῶντμημάτωντὰσυναμφότερα square on AC. But, the gnomon KLM and the squares
τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῆς ὅλης καὶ τοῦ BG and GD is (equivalent to) the whole of ADEB and
εἰρημένουτμήματοςπεριεχομένῳὀρθογωνίῳκαὶτῷἀπὸ CF , which are the squares on AB and BC (respectively).
τοῦλοιποῦτμήματοςτετραγώνῳ·ὅπερἔδειδεῖξαι. Thus, the (sum of the) squares on AB and BC is equal
to twice the rectangle contained by AB and BC, and the
square on AC.
Thus, if a straight-line is cut at random then the sum
of the squares on the whole (straight-line), and one of
the pieces (of the straight-line), is equal to twice the rectangle
contained by the whole, and the said piece, and the
square on the remaining piece. (Which is) the very thing
it was required to show.
† This proposition is a geometric version of the algebraic identity: (a + b) 2 + a 2 = 2(a + b) a + b 2 .
. Proposition 8 †
᾿Εὰνεὐθεῖαγραμμὴτμηθῇ,ὡςἔτυχεν,τὸτετράκιςὑπὸ If a straight-line is cut at random then four times the
τῆςὅληςκαὶἑνὸςτῶντμημάτωνπεριεχόμενονὀρθογώνιον rectangle contained by the whole (straight-line), and one
μετὰτοῦἀπὸτοῦλοιποῦτμήματοςτετραγώνουἴσονἐστὶ of the pieces (of the straight-line), plus the square on the
τῷἀπότετῆςὅληςκαὶτοῦεἰρημένουτμήματοςὡςἀπὸ remaining piece, is equal to the square described on the
μιᾶςἀναγραφέντιτετραγώνῳ.
whole and the former piece, as on one (complete straight-
ΕὐθεῖαγάρτιςἡΑΒτετμήσθω,ὡςἔτυχεν,κατὰτὸ line).
Γσημεῖον· λέγω, ὅτιτὸτετράκιςὑπὸτῶνΑΒ,ΒΓπε- For let any straight-line AB have been cut, at random,
ριεχόμενονὀρθογώνιονμετὰτοῦἀπὸτῆςΑΓτετραγώνου at point C. I say that four times the rectangle contained
ἴσονἐστὶτῷἀπὸτῆςΑΒ,ΒΓὡςἀπὸμιᾶςἀναγραφέντι by AB and BC, plus the square on AC, is equal to the
τετραγώνῳ.
square described on AB and BC, as on one (complete
᾿Εκβεβλήσθωγὰρἐπ᾿εὐθείας[τῇΑΒεὐθεῖα]ἡΒΔ, straight-line).
καὶκείσθωτῇΓΒἴσηἡΒΔ,καὶἀναγεγράφθωἀπὸτῆς For let BD have been produced in a straight-line
ΑΔτετράγωνοντὸΑΕΖΔ,καὶκαταγεγράφθωδιπλοῦντὸ [with the straight-line AB], and let BD be made equal
σχῆμα.
to CB [Prop. 1.3], and let the square AEFD have been
described on AD [Prop. 1.46], and let the (rest of the)
figure have been drawn double.
57

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 2
Α Γ Β
∆
A C B D
Μ
Η
Τ
Κ
Ν
M
G
T
K
N
Ξ
Σ
Υ
Π
Ρ
Ο
O
S
U
Q
R
P
Ε Θ Λ
Ζ
E H L F
᾿ΕπεὶοὖνἴσηἐστὶνἡΓΒτῇΒΔ,ἀλλὰἡμὲνΓΒτῇΗΚ Therefore, since CB is equal to BD, but CB is equal
ἐστινἴση,ἡδὲΒΔτῇΚΝ,καὶἡΗΚἄρατῇΚΝἐστινἴση. to GK [Prop. 1.34], and BD to KN [Prop. 1.34], GK is
διὰτὰαὐτὰδὴκαὶἡΠΡτῇΡΟἐστινἴση.καὶἐπεὶἴσηἐστὶν thus also equal to KN. So, for the same (reasons), QR is
ἡΒΓτῂΒΔ,ἡδὲΗΚτῇΚΝ,ἴσονἄραἐστὶκαὶτὸμὲν equal to RP . And since BC is equal to BD, and GK to
ΓΚτῷΚΔ,τὸδὲΗΡτῷΡΝ.ἀλλὰτὸΓΚτῷΡΝἐστιν KN, (square) CK is thus also equal to (square) KD, and
ἴσον·παραπληρώματαγὰρτοῦΓΟπαραλληλογράμμου·καὶ (square) GR to (square) RN [Prop. 1.36]. But, (square)
τὸΚΔἄρατῷΗΡἴσονἐστίν·τὰτέσσαραἄρατὰΔΚ,ΓΚ, CK is equal to (square) RN. For (they are) comple-
ΗΡ,ΡΝἴσαἀλλήλοιςἐστίν. τὰτέσσαραἄρατετραπλάσιά ments in the parallelogram CP [Prop. 1.43]. Thus,
ἐστιτοῦΓΚ.πάλινἐπεὶἴσηἐστὶνἡΓΒτῇΒΔ,ἀλλὰἡμὲν (square) KD is also equal to (square) GR. Thus, the
ΒΔτῇΒΚ,τουτέστιτῇΓΗἴση,ἡδὲΓΒτῇΗΚ,τουτέστι four (squares) DK, CK, GR, and RN are equal to one
τῇΗΠ,ἐστινἴση,καὶἡΓΗἄρατῇΗΠἴσηἐστίν.καὶἐπεὶ another. Thus, the four (taken together) are quadruple
ἴσηἐστὶνἡμὲνΓΗτῇΗΠ,ἡδὲΠΡτῇΡΟ,ἴσονἐστὶκαὶτὸ (square) CK. Again, since CB is equal to BD, but BD
μὲνΑΗτῷΜΠ,τὸδὲΠΛτῷΡΖ.ἀλλὰτὸΜΠτῷΠΛἐστιν (is) equal to BK—that is to say, CG—and CB is equal
ἴσον·παραπληρώματαγὰρτοῦΜΛπαραλληλογράμμου·καὶ to GK—that is to say, GQ—CG is thus also equal to GQ.
τὸΑΗἄρατῷΡΖἴσονἐστίν·τὰτέσσαραἄρατὰΑΗ,ΜΠ, And since CG is equal to GQ, and QR to RP , (rectan-
ΠΛ,ΡΖἴσαἀλλήλοιςἐστίν·τὰτέσσαραἄρατοῦΑΗἐστι gle) AG is also equal to (rectangle) MQ, and (rectangle)
τετραπλάσια.ἐδείχθηδὲκαὶτὰτέσσαρατὰΓΚ,ΚΔ,ΗΡ, QL to (rectangle) RF [Prop. 1.36]. But, (rectangle) MQ
ΡΝτοῦΓΚτετραπλάσια·τὰἄραὀκτώ,ἃπεριέχειτὸνΣΤΥ is equal to (rectangle) QL. For (they are) complements
γνώμονα,τετραπλάσιάἐστιτοῦΑΚ.καὶἐπεὶτὸΑΚτὸὑπὸ in the parallelogram ML [Prop. 1.43]. Thus, (rectangle)
τῶνΑΒ,ΒΔἐστιν·ἴσηγὰρἡΒΚτῇΒΔ·τὸἄρατετράκις AG is also equal to (rectangle) RF . Thus, the four (rect-
ὑπὸτῶνΑΒ,ΒΔτετραπλάσιόνἐστιτοῦΑΚ.ἐδείχθηδὲ angles) AG, MQ, QL, and RF are equal to one another.
τοῦΑΚτετραπλάσιοςκαὶὁΣΤΥγνώμων·τὸἄρατετράκις Thus, the four (taken together) are quadruple (rectan-
ὑπὸτῶνΑΒ,ΒΔἴσονἐστὶτῷΣΤΥγνώμονι.κοινὸνπρο- gle) AG. And it was also shown that the four (squares)
σκείσθωτὸΞΘ,ὅἐστινἴσοντῷἀπὸτῆςΑΓτετραγώνῳ·τὸ CK, KD, GR, and RN (taken together are) quadruple
ἄρατετράκιςὑπὸτῶνΑΒ,ΒΔπεριεχόμενονὀρθογώνιον (square) CK. Thus, the eight (figures taken together),
μετὰτοῦἀπὸΑΓτετραγώνουἴσονἐστὶτῷΣΤΥγνώμονι which comprise the gnomon STU, are quadruple (rectκαὶτῷΞΘ.ἀλλὰὁΣΤΥγνώμωνκαὶτὸΞΘὅλονἐστὶτὸ
angle) AK. And since AK is the (rectangle contained)
ΑΕΖΔτετράγωνον,ὅἐστινἀπὸτῆςΑΔ·τὸἄρατετράκις by AB and BD, for BK (is) equal to BD, four times the
ὑπὸτῶνΑΒ,ΒΔμετὰτοῦἀπὸΑΓἴσονἐστὶτῷἀπὸΑΔ (rectangle contained) by AB and BD is quadruple (rectτετραγώνῳ·ἴσηδὲἡΒΔτῇΒΓ.τὸἄρατετράκιςὑπὸτῶν
angle) AK. But the gnomon STU was also shown (to
ΑΒ,ΒΓπεριεχόμενονὀρθογώνιονμετὰτοῦἀπὸΑΓτε- be equal to) quadruple (rectangle) AK. Thus, four times
τραγώνουἴσονἐστὶτῷἀπὸτῆςΑΔ,τουτέστιτῷἀπὸτῆς the (rectangle contained) by AB and BD is equal to the
ΑΒκαὶΒΓὡςἀπὸμιᾶςἀναγραφέντιτετραγώνῳ. gnomon STU. Let OH, which is equal to the square on
᾿Εὰνἄραεὐθεῖαγραμμὴτμηθῇ,ὡςἔτυχεν,τὸτετράκις AC, have been added to both. Thus, four times the rect-
ὑπὸτῆςὅληςκαὶἑνὸςτῶντμημάτωνπεριεχόμενονὀρθογώ- angle contained by AB and BD, plus the square on AC,
νιονμετὰτοῦἀπὸτοῦλοιποῦτμήματοςτετραγώνουἴσου is equal to the gnomon STU, and the (square) OH. But,
58

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 2
ἐστὶτῷἀπότετῆςὅληςκαὶτοῦεἰρημένουτμήματοςὡς the gnomon STU and the (square) OH is (equivalent to)
ἀπὸμιᾶςἀναγραφέντιτετραγώνῳ·ὅπερἔδειδεῖξαι. the whole square AEFD, which is on AD. Thus, four
times the (rectangle contained) by AB and BD, plus the
(square) on AC, is equal to the square on AD. And BD
(is) equal to BC. Thus, four times the rectangle contained
by AB and BC, plus the square on AC, is equal to
the (square) on AD, that is to say the square described
on AB and BC, as on one (complete straight-line).
Thus, if a straight-line is cut at random then four times
the rectangle contained by the whole (straight-line), and
one of the pieces (of the straight-line), plus the square
on the remaining piece, is equal to the square described
on the whole and the former piece, as on one (complete
straight-line). (Which is) the very thing it was required
to show.
† This proposition is a geometric version of the algebraic identity: 4(a + b) a + b 2 = [(a + b) + a] 2 .
. Proposition 9 †
᾿Εὰνεὐθεῖαγραμμὴτμηθῇεἰςἴσακαὶἄνισα, τὰἀπὸ If a straight-line is cut into equal and unequal (pieces)
τῶνἀνίσωντῆςὅληςτμημάτωντετράγωναδιπλάσιάἐστι then the (sum of the) squares on the unequal pieces of the
τοῦτεἀπὸτῆςἡμισείαςκαὶτοῦἀπὸτῆςμεταξὺτῶντομῶν whole (straight-line) is double the (sum of the) square
τετραγώνου.
on half (the straight-line) and (the square) on the (difference)
between the (equal and unequal) pieces.
Ε
E
Η
Ζ
G
F
Α Γ ∆
Β
A C D B
ΕὐθεῖαγάρτιςἡΑΒτετμήσθωεἰςμὲνἴσακατὰτὸ For let any straight-line AB have been cut—equally at
Γ,εἱςδὲἄνισακατὰτὸΔ·λέγω,ὅτιτὰἀπὸτῶνΑΔ,ΔΒ C, and unequally at D. I say that the (sum of the) squares
τετράγωναδιπλάσιάἐστιτῶνἀπὸτῶνΑΓ,ΓΔτετραγώνων. on AD and DB is double the (sum of the squares) on AC
῎ΗχθωγὰρἀπὸτοῦΓτῇΑΒπρὸςὀρθὰςἡΓΕ,καὶ and CD.
κείσθωἴσηἑκατέρᾳτῶνΑΓ,ΓΒ,καὶἐπεζεύχθωσαναἱΕΑ, For let CE have been drawn from (point) C, at right-
ΕΒ,καὶδιὰμὲντοῦΔτῇΕΓπαράλληλοςἤχθωἡΔΖ,διὰ angles to AB [Prop. 1.11], and let it be made equal to
δὲτοῦΖτῇΑΒἡΖΗ,καὶἐπεζεύχθωἡΑΖ.καὶἐπεὶἴση each of AC and CB [Prop. 1.3], and let EA and EB
ἐστὶνἡΑΓτῇΓΕ,ἴσηἐστὶκαὶἡὑπὸΕΑΓγωνίατῇὑπὸ have been joined. And let DF have been drawn through
ΑΕΓ.καὶἐπεὶὀρθήἐστινἡπρὸςτῷΓ,λοιπαὶἄρααἱὑπὸ (point) D, parallel to EC [Prop. 1.31], and (let) FG
ΕΑΓ,ΑΕΓμιᾷὀρθῇἴσαιεἰσίν·καίεἰσινἴσαι·ἡμίσειαἄρα (have been drawn) through (point) F , (parallel) to AB
ὀρθῆςἐστινἑκατέρατῶνὑπὸΓΕΑ,ΓΑΕ.δὶατὰαὐτὰδὴ [Prop. 1.31]. And let AF have been joined. And since
καὶἑκατέρατῶνὑπὸΓΕΒ,ΕΒΓἡμίσειάἐστινὀρθῆς·ὅλη AC is equal to CE, the angle EAC is also equal to the
ἄραἡὑπὸΑΕΒὀρθήἐστιν. καὶἐπεὶἡὑπὸΗΕΖἡμίσειά (angle) AEC [Prop. 1.5]. And since the (angle) at C is
ἐστινὀρθῆς,ὀρθὴδὲἡὑπὸΕΗΖ·ἴσηγάρἐστιτῇἐντὸςκαὶ a right-angle, the (sum of the) remaining angles (of tri-
ἀπεναντίοντῇὑπὸΕΓΒ·λοιπὴἄραἡὑπὸΕΖΗἡμίσειάἐστιν angle AEC), EAC and AEC, is thus equal to one right-
59

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 2
ὀρθῆς·ἴσηἄρα[ἐστὶν]ἡὑπὸΗΕΖγωνίατῇὑπὸΕΖΗ·ὥστε angle [Prop. 1.32]. And they are equal. Thus, (angles)
καὶπλευρὰἡΕΗτῇΗΖἐστινἴση.πάλινἐπεὶἡπρὸςτῷΒ CEA and CAE are each half a right-angle. So, for the
γωνίαἡμίσειάἐστινὀρθῆς,ὀρθὴδὲἡὑπὸΖΔΒ·ἴσηγὰρ same (reasons), (angles) CEB and EBC are also each
πάλινἐστὶτῇἐντὸςκαὶἀπεναντίοντῇὑπὸΕΓΒ·λοιπὴἄρα half a right-angle. Thus, the whole (angle) AEB is a
ἡὑπὸΒΖΔἡμίσειάἐστινὀρθῆς·ἴσηἄραἡπρὸςτῷΒγωνία right-angle. And since GEF is half a right-angle, and
τῇὑπὸΔΖΒ·ὥστεκαὶπλευρὰἡΖΔπλευρᾷτῇΔΒἐστιν EGF (is) a right-angle—for it is equal to the internal and
ἴση.καὶἐπεὶἴσηἐστὶνἡΑΓτῇΓΕ,ἴσονἐστὶκαὶτὸἀπὸΑΓ opposite (angle) ECB [Prop. 1.29]—the remaining (anτῷἀπὸΓΕ·τὰἄραἀπὸτῶνΑΓ,ΓΕτετράγωναδιπλάσιά
gle) EFG is thus half a right-angle [Prop. 1.32]. Thus,
ἐστιτοῦἀπὸΑΓ.τοῖςδὲἀπὸτῶνΑΓ,ΓΕἴσονἐστὶτὸἀπὸ angle GEF [is] equal to EFG. So the side EG is also
τῆςΕΑτετράγωνον·ὀρθὴγὰρἡὑπὸΑΓΕγωνία·τὸἄρα equal to the (side) GF [Prop. 1.6]. Again, since the an-
ἀπὸτῆςΕΑδιπλάσιόνἐστιτοῦἀπὸτῆςΑΓ.πάλιν,ἐπεὶἴση gle at B is half a right-angle, and (angle) FDB (is) a
ἐστὶνἡΕΗτῇΗΖ,ἴσονκαὶτὸἀπὸτῆςΕΗτῷἀπὸτῆςΗΖ· right-angle—for again it is equal to the internal and opτὰἄραἀπὸτῶνΕΗ,ΗΖτετράγωναδιπλάσιάἐστιτοῦἀπὸ
posite (angle) ECB [Prop. 1.29]—the remaining (angle)
τῆςΗΖτετραγώνου.τοῖςδὲἀπὸτῶνΕΗ,ΗΖτετραγώνοις BFD is half a right-angle [Prop. 1.32]. Thus, the angle at
ἴσονἐστὶτὸἀπὸτῆςΕΖτετράγωνον·τὸἄραἀπὸτῆςΕΖ B (is) equal to DFB. So the side FD is also equal to the
τετράγωνονδιπλάσιόνἐστιτοῦἀπὸτῆςΗΖ.ἴσηδὲἡΗΖ side DB [Prop. 1.6]. And since AC is equal to CE, the
τῇΓΔ·τὸἄραἀπὸτῆςΕΖδιπλάσιόνἐστιτοῦἀπὸτῆςΓΔ. (square) on AC (is) also equal to the (square) on CE.
ἔστιδὲκαὶτὸἀπὸτῆςΕΑδιπλάσιοντοῦἀπὸτῆςΑΓ·τὰ Thus, the (sum of the) squares on AC and CE is dou-
ἄραἀπὸτῶνΑΕ,ΕΖτετράγωναδιπλάσιάἐστιτῶνἀπὸτῶν ble the (square) on AC. And the square on EA is equal
ΑΓ,ΓΔτετραγώνων. τοῖςδὲἀπὸτῶνΑΕ,ΕΖἴσονἐστὶ to the (sum of the) squares on AC and CE. For angle
τὸἀπὸτῆςΑΖτετράγωνον·ὀρθὴγάρἐστινἡὑπὸΑΕΖ ACE (is) a right-angle [Prop. 1.47]. Thus, the (square)
γωνία·τὸἄραἀπὸτῆςΑΖτετράγωνονδιπλάσιόνἐστιτῶν on EA is double the (square) on AC. Again, since EG
ἀπὸτῶνΑΓ,ΓΔ.τῷδὲἀπὸτῆςΑΖἴσατὰἀπὸτῶνΑΔ, is equal to GF , the (square) on EG (is) also equal to
ΔΖ·ὀρθὴγὰρἡπρὸςτῷΔγωνία·τὰἄραἀπὸτῶνΑΔ,ΔΖ the (square) on GF . Thus, the (sum of the squares) on
διπλάσιάἐστιτῶνἀπὸτῶνΑΓ,ΓΔτετραγώνων. ἴσηδὲἡ EG and GF is double the square on GF . And the square
ΔΖτῇΔΒ·τὰἄραἀπὸτῶνΑΔ,ΔΒτετράγωναδιπλάσιά on EF is equal to the (sum of the) squares on EG and
ἐστιτῶνἀπὸτῶνΑΓ,ΓΔτετράγώνων.
GF [Prop. 1.47]. Thus, the square on EF is double the
᾿Εὰνἄραεὐθεῖαγραμμὴτμηθῇεἰςἴσακαὶἄνισα,τὰἀπὸ (square) on GF . And GF (is) equal to CD [Prop. 1.34].
τῶνἀνίσωντῆςὅληςτμημάτωντετράγωναδιπλάσιάἐστι Thus, the (square) on EF is double the (square) on CD.
τοῦτεἀπὸτῆςἡμισείαςκαὶτοῦἀπὸτῆςμεταξὺτῶντομῶν And the (square) on EA is also double the (square) on
τετραγώνου·ὅπερἔδειδεῖξαι.
AC. Thus, the (sum of the) squares on AE and EF is
double the (sum of the) squares on AC and CD. And
the square on AF is equal to the (sum of the squares)
on AE and EF . For the angle AEF is a right-angle
[Prop. 1.47]. Thus, the square on AF is double the (sum
of the squares) on AC and CD. And the (sum of the
squares) on AD and DF (is) equal to the (square) on
AF . For the angle at D is a right-angle [Prop. 1.47].
Thus, the (sum of the squares) on AD and DF is double
the (sum of the) squares on AC and CD. And DF (is)
equal to DB. Thus, the (sum of the) squares on AD and
DB is double the (sum of the) squares on AC and CD.
Thus, if a straight-line is cut into equal and unequal
(pieces) then the (sum of the) squares on the unequal
pieces of the whole (straight-line) is double the (sum of
the) square on half (the straight-line) and (the square) on
the (difference) between the (equal and unequal) pieces.
(Which is) the very thing it was required to show.
† This proposition is a geometric version of the algebraic identity: a 2 + b 2 = 2[([a + b]/2) 2 + ([a + b]/2 − b) 2 ].
60

ËÌÇÁÉÁÏƺ ELEMENTS BOOK 2
. Proposition 10 †
᾿Εὰνεὐθεῖαγραμμὴτμηθῇδίχα,προστεθῇδέτιςαὐτῇ If a straight-line is cut in half, and any straight-line
εὐθεῖαἐπ᾿εὐθείας,τὸἀπὸτῆςὅληςσὺντῇπροσκειμένῃ added to it straight-on, then the sum of the square on
καὶτὸἀπὸτῆςπροσκειμένηςτὰσυναμφότερατετράγωνα the whole (straight-line) with the (straight-line) having
διπλάσιάἐστιτοῦτεἀπὸτῆςἡμισείαςκαὶτοῦἀπὸτῆςσυγ- been added, and the (square) on the (straight-line) havκειμένηςἔκτετῆςἡμισείαςκαὶτῆςπροσκειμένηςὡςἀπὸ
ing been added, is double the (sum of the square) on half
μιᾶςἀναγραφέντοςτετραγώνου.
(the straight-line), and the square described on the sum
of half (the straight-line) and (straight-line) having been
added, as on one (complete straight-line).
Ε
Ζ
E
F
Α Γ Β
∆
A C B
D
Η
ΕὐθεῖαγάρτιςἡΑΒτετμήσθωδίχακατὰτὸΓ,προ- For let any straight-line AB have been cut in half at
σκείσθωδέτιςαὐτῇεὐθεῖαἐπ᾿εὐθείαςἡΒΔ·λέγω,ὅτιτὰ (point) C, and let any straight-line BD have been added
ἀπὸτῶνΑΔ,ΔΒτετράγωναδιπλάσιάἐστιτῶνἀπὸτῶν to it straight-on. I say that the (sum of the) squares on
ΑΓ,ΓΔτετραγώνων.
AD and DB is double the (sum of the) squares on AC
῎ΗχθωγὰρἀπὸτοῦΓσημείουτῇΑΒπρὸςὀρθὰςἡΓΕ, and CD.
καὶκείσθωἴσηἑκατέρᾳτῶνΑΓ,ΓΒ,καὶἐπεζεύχθωσαν For let CE have been drawn from point C, at rightαἱΕΑ,ΕΒ·καὶδιὰμὲντοῦΕτῇΑΔπαράλληλοςἤχθωἡ
angles to AB [Prop. 1.11], and let it be made equal to
ΕΖ,διὰδὲτοῦΔτῇΓΕπαράλληλοςἤχθωἡΖΔ.καὶἐπεὶ each of AC and CB [Prop. 1.3], and let EA and EB have
εἰςπαραλλήλουςεὐθείαςτὰςΕΓ,ΖΔεὐθεῖάτιςἐνέπεσεν been joined. And let EF have been drawn through E,
ἡΕΖ,αἱὑπὸΓΕΖ,ΕΖΔἄραδυσὶνὀρθαῖςἴσαιεἰσίν· αἱ parallel to AD [Prop. 1.31], and let FD have been drawn
ἄρα ὑπὸ ΖΕΒ, ΕΖΔ δύο ὀρθῶν ἐλάσσονές εἰσιν· αἱ δὲ through D, parallel to CE [Prop. 1.31]. And since some
ἀπ᾿ἐλασσόνωνἢδύοὀρθῶνἐκβαλλόμεναισυμπίπτουσιν· straight-line EF falls across the parallel straight-lines EC
αἱ ἄρα ΕΒ, ΖΔ ἐκβαλλόμεναι ἐπὶ τὰ Β, Δ μέρη συμ- and FD, the (internal angles) CEF and EFD are thus
πεσοῦνται.ἐκβεβλήσθωσανκαὶσυμπιπτέτωσανκατὰτὸΗ, equal to two right-angles [Prop. 1.29]. Thus, FEB and
καὶἐπεζεύχθωἡΑΗ.καὶἐπεὶἴσηἐστὶνἡΑΓτῇΓΕ,ἴση EFD are less than two right-angles. And (straight-lines)
ἐστὶκαὶγωνίαἡὑπὸΕΑΓτῇὑπὸΑΕΓ·καὶὀρθὴἡπρὸςτῷ produced from (internal angles whose sum is) less than
Γ·ἡμίσειαἄραὀρθῆς[ἐστιν]ἑκατέρατῶνὑπὸΕΑΓ,ΑΕΓ. two right-angles meet together [Post. 5]. Thus, being proδιὰτὰαὐτὰδὴκαὶἑκατέρατῶνὑπὸΓΕΒ,ΕΒΓἡμίσειάἐστιν
duced in the direction of B and D, the (straight-lines)
ὀρθῆς·ὀρθὴἄραἐστὶνἡὑπὸΑΕΒ.καὶἐπεὶἡμίσειαὀρθῆς EB and FD will meet. Let them have been produced,
ἐστινἡὑπὸΕΒΓ,ἡμίσειαἄραὀρθῆςκαὶἡὑπὸΔΒΗ.ἔστι and let them meet together at G, and let AG have been
δὲκαὶἡὑπὸΒΔΗὀρθή·ἴσηγάρἐστιτῇὑπὸΔΓΕ·ἐναλλὰξ joined. And since AC is equal to CE, angle EAC is also
γάρ·λοιπὴἅραἡὑπὸΔΗΒἡμίσειάἐστινὀρθῆς·ἡἄραὑπὸ equal to (angle) AEC [Prop. 1.5]. And the (angle) at
ΔΗΒτῇὑπὸΔΒΗἐστινἴση·ὥστεκαὶπλευρὰἡΒΔπλευρᾷ C (is) a right-angle. Thus, EAC and AEC [are] each
τῇΗΔἐστινἴση. πάλιν, ἐπεὶἡὑπὸΕΗΖἡμίσειάἐστιν half a right-angle [Prop. 1.32]. So, for the same (rea-
ὀρθῆς,ὀρθὴδὲἡπρὸςτῷΖ·ἴσηγάρἐστιτῇἀπεναντίοντῇ sons), CEB and EBC are also each half a right-angle.
πρὸςτῷΓ·λοιπὴἄραἡὑπὸΖΕΗἡμίσειάἐστινὀρθῆς·ἴση Thus, (angle) AEB is a right-angle. And since EBC
ἄραἡὑπὸΕΗΖγωνίατῇὑπὸΖΕΗ·ὥστεκαὶπλευρὰἡΗΖ is half a right-angle, DBG (is) thus also half a rightπλευρᾷτῇΕΖἐστινἴση.
καὶἐπεὶ[ἴσηἐστὶνἡΕΓτῇΓΑ], angle [Prop. 1.15]. And BDG is also a right-angle. For
ἴσονἐστὶ[καὶ]τὸἀπὸτῆςΕΓτετράγωνοντῷἀπὸτῆςΓΑ it is equal to DCE. For (they are) alternate (angles)
G
61

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 2
τετραγώνῳ·τὰἄραἀπὸτῶνΕΓ,ΓΑτετράγωναδιπλάσιά [Prop. 1.29]. Thus, the remaining (angle) DGB is half
ἐστιτοῦἀπὸτῆςΓΑτετραγώνου.τοῖςδὲἀπὸτῶνΕΓ,ΓΑ a right-angle. Thus, DGB is equal to DBG. So side BD
ἴσονἐστὶτὸἀπὸτῆςΕΑ·τὸἄραἀπὸτῆςΕΑτετράγωνον is also equal to side GD [Prop. 1.6]. Again, since EGF is
διπλάσιόνἐστιτοῦἀπὸτῆςΑΓτετραγώνου.πάλιν,ἐπεὶἴση half a right-angle, and the (angle) at F (is) a right-angle,
ἐστὶνἡΖΗτῇΕΖ,ἴσονἐστὶκαὶτὸἀπὸτῆςΖΗτῷἀπὸτῆς for it is equal to the opposite (angle) at C [Prop. 1.34],
ΖΕ·τὰἄραἀπὸτῶνΗΖ,ΖΕδιπλάσιάἐστιτοῦἀπὸτῆςΕΖ. the remaining (angle) FEG is thus half a right-angle.
τοῖςδὲἀπὸτῶνΗΖ,ΖΕἴσονἐστὶτὸἀπὸτῆςΕΗ·τὸἄρα Thus, angle EGF (is) equal to FEG. So the side GF
ἀπὸτῆςΕΗδιπλάσιόνἐστιτοῦἀπὸτῆςΕΖ.ἴσηδὲἡΕΖτῇ is also equal to the side EF [Prop. 1.6]. And since [EC
ΓΔ·τὸἄραἀπὸτῆςΕΗτετράγωνονδιπλάσιόνἐστιτοῦἀπὸ is equal to CA] the square on EC is [also] equal to the
τῆςΓΔ.ἐδείχθηδὲκαὶτὸἀπὸτῆςΕΑδιπλάσιοντοῦἀπὸ square on CA. Thus, the (sum of the) squares on EC
τῆςΑΓ·τὰἄραἀπὸτῶνΑΕ,ΕΗτετράγωναδιπλάσιάἐστι and CA is double the square on CA. And the (square)
τῶνἀπὸτῶνΑΓ,ΓΔτετραγώνων. τοῖςδὲἀπὸτῶνΑΕ, on EA is equal to the (sum of the squares) on EC and
ΕΗτετραγώνοιςἴσονἐστὶτὸἀπὸτῆςΑΗτετράγωνον·τὸ CA [Prop. 1.47]. Thus, the square on EA is double the
ἄραἀπὸτῆςΑΗδιπλάσιόνἐστιτῶνἀπὸτῶνΑΓ,ΓΔ.τῷ square on AC. Again, since FG is equal to EF , the
δὲἀπὸτῆςΑΗἴσαἐστὶτὰἀπὸτῶνΑΔ,ΔΗ·τὰἄραἀπὸ (square) on FG is also equal to the (square) on FE.
τῶνΑΔ,ΔΗ[τετράγωνα]διπλάσιάἐστιτῶνἀπὸτῶνΑΓ, Thus, the (sum of the squares) on GF and FE is dou-
ΓΔ[τετραγώνων]. ἴσηδὲἡΔΗτῇΔΒ·τὰἄραἀπὸτῶν ble the (square) on EF . And the (square) on EG is equal
ΑΔ,ΔΒ[τετράγωνα]διπλάσιάἐστιτῶνἀπὸτῶνΑΓ,ΓΔ to the (sum of the squares) on GF and FE [Prop. 1.47].
τετραγώνων. Thus, the (square) on EG is double the (square) on EF .
᾿Εὰνἄραεὐθεῖαγραμμὴτμηθῇδίχα,προστεθῇδέτις And EF (is) equal to CD [Prop. 1.34]. Thus, the square
αὐτῇ εὐθεῖα ἐπ᾿ εὐθείας, τὸ ἀπὸ τῆς ὅλης σὺν τῇ προ- on EG is double the (square) on CD. But it was also
σκειμένῃ καὶ τὸ ἀπὸ τῆς προσκειμένης τὰ συναμφότερα shown that the (square) on EA (is) double the (square)
τετράγωναδιπλάσιάἐστιτοῦτεἀπὸτῆςἡμισείαςκαὶτοῦ on AC. Thus, the (sum of the) squares on AE and EG is
ἀπὸ τῆς συγκειμένης ἔκ τε τῆς ἡμισείας καὶ τῆς προ- double the (sum of the) squares on AC and CD. And the
σκειμένης ὡς ἀπὸμιᾶς ἀναγραφέντοςτετραγώνου· ὅπερ square on AG is equal to the (sum of the) squares on AE
ἔδειδεῖξαι.
and EG [Prop. 1.47]. Thus, the (square) on AG is double
the (sum of the squares) on AC and CD. And the (sum
of the squares) on AD and DG is equal to the (square)
on AG [Prop. 1.47]. Thus, the (sum of the) [squares] on
AD and DG is double the (sum of the) [squares] on AC
and CD. And DG (is) equal to DB. Thus, the (sum of
the) [squares] on AD and DB is double the (sum of the)
squares on AC and CD.
Thus, if a straight-line is cut in half, and any straightline
added to it straight-on, then the sum of the square
on the whole (straight-line) with the (straight-line) having
been added, and the (square) on the (straight-line)
having been added, is double the (sum of the square) on
half (the straight-line), and the square described on the
sum of half (the straight-line) and (straight-line) having
been added, as on one (complete straight-line). (Which
is) the very thing it was required to show.
† This proposition is a geometric version of the algebraic identity: (2 a + b) 2 + b 2 = 2[a 2 + (a + b) 2 ].
. Proposition 11 †
Τὴνδοθεῖσανεὐθεῖαντεμεῖνὥστετὸὑπὸτῆςὅληςκαὶ To cut a given straight-line such that the rectangle
τοῦἑτέρουτῶντμημάτωνπεριεχόμενονὀρθογώνιονἴσον contained by the whole (straight-line), and one of the
εἶναιτῷἀπὸτοῦλοιποῦτμήματοςτετραγώνῳ.
pieces (of the straight-line), is equal to the square on the
remaining piece.
62

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 2
Ζ
Η
F
G
Α
Θ
Β
A
H
B
Ε
E
Γ
Κ
∆
῎ΕστωἡδοθεῖσαεὐθεῖαἡΑΒ·δεῖδὴτὴνΑΒτεμεῖν Let AB be the given straight-line. So it is required to
ὥστε τὸ ὑπὸ τῆς ὅλης καὶ τοῦ ἑτέρου τῶν τμημάτων cut AB such that the rectangle contained by the whole
περιεχόμενον ὀρθογώνιον ἴσον εἶναι τῷ ἀπὸ τοῦ λοιποῦ (straight-line), and one of the pieces (of the straightτμήματοςτετραγώνῳ.
line), is equal to the square on the remaining piece.
ἈναγεγράφθωγὰρἀπὸτῆςΑΒτετράγωνοντὸΑΒΔΓ, For let the square ABDC have been described on AB
καὶτετμήσθωἡΑΓδίχακατὰτὸΕσημεῖον,καὶἐπεζεύχθω [Prop. 1.46], and let AC have been cut in half at point
ἡΒΕ,καὶδιήχθωἡΓΑἐπὶτὸΖ,καὶκείσθωτῇΒΕἴσηἡ E [Prop. 1.10], and let BE have been joined. And let
ΕΖ,καὶἀναγεγράφθωἀπὸτῆςΑΖτετράγωνοντὸΖΘ,καὶ CA have been drawn through to (point) F , and let EF
διήχθωἡΗΘἐπὶτὸΚ·λέγω,ὅτιἡΑΒτέτμηταικατὰτὸΘ, be made equal to BE [Prop. 1.3]. And let the square
ὥστετὸὑπὸτῶνΑΒ,ΒΘπεριεχόμενονὀρθογώνιονἴσον FH have been described on AF [Prop. 1.46], and let GH
ποιεῖντῷἀπὸτῆςΑΘτετραγώνῳ.
have been drawn through to (point) K. I say that AB has
᾿Επεὶ γὰρ εὐθεῖα ἡ ΑΓ τέτμηται δίχα κατὰ τὸ Ε, been cut at H such as to make the rectangle contained by
πρόσκειταιδὲ αὐτῇἡΖΑ, τὸ ἄρα ὑπὸτῶν ΓΖ, ΖΑ πε- AB and BH equal to the square on AH.
ριεχόμενονὀρθογώνιονμετὰτοῦἀπὸτῆςΑΕτετραγώνου For since the straight-line AC has been cut in half at
ἴσονἐστὶτῷἀπὸτῆςΕΖτετραγώνῳ. ἴσηδὲἡΕΖτῇΕΒ· E, and FA has been added to it, the rectangle contained
τὸἄραὑπὸτῶνΓΖ,ΖΑμετὰτοῦἀπὸτῆςΑΕἴσονἐστὶ by CF and FA, plus the square on AE, is thus equal to
τῷἀπὸΕΒ.ἀλλὰτῷἀπὸΕΒἴσαἐστὶτὰἀπὸτῶνΒΑ, the square on EF [Prop. 2.6]. And EF (is) equal to EB.
ΑΕ·ὀρθὴγὰρἡπρὸςτῷΑγωνία·τὸἄραὑπὸτῶνΓΖ, Thus, the (rectangle contained) by CF and FA, plus the
ΖΑμετὰτοῦἀπὸτῆςΑΕἴσονἐστὶτοῖςἀπὸτῶνΒΑ,ΑΕ. (square) on AE, is equal to the (square) on EB. But,
κοινὸνἀφῃρήσθωτὸἀπὸτῆςΑΕ·λοιπὸνἄρατὸὑπὸτῶν the (sum of the squares) on BA and AE is equal to the
ΓΖ,ΖΑπεριεχόμενονὀρθογώνιονἴσονἐστὶτῷἀπὸτῆςΑΒ (square) on EB. For the angle at A (is) a right-angle
τετραγώνῳ. καίἐστιτὸμὲνὑπὸτῶνΓΖ,ΖΑτὸΖΚ·ἴση [Prop. 1.47]. Thus, the (rectangle contained) by CF and
γὰρἡΑΖτῇΖΗ·τὸδὲἀπὸτῆςΑΒτὸΑΔ·τὸἄραΖΚἴσον FA, plus the (square) on AE, is equal to the (sum of
ἐστὶτῷΑΔ.κοινὸνἀρῃρήσθωτὸΑΚ·λοιπὸνἄρατὸΖΘ the squares) on BA and AE. Let the square on AE have
τῷΘΔἴσονἐστίν. καίἐστιτὸμὲνΘΔτὸὑπὸτῶνΑΒ, been subtracted from both. Thus, the remaining rectan-
ΒΘ·ἴσηγὰρἡΑΒτῇΒΔ·τὸδὲΖΘτὸἀπὸτῆςΑΘ·τὸ gle contained by CF and FA is equal to the square on
ἄραὑπὸτῶνΑΒ,ΒΘπεριεχόμενονὀρθογώνιονἴσονἐστὶ AB. And FK is the (rectangle contained) by CF and
τῷἀπὸΘΑτετραγώνῳ.
FA. For AF (is) equal to FG. And AD (is) the (square)
῾ΗἄραδοθεῖσαεὐθεῖαἡΑΒτέτμηταικατὰτὸΘὥστε on AB. Thus, the (rectangle) FK is equal to the (square)
τὸὑπὸτῶνΑΒ,ΒΘπεριεχόμενονὀρθογώνιονἴσονποιεῖν AD. Let (rectangle) AK have been subtracted from both.
τῷἀπὸτῆςΘΑτετραγώνῳ·ὅπερἔδειποιῆσαι.
Thus, the remaining (square) FH is equal to the (rectangle)
HD. And HD is the (rectangle contained) by AB
and BH. For AB (is) equal to BD. And FH (is) the
(square) on AH. Thus, the rectangle contained by AB
C
K
D
63

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 2
and BH is equal to the square on HA.
Thus, the given straight-line AB has been cut at
(point) H such as to make the rectangle contained by
AB and BH equal to the square on HA. (Which is) the
very thing it was required to do.
† This manner of cutting a straight-line—so that the ratio of the whole to the larger piece is equal to the ratio of the larger to the smaller piece—is
sometimes called the “Golden Section”.
. Proposition 12 †
᾿Εν τοῖς ἀμβλυγωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν In obtuse-angled triangles, the square on the side sub-
ἀμβλεῖανγωνίανὑποτεινούσηςπλευρᾶςτετράγωνονμεῖζόν tending the obtuse angle is greater than the (sum of the)
ἐστι τῶν ἀπὸ τῶν τὴν ἀμβλεῖαν γωνίαν περιεχουσῶν squares on the sides containing the obtuse angle by twice
πλευρῶντετραγώνωντῷπεριεχομένῳδὶςὑπὸτεμιᾶςτῶν the (rectangle) contained by one of the sides around the
περὶτὴνἀμβλεῖανγωνίαν,ἐφ᾿ἣνἡκάθετοςπίπτει,καὶτῆς obtuse angle, to which a perpendicular (straight-line)
ἀπολαμβανομένηςἐκτὸςὑπὸτῆςκαθέτουπρὸςτῇἀμβλείᾳ falls, and the (straight-line) cut off outside (the triangle)
γωνίᾳ.
by the perpendicular (straight-line) towards the obtuse
angle.
Β
B
∆ Α Γ
D A C
῎ΕστωἀμβλυγώνιοντρίγωνοντὸΑΒΓἀμβλεῖανἔχον Let ABC be an obtuse-angled triangle, having the anτὴνὑπὸΒΑΓ,καὶἤχθωἀπὸτοῦΒσημείουἐπὶτὴνΓΑ
gle BAC obtuse. And let BD be drawn from point B,
ἐκβληθεῖσαν κάθετος ἡ ΒΔ. λέγω, ὅτι τὸ ἀπὸ τῆς ΒΓ perpendicular to CA produced [Prop. 1.12]. I say that
τετράγωνονμεῖζόνἐστιτῶνἀπὸτῶνΒΑ,ΑΓτετραγώνων the square on BC is greater than the (sum of the) squares
τῷδὶςὑπὸτῶνΓΑ,ΑΔπεριεχομένῳὀρθογωνίῳ. on BA and AC, by twice the rectangle contained by CA
᾿ΕπεὶγὰρεὐθεῖαἡΓΔτέτμηται,ὡςἔτυχεν,κατὰτὸΑ and AD.
σημεῖον,τὸἄραἀπὸτῆςΔΓἴσονἐστὶτοῖςἀπὸτῶνΓΑ, For since the straight-line CD has been cut, at ran-
ΑΔτετραγώνοιςκαὶτῷδὶςὑπὸτῶνΓΑ,ΑΔπεριεχομένῳ dom, at point A, the (square) on DC is thus equal to
ὀρθογωνίῳ. κοινὸνπροσκείσθωτὸἀπὸτῆςΔΒ·τὰἄρα the (sum of the) squares on CA and AD, and twice the
ἀπὸτῶνΓΔ,ΔΒἴσαἐστὶτοῖςτεἀπὸτῶνΓΑ,ΑΔ,ΔΒτε- rectangle contained by CA and AD [Prop. 2.4]. Let the
τραγώνοιςκαὶτῷδὶςὑπὸτῶνΓΑ,ΑΔ[περιεχομένῳὀρθο- (square) on DB have been added to both. Thus, the (sum
γωνίῳ]. ἀλλὰτοῖςμὲνἀπὸτῶνΓΔ,ΔΒἴσονἐστὶτὸἀπὸ of the squares) on CD and DB is equal to the (sum of
τῆςΓΒ·ὀρθὴγὰρἡπροςτῷΔγωνία·τοῖςδὲἀπὸτῶνΑΔ, the) squares on CA, AD, and DB, and twice the [rect-
ΔΒἴσοντὸἀπὸτῆςΑΒ·τὸἄραἀπὸτῆςΓΒτετράγωνον angle contained] by CA and AD. But, the (square) on
ἴσονἐστὶτοῖςτεἀπὸτῶνΓΑ,ΑΒτετραγώνοιςκαὶτῷδὶς CB is equal to the (sum of the squares) on CD and DB.
ὑπὸτῶνΓΑ,ΑΔπεριεχομένῳὀρθογωνίῳ·ὥστετὸἀπὸτῆς For the angle at D (is) a right-angle [Prop. 1.47]. And
ΓΒτετράγωνοντῶνἀπὸτῶνΓΑ,ΑΒτετραγώνωνμεῖζόν the (square) on AB (is) equal to the (sum of the squares)
ἐστιτῷδὶςὑπὸτῶνΓΑ,ΑΔπεριεχομένῳὀρθογωνίῳ. on AD and DB [Prop. 1.47]. Thus, the square on CB
᾿Ενἄρατοῖςἀμβλυγωνίοιςτριγώνοιςτὸἀπὸτῆςτὴν is equal to the (sum of the) squares on CA and AB, and
ἀμβλεῖανγωνίανὑποτεινούσηςπλευρᾶςτετράγωνονμεῖζόν twice the rectangle contained by CA and AD. So the
ἐστι τῶν ἀπὸ τῶν τὴν ἀμβλεῖαν γωνίαν περιεχουσῶν square on CB is greater than the (sum of the) squares on
64

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 2
πλευρῶντετραγώνωντῷπεριχομένῳδὶςὑπότεμιᾶςτῶν CA and AB by twice the rectangle contained by CA and
περὶτὴνἀμβλεῖανγωνίαν,ἐφ᾿ἣνἡκάθετοςπίπτει,καὶτῆς AD.
ἀπολαμβανομένηςἐκτὸςὑπὸτῆςκαθέτουπρὸςτῇἀμβλείᾳ Thus, in obtuse-angled triangles, the square on the
γωνίᾳ·ὅπερἔδειδεῖξαι.
side subtending the obtuse angle is greater than the (sum
of the) squares on the sides containing the obtuse angle
by twice the (rectangle) contained by one of the
sides around the obtuse angle, to which a perpendicular
(straight-line) falls, and the (straight-line) cut off outside
(the triangle) by the perpendicular (straight-line) towards
the obtuse angle. (Which is) the very thing it was
required to show.
† This proposition is equivalent to the well-known cosine formula: BC 2 = AB 2 + AC 2 − 2 AB AC cos BAC, since cos BAC = −AD/AB.
. Proposition 13 †
᾿Εντοῖςὀξυγωνίοιςτριγώνοιςτὸἀπὸτῆςτὴνὀξεῖαν In acute-angled triangles, the square on the side subγωνίαν
ὑποτεινούσης πλευρᾶς τετράγωνον ἔλαττόν ἐστι tending the acute angle is less than the (sum of the)
τῶνἀπὸτῶντὴνὀξεῖανγωνίανπεριεχουσῶνπλευρῶντε- squares on the sides containing the acute angle by twice
τραγώνωντῷπεριεχομένῳδὶςὑπότεμιᾶςτῶνπερὶτὴν the (rectangle) contained by one of the sides around the
ὀξεῖανγωνίαν,ἐφ᾿ἣνἡκάθετοςπίπτει,καὶτῆςἀπολαμβα- acute angle, to which a perpendicular (straight-line) falls,
νομένηςἐντὸςὑπὸτῆςκαθέτουπρὸςτῇὀξείᾳγωνίᾳ. and the (straight-line) cut off inside (the triangle) by the
perpendicular (straight-line) towards the acute angle.
Α
A
Β
∆ Γ B D C
῎ΕστωὀξυγώνιοντρίγωνοντὸΑΒΓὀξεῖανἔχοντὴν Let ABC be an acute-angled triangle, having the anπρὸςτῷΒγωνίαν,καὶἤχθωἀπὸτοῦΑσημείουἐπὶτὴν
gle at (point) B acute. And let AD have been drawn from
ΒΓκάθετοςἡΑΔ·λέγω,ὅτιτὸἀπὸτῆςΑΓτετράγωνον point A, perpendicular to BC [Prop. 1.12]. I say that the
ἔλαττόνἐστιτῶνἀπὸτῶνΓΒ,ΒΑτετραγώνωντῷδὶςὑπὸ square on AC is less than the (sum of the) squares on
τῶνΓΒ,ΒΔπεριεχομένῳὀρθογωνίῳ.
CB and BA, by twice the rectangle contained by CB and
᾿ΕπεὶγὰρεὐθεῖαἡΓΒτέτμηται,ὡςἔτυχεν,κατὰτὸ BD.
Δ, τὰἄραἀπὸτῶνΓΒ,ΒΔτετράγωναἴσαἐστὶτῷτε For since the straight-line CB has been cut, at ranδὶςὑπὸτῶνΓΒ,ΒΔπεριεχομένῳὀρθογωνίῳκαὶτῷἀπὸ
dom, at (point) D, the (sum of the) squares on CB and
τῆςΔΓτετραγώνῳ. κοινὸνπροσκείσθωτὸἀπὸτῆςΔΑ BD is thus equal to twice the rectangle contained by CB
τετράγωνον·τὰἄραἀπὸτῶνΓΒ,ΒΔ,ΔΑτετράγωναἴσα and BD, and the square on DC [Prop. 2.7]. Let the
ἐστὶτῷτεδὶςὑπὸτῶνΓΒ,ΒΔπεριεχομένῳὀρθογωνίῳ square on DA have been added to both. Thus, the (sum
καὶτοῖςἀπὸτῶνΑΔ,ΔΓτετραγώνιος.ἀλλὰτοῖςμὲνἀπὸ of the) squares on CB, BD, and DA is equal to twice
τῶνΒΔ,ΔΑἴσοντὸἀπὸτῆςΑΒ·ὀρθὴγὰρἡπρὸςτῷΔ the rectangle contained by CB and BD, and the (sum of
γωνίᾳ·τοῖςδὲἀπὸτῶνΑΔ,ΔΓἴσοντὸἀπὸτῆςΑΓ·τὰ the) squares on AD and DC. But, the (square) on AB
ἄραἀπὸτῶνΓΒ,ΒΑἴσαἐστὶτῷτεἀπὸτῆςΑΓκαὶτῷδὶς (is) equal to the (sum of the squares) on BD and DA.
ὑπὸτῶνΓΒ,ΒΔ·ὥστεμόνοντὸἀπὸτῆςΑΓἔλαττόνἐστι For the angle at (point) D is a right-angle [Prop. 1.47].
65

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 2
τῶνἀπὸτῶνΓΒ,ΒΑτετραγώνωντῷδὶςὑπὸτῶνΓΒ,ΒΔ And the (square) on AC (is) equal to the (sum of the
περιεχομένῳὀρθογωνίῳ.
squares) on AD and DC [Prop. 1.47]. Thus, the (sum of
᾿Ενἄρατοῖςὀξυγωνίοιςτριγώνοιςτὸἀπὸτῆςτὴνὀξεῖαν the squares) on CB and BA is equal to the (square) on
γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον ἔλαττόν ἐστι AC, and twice the (rectangle contained) by CB and BD.
τῶνἀπὸτῶντὴνὀξεῖανγωνίανπεριεχουσῶνπλευρῶντε- So the (square) on AC alone is less than the (sum of the)
τραγώνωντῷπεριεχομένῳδὶςὑπότεμιᾶςτῶνπερὶτὴν squares on CB and BA by twice the rectangle contained
ὀξεῖανγωνίαν,ἐφ᾿ἣνἡκάθετοςπίπτει,καὶτῆςἀπολαμβα- by CB and BD.
νομένηςἐντὸςὑπὸτῆςκαθέτουπρὸςτῇὀξείᾳγωνίᾳ·ὅπερ Thus, in acute-angled triangles, the square on the side
ἔδειδεῖξαι.
subtending the acute angle is less than the (sum of the)
squares on the sides containing the acute angle by twice
the (rectangle) contained by one of the sides around the
acute angle, to which a perpendicular (straight-line) falls,
and the (straight-line) cut off inside (the triangle) by
the perpendicular (straight-line) towards the acute angle.
(Which is) the very thing it was required to show.
† This proposition is equivalent to the well-known cosine formula: AC 2 = AB 2 + BC 2 − 2 AB BC cos ABC, since cos ABC = BD/AB.
. Proposition 14
Τῷδοθέντιεὐθυγράμμῳἴσοντετράγωνονσυστήσας- To construct a square equal to a given rectilinear figθαι.
ure.
Α
Θ
A
H
Β
Η
Ε
Ζ
B
G
E
F
Γ
∆
῎Εστω τὸ δοθὲν εὐθύγραμμον τὸ Α· δεῖ δὴ τῷ Α Let A be the given rectilinear figure. So it is required
εὐθυγράμμῳἴσοντετράγωνονσυστήσασθαι. to construct a square equal to the rectilinear figure A.
Συνεστάτω γὰρ τῷ Α ἐυθυγράμμῳ ἴσον παραλληλό- For let the right-angled parallelogram BD, equal to
γραμμονὀρθογώνιοντὸΒΔ·εἰμὲνοὖνἴσηἐστὶνἡΒΕ the rectilinear figure A, have been constructed [Prop.
τῇΕΔ,γεγονὸςἂνεἴητὸἐπιταχθέν. συνέσταταιγὰρτῷ 1.45]. Therefore, if BE is equal to ED then that (which)
ΑεὐθυγράμμῳἴσοντετράγωνοντὸΒΔ·εἰδὲοὔ,μίατῶν was prescribed has taken place. For the square BD, equal
ΒΕ,ΕΔμείζωνἐστίν.ἔστωμείζωνἡΒΕ,καὶἐκβεβλήσθω to the rectilinear figure A, has been constructed. And if
ἐπὶτὸΖ,καὶκείσθωτῇΕΔἴσηἡΕΖ,καὶτετμήσθωἡΒΖ not, then one of the (straight-lines) BE or ED is greater
δίχακατὰτὸΗ,καὶκέντρῳτῷΗ,διαστήματιδὲἑνὶτῶν (than the other). Let BE be greater, and let it have
ΗΒ,ΗΖἡμικύκλιονγεγράφθωτὸΒΘΖ,καὶἐκβεβλήσθωἡ been produced to F , and let EF be made equal to ED
ΔΕἐπὶτὸΘ,καὶἐπεζεύχθωἡΗΘ.
[Prop. 1.3]. And let BF have been cut in half at (point)
᾿ΕπεὶοὖνεὐθεῖαἡΒΖτέτμηταιεἰςμὲνἴσακατὰτὸΗ,εἰς G [Prop. 1.10]. And, with center G, and radius one of
δὲἄνισακατὰτὸΕ,τὸἄραὑπὸτῶνΒΕ,ΕΖπεριεχόμενον the (straight-lines) GB or GF , let the semi-circle BHF
ὀρθογώνιονμετὰτοῦἀπὸτῆςΕΗτετραγώνουἴσονἐστὶ have been drawn. And let DE have been produced to H,
τῷἀπὸτῆςΗΖτετραγώνῳ. ἴσηδὲἡΗΖτῇΗΘ·τὸἄρα and let GH have been joined.
ὑπὸτῶνΒΕ,ΕΖμετὰτοῦἀπὸτῆςΗΕἴσονἐστὶτῷἀπὸ Therefore, since the straight-line BF has been cut—
τῆςΗΘ.τῷδὲἀπὸτῆςΗΘἴσαἐστὶτὰἀπὸτῶνΘΕ,ΕΗ equally at G, and unequally at E—the rectangle con-
C
D
66

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 2
τετράγωνα·τὸἄραὑπὸτῶνΒΕ,ΕΖμετὰτοῦἀπὸΗΕἴσα tained by BE and EF , plus the square on EG, is thus
ἐστὶτοῖςἀπὸτῶνΘΕ,ΕΗ.κοινὸνἀφῃρήσθωτὸἀπὸτῆςΗΕ equal to the square on GF [Prop. 2.5]. And GF (is) equal
τετράγωνον·λοιπὸνἄρατὸὑπὸτῶνΒΕ,ΕΖπεριεχόμενον to GH. Thus, the (rectangle contained) by BE and EF ,
ὄρθογώνιονἴσονἐστὶτῷἀπὸτῆςΕΘτετραγώνῳ.ἀλλὰτὸ plus the (square) on GE, is equal to the (square) on GH.
ὑπὸτῶνΒΕ,ΕΖτὸΒΔἐστιν·ἴσηγὰρἡΕΖτῇΕΔ·τὸ And the (sum of the) squares on HE and EG is equal to
ἄραΒΔπαραλληλόγραμμονἴσονἐστὶτῷἀπὸτῆςΘΕτε- the (square) on GH [Prop. 1.47]. Thus, the (rectangle
τραγώνῳ.ἴσονδὲτὸΒΔτῷΑεὐθυγράμμῳ.καὶτὸΑἄρα contained) by BE and EF , plus the (square) on GE, is
εὐθύγραμμονἴσονἐστὶτῷἀπὸτῆςΕΘἀναγραφησομένῳ equal to the (sum of the squares) on HE and EG. Let
τετραγώνῳ.
the square on GE have been taken from both. Thus, the
ΤῷἄραδοθέντιεὐθυγράμμῳτῷΑἴσοντετράγωνον remaining rectangle contained by BE and EF is equal to
συνέσταταιτὸἀπὸτῆςΕΘἀναγραφησόμενον· ὅπερἔδει the square on EH. But, BD is the (rectangle contained)
ποιῆσαι.
by BE and EF . For EF (is) equal to ED. Thus, the parallelogram
BD is equal to the square on HE. And BD
(is) equal to the rectilinear figure A. Thus, the rectilinear
figure A is also equal to the square (which) can be
described on EH.
Thus, a square—(namely), that (which) can be described
on EH—has been constructed, equal to the given
rectilinear figure A. (Which is) the very thing it was required
to do.
67

ELEMENTS BOOK 3
Fundamentals of Plane Geometry Involving
Circles
69

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
ÎÇÖÓ.
αʹ.῎Ισοικύκλοιεἰσίν,ὧναἱδιάμετροιἴσαιεἰσίν,ἢὧναἱ 1. Equal circles are (circles) whose diameters are
Definitions
ἐκτῶνκέντρωνἴσαιεἰσίν.
equal, or whose (distances) from the centers (to the cirβʹ.Εὐθεῖακύκλουἐφάπτεσθαιλέγεται,ἥτιςἁπτομένη
cumferences) are equal (i.e., whose radii are equal).
τοῦκύκλουκαὶἐκβαλλομένηοὐτέμνειτὸνκύκλον. 2. A straight-line said to touch a circle is any (straightγʹ.Κύκλοιἐφάπτεσθαιἀλλήλωνλέγονταιοἵτινεςἁπτό-
line) which, meeting the circle and being produced, does
μενοιἀλλήλωνοὐτέμνουσινἀλλήλους.
not cut the circle.
δʹ.᾿Εν κύκλῳἴσον ἀπέχειν ἀπὸ τοῦ κέντρου εὐθεῖαι 3. Circles said to touch one another are any (circles)
λέγονται, ὅταν αἱ ἀπὸ τοῦ κέντρου ἐπ᾿ αὐτὰς κάθετοι which, meeting one another, do not cut one another.
ἀγόμεναιἴσαιὦσιν.
4. In a circle, straight-lines are said to be equally far
εʹ.Μεῖζονδὲἀπέχεινλέγεται,ἐφ᾿ἣνἡμείζωνκάθετος from the center when the perpendiculars drawn to them
πίπτει.
from the center are equal.
ϛʹ.Τμῆμακύκλουἐστὶτὸπεριεχόμενονσχῆμαὑπότε 5. And (that straight-line) is said to be further (from
εὐθείαςκαὶκύκλουπεριφερείας.
the center) on which the greater perpendicular falls
ζʹ. Τμήματος δὲ γωνία ἐστὶν ἡ περιεχομένη ὑπό τε (from the center).
εὐθείαςκαὶκύκλουπεριφερείας.
6. A segment of a circle is the figure contained by a
ηʹ. ᾿Εν τμήματι δὲ γωνία ἐστίν, ὅταν ἐπὶ τῆς περι- straight-line and a circumference of a circle.
φερείαςτοῦτμήματοςληφθῇτισημεῖονκαὶἀπ᾿αὐτοῦἐπὶ 7. And the angle of a segment is that contained by a
τὰπέρατατῆςεὐθείας, ἥἐστιβάσιςτοῦτμήματος, ἐπι- straight-line and a circumference of a circle.
ζευχθῶσινεὐθεῖαι,ἡπεριεχομένηγωνίαὑπὸτῶνἐπιζευ- 8. And the angle in a segment is the angle contained
χθεισῶνεὐθειῶν.
by the joined straight-lines, when any point is taken on
θʹ.῞Οτανδὲ αἱπεριέχουσαιτὴν γωνίαν εὐθεῖαιἀπο- the circumference of a segment, and straight-lines are
λαμβάνωσίτιναπεριφέρειαν,ἐπ᾿ἐκείνηςλέγεταιβεβηκέναι joined from it to the ends of the straight-line which is
ἡγωνία.
the base of the segment.
ιʹ.Τομεὺςδὲκύκλουἐστίν,ὅτανπρὸςτῷκέντρῷτοῦ 9. And when the straight-lines containing an angle
κύκλουσυσταθῇγωνία,τὸπεριεχόμενονσχῆμαὑπότετῶν cut off some circumference, the angle is said to stand
τὴνγωνίανπεριεχουσῶνεὐθειῶνκαὶτῆςἀπολαμβανομένης upon that (circumference).
ὑπ᾿αὐτῶνπεριφερείας.
10. And a sector of a circle is the figure contained by
ιαʹ.῞Ομοίατμήματακύκλωνἐστὶτὰδεχόμεναγωνίας the straight-lines surrounding an angle, and the circum-
ἴσας,ἤἐνοἷςαἱγωνίαιἴσαιἀλλήλαιςεἰσίν.
ference cut off by them, when the angle is constructed at
the center of a circle.
11. Similar segments of circles are those accepting
equal angles, or in which the angles are equal to one another.
. Proposition 1
Τοῦδοθέντοςκύκλουτὸκέντρονεὑρεῖν.
To find the center of a given circle.
῎ΕστωὁδοθεὶςκύκλοςὁΑΒΓ·δεῖδὴτοῦΑΒΓκύκλου Let ABC be the given circle. So it is required to find
τὸκέντρονεὑρεῖν.
the center of circle ABC.
Διήχθωτιςεἰςαὐτόν, ὡς ἔτυχεν, εὐθεῖαἡΑΒ,καὶ Let some straight-line AB have been drawn through
τετμήσθωδίχακατὰτὸΔσημεῖον,καὶἀπὸτοῦΔτῇΑΒ (ABC), at random, and let (AB) have been cut in half at
πρὸςὀρθὰςἤχθωἡΔΓκαὶδιήχθωἐπὶτὸΕ,καὶτετμήσθω point D [Prop. 1.9]. And let DC have been drawn from
ἡΓΕδίχακατὰτὸΖ·λέγω,ὅτιτὸΖκέντρονἐστὶτοῦΑΒΓ D, at right-angles to AB [Prop. 1.11]. And let (CD) have
[κύκλου].
been drawn through to E. And let CE have been cut in
Μὴγάρ,ἀλλ᾿εἰδυνατόν,ἔστωτὸΗ,καὶἐπεζεύχθωσαν half at F [Prop. 1.9]. I say that (point) F is the center of
αἱΗΑ,ΗΔ,ΗΒ.καὶἐπεὶἴσηἐστὶνἡΑΔτῇΔΒ,κοινὴδὲἡ the [circle] ABC.
ΔΗ,δύοδὴαἱΑΔ,ΔΗδύοταῖςΗΔ,ΔΒἴσαιεἰσὶνἑκατέρα For (if) not then, if possible, let G (be the center of the
ἑκατέρᾳ·καὶβάσιςἡΗΑβάσειτῇΗΒἐστινἴση·ἐκκέντρου circle), and let GA, GD, and GB have been joined. And
γάρ·γωνίαἄραἡὑπὸΑΔΗγωνίᾳτῇὑπὸΗΔΒἴσηἐστίν. since AD is equal to DB, and DG (is) common, the two
70

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
ὅταν δὲ εὐθεῖαἐπ᾿ εὐθεῖανσταθεῖσατὰς ἐφεξῆςγωνίας (straight-lines) AD, DG are equal to the two (straight-
ἴσαςἀλλήλαιςποιῇ,ὀρθὴἑκατέρατῶνἴσωνγωνιῶνἐστιν· lines) BD, DG, † respectively. And the base GA is equal
ὀρθὴἄραἐστὶνἡὑπὸΗΔΒ.ἐστὶδὲκαὶἡὑπὸΖΔΒὀρθή· to the base GB. For (they are both) radii. Thus, angle
ἴσηἄραἡὑπὸΖΔΒτῇὑπὸΗΔΒ,ἡμείζωντῇἐλάττονι· ADG is equal to angle GDB [Prop. 1.8]. And when a
ὅπερἐστὶνἀδύνατον.οὐκἄρατὸΗκέντρονἐστὶτοῦΑΒΓ straight-line stood upon (another) straight-line make adκύκλου.ὁμοίωςδὴδείξομεν,ὅτιοὐδ᾿ἄλλοτιπλὴντοῦΖ.
jacent angles (which are) equal to one another, each of
the equal angles is a right-angle [Def. 1.10]. Thus, GDB
is a right-angle. And FDB is also a right-angle. Thus,
FDB (is) equal to GDB, the greater to the lesser. The
very thing is impossible. Thus, (point) G is not the center
of the circle ABC. So, similarly, we can show that neither
is any other (point) except F .
Γ
C
Ζ
Η
F
G
Α
∆
Β
A
D
B
Ε
ΤὸΖἄρασημεῖονκέντρονἐστὶτοῦΑΒΓ[κύκλου].
E
Thus, point F is the center of the [circle] ABC.
ÈÖ×Ñ.
᾿Εκδὴτούτουφανερόν, ὅτιἐὰνἐνκύκλῳεὐθεῖάτις So, from this, (it is) manifest that if any straight-line
Corollary
εὐθεῖάντιναδίχακαὶπρὸςὀρθὰςτέμνῃ,ἐπὶτῆςτεμνούσης in a circle cuts any (other) straight-line in half, and at
ἐστὶτὸκέντροντοῦκύκλου. —ὅπερἔδειποιῆσαι. right-angles, then the center of the circle is on the former
(straight-line). — (Which is) the very thing it was
required to do.
† The Greek text has “GD, DB”, which is obviously a mistake.
. Proposition 2
᾿Εὰνκύκλουἐπὶτῆςπεριφερείαςληφθῇδύοτυχόντα If two points are taken at random on the circumferσημεῖα,ἡἐπὶτὰσημεῖαἐπιζευγνυμένηεὐθεῖαἐντὸςπεσεῖται
ence of a circle then the straight-line joining the points
τοῦκύκλου.
will fall inside the circle.
῎ΕστωκύκλοςὁΑΒΓ,καὶἐπὶτῆςπεριφερείαςαὐτοῦ Let ABC be a circle, and let two points A and B have
εἰλήφθωδύοτυχόντασημεῖατὰΑ,Β·λέγω, ὅτιἡἀπὸ been taken at random on its circumference. I say that the
τοῦΑἐπὶτὸΒἐπιζευγνυμένηεὐθεῖαἐντὸςπεσεῖταιτοῦ straight-line joining A to B will fall inside the circle.
κύκλου.
For (if) not then, if possible, let it fall outside (the
Μὴγάρ,ἀλλ᾿εἰδυνατόν,πιπτέτωἐκτὸςὡςἡΑΕΒ,καὶ circle), like AEB (in the figure). And let the center of
εἰλήφθωτὸκέντροντοῦΑΒΓκύκλου,καὶἔστωτὸΔ,καὶ the circle ABC have been found [Prop. 3.1], and let it be
ἐπεζεύχθωσαναἱΔΑ,ΔΒ,καὶδιήχθωἡΔΖΕ.
(at point) D. And let DA and DB have been joined, and
᾿ΕπεὶοὖνἴσηἐστὶνἡΔΑτῇΔΒ,ἴσηἄρακαὶγωνίαἡ let DFE have been drawn through.
ὑπὸΔΑΕτῇὑπὸΔΒΕ·καὶἐπεὶτριγώνουτοῦΔΑΕμία Therefore, since DA is equal to DB, the angle DAE
71

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
πλευρὰπροσεκβέβληταιἡΑΕΒ,μείζωνἄραἡὑπὸΔΕΒ (is) thus also equal to DBE [Prop. 1.5]. And since in triγωνίατῆςὑπὸΔΑΕ.
ἴσηδὲἡὑπὸΔΑΕ τῇὑπὸΔΒΕ· angle DAE the one side, AEB, has been produced, anμείζωνἄραἡὑπὸΔΕΒτῆςὑπὸΔΒΕ.ὑπὸδὲτὴνμείζονα
gle DEB (is) thus greater than DAE [Prop. 1.16]. And
γωνίανἡμείζωνπλευρὰὑποτείνει·μείζωνἄραἡΔΒτῆς DAE (is) equal to DBE [Prop. 1.5]. Thus, DEB (is)
ΔΕ. ἴσηδὲἡΔΒτῇΔΖ.μείζων ἄραἡΔΖτῆςΔΕἡ greater than DBE. And the greater angle is subtended
ἐλάττωντῆςμείζονος· ὅπερἐστὶνἀδύνατον. οὐκἄραἡ by the greater side [Prop. 1.19]. Thus, DB (is) greater
ἀπὸτοῦΑἐπὶτὸΒἐπιζευγνυμένηεὐθεῖαἐκτὸςπεσεῖται than DE. And DB (is) equal to DF . Thus, DF (is)
τοῦκύκλου. ὁμοίωςδὴδείξομεν,ὅτιοὐδὲἐπ᾿αὐτῆςτῆς greater than DE, the lesser than the greater. The very
περιφερείας·ἐντὸςἄρα.
thing is impossible. Thus, the straight-line joining A to
B will not fall outside the circle. So, similarly, we can
show that neither (will it fall) on the circumference itself.
Thus, (it will fall) inside (the circle).
Γ
C
Α
∆
A
D
Ε
Ζ
Β
᾿Εὰνἄρακύκλουἐπὶτῆςπεριφερείαςληφθῇδύοτυχόντα Thus, if two points are taken at random on the cirσημεῖα,ἡἐπὶτὰσημεῖαἐπιζευγνυμένηεὐθεῖαἐντὸςπεσεῖται
cumference of a circle then the straight-line joining the
τοῦκύκλου·ὅπερἔδειδεῖξαι.
points will fall inside the circle. (Which is) the very thing
it was required to show.
. Proposition 3
᾿Εὰνἐνκύκλῳεὐθεῖάτιςδιὰτοῦκέντρουεὐθεῖάντινα In a circle, if any straight-line through the center cuts
μὴδιὰτοῦκέντρουδίχατέμνῃ,καὶπρὸςὀρθὰςαὐτὴντέμνει· in half any straight-line not through the center then it
καὶἐὰνπρὸςὀρθὰςαὐτὴντέμνῃ,καὶδίχααὐτὴντέμνει. also cuts it at right-angles. And (conversely) if it cuts it
῎ΕστωκύκλοςὁΑΒΓ,καὶἐναὐτῷεὐθεῖάτιςδιὰτοῦ at right-angles then it also cuts it in half.
κέντρουἡΓΔεὐθεῖάντιναμὴδιὰτοῦκέντρουτὴνΑΒδίχα Let ABC be a circle, and, within it, let some straightτεμνέτωκατὰτὸΖσημεῖον·λέγω,ὅτικαὶπρὸςὀρθὰςαὐτὴν
line through the center, CD, cut in half some straight-line
τέμνει.
not through the center, AB, at the point F . I say that
ΕἰλήφθωγὰρτὸκέντροντοῦΑΒΓκύκλου,καὶἔστω (CD) also cuts (AB) at right-angles.
τὸΕ,καὶἐπεζεύχθωσαναἱΕΑ,ΕΒ.
For let the center of the circle ABC have been found
ΚαὶἐπεὶἴσηἐστὶνἡΑΖτῇΖΒ,κοινὴδὲἡΖΕ,δύοδυσὶν [Prop. 3.1], and let it be (at point) E, and let EA and
ἴσαι[εἰσίν]·καὶβάσιςἡΕΑβάσειτῇΕΒἴση·γωνίαἄραἡ EB have been joined.
ὑπὸΑΖΕγωνίᾳτῇὑπὸΒΖΕἴσηἐστίν.ὅτανδὲεὐθεῖαἐπ᾿ And since AF is equal to FB, and FE (is) common,
εὐθεῖανσταθεῖσατὰςἐφεξῆςγωνίαςἴσαςἀλλήλαιςποιῇ, two (sides of triangle AFE) [are] equal to two (sides of
ὀρθὴἑκατέρατῶν ἴσωνγωνιῶνἐστιν· ἑκατέραἄρατῶν triangle BFE). And the base EA (is) equal to the base
ὑπὸΑΖΕ,ΒΖΕὀρθήἐστιν. ἡΓΔἄραδιὰτοῦκέντρου EB. Thus, angle AFE is equal to angle BFE [Prop. 1.8].
οὖσατὴνΑΒμὴδιὰτοῦκέντρουοὖσανδίχατέμνουσακαὶ And when a straight-line stood upon (another) straightπρὸςὀρθὰςτέμνει.
line makes adjacent angles (which are) equal to one another,
each of the equal angles is a right-angle [Def. 1.10].
Thus, AFE and BFE are each right-angles. Thus, the
E
F
B
72

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
Γ
(straight-line) CD, which is through the center and cuts
in half the (straight-line) AB, which is not through the
center, also cuts (AB) at right-angles.
C
Ε
E
Α
Ζ
Β
A
F
B
∆
ἈλλὰδὴἡΓΔτὴνΑΒπρὸςὀρθὰςτεμνέτω·λέγω,ὅτι And so let CD cut AB at right-angles. I say that it
καὶδίχααὐτὴντέμνει,τουτέστιν,ὅτιἴσηἐστὶνἡΑΖτῇΖΒ. also cuts (AB) in half. That is to say, that AF is equal to
Τῶνγὰραὐτῶνκατασκευασθέντων,ἐπεὶἴσηἐστὶνἡ FB.
ΕΑτῇΕΒ,ἴσηἐστὶκαὶγωνίαἡὑπὸΕΑΖτῇὑπὸΕΒΖ. For, with the same construction, since EA is equal
ἐστὶδὲκαὶὀρθὴἡὑπὸΑΖΕὀρθῇτῇὑπὸΒΖΕἴση·δύο to EB, angle EAF is also equal to EBF [Prop. 1.5].
ἄρατρίγωνάἐστιΕΑΖ,ΕΖΒτὰςδύογωνίαςδυσὶγωνίαις And the right-angle AFE is also equal to the right-angle
ἴσαςἔχοντακαὶμίανπλευρὰνμιᾷπλευρᾷἴσηνκοινὴναὐτῶν BFE. Thus, EAF and EFB are two triangles having
τὴνΕΖὑποτείνουσανὑπὸμίαντῶνἴσωνγωνιῶν·καὶτὰς two angles equal to two angles, and one side equal to
λοιπὰςἄραπλευρὰςταῖςλοιπαῖςπλευραῖςἴσαςἕξει·ἴσηἄρα one side—(namely), their common (side) EF , subtend-
ἡΑΖτῇΖΒ.
ing one of the equal angles. Thus, they will also have the
᾿Εὰνἄραἐνκύκλῳεὐθεῖάτιςδιὰτοῦκέντρουεὐθεῖάν remaining sides equal to the (corresponding) remaining
τιναμὴδιὰτοῦκέντρουδίχατέμνῃ,καὶπρὸςὀρθὰςαὐτὴν sides [Prop. 1.26]. Thus, AF (is) equal to FB.
τέμνει· καὶἐὰνπρὸςὀρθὰςαὐτὴντέμνῃ,καὶδίχααὐτὴν Thus, in a circle, if any straight-line through the cenτέμνει·ὅπερἔδειδεῖξαι.
ter cuts in half any straight-line not through the center
then it also cuts it at right-angles. And (conversely) if it
cuts it at right-angles then it also cuts it in half. (Which
is) the very thing it was required to show.
. Proposition 4
᾿Εὰνἐνκύκλῳδύοεὐθεῖαιτέμνωσινἀλλήλαςμὴδὶατοῦ In a circle, if two straight-lines, which are not through
κέντρουοὖσαι,οὐτέμνουσινἀλλήλαςδίχα.
the center, cut one another then they do not cut one an-
῎ΕστωκύκλοςὁΑΒΓΔ,καὶἐναὐτῷδύοεὐθεῖαιαἱΑΓ, other in half.
ΒΔτεμνέτωσανἀλλήλαςκατὰτὸΕμὴδιὰτοῦκέντρου Let ABCD be a circle, and within it, let two straightοὖσαι·λέγω,ὅτιοὐτέμνουσινἀλλήλαςδίχα.
lines, AC and BD, which are not through the center, cut
Εἰγὰρδυνατόν,τεμνέτωσανἀλλήλαςδίχαὥστεἴσην one another at (point) E. I say that they do not cut one
εἶναιτὴνμὲνΑΕτῇΕΓ,τὴνδὲΒΕτῇΕΔ·καὶεἰλήφθωτὸ another in half.
κέντροντοῦΑΒΓΔκύκλου,καὶἔστωτὸΖ,καὶἐπεζεύχθω For, if possible, let them cut one another in half, such
ἡΖΕ.
that AE is equal to EC, and BE to ED. And let the
᾿ΕπεὶοὖνεὐθεῖάτιςδιὰτοῦκέντρουἡΖΕεὐθεῖάντινα center of the circle ABCD have been found [Prop. 3.1],
μὴδιὰτοῦκέντρουτὴνΑΓδίχατέμνει,καὶπρὸςὀρθὰς and let it be (at point) F , and let FE have been joined.
αὐτὴντέμνει·ὀρθὴἄραἐστὶνἡὑπὸΖΕΑ·πάλιν,ἐπεὶεὐθεῖά Therefore, since some straight-line through the center,
τιςἡΖΕεὐθεῖάντινατὴνΒΔδίχατέμνει,καὶπρὸςὀρθὰς FE, cuts in half some straight-line not through the cenαὐτὴντέμνει·ὀρθὴἄραἡὑπὸΖΕΒ.ἐδείχθηδὲκαὶἡὑπὸ
ter, AC, it also cuts it at right-angles [Prop. 3.3]. Thus,
ΖΕΑὀρθή·ἴσηἄραἡὑπὸΖΕΑτῇὑπὸΖΕΒἡἐλάττωντῇ FEA is a right-angle. Again, since some straight-line FE
D
73

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
μείζονι·ὅπερἐστὶνἀδύνατον.οὐκἄρααἱΑΓ,ΒΔτέμνουσιν cuts in half some straight-line BD, it also cuts it at right-
ἀλλήλαςδίχα.
angles [Prop. 3.3]. Thus, FEB (is) a right-angle. But
FEA was also shown (to be) a right-angle. Thus, FEA
(is) equal to FEB, the lesser to the greater. The very
thing is impossible. Thus, AC and BD do not cut one
another in half.
Α
Ζ
∆
A
F
D
Β
Ε
Γ
᾿Εὰνἄραἐνκύκλῳδύοεὐθεῖαιτέμνωσινἀλλήλαςμὴδὶα Thus, in a circle, if two straight-lines, which are not
τοῦκέντρουοὖσαι,οὐτέμνουσινἀλλήλαςδίχα·ὅπερἔδει through the center, cut one another then they do not cut
δεῖξαι.
one another in half. (Which is) the very thing it was required
to show.
. Proposition 5
᾿Εὰνδύοκύκλοιτέμνωσινἀλλήλους,οὐκἔσταιαὐτῶν If two circles cut one another then they will not have
τὸαὐτὸκέντρον.
the same center.
Α
∆
Γ
A
B
D
E
C
C
Ε
E
Β
Ζ
B
F
Η
ΔύογὰρκύκλοιοἱΑΒΓ,ΓΔΗτεμνέτωσανἀλλήλους For let the two circles ABC and CDG cut one another
κατὰτὰΒ,Γσημεῖα. λέγω,ὅτιοὐκἔσταιαὐτῶντὸαὐτὸ at points B and C. I say that they will not have the same
κέντρον.
center.
Εἰγὰρδυνατόν,ἔστωτὸΕ,καὶἐπεζεύχθωἡΕΓ,καὶ For, if possible, let E be (the common center), and
διήχθωἡΕΖΗ,ὡςἔτυχεν.καὶἐπεὶτὸΕσημεῖονκέντρον let EC have been joined, and let EFG have been drawn
ἐστὶτοῦΑΒΓκύκλου,ἵσηἐστὶνἡΕΓτῇΕΖ.πάλιν,ἐπεὶτὸ through (the two circles), at random. And since point
ΕσημεῖονκέντρονἐστὶτοῦΓΔΗκύκλου,ἴσηἐστὶνἡΕΓ E is the center of the circle ABC, EC is equal to EF .
τῇΕΗ·ἐδείχθηδὲἡΕΓκαὶτῇΕΖἴση·καὶἡΕΖἄρατῇΕΗ Again, since point E is the center of the circle CDG, EC
ἐστινἴσηἡἐλάσσωντῇμείζονι·ὅπερἐστὶνἀδύνατον.οὐκ is equal to EG. But EC was also shown (to be) equal
ἄρατὸΕσημεῖονκέντρονἐστὶτῶνΑΒΓ,ΓΔΗκύκλων. to EF . Thus, EF is also equal to EG, the lesser to the
᾿Εὰν ἄρα δύο κύκλοι τέμνωσιν ἀλλήλους, οὐκ ἔστιν greater. The very thing is impossible. Thus, point E is not
G
74

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
αὐτῶντὸαὐτὸκέντρον·ὅπερἔδειδεῖξαι.
the (common) center of the circles ABC and CDG.
Thus, if two circles cut one another then they will not
have the same center. (Which is) the very thing it was
required to show.
¦. Proposition 6
᾿Εὰνδύοκύκλοιἐφάπτωνταιἀλλήλων,οὐκἔσταιαὐτῶν If two circles touch one another then they will not
τὸαὐτὸκέντρον.
have the same center.
Γ
C
Ζ
Ε
Β
F
E
B
∆
D
Α
A
ΔύογὰρκύκλοιοἱΑΒΓ,ΓΔΕἐφαπτέσθωσανἀλλήλων For let the two circles ABC and CDE touch one anκατὰτὸΓσημεῖον·
λέγω, ὅτιοὐκἔσταιαὐτῶντὸαὐτὸ other at point C. I say that they will not have the same
κέντρον.
center.
Εἰγὰρδυνατόν,ἔστωτὸΖ,καὶἐπεζεύχθωἡΖΓ,καὶ For, if possible, let F be (the common center), and
διήχθω,ὡςἔτυχεν,ἡΖΕΒ.
let FC have been joined, and let FEB have been drawn
᾿ΕπεὶοὖντὸΖσημεῖονκέντρονἐστὶτοῦΑΒΓκύκλου, through (the two circles), at random.
ἴσηἐστὶνἡΖΓτῇΖΒ.πάλιν,ἐπεὶτὸΖσημεῖονκέντρον Therefore, since point F is the center of the circle
ἐστὶτοῦΓΔΕκύκλου,ἴσηἐστὶνἡΖΓτῇΖΕ.ἐδείχθηδὲἡ ABC, FC is equal to FB. Again, since point F is the
ΖΓτῇΖΒἴση·καὶἡΖΕἄρατῇΖΒἐστινἴση,ἡἐλάττων center of the circle CDE, FC is equal to FE. But FC
τῇμείζονι· ὅπερἐστὶνἀδύνατον. οὐκἄρατὸΖσημεῖον was shown (to be) equal to FB. Thus, FE is also equal
κέντρονἐστὶτῶνΑΒΓ,ΓΔΕκύκλων.
to FB, the lesser to the greater. The very thing is impos-
᾿Εὰνἄραδύοκύκλοιἐφάπτωνταιἀλλήλων,οὐκἔσται sible. Thus, point F is not the (common) center of the
αὐτῶντὸαὐτὸκέντρον·ὅπερἔδειδεῖξαι.
circles ABC and CDE.
Thus, if two circles touch one another then they will
not have the same center. (Which is) the very thing it was
required to show.
Þ. Proposition 7
᾿Εὰνκύκλουἐπὶτῆςδιαμέτρουληφθῇτισημεῖον,ὃμή If some point, which is not the center of the circle,
ἐστικέντροντοῦκύκλου,ἀπὸδὲτοῦσημείουπρὸςτὸν is taken on the diameter of a circle, and some straightκύκλονπροσπίπτωσινεὐθεῖαίτινες,μεγίστημὲνἔσται,ἐφ᾿
lines radiate from the point towards the (circumference
ἧςτὸκέντρον,ἐλαχίστηδὲἡλοιπή,τῶνδὲἄλλωνἀεὶἡ of the) circle, then the greatest (straight-line) will be that
ἔγγιοντῆςδὶατοῦκέντρουτῆςἀπώτερονμείζων ἐστίν, on which the center (lies), and the least the remainder
δύοδὲμόνονἴσαιἀπὸτοῦσημείουπροσπεσοῦνταιπρὸς (of the same diameter). And for the others, a (straightτὸνκύκλονἐφ᾿ἑκάτερατῆςἐλαχίστης.
line) nearer † to the (straight-line) through the center is
always greater than a (straight-line) further away. And
only two equal (straight-lines) will radiate from the point
towards the (circumference of the) circle, (one) on each
75

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
Γ
Η
(side) of the least (straight-line).
C
G
Β
B
Α
Ε
Ζ
∆
A
E
F
D
Κ Θ
῎ΕστωκύκλοςὁΑΒΓΔ,διάμετροςδὲαὐτοῦἔστωἡΑΔ, Let ABCD be a circle, and let AD be its diameter, and
καὶἐπὶτῆςΑΔεἰλήφθωτισημεῖοντὸΖ,ὃμήἐστικέντρον let some point F , which is not the center of the circle,
τοῦκύκλου,κέντρονδὲτοῦκύκλουἔστωτὸΕ,καὶἀπὸτοῦ have been taken on AD. Let E be the center of the circle.
ΖπρὸςτὸνΑΒΓΔκύκλονπροσπιπτέτωσανεὐθεῖαίτινεςαἱ And let some straight-lines, FB, FC, and FG, radiate
ΖΒ,ΖΓ,ΖΗ·λέγω,ὅτιμεγίστημένἐστινἡΖΑ,ἐλαχίστη from F towards (the circumference of) circle ABCD. I
δὲἡΖΔ,τῶνδὲἄλλωνἡμὲνΖΒτῆςΖΓμείζων,ἡδὲΖΓ say that FA is the greatest (straight-line), FD the least,
τῆςΖΗ.
and of the others, FB (is) greater than FC, and FC than
᾿ΕπεζεύχθωσανγὰραἱΒΕ,ΓΕ,ΗΕ.καὶἐπεὶπαντὸς FG.
τριγώνουαἱδύοπλευραὶτῆςλοιπῆςμείζονέςεἰσιν,αἱἄρα For let BE, CE, and GE have been joined. And since
ΕΒ,ΕΖτῆςΒΖμείζονέςεἰσιν.ἴσηδὲἡΑΕτῇΒΕ[αἱἄρα for every triangle (any) two sides are greater than the
ΒΕ,ΕΖἴσαιεἰσὶτῇΑΖ]·μείζωνἄραἡΑΖτῆςΒΖ.πάλιν, remaining (side) [Prop. 1.20], EB and EF is thus greater
ἐπεὶἴσηἐστὶνἡΒΕτῇΓΕ,κοινὴδὲἡΖΕ,δύοδὴαἱΒΕ, than BF . And AE (is) equal to BE [thus, BE and EF
ΕΖδυσὶταῖςΓΕ,ΕΖἴσαιεἰσίν.ἀλλὰκαὶγωνίαἡὑπὸΒΕΖ is equal to AF ]. Thus, AF (is) greater than BF . Again,
γωνίαςτῆςὑπὸΓΕΖμείζων·βάσιςἄραἡΒΖβάσεωςτῆς since BE is equal to CE, and FE (is) common, the two
ΓΖμείζωνἐστίν. διὰτὰαὐτὰδὴκαὶἡΓΖτῆςΖΗμείζων (straight-lines) BE, EF are equal to the two (straight-
ἐστίν.
lines) CE, EF (respectively). But, angle BEF (is) also
Πάλιν,ἐπεὶαἱΗΖ,ΖΕτῆςΕΗμείζονέςεἰσιν,ἴσηδὲἡ greater than angle CEF . ‡ Thus, the base BF is greater
ΕΗτῇΕΔ,αἱἄραΗΖ,ΖΕτῆςΕΔμείζονέςεἰσιν. κοινὴ than the base CF . Thus, the base BF is greater than the
ἀφῃρήσθωἡΕΖ·λοιπὴἄραἡΗΖλοιπῆςτῆςΖΔμείζων base CF [Prop. 1.24]. So, for the same (reasons), CF is
ἐστίν.μεγίστημὲνἄραἡΖΑ,ἐλαχίστηδὲἡΖΔ,μείζωνδὲ also greater than FG.
ἡμὲνΖΒτῆςΖΓ,ἡδὲΖΓτῆςΖΗ.
Again, since GF and FE are greater than EG
Λέγω,ὅτικαὶἀπὸτοῦΖσημείουδύομόνονἴσαιπρο- [Prop. 1.20], and EG (is) equal to ED, GF and FE
σπεσοῦνταιπρὸςτὸνΑΒΓΔκύκλονἐφ᾿ἑκάτερατῆςΖΔ are thus greater than ED. Let EF have been taken from
ἐλαχίστης.συνεστάτωγὰρπρὸςτῇΕΖεὐθείᾳκαὶτῷπρὸς both. Thus, the remainder GF is greater than the reαὐτῇσημείῳτῷΕτῇὑπὸΗΕΖγωνίᾳἴσηἡὑπὸΖΕΘ,καὶ
mainder FD. Thus, FA (is) the greatest (straight-line),
ἐπεζεύχθωἡΖΘ.ἐπεὶοὖνἴσηἐστὶνἡΗΕτῇΕΘ,κοινὴ FD the least, and FB (is) greater than FC, and FC than
δὲἡΕΖ,δύοδὴαἱΗΕ,ΕΖδυσὶταῖςΘΕ,ΕΖἴσαιεἰσίν· FG.
καὶγωνίαἡὑπὸΗΕΖγωνίᾳτῇὑπὸΘΕΖἴση·βάσιςἄρα I also say that from point F only two equal (straight-
ἡΖΗβάσειτῇΖΘἴσηἐστίν. λέγωδή,ὅτιτῇΖΗἄλλη lines) will radiate towards (the circumference of) circle
ἴσηοὐπροσπεσεῖταιπρὸςτὸνκύκλονἀπὸτοῦΖσημείου. ABCD, (one) on each (side) of the least (straight-line)
εἰγὰρδυνατόν,προσπιπτέτωἡΖΚ.καὶἐπεὶἡΖΚτῇΖΗ FD. For let the (angle) FEH, equal to angle GEF , have
ἴσηἐστίν,ἀλλὰἡΖΘτῇΖΗ[ἴσηἐστίν],καὶἡΖΚἄρατῇ been constructed on the straight-line EF , at the point E
ΖΘἐστινἴση,ἡἔγγιοντῆςδιὰτοῦκέντρουτῇἀπώτερον on it [Prop. 1.23], and let FH have been joined. There-
ἴση·ὅπερἀδύνατον. οὐκἄραἀπὸτοῦΖσημείουἑτέρατις fore, since GE is equal to EH, and EF (is) common,
K
H
76

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
προσπεσεῖταιπρὸςτὸνκύκλονἴσητῇΗΖ·μίαἄραμόνη. the two (straight-lines) GE, EF are equal to the two
᾿Εὰνἄρακύκλουἐπὶτῆςδιαμέτρουληφθῇτισημεῖον, (straight-lines) HE, EF (respectively). And angle GEF
ὃμήἐστικέντροντοῦκύκλου,ἀπὸδὲτοῦσημείουπρὸς (is) equal to angle HEF . Thus, the base FG is equal to
τὸνκύκλονπροσπίπτωσινεὐθεῖαίτινες,μεγίστημὲνἔσται, the base FH [Prop. 1.4]. So I say that another (straight-
ἐφ᾿ἧςτὸκέντρον,ἐλαχίστηδὲἡλοιπή,τῶνδὲἄλλωνἀεὶἡ line) equal to FG will not radiate towards (the circumfer-
ἔγγιοντῆςδὶατοῦκέντρουτῆςἀπώτερονμείζωνἐστίν,δύο ence of) the circle from point F . For, if possible, let FK
δὲμόνονἴσαιἀπὸτοῦαὐτοῦσημείουπροσπεσοῦνταιπρὸς (so) radiate. And since FK is equal to FG, but FH [is
τὸνκύκλονἐφ᾿ἑκάτερατῆςἐλαχίστης·ὅπερἔδειδεῖξαι. equal] to FG, FK is thus also equal to FH, the nearer
to the (straight-line) through the center equal to the further
away. The very thing (is) impossible. Thus, another
(straight-line) equal to GF will not radiate from the point
F towards (the circumference of) the circle. Thus, (there
is) only one (such straight-line).
Thus, if some point, which is not the center of the
circle, is taken on the diameter of a circle, and some
straight-lines radiate from the point towards the (circumference
of the) circle, then the greatest (straight-line)
will be that on which the center (lies), and the least
the remainder (of the same diameter). And for the others,
a (straight-line) nearer to the (straight-line) through
the center is always greater than a (straight-line) further
away. And only two equal (straight-lines) will radiate
from the same point towards the (circumference of the)
circle, (one) on each (side) of the least (straight-line).
(Which is) the very thing it was required to show.
† Presumably, in an angular sense.
‡ This is not proved, except by reference to the figure.
. Proposition 8
᾿Εὰν κύκλου ληφθῇ τι σημεῖον ἐκτός, ἀπὸ δὲ τοῦ If some point is taken outside a circle, and some
σημείουπρὸςτὸνκύκλονδιαχθῶσινεὐθεῖαίτινες,ὧνμία straight-lines are drawn from the point to the (circumμὲνδιὰτοῦκέντρου,αἱδὲλοιπαί,ὡςἔτυχεν,τῶνμὲνπρὸς
ference of the) circle, one of which (passes) through
τὴν κοίλην περιφέρειαν προσπιπτουσῶν εὐθειῶν μεγίστη the center, the remainder (being) random, then for the
μένἐστινἡδιὰτοῦκέντρου,τῶνδὲἄλλωνἀεὶἡἔγγιοντῆς straight-lines radiating towards the concave (part of the)
διὰτοῦκέντρουτῆςἀπώτερονμείζωνἐστίν,τῶνδὲπρὸς circumference, the greatest is that (passing) through the
τὴνκυρτὴνπεριφέρειανπροσπιπτουσῶνεὐθειῶνἐλαχίστη center. For the others, a (straight-line) nearer † to the
μένἐστινἡμεταξὺτοῦτεσημείουκαὶτῆςδιαμέτρου,τῶν (straight-line) through the center is always greater than
δὲἄλλωνἀεὶἡἔγγιοντῆςἐλαχίστηςτῆςἀπώτερόνἐστιν one further away. For the straight-lines radiating towards
ἐλάττων,δύοδὲμόνονἴσαιἀπὸτοῦσημείουπροσπεσοῦνται the convex (part of the) circumference, the least is that
πρὸςτὸνκύκλονἐφ᾿ἑκάτερατῆςἐλαχίστης.
between the point and the diameter. For the others, a
῎ΕστωκύκλοςὁΑΒΓ,καὶτοῦΑΒΓεἰλήφθωτισημεῖον (straight-line) nearer to the least (straight-line) is always
ἐκτὸςτὸΔ,καὶἀπ᾿αὐτοῦδιήχθωσανεὐθεῖαίτινεςαἱΔΑ, less than one further away. And only two equal (straight-
ΔΕ, ΔΖ, ΔΓ, ἔστω δὲ ἡ ΔΑ διὰ τοῦ κέντρου. λέγω, lines) will radiate from the point towards the (circum-
ὅτιτῶνμὲνπρὸςτὴνΑΕΖΓκοίληνπεριφέρειανπροσπι- ference of the) circle, (one) on each (side) of the least
πτουσῶνεὐθειῶνμεγίστημένἐστινἡδιὰτοῦκέντρουἡ (straight-line).
ΔΑ,μείζωνδὲἡμὲνΔΕτῆςΔΖἡδὲΔΖτῆςΔΓ,τῶν Let ABC be a circle, and let some point D have been
δὲ πρὸς τὴν ΘΛΚΗ κυρτὴν περιφέρειαν προσπιπτουσῶν taken outside ABC, and from it let some straight-lines,
εὐθειῶνἐλαχίστημένἐστινἡΔΗἡμεταξὺτοῦσημείουκαὶ DA, DE, DF , and DC, have been drawn through (the
τῆςδιαμέτρουτῆςΑΗ,ἀεὶδὲἡἔγγιοντῆςΔΗἐλαχίστης circle), and let DA be through the center. I say that for
ἐλάττωνἐστὶτῆςἀπώτερον,ἡμὲνΔΚτῆςΔΛ,ἡδὲΔΛ the straight-lines radiating towards the concave (part of
77

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
τῆςΔΘ.
∆
the) circumference, AEFC, the greatest is the one (passing)
through the center, (namely) AD, and (that) DE (is)
greater than DF , and DF than DC. For the straight-lines
radiating towards the convex (part of the) circumference,
HLKG, the least is the one between the point and the diameter
AG, (namely) DG, and a (straight-line) nearer to
the least (straight-line) DG is always less than one farther
away, (so that) DK (is less) than DL, and DL than
than DH.
D
Θ
Λ ΚΗ
H
L K
G
B
N
Ζ
Ε
Γ
Μ
Β
Ν
F
E
C
M
Α
A
ΕἰλήφθωγὰρτὸκέντροντοῦΑΒΓκύκλουκαὶἔστωτὸ For let the center of the circle have been found
Μ·καὶἐπεζεύχθωσαναἱΜΕ,ΜΖ,ΜΓ,ΜΚ,ΜΛ,ΜΘ. [Prop. 3.1], and let it be (at point) M [Prop. 3.1]. And let
ΚαὶἐπεὶἴσηἐστὶνἡΑΜτῇΕΜ,κοινὴπροσκείσθωἡ ME, MF , MC, MK, ML, and MH have been joined.
ΜΔ·ἡἄραΑΔἴσηἐστὶταῖςΕΜ,ΜΔ.ἀλλ᾿αἱΕΜ,ΜΔ And since AM is equal to EM, let MD have been
τῆςΕΔμείζονέςεἰσιν·καὶἡΑΔἄρατῆςΕΔμείζωνἐστίν. added to both. Thus, AD is equal to EM and MD. But,
πάλιν,ἐπεὶἴσηἐστὶνἡΜΕτῇΜΖ,κοινὴδὲἡΜΔ,αἱΕΜ, EM and MD is greater than ED [Prop. 1.20]. Thus,
ΜΔἄραταῖςΖΜ,ΜΔἴσαιεἰσίν·καὶγωνίαἡὑπὸΕΜΔ AD is also greater than ED. Again, since ME is equal
γωνίαςτῆςὑπὸΖΜΔμείζωνἐστίν.βάσιςἄραἡΕΔβάσεως to MF , and MD (is) common, the (straight-lines) EM,
τῆςΖΔμείζωνἐστίν·ὁμοίωςδὴδείξομεν,ὅτικαὶἡΖΔτῆς MD are thus equal to FM, MD. And angle EMD is
ΓΔμείζωνἐστίν·μεγίστημὲνἄραἡΔΑ,μείζωνδὲἡμὲν greater than angle FMD. ‡ Thus, the base ED is greater
ΔΕτῆςΔΖ,ἡδὲΔΖτῆςΔΓ. than the base FD [Prop. 1.24]. So, similarly, we can
ΚαὶἐπεὶαἱΜΚ,ΚΔτῆςΜΔμείζονέςεἰσιν,ἴσηδὲἡ show that FD is also greater than CD. Thus, AD (is) the
ΜΗτῇΜΚ,λοιπὴἄραἡΚΔλοιπῆςτῆςΗΔμείζωνἐστίν· greatest (straight-line), and DE (is) greater than DF ,
ὥστεἡΗΔτῆςΚΔἐλάττωνἐστίν·καὶἐπεὶτριγώνουτοῦ and DF than DC.
ΜΛΔἐπὶμιᾶςτῶνπλευρῶντῆςΜΔδύοεὐθεῖαιἐντὸς And since MK and KD is greater than MD [Prop.
συνεστάθησαναἱΜΚ,ΚΔ,αἱἄραΜΚ,ΚΔτῶνΜΛ,ΛΔ 1.20], and MG (is) equal to MK, the remainder KD
ἐλάττονέςεἰσιν· ἴσηδὲἡΜΚτῇΜΛ·λοιπὴἄραἡΔΚ is thus greater than the remainder GD. So GD is less
λοιπῆςτῆςΔΛἐλάττωνἐστίν. ὁμοίωςδὴδείξομεν, ὅτι than KD. And since in triangle MLD, the two interκαὶἡΔΛτῆςΔΘἐλάττωνἐστίν·ἐλαχίστημὲνἄραἡΔΗ,
nal straight-lines MK and KD were constructed on one
ἐλάττωνδὲἡμὲνΔΚτῆςΔΛἡδὲΔΛτῆςΔΘ. of the sides, MD, then MK and KD are thus less than
Λέγω, ὅτι καὶ δύο μόνον ἴσαι ἀπὸ τοῦ Δ σημείου ML and LD [Prop. 1.21]. And MK (is) equal to ML.
προσπεσοῦνται πρὸς τὸν κύκλον ἐφ᾿ ἑκάτερα τῆς ΔΗ Thus, the remainder DK is less than the remainder DL.
ἐλαχίστης· συνεστάτωπρὸςτῇΜΔεὐθείᾳκαὶτῷπρὸς So, similarly, we can show that DL is also less than DH.
αὐτῇσημείῳτῷΜτῇὑπὸΚΜΔγωνίᾳἴσηγωνίαἡὑπὸ Thus, DG (is) the least (straight-line), and DK (is) less
ΔΜΒ,καὶἐπεζεύχθωἡΔΒ.καὶἐπεὶἴσηἐστὶνἡΜΚτῇ than DL, and DL than DH.
ΜΒ,κοινὴδὲἡΜΔ,δύοδὴαἱΚΜ,ΜΔδύοταῖςΒΜ,ΜΔ I also say that only two equal (straight-lines) will radi-
78

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
ἴσαιεἰσὶνἑκατέραἑκατέρᾳ·καὶγωνίαἡὑπὸΚΜΔγωνίᾳ ate from point D towards (the circumference of) the cirτῇὑπὸΒΜΔἴση·βάσιςἄραἡΔΚβάσειτῇΔΒἴσηἐστίν.
cle, (one) on each (side) on the least (straight-line), DG.
λέγω [δή], ὅτι τῇ ΔΚ εὐθείᾳἄλληἴση οὐπροσπεσεῖται Let the angle DMB, equal to angle KMD, have been
πρὸςτὸνκύκλονἀπὸτοῦΔσημείου.εἰγὰρδυνατόν,προ- constructed on the straight-line MD, at the point M on
σπιπτέτωκαὶἔστωἡΔΝ.ἐπεὶοὖνἡΔΚτῇΔΝἐστινἴση, it [Prop. 1.23], and let DB have been joined. And since
ἀλλ᾿ἡΔΚτῇΔΒἐστινἴση,καὶἡΔΒἄρατῇΔΝἐστιν MK is equal to MB, and MD (is) common, the two
ἴση,ἡἔγγιοντῆςΔΗἐλαχίστηςτῇἀπώτερον[ἐστιν]ἴση· (straight-lines) KM, MD are equal to the two (straight-
ὅπερἀδύνατονἐδείχθη. οὐκἄραπλείουςἢδύοἴσαιπρὸς lines) BM, MD, respectively. And angle KMD (is)
τὸνΑΒΓκύκλονἀπὸτοῦΔσημείουἐφ᾿ἑκάτερατῆςΔΗ equal to angle BMD. Thus, the base DK is equal to the
ἐλαχίστηςπροσπεσοῦνται.
base DB [Prop. 1.4]. [So] I say that another (straight-
᾿Εὰνἄρακύκλουληφθῇτισημεῖονἐκτός,ἀπὸδὲτοῦ line) equal to DK will not radiate towards the (circumσημείουπρὸςτὸνκύκλονδιαχθῶσινεὐθεῖαίτινες,ὧνμία
ference of the) circle from point D. For, if possible, let
μὲνδιὰτοῦκέντρουαἱδὲλοιπαί,ὡςἔτυχεν,τῶνμὲνπρὸς (such a straight-line) radiate, and let it be DN. Thereτὴν
κοίλην περιφέρειαν προσπιπτουσῶν εὐθειῶν μεγίστη fore, since DK is equal to DN, but DK is equal to DB,
μένἐστινἡδιὰτοῦκέντου,τῶνδὲἄλλωνἀεὶἡἔγγιοντῆς then DB is thus also equal to DN, (so that) a (straightδιὰτοῦκέντρουτῆςἀπώτερονμείζωνἐστίν,τῶνδὲπρὸς
line) nearer to the least (straight-line) DG [is] equal to
τὴνκυρτὴνπεριφέρειανπροσπιπτουσῶνεὐθειῶνἐλαχίστη one further away. The very thing was shown (to be) imμένἐστινἡμεταξὺτοῦτεσημείουκαὶτῆςδιαμέτρου,τῶν
possible. Thus, not more than two equal (straight-lines)
δὲἄλλωνἀεὶἡἔγγιοντῆςἐλαχίστηςτῆςἀπώτερόνἐστιν will radiate towards (the circumference of) circle ABC
ἐλάττων,δύοδὲμόνονἴσαιἀπὸτοῦσημείουπροσπεσοῦνται from point D, (one) on each side of the least (straightπρὸς
τὸν κύκλον ἐφ᾿ ἑκάτερα τῆς ἐλαχίστης· ὅπερ ἔδει line) DG.
δεῖξαι.
Thus, if some point is taken outside a circle, and some
straight-lines are drawn from the point to the (circumference
of the) circle, one of which (passes) through the center,
the remainder (being) random, then for the straightlines
radiating towards the concave (part of the) circumference,
the greatest is that (passing) through the center.
For the others, a (straight-line) nearer to the (straightline)
through the center is always greater than one further
away. For the straight-lines radiating towards the
convex (part of the) circumference, the least is that between
the point and the diameter. For the others, a
(straight-line) nearer to the least (straight-line) is always
less than one further away. And only two equal (straightlines)
will radiate from the point towards the (circumference
of the) circle, (one) on each (side) of the least
(straight-line). (Which is) the very thing it was required
to show.
† Presumably, in an angular sense.
‡ This is not proved, except by reference to the figure.
. Proposition 9
᾿Εὰν κύκλου ληφθῇ τι σημεῖον ἐντός, ἀπο δὲ τοῦ If some point is taken inside a circle, and more than
σημείουπρὸςτὸνκύκλονπροσπίπτωσιπλείουςἢδύοἴσαι two equal straight-lines radiate from the point towards
εὐθεῖαι,τὸληφθὲνσημεῖονκέντρονἐστὶτοῦκύκλου. the (circumference of the) circle, then the point taken is
῎ΕστωκύκλοςὁΑΒΓ,ἐντὸςδὲαὐτοῦσημεῖοντὸΔ,καὶ the center of the circle.
ἀπὸτοῦΔπρὸςτὸνΑΒΓκύκλονπροσπιπτέτωσανπλείους Let ABC be a circle, and D a point inside it, and let
ἢδύοἴσαιεὐθεῖαιαἱΔΑ,ΔΒ,ΔΓ·λέγω,ὅτιτὸΔσημεῖον more than two equal straight-lines, DA, DB, and DC, raκέντρονἐστὶτοῦΑΒΓκύκλου.
diate from D towards (the circumference of) circle ABC.
79

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
Λ
I say that point D is the center of circle ABC.
L
Β
Ζ
Γ
B
F
C
Κ
Ε
∆
Η
K
E
D
G
Α
Θ
H
᾿Επεζεύχθωσαν γὰρ αἱ ΑΒ, ΒΓ καὶ τετμήσθωσαν For let AB and BC have been joined, and (then)
δίχακατὰτὰΕ,Ζσημεῖα,καὶἐπιζευχθεῖσαιαἱΕΔ,ΖΔ have been cut in half at points E and F (respectively)
διήχθωσανἐπὶτὰΗ,Κ,Θ,Λσημεῖα.
[Prop. 1.10]. And ED and FD being joined, let them
᾿ΕπεὶοὖνἴσηἐστὶνἡΑΕτῇΕΒ,κοινὴδὲἡΕΔ,δύο have been drawn through to points G, K, H, and L.
δὴαἱΑΕ,ΕΔδύοταῖςΒΕ,ΕΔἴσαιεἰσίν·καὶβάσιςἡΔΑ Therefore, since AE is equal to EB, and ED (is) comβάσειτῇΔΒἴση·γωνίαἄραἡὑπὸΑΕΔγωνίᾳτῇὑπὸΒΕΔ
mon, the two (straight-lines) AE, ED are equal to the
ἴσηἐστίν·ὀρθὴἄραἑκατέρατῶνὑπὸΑΕΔ,ΒΕΔγωνιῶν·ἡ two (straight-lines) BE, ED (respectively). And the base
ΗΚἄρατὴνΑΒτέμνειδίχακαὶπρὸςὀρθάς.καὶἐπεί,ἐὰν DA (is) equal to the base DB. Thus, angle AED is equal
ἐνκύκλῳεὐθεῖάτιςεὐθεῖάντιναδίχατεκαὶπρὸςὀρθὰς to angle BED [Prop. 1.8]. Thus, angles AED and BED
τέμνῃ,ἐπὶτῆςτεμνούσηςἐστὶτὸκέντροντοῦκύκλου,ἐπὶ (are) each right-angles [Def. 1.10]. Thus, GK cuts AB in
τῆςΗΚἄραἐστὶτὸκέντροντοῦκύκλου.διὰτὰαὐτὰδὴκαὶ half, and at right-angles. And since, if some straight-line
ἐπὶτῆςΘΛἐστιτὸκέντροντοῦΑΒΓκύκλου. καὶοὐδὲν in a circle cuts some (other) straight-line in half, and at
ἕτερονκοινὸνἔχουσιναἱΗΚ,ΘΛεὐθεῖαιἢτὸΔσημεῖον· right-angles, then the center of the circle is on the former
τὸΔἄρασημεῖονκέντρονἐστὶτοῦΑΒΓκύκλου. (straight-line) [Prop. 3.1 corr.], the center of the circle is
᾿Εὰνἄρακύκλουληφθῇτισημεῖονἐντός,ἀπὸδὲτοῦ thus on GK. So, for the same (reasons), the center of
σημείουπρὸςτὸνκύκλονπροσπίπτωσιπλείουςἢδύοἴσαι circle ABC is also on HL. And the straight-lines GK and
εὐθεῖαι,τὸληφθὲνσημεῖονκέντρονἐστὶτοῦκύκλου·ὅπερ HL have no common (point) other than point D. Thus,
ἔδειδεῖξαι.
point D is the center of circle ABC.
Thus, if some point is taken inside a circle, and more
than two equal straight-lines radiate from the point towards
the (circumference of the) circle, then the point
taken is the center of the circle. (Which is) the very thing
it was required to show.
. Proposition 10
Κύκλοςκύκλονοὐτέμνεικατὰπλείονασημεῖαἢδύο. A circle does not cut a(nother) circle at more than two
Εἰ γὰρ δυνατόν, κύκλος ὁ ΑΒΓ κύκλον τὸν ΔΕΖ points.
τεμνέτωκατὰπλείονασημεῖαἢδύοτὰΒ,Η,Ζ,Θ,καὶ For, if possible, let the circle ABC cut the circle DEF
ἐπιζευχθεῖσαιαἱΒΘ,ΒΗδίχατεμνέσθωσανκατὰτὰΚ,Λ at more than two points, B, G, F , and H. And BH and
σημεῖα·καὶἀπὸτῶνΚ,ΛταῖςΒΘ,ΒΗπρὸςὀρθὰςἀχθεῖσαι BG being joined, let them (then) have been cut in half
αἱΚΓ,ΛΜδιήχθωσανἐπὶτὰΑ,Εσημεῖα.
at points K and L (respectively). And KC and LM being
drawn at right-angles to BH and BG from K and
L (respectively) [Prop. 1.11], let them (then) have been
drawn through to points A and E (respectively).
A
80

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
Μ
Θ
Ν
Ζ
Α
Κ
∆
Ο
Λ
Β
Η
Ξ Ε
Γ
C
᾿ΕπεὶοὖνἐνκύκλῳτῷΑΒΓεὐθεῖάτιςἡΑΓεὐθεῖάν Therefore, since in circle ABC some straight-line
τινατὴνΒΘδίχακαὶπρὸςὀρθὰςτέμνει,ἐπὶτῆςΑΓἄρα AC cuts some (other) straight-line BH in half, and at
ἐστὶτὸκέντροντοῦΑΒΓκύκλου.πάλιν,ἐπεὶἐνκύκλῳτῷ right-angles, the center of circle ABC is thus on AC
αὐτῷτῷΑΒΓεὐθεῖάτιςἡΝΞεὐθεῖάντινατὴνΒΗδίχα [Prop. 3.1 corr.]. Again, since in the same circle ABC
καὶπρὸςὀρθὰςτέμνει,ἐπὶτῆςΝΞἄραἐστὶτὸκέντρον some straight-line NO cuts some (other straight-line) BG
τοῦΑΒΓκύκλου. ἐδείχθηδὲκαὶἐπὶτῆςΑΓ,καὶκατ᾿ in half, and at right-angles, the center of circle ABC is
οὐδὲνσυμβάλλουσιναἱΑΓ,ΝΞεὐθεῖαιἢκατὰτὸΟ·τὸ thus on NO [Prop. 3.1 corr.]. And it was also shown (to
ΟἄρασημεῖονκέντρονἐστὶτοῦΑΒΓκύκλου. ὁμοίωςδὴ be) on AC. And the straight-lines AC and NO meet at
δείξομεν,ὅτικαὶτοῦΔΕΖκύκλουκέντρονἐστὶτὸΟ·δύο no other (point) than P . Thus, point P is the center of
ἄρακύκλωντεμνόντωνἀλλήλουςτῶνΑΒΓ,ΔΕΖτὸαὐτό circle ABC. So, similarly, we can show that P is also the
ἐστικέντροντὸΟ·ὅπερἐστὶνἀδύνατον.
center of circle DEF . Thus, two circles cutting one an-
Οὐκἄρακύκλοςκύκλοντέμνεικατὰπλείονασημεῖαἢ other, ABC and DEF , have the same center P . The very
δύο·ὅπερἔδειδεῖξαι. thing is impossible [Prop. 3.5].
Thus, a circle does not cut a(nother) circle at more
than two points. (Which is) the very thing it was required
to show.
. Proposition 11
᾿Εὰνδύοκύκλοιἐφάπτωνταιἀλλήλωνἐντός,καὶληφθῇ If two circles touch one another internally, and their
αὐτῶντὰκέντρα, ἡἐπὶτὰκέντρααὐτῶνἐπιζευγνυμένη centers are found, then the straight-line joining their cenεὐθεῖα
καὶ ἐκβαλλομένη ἐπὶ τὴν συναφὴν πεσεῖται τῶν ters, being produced, will fall upon the point of union of
κύκλων.
the circles.
ΔύογὰρκύκλοιοἱΑΒΓ,ΑΔΕἐφαπτέσθωσανἀλλήλων For let two circles, ABC and ADE, touch one another
ἐντὸςκατὰτὸΑσημεῖον,καὶεἰλήφθωτοῦμὲνΑΒΓκύκλου internally at point A, and let the center F of circle ABC
κέντροντὸΖ,τοῦδὲΑΔΕτὸΗ·λέγω,ὅτιἡἀπὸτοῦΗἐπὶ have been found [Prop. 3.1], and (the center) G of (cirτὸΖἐπιζευγνυμένηεὐθεῖαἐκβαλλομένηἐπὶτὸΑπεσεῖται.
cle) ADE [Prop. 3.1]. I say that the straight-line joining
Μὴ γάρ, ἀλλ᾿ εἰ δυνατόν, πιπτέτω ὡς ἡ ΖΗΘ, καὶ G to F , being produced, will fall on A.
ἐπεζεύχθωσαναἱΑΖ,ΑΗ.
For (if) not then, if possible, let it fall like FGH (in
᾿ΕπεὶοὖναἱΑΗ,ΗΖτῆςΖΑ,τουτέστιτῆςΖΘ,μείζονές the figure), and let AF and AG have been joined.
εἰσιν,κοινὴἀφῃρήσθωἡΖΗ·λοιπὴἄραἡΑΗλοιπῆςτῆς Therefore, since AG and GF is greater than FA, that
ΗΘμείζωνἐστίν. ἴσηδὲἡΑΗτῇΗΔ·καὶἡΗΔἄρα is to say FH [Prop. 1.20], let FG have been taken from
τῆςΗΘμείζωνἐστὶνἡἐλάττωντῆςμείζονος·ὅπερἐστὶν both. Thus, the remainder AG is greater than the re-
ἀδύνατον·οὐκἄραἡἀπὸτοῦΖἐπὶτὸΗἐπιζευγνυμένη mainder GH. And AG (is) equal to GD. Thus, GD is
εὐθεὶα ἐκτὸς πεσεῖται· κατὰ τὸ Α ἄρα ἐπὶ τῆς συναφῆς also greater than GH, the lesser than the greater. The
πεσεῖται.
very thing is impossible. Thus, the straight-line joining F
to G will not fall outside (one circle but inside the other).
Thus, it will fall upon the point of union (of the circles)
M
H
N
F
A
K
D
P
B
L
G
O
E
81

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
Θ
at point A.
H
Α
∆
A
D
Η
G
Ζ
Β
F
B
Ε
E
Γ
᾿Εὰνἄραδύοκύκλοιἐφάπτωνταιἀλλήλωνἐντός,[καὶ Thus, if two circles touch one another internally, [and
ληφθῇαὐτῶντὰκέντρα],ἡἐπὶτὰκέντρααὐτῶνἐπιζευ- their centers are found], then the straight-line joining
γνυμένηεὐθεῖα[καὶἐκβαλλομένη]ἐπὶτὴνσυναφὴνπεσεῖται their centers, [being produced], will fall upon the point
τῶνκύκλων·ὅπερἔδειδεῖξαι.
of union of the circles. (Which is) the very thing it was
required to show.
. Proposition 12
᾿Εὰνδύοκύκλοιἐφάπτωνταιἀλλήλωνἐκτός,ἡἐπὶτὰ If two circles touch one another externally then the
κέντρααὐτῶνἐπιζευγνυμένηδιὰτῆςἐπαφῆςἐλεύσεται. (straight-line) joining their centers will go through the
point of union.
Β
B
C
Ζ
Α
Γ
∆
F
A
C
D
Η
G
Ε
ΔύογὰρκύκλοιοἱΑΒΓ,ΑΔΕἐφαπτέσθωσανἀλλήλων For let two circles, ABC and ADE, touch one an-
ἐκτὸς κατὰ τὸ Α σημεῖον, καὶ εἰλήφθω τοῦ μὲν ΑΒΓ other externally at point A, and let the center F of ABC
κέντροντὸΖ,τοῦδὲΑΔΕτὸΗ·λέγω, ὅτιἡἀπὸτοῦ have been found [Prop. 3.1], and (the center) G of ADE
ΖἐπὶτὸΗἐπιζευγνυμένηεὐθεῖαδιὰτῆςκατὰτὸΑἐπαφῆς [Prop. 3.1]. I say that the straight-line joining F to G will
ἐλεύσεται. go through the point of union at A.
Μὴγάρ, ἀλλ᾿εἰδυνατόν, ἐρχέσθωὡςἡΖΓΔΗ,καὶ For (if) not then, if possible, let it go like FCDG (in
ἐπεζεύχθωσαναἱΑΖ,ΑΗ.
the figure), and let AF and AG have been joined.
᾿ΕπεὶοὖντὸΖσημεῖονκέντρονἐστὶτοῦΑΒΓκύκλου, Therefore, since point F is the center of circle ABC,
ἴσηἐστὶνἡΖΑτῇΖΓ.πάλιν,ἐπεὶτὸΗσημεῖονκέντρον FA is equal to FC. Again, since point G is the center of
ἐστὶτοῦΑΔΕκύκλου,ἴσηἐστὶνἡΗΑτῇΗΔ.ἐδείχθη circle ADE, GA is equal to GD. And FA was also shown
E
82

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
δὲκαὶἡΖΑτῇΖΓἴση·αἱἄραΖΑ,ΑΗταῖςΖΓ,ΗΔἴσαι (to be) equal to FC. Thus, the (straight-lines) FA and
εἰσίν·ὥστεὅληἡΖΗτῶνΖΑ,ΑΗμείζωνἐστίν·ἀλλὰκαὶ AG are equal to the (straight-lines) FC and GD. So the
ἐλάττων·ὅπερἐστὶνἀδύνατον. οὐκἄραἡἀπὸτοῦΖἐπὶ whole of FG is greater than FA and AG. But, (it is) also
τὸΗἐπιζευγνυμένηεὐθεῖαδιὰτῆςκατὰτὸΑἐπαφῆςοὐκ less [Prop. 1.20]. The very thing is impossible. Thus, the
ἐλεύσεται·δι᾿αὐτῆςἄρα.
straight-line joining F to G cannot not go through the
᾿Εὰνἄραδύοκύκλοιἐφάπτωνταιἀλλήλωνἐκτός,ἡἐπὶ point of union at A. Thus, (it will go) through it.
τὰ κέντρααὐτῶνἐπιζευγνυμένη[εὐθεῖα]διὰτῆςἐπαφῆς Thus, if two circles touch one another externally then
ἐλεύσεται·ὅπερἔδειδεῖξαι.
the [straight-line] joining their centers will go through
the point of union. (Which is) the very thing it was required
to show.
. Proposition 13
Κύκλοςκύκλουοὐκἐφάπτεταικατὰπλείονασημεῖαἢ A circle does not touch a(nother) circle at more than
καθ᾿ἕν,ἐάντεἐντὸςἐάντεἐκτὸςἐφάπτηται.
one point, whether they touch internally or externally.
Κ
K
Α
A
Ε
Γ
E
C
Η Θ
G H
Β ∆ B
D
Ζ
F
Εἰγὰρδυνατόν,κύκλοςὁΑΒΓΔκύκλουτοῦΕΒΖΔ For, if possible, let circle ABDC † touch circle EBFD—
ἐφαπτέσθωπρότερονἐντὸςκατὰπλείονασημεῖαἢἓντὰΔ, first of all, internally—at more than one point, D and B.
Β. And let the center G of circle ABDC have been found
ΚαὶεἰλήφθωτοῦμὲνΑΒΓΔκύκλουκέντροντὸΗ,τοῦ [Prop. 3.1], and (the center) H of EBFD [Prop. 3.1].
δὲΕΒΖΔτὸΘ.
Thus, the (straight-line) joining G and H will fall on
῾ΗἄραἀπὸτοῦΗἐπὶτὸΘἐπιζευγνυμένηἐπὶτὰΒ, B and D [Prop. 3.11]. Let it fall like BGHD (in the
Δπεσεῖται. πιπτέτωὡςἡΒΗΘΔ.καὶἐπεὶτὸΗσημεῖον figure). And since point G is the center of circle ABDC,
κέντρονἐστὶτοῦΑΒΓΔκύκλου,ἴσηἐστὶνἡΒΗτῇΗΔ· BG is equal to GD. Thus, BG (is) greater than HD.
μείζωνἄραἡΒΗτῆςΘΔ·πολλῷἄραμείζωνἡΒΘτῆςΘΔ. Thus, BH (is) much greater than HD. Again, since point
πάλιν,ἐπεὶτὸΘσημεῖονκέντρονἐστὶτοῦΕΒΖΔκύκλου, H is the center of circle EBFD, BH is equal to HD.
ἴσηἐστὶνἡΒΘτῇΘΔ·ἐδείχθηδὲαὐτῆςκαὶπολλῷμείζων· But it was also shown (to be) much greater than it. The
ὅπερἀδύνατον·οὐκἄρακύκλοςκύκλουἐφάπτεταιἐντὸς very thing (is) impossible. Thus, a circle does not touch
κατὰπλείονασημεῖαἢἕν.
a(nother) circle internally at more than one point.
Λέγωδή,ὅτιοὐδὲἐκτός.
So, I say that neither (does it touch) externally (at
Εἰ γὰρ δυνατόν, κύκλοςὁΑΓΚ κύκλουτοῦΑΒΓΔ more than one point).
ἐφαπτέσθωἐκτὸςκατὰπλείονασημεῖαἢἓντὰΑ,Γ,καὶ For, if possible, let circle ACK touch circle ABDC
ἐπεζεύχθωἡΑΓ.
externally at more than one point, A and C. And let AC
῎ΕπεὶοὖνκύκλωντῶνΑΒΓΔ,ΑΓΚεἴληπταιἐπὶτῆς have been joined.
περιφερείαςἑκατέρουδύοτυχόντασημεῖατὰΑ,Γ,ἡἐπὶ Therefore, since two points, A and C, have been taken
τὰσημεῖαἐπιζευγνυμένηεὐθεῖαἐντὸςἑκατέρουπεσεῖται· at random on the circumference of each of the circles
ἀλλὰτοῦμὲνΑΒΓΔἐντὸςἔπεσεν, τοῦδὲΑΓΚἐκτός· ABDC and ACK, the straight-line joining the points will
ὅπερἄτοπον·οὐκἄρακύκλοςκύκλουἐφάπτεταιἐκτὸςκατὰ fall inside each (circle) [Prop. 3.2]. But, it fell inside
πλείονασημεῖαἢἕν.ἐδείχθηδέ,ὅτιοὐδὲἐντός. ABDC, and outside ACK [Def. 3.3]. The very thing
83

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
Κύκλοςἄρακύκλουοὐκἐφάπτεταικατὰπλείονασημεῖα (is) absurd. Thus, a circle does not touch a(nother) circle
ἢ[καθ᾿]ἕν,ἐάντεἐντὸςἐάντεἐκτὸςἐφάπτηται·ὅπερἔδει externally at more than one point. And it was shown that
δεῖξαι.
neither (does it) internally.
Thus, a circle does not touch a(nother) circle at more
than one point, whether they touch internally or externally.
(Which is) the very thing it was required to show.
† The Greek text has “ABCD”, which is obviously a mistake.
. Proposition 14
᾿Εν κύκλῳ αἱ ἴσαι εὐθεῖαι ἴσον ἀπέχουσιν ἀπὸ τοῦ In a circle, equal straight-lines are equally far from the
κέντρου, καὶ αἱ ἴσον ἀπέχουσαι ἀπὸ τοῦ κέντρου ἴσαι center, and (straight-lines) which are equally far from the
ἀλλήλαιςεἰσίν.
center are equal to one another.
∆
D
Β
Ε
Η
B
E
G
Ζ
F
Α
Γ
῎ΕστωκύκλοςὁΑΒΓΔ,καὶἐναὐτῷἴσαιεὐθεῖαιἔστω- Let ABDC † be a circle, and let AB and CD be equal
σαναἱΑΒ,ΓΔ·λέγω,ὅτιαἱΑΒ,ΓΔἴσονἀπέχουσινἀπὸ straight-lines within it. I say that AB and CD are equally
τοῦκέντρου.
far from the center.
ΕἰλήφθωγὰρτὸκέντοντοῦΑΒΓΔκύκλουκαὶἔστωτὸ For let the center of circle ABDC have been found
Ε,καὶἀπὸτοῦΕἐπὶτὰςΑΒ,ΓΔκάθετοιἤχθωσαναἱΕΖ, [Prop. 3.1], and let it be (at) E. And let EF and EG
ΕΗ,καὶἐπεζεύχθωσαναἱΑΕ,ΕΓ.
have been drawn from (point) E, perpendicular to AB
᾿ΕπεὶοὖνεὐθεῖάτιςδὶατοῦκέντρουἡΕΖεὐθεῖάντινα and CD (respectively) [Prop. 1.12]. And let AE and EC
μὴδιὰτοῦκέντρουτὴνΑΒπρὸςὀρθὰςτέμνει,καὶδίχα have been joined.
αὐτὴντέμνει. ἴσηἄραἡΑΖτῇΖΒ·διπλῆἄραἡΑΒτῆς Therefore, since some straight-line, EF , through the
ΑΖ.διὰτὰαὐτὰδὴκαὶἡΓΔτῆςΓΗἐστιδιπλῆ·καίἐστιν center (of the circle), cuts some (other) straight-line, AB,
ἴσηἡΑΒτῇΓΔ·ἴσηἄρακαὶἡΑΖτῇΓΗ.καὶἐπεὶἴσηἐστὶν not through the center, at right-angles, it also cuts it in
ἡΑΕτῇΕΓ,ἴσονκαὶτὸἀπὸτῆςΑΕτῷἀπὸτῆςΕΓ.ἀλλὰ half [Prop. 3.3]. Thus, AF (is) equal to FB. Thus, AB
τῷμὲνἀπὸτῆςΑΕἴσατὰἀπὸτῶνΑΖ,ΕΖ·ὀρθὴγὰρἡ (is) double AF . So, for the same (reasons), CD is also
πρὸςτῷΖγωνία·τῷδὲἀπὸτῆςΕΓἴσατὰἀπὸτῶνΕΗ,ΗΓ· double CG. And AB is equal to CD. Thus, AF (is)
ὀρθὴγὰρἡπρὸςτῷΗγωνία·τὰἄραἀπὸτῶνΑΖ,ΖΕἴσα also equal to CG. And since AE is equal to EC, the
ἐστὶτοῖςἀπὸτῶνΓΗ,ΗΕ,ὧντὸἀπὸτῆςΑΖἴσονἐστὶτῷ (square) on AE (is) also equal to the (square) on EC.
ἀπὸτῆςΓΗ·ἴσηγάρἐστινἡΑΖτῇΓΗ·λοιπὸνἄρατὸἀπὸ But, the (sum of the squares) on AF and EF (is) equal
τῆςΖΕτῷἀπὸτῆςΕΗἴσονἐστίν·ἴσηἄραἡΕΖτῇΕΗ.ἐν to the (square) on AE. For the angle at F (is) a rightδὲκύκλῳἴσονἀπέχεινἀπὸτοῦκέντρουεὐθεῖαιλέγονται,
angle [Prop. 1.47]. And the (sum of the squares) on EG
ὅταναἱἀπὸτοῦκέντρουἐπ᾿αὐτὰςκάθετοιἀγόμεναιἴσαι and GC (is) equal to the (square) on EC. For the angle
ὦσιν·αἱἄραΑΒ,ΓΔἴσονἀπέχουσινἀπὸτοῦκέντρου. at G (is) a right-angle [Prop. 1.47]. Thus, the (sum of
ἈλλὰδὴαἱΑΒ,ΓΔεὐθεῖαιἴσονἀπεχέτωσανἀπὸτοῦ the squares) on AF and FE is equal to the (sum of the
κέντρου,τουτέστινἴσηἔστωἡΕΖτῇΕΗ.λέγω,ὅτιἴση squares) on CG and GE, of which the (square) on AF
ἐστὶκαὶἡΑΒτῇΓΔ.
is equal to the (square) on CG. For AF is equal to CG.
A
C
84

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
Τῶνγὰραὐτῶνκατασκευασθέντωνὁμοίωςδείξομεν, Thus, the remaining (square) on FE is equal to the (re-
ὅτιδιπλῆἐστινἡμὲνΑΒτῆςΑΖ,ἡδὲΓΔτῆςΓΗ·καὶἐπεὶ maining square) on EG. Thus, EF (is) equal to EG. And
ἴσηἐστὶνἡΑΕτῇΓΕ,ἴσονἐστὶτὸἀπὸτῆςΑΕτῷἀπὸ straight-lines in a circle are said to be equally far from
τῆςΓΕ·ἀλλὰτῷμὲνἀπὸτῆςΑΕἴσαἐστὶτὰἀπὸτῶνΕΖ, the center when perpendicular (straight-lines) which are
ΖΑ,τῷδὲἀπὸτῆςΓΕἴσατὰἀπὸτῶνΕΗ,ΗΓ.τὰἄραἀπὸ drawn to them from the center are equal [Def. 3.4]. Thus,
τῶνΕΖ,ΖΑἴσαἐστὶτοῖςἀπὸτῶνΕΗ,ΗΓ·ὧντὸἀπὸτῆς AB and CD are equally far from the center.
ΕΖτῷἀπὸτῆςΕΗἐστινἴσον·ἴσηγὰρἡΕΖτῇΕΗ·λοιπὸν So, let the straight-lines AB and CD be equally far
ἄρατὸἀπὸτῆςΑΖἴσονἐστὶτῷἀπὸτῆςΓΗ·ἴσηἄραἡΑΖ from the center. That is to say, let EF be equal to EG. I
τῇΓΗ·καίἐστιτῆςμὲνΑΖδιπλῆἡΑΒ,τῆςδὲΓΗδιπλῆ say that AB is also equal to CD.
ἡΓΔ·ἴσηἄραἡΑΒτῇΓΔ.
For, with the same construction, we can, similarly,
᾿Εν κύκλῳ ἄρα αἱ ἴσαι εὐθεῖαι ἴσον ἀπέχουσιν ἀπὸ show that AB is double AF , and CD (double) CG. And
τοῦκέντρου,καὶαἱἴσονἀπέχουσαιἀπὸτοῦκέντρουἴσαι since AE is equal to CE, the (square) on AE is equal to
ἀλλήλαιςεἰσίν·ὅπερἔδειδεῖξαι.
the (square) on CE. But, the (sum of the squares) on
EF and FA is equal to the (square) on AE [Prop. 1.47].
And the (sum of the squares) on EG and GC (is) equal
to the (square) on CE [Prop. 1.47]. Thus, the (sum of
the squares) on EF and FA is equal to the (sum of the
squares) on EG and GC, of which the (square) on EF is
equal to the (square) on EG. For EF (is) equal to EG.
Thus, the remaining (square) on AF is equal to the (remaining
square) on CG. Thus, AF (is) equal to CG. And
AB is double AF , and CD double CG. Thus, AB (is)
equal to CD.
Thus, in a circle, equal straight-lines are equally far
from the center, and (straight-lines) which are equally far
from the center are equal to one another. (Which is) the
very thing it was required to show.
† The Greek text has “ABCD”, which is obviously a mistake.
. Proposition 15
᾿Ενκύκλῳμεγίστημὲνἡδιάμετρος,τῶνδὲἄλλωνἀεὶ In a circle, a diameter (is) the greatest (straight-line),
ἡἔγγιοντοῦκέντρουτῆςἀπώτερονμείζωνἐστίν. and for the others, a (straight-line) nearer to the center
῎ΕστωκύκλοςὁΑΒΓΔ,διάμετροςδὲαὐτοῦἔστωἡΑΔ, is always greater than one further away.
κέντρονδὲτὸΕ,καὶἔγγιονμὲντῆςΑΔδιαμέτρουἔστωἡ Let ABCD be a circle, and let AD be its diameter,
ΒΓ,ἀπώτερονδὲἡΖΗ·λέγω,ὅτιμεγίστημένἐστινἡΑΔ, and E (its) center. And let BC be nearer to the diameter
μείζωνδὲἡΒΓτῆςΖΗ.
AD, † and FG further away. I say that AD is the greatest
῎Ηχθωσαν γὰρ ἀπὸ τοῦ Ε κέντρου ἐπὶ τὰς ΒΓ, ΖΗ (straight-line), and BC (is) greater than FG.
κάθετοιαἱΕΘ,ΕΚ.καὶἐπεὶἔγγιονμὲντοῦκέντρουἐστὶν For let EH and EK have been drawn from the cen-
ἡΒΓ,ἀπώτερονδὲἡΖΗ,μείζωνἄραἡΕΚτῆςΕΘ.κείσθω ter E, at right-angles to BC and FG (respectively)
τῇΕΘἴσηἡΕΛ,καὶδιὰτοῦΛτῇΕΚπρὸςὀρθὰςἀχθεῖσα [Prop. 1.12]. And since BC is nearer to the center,
ἡΛΜδιήχθωἐπὶτὸΝ,καὶἐπεζεύχθωσαναἱΜΕ,ΕΝ,ΖΕ, and FG further away, EK (is) thus greater than EH
ΕΗ. [Def. 3.5]. Let EL be made equal to EH [Prop. 1.3].
ΚαὶἐπεὶἴσηἐστὶνἡΕΘτῇΕΛ,ἴσηἐστὶκαὶἡΒΓτῇ And LM being drawn through L, at right-angles to EK
ΜΝ.πάλιν,ἐπεὶἴσηἐστὶνἡμὲνΑΕτῇΕΜ,ἡδὲΕΔτῇ [Prop. 1.11], let it have been drawn through to N. And
ΕΝ,ἡἄραΑΔταῖςΜΕ,ΕΝἴσηἐστίν.ἀλλ᾿αἱμὲνΜΕ,ΕΝ let ME, EN, FE, and EG have been joined.
τῆςΜΝμείζονέςεἰσιν[καὶἡΑΔτῆςΜΝμείζωνἐστίν], And since EH is equal to EL, BC is also equal to
ἴσηδὲἡΜΝτῇΒΓ·ἡΑΔἄρατῆςΒΓμείζωνἐστίν. καὶ MN [Prop. 3.14]. Again, since AE is equal to EM, and
ἐπεὶδύοαἱΜΕ,ΕΝδύοταῖςΖΕ,ΕΗἴσαιεἰσίν,καὶγωνία ED to EN, AD is thus equal to ME and EN. But, ME
ἡὑπὸΜΕΝγωνίαςτῆςὑπὸΖΕΗμείζων[ἐστίν],βάσιςἄρα and EN is greater than MN [Prop. 1.20] [also AD is
85

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
ἡΜΝβάσεωςτῆςΖΗμείζωνἐστίν. ἀλλὰἡΜΝτῇΒΓ greater than MN], and MN (is) equal to BC. Thus, AD
ἐδείχθηἴση[καὶἡΒΓτῆςΖΗμείζωνἐστίν]. μεγίστημὲν is greater than BC. And since the two (straight-lines)
ἄραἡΑΔδιάμετρος,μείζωνδὲἡΒΓτῆςΖΗ. ME, EN are equal to the two (straight-lines) FE, EG
(respectively), and angle MEN [is] greater than angle
FEG, ‡ the base MN is thus greater than the base FG
[Prop. 1.24]. But, MN was shown (to be) equal to BC
[(so) BC is also greater than FG]. Thus, the diameter
AD (is) the greatest (straight-line), and BC (is) greater
than FG.
Α
A
Μ
M
Β
B
Ζ
F
Κ
Λ
Ε
Θ
K
L
E
H
Η
G
Ν
Γ
N C
∆
D
᾿Ενκύκλῳἄραμεγίστημὲνέστινἡδιάμετρος,τῶνδὲ Thus, in a circle, a diameter (is) the greatest (straight-
ἄλλωνἀεὶἡἔγγιοντοῦκέντρουτῆςἀπώτερονμείζωνἐστίν· line), and for the others, a (straight-line) nearer to the
ὅπερἔδειδεῖξαι.
center is always greater than one further away. (Which
is) the very thing it was required to show.
† Euclid should have said “to the center”, rather than ”to the diameter AD”, since BC, AD and FG are not necessarily parallel.
‡ This is not proved, except by reference to the figure.
¦. Proposition 16
῾Η τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπ᾿ ἄκρας A (straight-line) drawn at right-angles to the diameter
ἀγομένηἐκτὸςπεσεῖταιτοῦκύκλου, καὶεἰςτὸν μεταξὺ of a circle, from its end, will fall outside the circle. And
τόποντῆςτεεὐθείαςκαὶτῆςπεριφερείαςἑτέραεὐθεῖαοὐ another straight-line cannot be inserted into the space beπαρεμπεσεῖται,
καὶ ἡ μὲν τοῦ ἡμικυκλίου γωνία ἁπάσης tween the (aforementioned) straight-line and the circumγωνίας
ὀξείας εὐθυγράμμου μείζων ἐστίν, ἡ δὲ λοιπὴ ference. And the angle of the semi-circle is greater than
ἐλάττων.
any acute rectilinear angle whatsoever, and the remain-
῎ΕστωκύκλοςὁΑΒΓπερὶκέντροντὸΔκαὶδιάμετρον ing (angle is) less (than any acute rectilinear angle).
τὴνΑΒ·λέγω,ὅτιἡἀπὸτοῦΑτῇΑΒπρὸςὀρθὰςἀπ᾿ Let ABC be a circle around the center D and the di-
ἄκραςἀγομένηἐκτὸςπεσεῖταιτοῦκύκλου. ameter AB. I say that the (straight-line) drawn from A,
Μὴγάρ,ἀλλ᾿εἰδυνατόν,πιπτέτωἐντὸςὡςἡΓΑ,καὶ at right-angles to AB [Prop 1.11], from its end, will fall
ἐπεζεύχθωἡΔΓ.
outside the circle.
᾿ΕπεὶἴσηἐστὶνἡΔΑτῇΔΓ,ἴσηἐστὶκαὶγωνίαἡὑπὸ For (if) not then, if possible, let it fall inside, like CA
ΔΑΓγωνίᾳτῇὑπὸΑΓΔ.ὀρθὴδὲἡὑπὸΔΑΓ·ὀρθὴἄρα (in the figure), and let DC have been joined.
καὶἡὑπὸΑΓΔ·τριγώνουδὴτοῦΑΓΔαἱδύογωνίαιαἱ Since DA is equal to DC, angle DAC is also equal
ὑπὸΔΑΓ,ΑΓΔδύοὀρθαῖςἴσαιεἰσίν·ὅπερἐστὶνἀδύνατον. to angle ACD [Prop. 1.5]. And DAC (is) a right-angle.
οὐκἄραἡἀπὸτοῦΑσημείουτῇΒΑπρὸςὀρθὰςἀγομένη Thus, ACD (is) also a right-angle. So, in triangle ACD,
ἐντὸςπεσεῖταιτοῦκύκλου. ὁμοίωςδὴδεῖξομεν,ὅτιοὐδ᾿ the two angles DAC and ACD are equal to two right-
ἐπὶτῆςπεριφερείας·ἐκτὸςἄρα.
angles. The very thing is impossible [Prop. 1.17]. Thus,
the (straight-line) drawn from point A, at right-angles
86

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
Β
to BA, will not fall inside the circle. So, similarly, we
can show that neither (will it fall) on the circumference.
Thus, (it will fall) outside (the circle).
B
Γ
∆
C
D
Ζ
Ε
Η
Θ
Α
ΠιπτέτωὡςἡΑΕ·λέγωδή,ὅτιεἰςτὸνμεταξὺτόπον Let it fall like AE (in the figure). So, I say that another
τῆςτεΑΕεὐθείαςκαὶτῆςΓΘΑπεριφερείαςἑτέραεὐθεῖα straight-line cannot be inserted into the space between
οὐπαρεμπεσεῖται.
the straight-line AE and the circumference CHA.
Εἰγὰρδυνατόν,παρεμπιπτέτωὡςἡΖΑ,καὶἤχθωἀπὸ For, if possible, let it be inserted like FA (in the figτοῦΔσημείουἐπὶτῆνΖΑκάθετοςἡΔΗ.καὶἐπεὶὀρθή
ure), and let DG have been drawn from point D, perpen-
ἐστινἡὑπὸΑΗΔ,ἐλάττωνδὲὀρθῆςἡὑπὸΔΑΗ,μείζων dicular to FA [Prop. 1.12]. And since AGD is a right-
ἄραἡΑΔτῆςΔΗ.ἴσηδὲἡΔΑτῇΔΘ·μείζωνἄραἡΔΘ angle, and DAG (is) less than a right-angle, AD (is)
τῆςΔΗ,ἡἐλάττωντῆςμείζονος·ὅπερἐστὶνἀδύνατον.οὐκ thus greater than DG [Prop. 1.19]. And DA (is) equal
ἄραεἰςτὸνμεταξὺτόποντῆςτεεὐθείαςκαὶτῆςπεριφερείας to DH. Thus, DH (is) greater than DG, the lesser than
ἑτέραεὐθεῖαπαρεμπεσεῖται.
the greater. The very thing is impossible. Thus, another
Λέγω,ὅτικαὶἡμὲντοῦἡμικυκλίουγωνίαἡπεριεχομένη straight-line cannot be inserted into the space between
ὑπότετῆςΒΑεὐθείαςκαὶτῆςΓΘΑπεριφερείαςἁπάσης the straight-line (AE) and the circumference.
γωνίαςὀξείαςεὐθυγράμμουμείζωνἐστίν,ἡδὲλοιπὴἡπε- And I also say that the semi-circular angle contained
ριεχομένηὑπότετῆςΓΘΑπεριφερείαςκαὶτῆςΑΕεὐθείας by the straight-line BA and the circumference CHA is
ἁπάσηςγωνίαςὀξείαςεὐθυγράμμουἐλάττωνἐστίν. greater than any acute rectilinear angle whatsoever, and
Εἰ γὰρ ἐστί τις γωνία εὐθύγραμμος μείζων μὲν τῆς the remaining (angle) contained by the circumference
περιεχομένηςὑπότετῆςΒΑεὐθείαςκαὶτῆςΓΘΑπερι- CHA and the straight-line AE is less than any acute recφερείας,
ἐλάττων δὲ τῆς περιεχομένης ὑπό τε τῆς ΓΘΑ tilinear angle whatsoever.
περιφερείαςκαὶτὴςΑΕεὐθείας,εἰςτὸνμεταξὺτόποντῆς For if any rectilinear angle is greater than the (anτε
ΓΘΑ περιφερείας καὶ τῆς ΑΕ εὐθείας εὐθεῖα παρεμ- gle) contained by the straight-line BA and the circumπεσεῖται,
ἥτιςποιήσειμείζοναμὲντῆςπεριεχομένηςὑπὸ ference CHA, or less than the (angle) contained by the
τετῆςΒΑεὐθείαςκαὶτῆςΓΘΑπεριφερείαςὑπὸεὐθειῶν circumference CHA and the straight-line AE, then a
περιεχομένην, ἐλάττοναδὲτῆςπεριεχομένηςὑπότετῆς straight-line can be inserted into the space between the
ΓΘΑπεριφερείαςκαὶτῆςΑΕεὐθείας. οὐπαρεμπίπτειδέ· circumference CHA and the straight-line AE—anything
οὐκἄρατῆςπεριεχομένηςγωνίαςὑπότετῆςΒΑεὐθείας which will make (an angle) contained by straight-lines
καὶτῆςΓΘΑπεριφερείαςἔσταιμείζωνὀξεῖαὑπὸεὐθειῶν greater than the angle contained by the straight-line BA
περιεχομένη,οὐδὲμὴνἐλάττωντῆςπεριεχομένηςὑπότε and the circumference CHA, or less than the (angle)
τῆςΓΘΑπεριφερείαςκαὶτῆςΑΕεὐθείας.
contained by the circumference CHA and the straightline
AE. But (such a straight-line) cannot be inserted.
Thus, an acute (angle) contained by straight-lines cannot
be greater than the angle contained by the straight-line
BA and the circumference CHA, neither (can it be) less
than the (angle) contained by the circumference CHA
and the straight-line AE.
F
E
G
H
A
87

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
ÈÖ×Ñ.
᾿Εκδὴτούτουφανερόν,ὅτιἡτῇδιαμέτρῳτοῦκύκλου So, from this, (it is) manifest that a (straight-line)
Corollary
πρὸς ὀρθὰς ἀπ᾿ ἄκρας ἀγομένη ἐφάπτεται τοῦ κύκλου drawn at right-angles to the diameter of a circle, from
[καὶὅτιεὐθεῖακύκλουκαθ᾿ἓνμόνονἐφάπτεταισημεῖον, its extremity, touches the circle [and that the straight-line
ἐπειδήπερκαὶἡκατὰδύοαὐτῷσυμβάλλουσαἐντὸςαὐτοῦ touches the circle at a single point, inasmuch as it was
πίπτουσαἐδείχθη]·ὅπερἔδειδεῖξαι.
also shown that a (straight-line) meeting (the circle) at
two (points) falls inside it [Prop. 3.2] ]. (Which is) the
very thing it was required to show.
Þ. Proposition 17
Ἀπὸτοῦδοθέντοςσημείουτοῦδοθέντοςκύκλουἐφα- To draw a straight-line touching a given circle from a
πτομένηνεὐθεῖανγραμμὴνἀγαγεῖν.
given point.
Α
A
Ζ
Β
∆
Ε
Γ
Η
F
B
D
E
C
G
῎ΕστωτὸμὲνδοθὲνσημεῖοντὸΑ,ὁδὲδοθεὶςκύκλος Let A be the given point, and BCD the given circle.
ὁΒΓΔ·δεῖδὴἀπὸτοῦΑσημείουτοῦΒΓΔκύκλουἐφα- So it is required to draw a straight-line touching circle
πτομένηνεὐθεῖανγραμμὴνἀγαγεῖν. BCD from point A.
Εἰλήφθω γὰρ τὸ κέντρον τοῦ κύκλου τὸ Ε, καὶ For let the center E of the circle have been found
ἐπεζεύχθωἡΑΕ,καὶκέντρῳμὲντῷΕδιαστήματιδὲτῷ [Prop. 3.1], and let AE have been joined. And let (the
ΕΑκύκλοςγεγράφθωὁΑΖΗ,καὶἀπὸτοῦΔτῇΕΑπρὸς circle) AFG have been drawn with center E and radius
ὀρθὰςἤχθωἡΔΖ,καὶἐπεζεύχθωσαναἱΕΖ,ΑΒ·λέγω, EA. And let DF have been drawn from from (point) D,
ὅτιἀπὸτοῦΑσημείουτοῦΒΓΔκύκλουἐφαπτομένηἦκται at right-angles to EA [Prop. 1.11]. And let EF and AB
ἡΑΒ.
have been joined. I say that the (straight-line) AB has
᾿ΕπεὶγὰρτὸΕκέντρονἐστὶτῶνΒΓΔ,ΑΖΗκύκλων, been drawn from point A touching circle BCD.
ἴσηἄραἐστὶνἡμὲνΕΑτῇΕΖ,ἡδὲΕΔτῇΕΒ·δύοδὴ For since E is the center of circles BCD and AFG,
αἱΑΕ,ΕΒδύοταῖςΖΕ,ΕΔἴσαιεἰσίν·καὶγωνίανκοινὴν EA is thus equal to EF , and ED to EB. So the two
περιέχουσιτὴνπρὸςτῷΕ·βάσιςἄραἡΔΖβάσειτῇΑΒ (straight-lines) AE, EB are equal to the two (straight-
ἴσηἐστίν, καὶτὸΔΕΖτρίγωνοντῷΕΒΑτριγώνῳἴσον lines) FE, ED (respectively). And they contain a com-
ἐστίν,καὶαἱλοιπαὶγωνίαιταῖςλοιπαῖςγωνίαις·ἴσηἄραἡ mon angle at E. Thus, the base DF is equal to the
ὑπὸΕΔΖτῇὑπὸΕΒΑ.ὀρθὴδὲἡὑπὸΕΔΖ·ὀρθὴἄρακαὶἡ base AB, and triangle DEF is equal to triangle EBA,
ὑπὸΕΒΑ.καίἐστινἡΕΒἐκτοῦκέντρου·ἡδὲτῇδιαμέτρῳ and the remaining angles (are equal) to the (correτοῦκύκλουπρὸςὀρθὰςἀπ᾿ἄκραςἀγομένηἐφάπτεταιτοῦ
sponding) remaining angles [Prop. 1.4]. Thus, (angle)
κύκλου·ἡΑΒἄραἐφάπτεταιτοῦΒΓΔκύκλου. EDF (is) equal to EBA. And EDF (is) a right-angle.
ἈπὸτοῦἄραδοθέντοςσημείουτοῦΑτοῦδοθέντος Thus, EBA (is) also a right-angle. And EB is a raκύκλουτοῦΒΓΔἐφαπτομένηεὐθεῖαγραμμὴἦκταιἡΑΒ·
dius. And a (straight-line) drawn at right-angles to the
ὅπερἔδειποιῆσαι.
diameter of a circle, from its extremity, touches the circle
[Prop. 3.16 corr.]. Thus, AB touches circle BCD.
Thus, the straight-line AB has been drawn touching
88

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
the given circle BCD from the given point A. (Which is)
the very thing it was required to do.
. Proposition 18
᾿Εὰνκύκλουἐφάπτηταίτιςεὐθεῖα,ἀπὸδὲτοῦκέντρου If some straight-line touches a circle, and some
ἐπὶτὴνἁφὴνἐπιζευχθῇτιςεὐθεῖα,ἡἐπιζευχθεῖσακάθετος (other) straight-line is joined from the center (of the cir-
ἔσταιἐπὶτὴνἐφαπτομένην.
cle) to the point of contact, then the (straight-line) so
joined will be perpendicular to the tangent.
Α
A
Ζ
Β
Η
∆
F
B
G
D
Γ
C
Ε
ΚύκλουγὰρτοῦΑΒΓἐφαπτέσθωτιςεὐθεῖαἡΔΕκατὰ For let some straight-line DE touch the circle ABC at
τὸΓσημεῖον,καὶεἰλήφθωτὸκέντροντοῦΑΒΓκύκλουτὸ point C, and let the center F of circle ABC have been
Ζ,καὶἀπὸτοῦΖἐπὶτὸΓἐπεζεύχθωἡΖΓ·λέγω,ὅτιἡΖΓ found [Prop. 3.1], and let FC have been joined from F
κάθετόςἐστινἐπὶτὴνΔΕ.
to C. I say that FC is perpendicular to DE.
Εἰγὰρμή,ἤχθωἀπὸτοῦΖἐπὶτὴνΔΕκάθετοςἡΖΗ. For if not, let FG have been drawn from F , perpen-
᾿ΕπεὶοὖνἡὑπὸΖΗΓγωνίαὀρθήἐστιν,ὀξεῖαἄραἐστὶν dicular to DE [Prop. 1.12].
ἡὑπὸΖΓΗ·ὑπὸδὲτὴνμείζοναγωνίανἡμείζωνπλευρὰ Therefore, since angle FGC is a right-angle, (angle)
ὑποτείνει·μείζωνἄραἡΖΓτῆςΖΗ·ἴσηδὲἡΖΓτῇΖΒ· FCG is thus acute [Prop. 1.17]. And the greater angle is
μείζωνἄρακαὶἡΖΒτῆςΖΗἡἐλάττωντῆςμείζονος·ὅπερ subtended by the greater side [Prop. 1.19]. Thus, FC (is)
ἐστὶνἀδύνατον.οὐκἄραἡΖΗκάθετόςἐστινἐπὶτὴνΔΕ. greater than FG. And FC (is) equal to FB. Thus, FB
ὁμοίωςδὴδεῖξομεν,ὅτιοὐδ᾿ἄλλητιςπλὴντῆςΖΓ·ἡΖΓ (is) also greater than FG, the lesser than the greater. The
ἄρακάθετόςἐστινἐπὶτὴνΔΕ.
very thing is impossible. Thus, FG is not perpendicular to
᾿Εὰν ἄρα κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τοῦ DE. So, similarly, we can show that neither (is) any other
κέντρουἐπὶτὴνἁφὴνἐπιζευχθῇτιςεὐθεῖα,ἡἐπιζευχθεῖσα (straight-line) except FC. Thus, FC is perpendicular to
κάθετοςἔσταιἐπὶτὴνἐφαπτομένην·ὅπερἔδειδεῖξαι. DE.
Thus, if some straight-line touches a circle, and some
(other) straight-line is joined from the center (of the circle)
to the point of contact, then the (straight-line) so
joined will be perpendicular to the tangent. (Which is)
the very thing it was required to show.
. Proposition 19
᾿Εὰνκύκλουἐφάπτηταίτιςεὐθεῖα,ἀπὸδὲτῆςἁφῆςτῇ If some straight-line touches a circle, and a straight-
ἐφαπτομένῃπρὸςὀρθὰς[γωνίας]εὐθεῖαγραμμὴἀχθῇ,ἐπὶ line is drawn from the point of contact, at right-[angles]
τῆςἀχθείσηςἔσταιτὸκέντροντοῦκύκλου.
to the tangent, then the center (of the circle) will be on
ΚύκλουγὰρτοῦΑΒΓἐφαπτέσθωτιςεὐθεῖαἡΔΕκατὰ the (straight-line) so drawn.
τὸΓσημεῖον,καὶἀπὸτοῦΓτῇΔΕπρὸςὀρθὰςἤχθωἡ For let some straight-line DE touch the circle ABC at
ΓΑ·λέγω,ὅτιἐπὶτῆςΑΓἐστιτὸκέντροντοῦκύκλου. point C. And let CA have been drawn from C, at right-
E
89

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
Α
angles to DE [Prop. 1.11]. I say that the center of the
circle is on AC.
A
Β
Ζ
B
F
∆
Γ
Ε
Μὴγάρ,ἀλλ᾿εἰδυνατόν,ἔστωτὸΖ,καὶἐπεζεύχθωἡ For (if) not, if possible, let F be (the center of the
ΓΖ.
circle), and let CF have been joined.
᾿Επεὶ[οὖν]κύκλουτοῦΑΒΓἐφάπτεταίτιςεὐθεῖαἡΔΕ, [Therefore], since some straight-line DE touches the
ἀπὸδὲτοῦκέντρουἐπὶτὴνἁφὴνἐπέζευκταιἡΖΓ,ἡΖΓἄρα circle ABC, and FC has been joined from the center to
κάθετόςἐστινἐπὶτὴνΔΕ·ὀρθὴἄραἐστὶνἡὑπὸΖΓΕ.ἐστὶ the point of contact, FC is thus perpendicular to DE
δὲκαὶἡὑπὸΑΓΕὀρθή·ἴσηἄραἐστὶνἡὑπὸΖΓΕτῇὑπὸ [Prop. 3.18]. Thus, FCE is a right-angle. And ACE
ΑΓΕἡἐλάττωντῇμείζονι·ὅπερἐστὶνἀδύνατον. οὐκἄρα is also a right-angle. Thus, FCE is equal to ACE, the
τὸΖκέντρονἐστὶτοῦΑΒΓκύκλου. ὁμοίωςδὴδείξομεν, lesser to the greater. The very thing is impossible. Thus,
ὅτιοὐδ᾿ἄλλοτιπλὴνἐπὶτῆςΑΓ.
F is not the center of circle ABC. So, similarly, we can
᾿Εὰνἄρακύκλουἐφάπτηταίτιςεὐθεῖα,ἀπὸδὲτῆςἁφῆς show that neither is any (point) other (than one) on AC.
τῇἐφαπτομένῃπρὸςὀρθὰςεὐθεῖαγραμμὴἀχθῇ,ἐπὶτῆς
ἀχθείσηςἔσταιτὸκέντροντοῦκύκλου·ὅπερἔδειδεῖξαι. Thus, if some straight-line touches a circle, and a straightline
is drawn from the point of contact, at right-angles to
the tangent, then the center (of the circle) will be on the
(straight-line) so drawn. (Which is) the very thing it was
required to show.
. Proposition 20
᾿Ενκύκλῳἡπρὸςτῷκέντρῳγωνίαδιπλασίωνἐστὶτῆς In a circle, the angle at the center is double that at the
πρὸςτῇπεριφερείᾳ,ὅταντὴναὐτὴνπεριφέρειανβάσινἔχω- circumference, when the angles have the same circumferσιναἱγωνίαι.
ence base.
῎ΕστωκύκλοςὁΑΒΓ,καὶπρὸςμὲντῷκέντρῳαὐτοῦ Let ABC be a circle, and let BEC be an angle at its
γωνίαἔστωἡὑπὸΒΕΓ,πρὸςδὲτῇπεριφερείᾳἡὑπὸΒΑΓ, center, and BAC (one) at (its) circumference. And let
ἐχέτωσανδὲτὴναὐτὴνπεριφέρειανβάσιντὴνΒΓ·λέγω, them have the same circumference base BC. I say that
ὅτιδιπλασίωνἐστὶνἡὑπὸΒΕΓγωνίατῆςὑπὸΒΑΓ. angle BEC is double (angle) BAC.
᾿ΕπιζευχθεῖσαγὰρἡΑΕδιήχθωἐπὶτὸΖ.
For being joined, let AE have been drawn through to
᾿ΕπεὶοὖνἴσηἐστὶνἡΕΑτῇΕΒ,ἴσηκαὶγωνίαἡὑπὸ F .
ΕΑΒτῇὑπὸΕΒΑ·αἱἄραὑπὸΕΑΒ,ΕΒΑγωνίαιτῆςὑπὸ Therefore, since EA is equal to EB, angle EAB (is)
ΕΑΒδιπλασίουςεἰσίν. ἴσηδὲἡὑπὸΒΕΖταῖςὑπὸΕΑΒ, also equal to EBA [Prop. 1.5]. Thus, angle EAB and
ΕΒΑ·καὶἡὑπὸΒΕΖἄρατῆςὑπὸΕΑΒἐστιδιπλῆ.διὰτὰ EBA is double (angle) EAB. And BEF (is) equal to
αὐτὰδὴκαὶἡὑπὸΖΕΓτῆςὑπὸΕΑΓἐστιδιπλῆ. ὅληἄρα EAB and EBA [Prop. 1.32]. Thus, BEF is also double
ἡὑπὸΒΕΓὅληςτῆςὑπὸΒΑΓἐστιδιπλῆ.
EAB. So, for the same (reasons), FEC is also double
EAC. Thus, the whole (angle) BEC is double the whole
(angle) BAC.
D
C
E
90

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
Α
∆
A
D
Ε
E
Γ
C
Η
Β
Ζ
Κεκλάσθωδὴπάλιν,καὶἔστωἑτέραγωνίαἡὑπὸΒΔΓ, So let another (straight-line) have been inflected, and
καὶἐπιζευχθεῖσαἡΔΕἐκβεβλήσθωἐπὶτὸΗ.ὁμοίωςδὴ let there be another angle, BDC. And DE being joined,
δείξομεν,ὅτιδιπλῆἐστινἡὑπὸΗΕΓγωνίατῆςὑπὸΕΔΓ, let it have been produced to G. So, similarly, we can show
ὧνἡὑπὸΗΕΒδιπλῆἐστιτῆςὑπὸΕΔΒ·λοιπὴἄραἡὑπὸ that angle GEC is double EDC, of which GEB is double
ΒΕΓδιπλῆἐστιτῆςὑπὸΒΔΓ.
EDB. Thus, the remaining (angle) BEC is double the
᾿Ενκύκλῳἄραἡπρὸςτῷκέντρῳγωνίαδιπλασίωνἐστὶ (remaining angle) BDC.
τῆςπρὸςτῇπεριφερείᾳ,ὅταντὴναὐτὴνπεριφέρειανβάσιν Thus, in a circle, the angle at the center is double that
ἔχωσιν[αἱγωνίαι]·ὅπερἔδειδεῖξαι.
at the circumference, when [the angles] have the same
circumference base. (Which is) the very thing it was required
to show.
. Proposition 21
᾿Ενκύκλῳαἱἐντῷαὐτῷτμήματιγωνίαιἴσαιἀλλήλαις In a circle, angles in the same segment are equal to
εἰσίν.
one another.
Α
G
A
B
F
Ζ
Ε
F
E
Β ∆ B
D
Γ
῎ΕστωκύκλοςὁΑΒΓΔ,καὶἐντῷαὐτῷτμήματιτῷ Let ABCD be a circle, and let BAD and BED be
ΒΑΕΔγωνίαιἔστωσαναἱὑπὸΒΑΔ,ΒΕΔ·λέγω,ὅτιαἱ angles in the same segment BAED. I say that angles
ὑπὸΒΑΔ,ΒΕΔγωνίαιἴσαιἀλλήλαιςεἰσίν.
BAD and BED are equal to one another.
ΕἰλήφθωγὰρτοῦΑΒΓΔκύκλουτὸκέντρον,καὶἔστω For let the center of circle ABCD have been found
τὸΖ,καὶἐπεζεύχθωσαναἱΒΖ,ΖΔ.
[Prop. 3.1], and let it be (at point) F . And let BF and
ΚαὶἐπεὶἡμὲνὑπὸΒΖΔγωνίαπρὸςτῷκέντρῳἐστίν,ἡ FD have been joined.
δὲὑπὸΒΑΔπρὸςτῇπεριφερείᾳ,καὶἔχουσιτὴναὐτὴνπε- And since angle BFD is at the center, and BAD at
ριφέρειανβάσιντὴνΒΓΔ,ἡἄραὑπὸΒΖΔγωνίαδιπλασίων the circumference, and they have the same circumference
ἐστὶτῆςὑπὸΒΑΔ.διὰτὰαὐτὰδὴἡὑπὸΒΖΔκαὶτῆςὑπὸ base BCD, angle BFD is thus double BAD [Prop. 3.20].
C
91

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
ΒΕΔἐστιδιπλσίων·ἴσηἄραἡὑπὸΒΑΔτῇὑπὸΒΕΔ. So, for the same (reasons), BFD is also double BED.
᾿Εν κύκλῳ ἄρα αἱ ἐν τῷ αὐτῷ τμήματι γωνίαι ἴσαι Thus, BAD (is) equal to BED.
ἀλλήλαιςεἰσίν·ὅπερἔδειδεῖξαι.
Thus, in a circle, angles in the same segment are equal
to one another. (Which is) the very thing it was required
to show.
. Proposition 22
Τῶνἐντοῖςκύκλοιςτετραπλεύρωναἱἀπεναντίονγωνίαι For quadrilaterals within circles, the (sum of the) opδυσὶνὀρθαῖςἴσαιεἰσίν.
posite angles is equal to two right-angles.
Β
B
Α
A
Γ
C
∆
῎ΕστωκύκλοςὁΑΒΓΔ,καὶἐναὐτῷτετράπλευρονἔστω Let ABCD be a circle, and let ABCD be a quadrilatτὸΑΒΓΔ·λέγω,ὅτιαἱἀπεναντίονγωνίαιδυσὶνὀρθαῖςἴσαι
eral within it. I say that the (sum of the) opposite angles
εἰσίν.
is equal to two right-angles.
᾿ΕπεζεύχθωσαναἱΑΓ,ΒΔ.
Let AC and BD have been joined.
᾿Επεὶοὖνπαντὸςτριγώνουαἱτρεῖςγωνίαιδυσὶνὀρθαῖς Therefore, since the three angles of any triangle are
ἴσαιεἰσίν,τοῦΑΒΓἄρατριγώνουαἱτρεῖςγωνίαιαἱὑπὸ equal to two right-angles [Prop. 1.32], the three angles
ΓΑΒ,ΑΒΓ,ΒΓΑδυσὶνὀρθαῖςἴσαιεἰσίν.ἴσηδὲἡμὲνὑπὸ CAB, ABC, and BCA of triangle ABC are thus equal
ΓΑΒτῇὑπὸΒΔΓ·ἐνγὰρτῷαὐτῷτμήματίεἰσιτῷΒΑΔΓ· to two right-angles. And CAB (is) equal to BDC. For
ἡδὲὑπὸΑΓΒτῇὑπὸΑΔΒ·ἐνγὰρτῷαὐτῷτμήματίεἰσι they are in the same segment BADC [Prop. 3.21]. And
τῷΑΔΓΒ·ὅληἄραἡὑπὸΑΔΓταῖςὑπὸΒΑΓ,ΑΓΒἴση ACB (is equal) to ADB. For they are in the same seg-
ἐστίν. κοινὴπροσκείσθωἡὑπὸΑΒΓ·αἱἄραὑπὸΑΒΓ, ment ADCB [Prop. 3.21]. Thus, the whole of ADC is
ΒΑΓ,ΑΓΒταῖςὑπὸΑΒΓ,ΑΔΓἴσαιεἰσίν. ἀλλ᾿αἱὑπὸ equal to BAC and ACB. Let ABC have been added to
ΑΒΓ,ΒΑΓ,ΑΓΒδυσὶνὀρθαῖςἴσαιεἰσίν.καὶαἱὑπὸΑΒΓ, both. Thus, ABC, BAC, and ACB are equal to ABC
ΑΔΓἄραδυσὶνὀρθαῖςἴσαιεἰσίν. ὁμοίωςδὴδείξομεν,ὅτι and ADC. But, ABC, BAC, and ACB are equal to two
καὶαἱὑπὸΒΑΔ,ΔΓΒγωνίαιδυσὶνὀρθαῖςἴσαιεἰσίν. right-angles. Thus, ABC and ADC are also equal to two
Τῶνἄραἐντοῖςκύκλοιςτετραπλεύρωναἱἀπεναντίον right-angles. Similarly, we can show that angles BAD
γωνίαιδυσὶνὀρθαῖςἴσαιεἰσίν·ὅπερἔδειδεῖξαι.
and DCB are also equal to two right-angles.
Thus, for quadrilaterals within circles, the (sum of
the) opposite angles is equal to two right-angles. (Which
is) the very thing it was required to show.
. Proposition 23
᾿Επὶτῆςαὐτῆςεὐθείαςδύοτμήματακύκλωνὅμοιακαὶ Two similar and unequal segments of circles cannot be
ἄνισαοὐσυσταθήσεταιἐπὶτὰαὐτὰμέρη.
constructed on the same side of the same straight-line.
Εἰ γὰρ δυνατόν, ἐπὶ τῆς αὐτῆςεὐθείαςτῆς ΑΒ δύο For, if possible, let the two similar and unequal segτμήματακύκλωνὅμοιακαὶἄνισασυνεστάτωἐπὶτὰαὐτὰ
ments of circles, ACB and ADB, have been constructed
μέρητὰΑΓΒ,ΑΔΒ,καὶδιήχθωἡΑΓΔ,καὶἐπεζεύχθωσαν on the same side of the same straight-line AB. And let
D
92

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
αἱΓΒ,ΔΒ.
ACD have been drawn through (the segments), and let
CB and DB have been joined.
Γ
∆
C
D
Α
Β
᾿ΕπεὶοὖνὅμοιόνἐστιτὸΑΓΒτμῆματῷΑΔΒτμήματι, Therefore, since segment ACB is similar to segment
ὅμοιαδὲτμήματακύκλωνἐστὶτὰδεχόμεναγωνίαςἴσας, ADB, and similar segments of circles are those accept-
ἴσηἄραἐστὶνἡὑπὸΑΓΒγωνίατῇὑπὸΑΔΒἡἐκτὸςτῇ ing equal angles [Def. 3.11], angle ACB is thus equal
ἐντός·ὅπερἐστὶνἀδύνατον.
to ADB, the external to the internal. The very thing is
Οὐκἄραἐπὶτῆςαὐτῆςεὐθείαςδύοτμήματακύκλων impossible [Prop. 1.16].
ὅμοιακαὶἄνισασυσταθήσεταιἐπὶτὰαὐτὰμέρη·ὅπερἔδει Thus, two similar and unequal segments of circles
δεῖξαι.
cannot be constructed on the same side of the same
straight-line.
. Proposition 24
Τὰἐπὶἴσωνεὐθειῶνὅμοιατμήματακύλωνἴσαἀλλήλοις Similar segments of circles on equal straight-lines are
ἐστίν.
equal to one another.
Ε
A
E
B
Α
Β
A
B
Ζ
Η
F
G
Γ
∆
῎Εστωσαν γὰρ ἐπὶ ἴσων εὐθειῶν τῶν ΑΒ, ΓΔ ὅμοια For let AEB and CFD be similar segments of circles
τμήματακύκλωντὰΑΕΒ,ΓΖΔ·λέγω, ὅτιἴσονἐστὶτὸ on the equal straight-lines AB and CD (respectively). I
ΑΕΒτμῆματῷΓΖΔτμήματι.
say that segment AEB is equal to segment CFD.
᾿ΕφαρμοζομένουγὰρτοῦΑΕΒτμήματοςἐπὶτὸΓΖΔκαὶ For if the segment AEB is applied to the segment
τιθεμένουτοῦμὲνΑσημείουἐπὶτὸΓτῆςδὲΑΒεὐθείας CFD, and point A is placed on (point) C, and the
ἐπὶτὴνΓΔ,ἐφαρμόσεικαὶτὸΒσημεῖονἐπὶτὸΔσημεῖον straight-line AB on CD, then point B will also coincide
διὰτὸἴσηνεἶναιτὴνΑΒτῇΓΔ·τῆςδὲΑΒἐπὶτὴνΓΔἐφαρ- with point D, on account of AB being equal to CD. And
μοσάσηςἐφαρμόσεικαὶτὸΑΕΒτμῆμαἐπὶτὸΓΖΔ.εἰγὰρ if AB coincides with CD then the segment AEB will also
ἡΑΒεὐθεῖαἐπὶτὴνΓΔἐφαρμόσει,τὸδὲΑΕΒτμῆμαἐπὶ coincide with CFD. For if the straight-line AB coincides
τὸΓΖΔμὴἐφαρμόσει,ἤτοιἐντὸςαὐτοῦπεσεῖταιἢἐκτὸς with CD, and the segment AEB does not coincide with
ἢπαραλλάξει,ὡςτὸΓΗΔ,καὶκύκλοςκύκλοντέμνεικατὰ CFD, then it will surely either fall inside it, outside (it), †
πλείονασημεῖαἢδύο·ὅπερἐστίνἀδύνατον.οὐκἄραἐφαρ- or it will miss like CGD (in the figure), and a circle (will)
μοζομένηςτῆςΑΒεὐθείαςἐπὶτὴνΓΔοὐκἐφαρμόσεικαὶ cut (another) circle at more than two points. The very
C
D
93

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
τὸΑΕΒτμῆμαἐπὶτὸΓΖΔ·ἐφαρμόσειἄρα,καὶἴσοναὐτῷ thing is impossible [Prop. 3.10]. Thus, if the straight-line
ἔσται.
AB is applied to CD, the segment AEB cannot not also
Τὰἄραἐπὶἴσωνεὐθειῶνὅμοιατμήματακύκλωνἴσα coincide with CFD. Thus, it will coincide, and will be
ἀλλήλοιςἐστίν·ὅπερἔδειδεῖξαι. equal to it [C.N. 4].
Thus, similar segments of circles on equal straightlines
are equal to one another. (Which is) the very thing
it was required to show.
† Both this possibility, and the previous one, are precluded by Prop. 3.23.
. Proposition 25
Κύκλουτμήματοςδοθέντοςπροσαναγράψαιτὸνκύκλον, For a given segment of a circle, to complete the circle,
οὗπέρἐστιτμῆμα.
the very one of which it is a segment.
Α
Α
Α
A
A
A
Β
∆
Ε
Β
∆
Β
Ε
∆
B
D
E
B
D
B
E
D
Γ
Γ Γ
C
C
C
῎ΕστωτὸδοθὲντμῆμακύκλουτὸΑΒΓ·δεῖδὴτοῦΑΒΓ Let ABC be the given segment of a circle. So it is reτμήματοςπροσαναγράψαιτὸνκύκλον,οὖπέρἐστιτμῆμα.
quired to complete the circle for segment ABC, the very
ΤετμήσθωγὰρἡΑΓδίχακατὰτὸΔ,καὶἤχθωἀπὸτοῦ one of which it is a segment.
ΔσημείουτῇΑΓπρὸςὀρθὰςἡΔΒ,καὶἐπεζεύχθωἡΑΒ· For let AC have been cut in half at (point) D
ἡὑπὸΑΒΔγωνίαἄρατῆςὑπὸΒΑΔἤτοιμείζωνἐστὶνἢ [Prop. 1.10], and let DB have been drawn from point
ἴσηἢἐλάττων.
D, at right-angles to AC [Prop. 1.11]. And let AB have
῎Εστω πρότερονμείζων, καὶσυνεστάτωπρὸςτῇΒΑ been joined. Thus, angle ABD is surely either greater
εὐθείᾳκαὶτῷπρὸςαὐτῇσημείῳτῷΑτῇὑπὸΑΒΔγωνίᾳ than, equal to, or less than (angle) BAD.
ἴσηἡὑπὸΒΑΕ,καὶδιήχθωἡΔΒἐπὶτὸΕ,καὶἐπεζεύχθω First of all, let it be greater. And let (angle) BAE,
ἡΕΓ.ἐπεὶοὖνἴσηἐστὶνἡὑπὸΑΒΕγωνίατῇὑπὸΒΑΕ, equal to angle ABD, have been constructed on the
ἴσηἄραἐστὶκαὶἡΕΒεὐθεῖατῇΕΑ.καὶἐπεὶἴσηἐστὶνἡ straight-line BA, at the point A on it [Prop. 1.23]. And
ΑΔτῇΔΓ,κοινὴδὲἡΔΕ,δύοδὴαἱΑΔ,ΔΕδύοταῖς let DB have been drawn through to E, and let EC have
ΓΔ,ΔΕἴσαιεἰσὶνἑκατέραἑκατέρᾳ·καὶγωνίαἡὑπὸΑΔΕ been joined. Therefore, since angle ABE is equal to
γωνίᾳτῇὑπὸΓΔΕἐστινἴση·ὀρθὴγὰρἑκατέρα·βάσιςἄρα BAE, the straight-line EB is thus also equal to EA
ἡΑΕβάσειτῇΓΕἐστινἴση. ἀλλὰἡΑΕτῇΒΕἐδείχθη [Prop. 1.6]. And since AD is equal to DC, and DE (is)
ἴση·καὶἡΒΕἄρατῇΓΕἐστινἴση·αἱτρεῖςἄρααἱΑΕ,ΕΒ, common, the two (straight-lines) AD, DE are equal to
ΕΓἴσαιἀλλήλαιςεἰσίν·ὁἄρακέντρῷτῷΕδιαστήματιδὲ the two (straight-lines) CD, DE, respectively. And angle
ἑνὶτῶνΑΕ,ΕΒ,ΕΓκύκλοςγραφόμενοςἥξεικαὶδιὰτῶν ADE is equal to angle CDE. For each (is) a right-angle.
λοιπῶνσημείωνκαὶἔσταιπροσαναγεγραμμένος. κύκλου Thus, the base AE is equal to the base CE [Prop. 1.4].
ἄρατμήματοςδοθέντοςπροσαναγέγραπταιὁκύκλος. καὶ But, AE was shown (to be) equal to BE. Thus, BE is
δῆλον,ὡςτὸΑΒΓτμῆμαἔλαττόνἐστινἡμικυκλίουδιὰτὸ also equal to CE. Thus, the three (straight-lines) AE,
τὸΕκέντρονἐκτὸςαὐτοῦτυγχάνειν.
EB, and EC are equal to one another. Thus, if a cir-
῾Ομοίως[δὲ]κἂνᾖἡὑπὸΑΒΔγωνίαἴσητῇὑπὸΒΑΔ, cle is drawn with center E, and radius one of AE, EB,
τῆςΑΔἴσηςγενομένηςἑκατέρᾳτῶνΒΔ,ΔΓαἱτρεῖςαἱ or EC, it will also go through the remaining points (of
ΔΑ,ΔΒ,ΔΓἴσαιἀλλήλαιςἔσονται,καὶἔσταιτὸΔκέντρον the segment), and the (associated circle) will have been
τοῦπροσαναπεπληρωμένουκύκλου, καὶδηλαδὴἔσταιτὸ completed [Prop. 3.9]. Thus, a circle has been completed
ΑΒΓἡμικύκλιον.
from the given segment of a circle. And (it is) clear that
᾿ΕὰνδὲἡὑπὸΑΒΔἐλάττωνᾖτῆςὑπὸΒΑΔ,καὶσυ- the segment ABC is less than a semi-circle, because the
στησώμεθαπρὸςτῇΒΑεὐθείᾳκαὶτῷπρὸςαὐτῇσημείῳ center E happens to lie outside it.
94

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
τῷΑτῇὑπὸΑΒΔγωνίᾳἴσην,ἐντὸςτοῦΑΒΓτμήματος [And], similarly, even if angle ABD is equal to BAD,
πεσεῖταιτὸκέντρονἐπὶτῆςΔΒ,καὶἔσταιδηλαδὴτὸΑΒΓ (since) AD becomes equal to each of BD [Prop. 1.6] and
τμῆμαμεῖζονἡμικυκλίου.
DC, the three (straight-lines) DA, DB, and DC will be
Κύκλουἄρατμήματοςδοθέντοςπροσαναγέγραπταιὁ equal to one another. And point D will be the center
κύκλος·ὅπερἔδειποιῆσαι.
of the completed circle. And ABC will manifestly be a
semi-circle.
And if ABD is less than BAD, and we construct (angle
BAE), equal to angle ABD, on the straight-line BA,
at the point A on it [Prop. 1.23], then the center will fall
on DB, inside the segment ABC. And segment ABC will
manifestly be greater than a semi-circle.
Thus, a circle has been completed from the given segment
of a circle. (Which is) the very thing it was required
to do.
¦. Proposition 26
᾿Εντοῖςἴσοιςκύκλοιςαἱἴσαιγωνίαιἐπὶἴσωνπεριφε- In equal circles, equal angles stand upon equal cirρειῶνβεβήκασιν,ἐάντεπρὸςτοῖςκέντροιςἐάντεπρὸς
cumferences whether they are standing at the center or
ταῖςπεριφερείαιςὦσιβεβηκυῖαι.
at the circumference.
Α
A
Η
Θ
∆
G
H
D
Β
Γ
Ε
Ζ
B
C
E
F
Κ Λ
K
L
῎ΕστωσανἴσοικύκλοιοἱΑΒΓ,ΔΕΖκαὶἐναὐτοῖςἴσαι Let ABC and DEF be equal circles, and within them
γωνίαιἔστωσανπρὸςμὲντοῖςκέντροιςαἱὑπὸΒΗΓ,ΕΘΖ, let BGC and EHF be equal angles at the center, and
πρὸςδὲταῖςπεριφερείαιςαἱὑπὸΒΑΓ,ΕΔΖ·λέγω,ὅτιἴση BAC and EDF (equal angles) at the circumference. I
ἐστὶνἡΒΚΓπεριφέρειατῇΕΛΖπεριφερείᾳ.
say that circumference BKC is equal to circumference
᾿ΕπεζεύχθωσανγὰραἱΒΓ,ΕΖ. ELF .
ΚαὶἐπεὶἴσοιεἰσὶνοἱΑΒΓ,ΔΕΖκύκλοι,ἴσαιεἰσὶναἱ For let BC and EF have been joined.
ἐκτῶνκέντρων·δύοδὴαἱΒΗ,ΗΓδύοταῖςΕΘ,ΘΖἴσαι· And since circles ABC and DEF are equal, their radii
καὶγωνίαἡπρὸςτῷΗγωνίᾳτῇπρὸςτῷΘἴση·βάσιςἄρα are equal. So the two (straight-lines) BG, GC (are) equal
ἡΒΓβάσειτῇΕΖἐστινἴση. καὶἐπεὶἴσηἐστὶνἡπρὸςτῷ to the two (straight-lines) EH, HF (respectively). And
ΑγωνίατῇπρὸςτῷΔ,ὅμοιονἄραἐστὶτὸΒΑΓτμῆματῷ the angle at G (is) equal to the angle at H. Thus, the base
ΕΔΖτμήματι·καίεἰσινἐπὶἴσωνεὐθειῶν[τῶνΒΓ,ΕΖ]·τὰ BC is equal to the base EF [Prop. 1.4]. And since the
δὲἐπὶἴσωνεὐθειῶνὅμοιατμήματακύκλωνἴσαἀλλήλοις angle at A is equal to the (angle) at D, the segment BAC
ἐστίν·ἴσονἄρατὸΒΑΓτμῆματῷΕΔΖ.ἔστιδὲκαὶὅλοςὁ is thus similar to the segment EDF [Def. 3.11]. And
ΑΒΓκύκλοςὅλῳτῷΔΕΖκύκλῳἴσος·λοιπὴἄραἡΒΚΓ they are on equal straight-lines [BC and EF ]. And simiπεριφέρειατῇΕΛΖπεριφερείᾳἐστὶνἴση.
lar segments of circles on equal straight-lines are equal to
᾿Ενἄρατοῖςἴσοιςκύκλοιςαἱἴσαιγωνίαιἐπὶἴσωνπερι- one another [Prop. 3.24]. Thus, segment BAC is equal to
φερειῶνβεβήκασιν,ἐάντεπρὸςτοῖςκέντροιςἐάντεπρὸς (segment) EDF . And the whole circle ABC is also equal
ταῖςπεριφερείαςὦσιβεβηκυῖαι·ὅπερἔδειδεῖξαι. to the whole circle DEF . Thus, the remaining circumference
BKC is equal to the (remaining) circumference
ELF .
Thus, in equal circles, equal angles stand upon equal
circumferences, whether they are standing at the center
95

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
or at the circumference. (Which is) the very thing which
it was required to show.
Þ. Proposition 27
᾿Εντοῖςἴσοιςκύκλοιςαἱἐπὶἴσωνπεριφερειῶνβεβηκυῖαι In equal circles, angles standing upon equal circumγωνίαιἴσαιἀλλήλαιςεἰσίν,ἐάντεπρὸςτοῖςκέντροιςἐάντε
ferences are equal to one another, whether they are
πρὸςταῖςπεριφερείαιςὦσιβεβηκυῖαι.
standing at the center or at the circumference.
Α
∆
A
D
Η
Θ
G
H
Β
Γ
Κ
Ε
Ζ
᾿ΕνγὰρἴσοιςκύκλοιςτοῖςΑΒΓ,ΔΕΖἐπὶἴσωνπερι- For let the angles BGC and EHF at the centers G
φερειῶντῶνΒΓ,ΕΖπρὸςμὲντοῖςΗ,Θκέντροιςγωνίαι and H, and the (angles) BAC and EDF at the circumβεβηκέτωσαναἱὑπὸΒΗΓ,ΕΘΖ,πρὸςδὲταῖςπεριφερείαις
ferences, stand upon the equal circumferences BC and
αἱὑπὸΒΑΓ,ΕΔΖ·λέγω,ὅτιἡμὲνὑπὸΒΗΓγωνίατῇὑπὸ EF , in the equal circles ABC and DEF (respectively). I
ΕΘΖἐστινἴση,ἡδὲὑπὸΒΑΓτῇὑπὸΕΔΖἐστινἴση. say that angle BGC is equal to (angle) EHF , and BAC
ΕἰγὰρἄνισόςἐστινἡὑπὸΒΗΓτῇὑπὸΕΘΖ,μίααὐτῶν is equal to EDF .
μείζων ἐστίν. ἔστωμείζωνἡὑπὸΒΗΓ,καὶσυνεστάτω For if BGC is unequal to EHF , one of them is greater.
πρὸςτῇΒΗεὐθείᾳκαὶτῷπρὸςαὐτῇσημείῳτῷΗτῇὑπὸ Let BGC be greater, and let the (angle) BGK, equal to
ΕΘΖγωνίᾳἴσηἡὑπὸΒΗΚ·αἱδὲἴσαιγωνίαιἐπὶἴσων angle EHF , have been constructed on the straight-line
περιφερειῶνβεβήκασιν,ὅτανπρὸςτοῖςκέντροιςὦσιν·ἴση BG, at the point G on it [Prop. 1.23]. But equal angles
ἄραἡΒΚπεριφέρειατῇΕΖπεριφερείᾳ.ἀλλὰἡΕΖτῇΒΓ (in equal circles) stand upon equal circumferences, when
ἐστινἴση· καὶἡΒΚἄρατῇΒΓἐστινἴσηἡἐλάττωντῇ they are at the centers [Prop. 3.26]. Thus, circumference
μείζονι·ὅπερἐστὶνἀδύνατον. οὐκἄραἄνισόςἐστινἡὑπὸ BK (is) equal to circumference EF . But, EF is equal
ΒΗΓγωνίατῇὑπὸΕΘΖ·ἴσηἄρα. καίἐστιτῆςμὲνὑπὸ to BC. Thus, BK is also equal to BC, the lesser to the
ΒΗΓἡμίσειαἡπρὸςτῷΑ,τῆςδὲὑπὸΕΘΖἡμίσειαἡπρὸς greater. The very thing is impossible. Thus, angle BGC
τῷΔ·ἴσηἄρακαὶἡπρὸςτῷΑγωνίατῇπρὸςτῷΔ. is not unequal to EHF . Thus, (it is) equal. And the
᾿Ενἄρατοῖςἴσοιςκύκλοιςαἱἐπὶἴσωνπεριφερειῶνβε- (angle) at A is half BGC, and the (angle) at D half EHF
βηκυῖαιγωνίαιἴσαιἀλλήλαιςεἰσίν,ἐάντεπρὸςτοῖςκέντροις [Prop. 3.20]. Thus, the angle at A (is) also equal to the
ἐάντεπρὸςταῖςπεριφερείαιςὦσιβεβηκυῖαι·ὅπερἔδειδεῖξαι. (angle) at D.
Thus, in equal circles, angles standing upon equal circumferences
are equal to one another, whether they are
standing at the center or at the circumference. (Which is)
the very thing it was required to show.
. Proposition 28
᾿Εντοῖςἴσοιςκύκλοιςαἱἴσαιεὐθεῖαιἴσαςπεριφερείας In equal circles, equal straight-lines cut off equal cir-
ἀφαιροῦσιτὴνμὲνμείζονατῇμείζονιτὴνδὲἐλάττονατῇ cumferences, the greater (circumference being equal) to
ἐλάττονι.
the greater, and the lesser to the lesser.
῎ΕστωσανἴσοικύκλοιοἱΑΒΓ,ΔΕΖ,καὶἐντοῖςκύκλοις Let ABC and DEF be equal circles, and let AB
ἴσαιεὐθεῖαιἔστωσαναἱΑΒ,ΔΕτὰςμὲνΑΓΒ,ΑΖΕπερι- and DE be equal straight-lines in these circles, cutting
φερείαςμείζοναςἀφαιροῦσαιτὰςδὲΑΗΒ,ΔΘΕἐλάττονας· off the greater circumferences ACB and DFE, and the
λέγω,ὅτιἡμὲνΑΓΒμείζωνπεριφέρειαἴσηἐστὶτῇΔΖΕ lesser (circumferences) AGB and DHE (respectively). I
μείζονιπεριφερείᾳἡδὲΑΗΒἐλάττωνπεριφέρειατῇΔΘΕ. say that the greater circumference ACB is equal to the
greater circumference DFE, and the lesser circumfer-
B
K
C
E
F
96

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
Γ
Ζ
ence AGB to (the lesser) DHE.
C
F
Κ
Λ
K
L
Α
Β ∆ Ε
A
B
D
E
Η
Θ
G
H
Εἰλήφθω γὰρ τὰ κέντρα τῶν κύκλων τὰ Κ, Λ, καὶ For let the centers of the circles, K and L, have been
ἐπεζεύχθωσαναἱΑΚ,ΚΒ,ΔΛ,ΛΕ.
found [Prop. 3.1], and let AK, KB, DL, and LE have
Καὶ ἐπεὶ ἴσοι κύκλοι εἰσίν, ἴσαι εἰσὶ καὶ αἱ ἐκ τῶν been joined.
κέντρων·δύοδὴαἱΑΚ,ΚΒδυσὶταῖςΔΛ,ΛΕἴσαιεἰσίν· And since (ABC and DEF ) are equal circles, their
καὶβάσιςἡΑΒβάσειτῇΔΕἴση·γωνίαἄραἡὑπὸΑΚΒ radii are also equal [Def. 3.1]. So the two (straightγωνίᾳτῇὑπὸΔΛΕἴσηἐστίν.
αἱδὲἴσαιγωνίαιἐπὶἴσων lines) AK, KB are equal to the two (straight-lines) DL,
περιφερειῶνβεβήκασιν,ὅτανπρὸςτοῖςκέντροιςὦσιν·ἴση LE (respectively). And the base AB (is) equal to the
ἄραἡΑΗΒπεριφέρειατῇΔΘΕ.ἐστὶδὲκαὶὅλοςὁΑΒΓ base DE. Thus, angle AKB is equal to angle DLE
κύκλοςὅλῳτῷΔΕΖκύκλῳἴσος·καὶλοιπὴἄραἡΑΓΒ [Prop. 1.8]. And equal angles stand upon equal circumπεριφέρειαλοιπῇτῇΔΖΕπεριφερείᾳἴσηἐστίν.
ferences, when they are at the centers [Prop. 3.26]. Thus,
᾿Εν ἄρα τοῖς ἴσοις κύκλοις αἱ ἴσαι εὐθεῖαι ἴσας πε- circumference AGB (is) equal to DHE. And the whole
ριφερείας ἀφαιροῦσι τὴν μὲν μείζονα τῇ μείζονι τὴν δὲ circle ABC is also equal to the whole circle DEF . Thus,
ἐλάττονατῇἐλάττονι·ὅπερἔδειδεῖξαι.
the remaining circumference ACB is also equal to the
remaining circumference DFE.
Thus, in equal circles, equal straight-lines cut off
equal circumferences, the greater (circumference being
equal) to the greater, and the lesser to the lesser. (Which
is) the very thing it was required to show.
. Proposition 29
᾿Εντοῖςἴσοιςκύκλοιςτὰςἴσαςπεριφερείαςἴσαιεὐθεῖαι In equal circles, equal straight-lines subtend equal cir-
ὑποτείνουσιν.
cumferences.
Α
∆
A
D
Κ
Λ
K
L
Β Γ Ε Ζ B
C E
F
Η
Θ
G
H
῎ΕστωσανἴσοικύκλοιοἱΑΒΓ,ΔΕΖ,καὶἐναὐτοῖςἴσαι Let ABC and DEF be equal circles, and within them
περιφέρειαιἀπειλήφθωσαναἱΒΗΓ,ΕΘΖ,καὶἐπεζεύχθωσαν let the equal circumferences BGC and EHF have been
αἱΒΓ,ΕΖεὐθεῖαι·λέγω,ὅτιἴσηἐστὶνἡΒΓτῇΕΖ. cut off. And let the straight-lines BC and EF have been
Εἰλήφθωγὰρτὰκέντρατῶνκύκλων,καὶἔστωτὰΚ, joined. I say that BC is equal to EF .
Λ,καὶἐπεζεύχθωσαναἱΒΚ,ΚΓ,ΕΛ,ΛΖ.
For let the centers of the circles have been found
ΚαὶἐπεὶἴσηἐστὶνἡΒΗΓπεριφέρειατῇΕΘΖπεριφερείᾳ, [Prop. 3.1], and let them be (at) K and L. And let BK,
97

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
ἴσηἐστὶκαὶγωνίαἡὑπὸΒΚΓτῇὑπὸΕΛΖ.καὶἐπεὶἴσοι KC, EL, and LF have been joined.
εἰσὶνοἱΑΒΓ,ΔΕΖκύκλοι,ἴσαιεἰσὶκαὶαἱἐκτῶνκέντρων· And since the circumference BGC is equal to the cirδύοδὴαἱΒΚ,ΚΓδυσὶταῖςΕΛ,ΛΖἴσαιεἰσίν·καὶγωνίας
cumference EHF , the angle BKC is also equal to (an-
ἴσαςπεριέχουσιν·βάσιςἄραἡΒΓβάσειτῇΕΖἴσηἐστίν· gle) ELF [Prop. 3.27]. And since the circles ABC and
᾿Εν ἄρα τοῖς ἴσοις κύκλοιςτὰς ἴσας περιφερείαςἴσαι DEF are equal, their radii are also equal [Def. 3.1]. So
εὐθεῖαιὑποτείνουσιν·ὅπερἔδειδεῖξαι.
the two (straight-lines) BK, KC are equal to the two
(straight-lines) EL, LF (respectively). And they contain
equal angles. Thus, the base BC is equal to the base EF
[Prop. 1.4].
Thus, in equal circles, equal straight-lines subtend
equal circumferences. (Which is) the very thing it was
required to show.
Ð. Proposition 30
Τὴνδοθεῖσανπεριφέρειανδίχατεμεῖν.
To cut a given circumference in half.
∆
D
Α
Γ
Β
῎ΕστωἡδοθεῖσαπεριφέρειαἡΑΔΒ·δεῖδὴτὴνΑΔΒ Let ADB be the given circumference. So it is required
περιφέρειανδίχατεμεῖν.
to cut circumference ADB in half.
᾿ΕπεζεύχθωἡΑΒ,καὶτετμήσθωδίχακατὰτὸΓ,καὶ Let AB have been joined, and let it have been cut in
ἀπὸτοῦΓσημείουτῇΑΒεὐθείᾳπρὸςὀρθὰςἤχθωἡΓΔ, half at (point) C [Prop. 1.10]. And let CD have been
καὶἐπεζεύχθωσαναἱΑΔ,ΔΒ. drawn from point C, at right-angles to AB [Prop. 1.11].
ΚαὶἐπεὶἴσηἐστὶνἡΑΓτῇΓΒ,κοινὴδὲἡΓΔ,δύο And let AD, and DB have been joined.
δὴαἱΑΓ,ΓΔδυσὶταῖςΒΓ,ΓΔἴσαιεἰσίν· καὶγωνίαἡ And since AC is equal to CB, and CD (is) com-
ὑπὸΑΓΔγωνίᾳτῇὑπὸΒΓΔἴση·ὀρθὴγὰρἑκατέρα·βάσις mon, the two (straight-lines) AC, CD are equal to the
ἄραἡΑΔβάσειτῇΔΒἴσηἐστίν. αἱδὲἴσαιεὐθεῖαιἴσας two (straight-lines) BC, CD (respectively). And angle
περιφερείαςἀφαιροῦσιτὴνμὲνμείζονατῇμείζονιτὴνδὲ ACD (is) equal to angle BCD. For (they are) each right-
ἐλάττονατῇἐλάττονι·κάιἐστινἑκατέρατῶνΑΔ,ΔΒπε- angles. Thus, the base AD is equal to the base DB
ριφερειῶνἐλάττωνἡμικυκλίου·ἴσηἄραἡΑΔπεριφέρειατῇ [Prop. 1.4]. And equal straight-lines cut off equal circum-
ΔΒπεριφερείᾳ.
ferences, the greater (circumference being equal) to the
῾Η ἄρα δοθεῖσα περιφέρειαδίχα τέτμηταικατὰ τὸ Δ greater, and the lesser to the lesser [Prop. 1.28]. And the
σημεῖον·ὅπερἔδειποιῆσαι.
circumferences AD and DB are each less than a semicircle.
Thus, circumference AD (is) equal to circumference
DB.
Thus, the given circumference has been cut in half at
point D. (Which is) the very thing it was required to do.
Ð. Proposition 31
᾿Ενκύκλῳἡμὲνἐντῷἡμικυκλίῳγωνίαὀρθήἐστιν,ἡδὲ In a circle, the angle in a semi-circle is a right-angle,
ἐντῷμείζονιτμήματιἐλάττωνὀρθῆς,ἡδὲἐντῷἐλάττονι and that in a greater segment (is) less than a right-angle,
τμήματιμείζωνὀρθῆς·καὶἔπιἡμὲντοῦμείζονοςτμήματος and that in a lesser segment (is) greater than a rightγωνίαμείζωνἐστὶνὀρθῆς,ἡδὲτοῦἐλάττονοςτμήματος
angle. And, further, the angle of a segment greater (than
γωνίαἐλάττωνὀρθῆς.
a semi-circle) is greater than a right-angle, and the an-
A
C
B
98

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
Ζ
∆
Γ
gle of a segment less (than a semi-circle) is less than a
right-angle.
F
D
C
Α
A
Ε
E
Β
῎ΕστωκύκλοςὁΑΒΓΔ,διάμετροςδὲαὐτοῦἔστωἡΒΓ, Let ABCD be a circle, and let BC be its diameter, and
κέντρονδὲτὸΕ,καὶἐπεζεύχθωσαναἱΒΑ,ΑΓ,ΑΔ,ΔΓ· E its center. And let BA, AC, AD, and DC have been
λέγω,ὅτιἡμὲνἐντῷΒΑΓἡμικυκλίῳγωνίαἡὑπὸΒΑΓ joined. I say that the angle BAC in the semi-circle BAC
ὀρθήἐστιν,ἡδὲἐντῷΑΒΓμείζονιτοῦἡμικυκλίουτμήματι is a right-angle, and the angle ABC in the segment ABC,
γωνίαἡὑπὸΑΒΓἐλάττωνἐστὶνὀρθῆς,ἡδὲἐντῷΑΔΓ (which is) greater than a semi-circle, is less than a right-
ἐλάττονιτοῦἡμικυκλίουτμήματιγωνίαἡὑπὸΑΔΓμείζων angle, and the angle ADC in the segment ADC, (which
ἐστὶνὀρθῆς.
is) less than a semi-circle, is greater than a right-angle.
᾿ΕπεζεύχθωἡΑΕ,καὶδιήχθωἡΒΑἐπὶτὸΖ.
Let AE have been joined, and let BA have been
ΚαὶἐπεὶἴσηἐστὶνἡΒΕτῇΕΑ,ἴσηἐστὶκαὶγωνίαἡ drawn through to F .
ὑπὸΑΒΕτῇὑπὸΒΑΕ.πάλιν,ἐπεὶἴσηἐστὶνἡΓΕτῇΕΑ, And since BE is equal to EA, angle ABE is also
ἴσηἐστὶκαὶἡὑπὸΑΓΕτῇὑπὸΓΑΕ·ὅληἄραἡὑπὸΒΑΓ equal to BAE [Prop. 1.5]. Again, since CE is equal to
δυσὶταῖςὑπὸΑΒΓ,ΑΓΒἴσηἐστίν.ἐστὶδὲκαὶἡὑπὸΖΑΓ EA, ACE is also equal to CAE [Prop. 1.5]. Thus, the
ἐκτὸςτοῦΑΒΓτριγώνουδυσὶταῖςὑπὸΑΒΓ,ΑΓΒγωνίαις whole (angle) BAC is equal to the two (angles) ABC
ἴση·ἴσηἄρακαὶἡὑπὸΒΑΓγωνίατῇὑπὸΖΑΓ·ὀρθὴἄρα and ACB. And FAC, (which is) external to triangle
ἑκατέρα·ἡἄραἐντῷΒΑΓἡμικυκλίῳγωνίαἡὑπὸΒΑΓ ABC, is also equal to the two angles ABC and ACB
ὀρθήἐστιν.
[Prop. 1.32]. Thus, angle BAC (is) also equal to FAC.
ΚαὶἐπεὶτοῦΑΒΓτρίγωνουδύογωνίαιαἱὑπὸΑΒΓ, Thus, (they are) each right-angles. [Def. 1.10]. Thus, the
ΒΑΓ δύο ὀρθῶν ἐλάττονές εἰσιν, ὀρθὴ δὲ ἡ ὑπὸ ΒΑΓ, angle BAC in the semi-circle BAC is a right-angle.
ἐλάττωνἄραὀρθῆςἐστινἡὑπὸΑΒΓγωνία·καίἐστινἐν And since the two angles ABC and BAC of trianτῷΑΒΓμείζονιτοῦἡμικυκλίουτμήματι.
gle ABC are less than two right-angles [Prop. 1.17], and
ΚαὶἐπεὶἐνκύκλῳτετράπλευρόνἐστιτὸΑΒΓΔ,τῶνδὲ BAC is a right-angle, angle ABC is thus less than a right-
ἐντοῖςκύκλοιςτετραπλεύρωναἱἀπεναντίονγωνίαιδυσὶν angle. And it is in segment ABC, (which is) greater than
ὀρθαῖςἴσαιεἰσίν[αἱἄραὑπὸΑΒΓ,ΑΔΓγωνίαιδυσὶνὀρθαῖς a semi-circle.
ἴσαςεἰσίν],καίἐστινἡὑπὸΑΒΓἐλάττωνὀρθῆς·λοιπὴἄρα And since ABCD is a quadrilateral within a circle,
ἡὑπὸΑΔΓγωνίαμείζωνὀρθῆςἐστιν·καίἐστινἐντῷΑΔΓ and for quadrilaterals within circles the (sum of the) op-
ἐλάττονιτοῦἡμικυκλίουτμήματι. posite angles is equal to two right-angles [Prop. 3.22]
Λέγω,ὅτικαὶἡμὲντοῦμείζονοςτμήματοςγωνίαἡπε- [angles ABC and ADC are thus equal to two rightριεχομένηὑπό[τε]τῆςΑΒΓπεριφερείαςκαὶτῆςΑΓεὐθείας
angles], and (angle) ABC is less than a right-angle. The
μείζωνἐστὶνὀρθῆς,ἡδὲτοῦἐλάττονοςτμήματοςγωνίαἡ remaining angle ADC is thus greater than a right-angle.
περιεχομένηὑπό[τε]τῆςΑΔ[Γ]περιφερείαςκαὶτῆςΑΓ And it is in segment ADC, (which is) less than a semiεὐθείαςἐλάττωνἐστὶνὀρθῆς.
καίἐστιναὐτόθενφανερόν. circle.
ἐπεὶγὰρἡὑπὸτῶνΒΑ,ΑΓεὐθειῶνὀρθήἐστιν, ἡἄρα I also say that the angle of the greater segment,
ὑπὸτῆςΑΒΓπεριφερείαςκαὶτῆςΑΓεὐθείαςπεριεχομένη (namely) that contained by the circumference ABC and
μείζωνἐστὶνὀρθῆς.πάλιν,ἐπεὶἡὑπὸτῶνΑΓ,ΑΖεὐθειῶν the straight-line AC, is greater than a right-angle. And
ὀρθήἐστιν,ἡἄραὑπὸτῆςΓΑεὐθείαςκαὶτῆςΑΔ[Γ]περι- the angle of the lesser segment, (namely) that contained
B
99

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
φερείαςπεριεχομένηἐλάττωνἐστὶνὀρθῆς.
by the circumference AD[C] and the straight-line AC, is
᾿Ενκύκλῳἄραἡμὲνἐντῷἡμικυκλίῳγωνίαὀρθήἐστιν, less than a right-angle. And this is immediately apparent.
ἡ δὲ ἐντῷ μείζονι τμήματιἐλάττωνὀρθῆς, ἡδὲ ἐν τῷ For since the (angle contained by) the two straight-lines
ἐλάττονι[τμήματι]μείζωνὀρθῆς·καὶἔπιἡμὲντοῦμείζονος BA and AC is a right-angle, the (angle) contained by
τμήματος[γωνία]μείζων[ἐστὶν]ὀρθῆς,ἡδὲτοῦἐλάττονος the circumference ABC and the straight-line AC is thus
τμήματος[γωνία]ἐλάττωνὀρθῆς·ὅπερἔδειδεῖξαι. greater than a right-angle. Again, since the (angle contained
by) the straight-lines AC and AF is a right-angle,
the (angle) contained by the circumference AD[C] and
the straight-line CA is thus less than a right-angle.
Thus, in a circle, the angle in a semi-circle is a rightangle,
and that in a greater segment (is) less than a
right-angle, and that in a lesser [segment] (is) greater
than a right-angle. And, further, the [angle] of a segment
greater (than a semi-circle) [is] greater than a rightangle,
and the [angle] of a segment less (than a semicircle)
is less than a right-angle. (Which is) the very thing
it was required to show.
Ð. Proposition 32
᾿Εὰνκύκλουἐφάπτηταίτιςεὐθεῖα,ἀπὸδὲτῆςἁφῆςεἰς If some straight-line touches a circle, and some
τὸν κύκλονδιαχθῇτιςεὐθεῖατέμνουσατὸνκύκλον, ἃς (other) straight-line is drawn across, from the point of
ποιεῖγωνίαςπρὸςτῇἐφαπτομένῃ,ἴσαιἔσονταιταῖςἐντοῖς contact into the circle, cutting the circle (in two), then
ἐναλλὰξτοῦκύκλουτμήμασιγωνίαις.
those angles the (straight-line) makes with the tangent
will be equal to the angles in the alternate segments of
the circle.
A
A
D
D
C
C
E
F
E
F
B
B
ΚύκλουγὰρτοῦΑΒΓΔἐφαπτέσθωτιςεὐθεῖαἡΕΖ For let some straight-line EF touch the circle ABCD
κατὰ τὸ Β σημεῖον, καὶ ἀπὸ τοῦ Β σημείουδιήχθω τις at the point B, and let some (other) straight-line BD
εὐθεῖαεἰςτὸνΑΒΓΔκύκλοντέμνουσααὐτὸνἡΒΔ.λέγω, have been drawn from point B into the circle ABCD,
ὅτιἃςποιεῖγωνίαςἡΒΔμετὰτῆςΕΖἐφαπτομένης,ἴσας cutting it (in two). I say that the angles BD makes with
ἔσονταιταῖςἐντοῖςἐναλλὰξτμήμασιτοῦκύκλουγωνίαις, the tangent EF will be equal to the angles in the alterτουτέστιν,ὅτιἡμὲνὑπὸΖΒΔγωνίαἴσηἐστὶτῇἐντῷΒΑΔ
nate segments of the circle. That is to say, that angle
τμήματισυνισταμένῃγωνίᾳ,ἡδὲὑπὸΕΒΔγωνίαἴσηἐστὶ FBD is equal to the angle constructed in segment BAD,
τῇἐντῷΔΓΒτμήματισυνισταμένῃγωνίᾳ.
and angle EBD is equal to the angle constructed in seg-
῎ΗχθωγὰρἀπὸτοῦΒτῇΕΖπρὸςὀρθὰςἡΒΑ,καὶ ment DCB.
εἰλήφθωἐπὶτῆςΒΔπεριφερείαςτυχὸνσημεῖοντὸΓ,καὶ For let BA have been drawn from B, at right-angles
ἐπεζεύχθωσαναἱΑΔ,ΔΓ,ΓΒ.
to EF [Prop. 1.11]. And let the point C have been taken
ΚαὶἐπεὶκύκλουτοῦΑΒΓΔἐφάπτεταίτιςεὐθεῖαἡΕΖ at random on the circumference BD. And let AD, DC,
100

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
κατὰτὸΒ,καὶἀπὸτῆςἁφῆςἦκταιτῇἐφαπτομένῃπρὸς and CB have been joined.
ὀρθὰςἡΒΑ,ἐπὶτῆςΒΑἄρατὸκέντρονἐστὶτοῦΑΒΓΔ And since some straight-line EF touches the circle
κύκλου.ἡΒΑἄραδιάμετόςἐστιτοῦΑΒΓΔκύκλου·ἡἄρα ABCD at point B, and BA has been drawn from the
ὑπὸΑΔΒγωνίαἐνἡμικυκλίῳοὖσαὀρθήἐστιν.λοιπαὶἄρα point of contact, at right-angles to the tangent, the center
αἱὑπὸΒΑΔ,ΑΒΔμιᾷὀρθῇἴσαιεἰσίν. ἐστὶδὲκαὶἡὑπὸ of circle ABCD is thus on BA [Prop. 3.19]. Thus, BA
ΑΒΖὀρθή·ἡἄραὑπὸΑΒΖἴσηἐστὶταῖςὑπὸΒΑΔ,ΑΒΔ. is a diameter of circle ABCD. Thus, angle ADB, being
κοινὴἀφῃρήσθωἡὑπὸΑΒΔ·λοιπὴἄραἡὑπὸΔΒΖγωνία in a semi-circle, is a right-angle [Prop. 3.31]. Thus, the
ἴσηἐστὶτῇἐντῷἐναλλὰξτμήματιτοῦκύκλουγωνίᾳτῇ remaining angles (of triangle ADB) BAD and ABD are
ὑπὸΒΑΔ.καὶἐπεὶἐνκύκλῳτετράπλευρόνἐστιτὸΑΒΓΔ, equal to one right-angle [Prop. 1.32]. And ABF is also a
αἱἀπεναντίοναὐτοῦγωνίαιδυσὶνὀρθαῖςἴσαιεἰσίν. εἰσὶδὲ right-angle. Thus, ABF is equal to BAD and ABD. Let
καὶαἱὑπὸΔΒΖ,ΔΒΕδυσὶνὀρθαῖςἴσαι·αἱἄραὑπὸΔΒΖ, ABD have been subtracted from both. Thus, the remain-
ΔΒΕταῖςὑπὸΒΑΔ,ΒΓΔἴσαιεἰσίν,ὧνἡὑπὸΒΑΔτῇὑπὸ ing angle DBF is equal to the angle BAD in the alternate
ΔΒΖἐδείχθηἴση·λοιπὴἄραἡὑπὸΔΒΕτῇἐντῷἐναλλὰξ segment of the circle. And since ABCD is a quadrilateral
τοῦκύκλουτμήματιτῷΔΓΒτῇὑπὸΔΓΒγωνίᾳἐστὶνἴση. in a circle, (the sum of) its opposite angles is equal to
᾿Εὰνἄρακύκλουἐφάπτηταίτιςεὐθεῖα,ἀπὸδὲτῆςἁφῆς two right-angles [Prop. 3.22]. And DBF and DBE is
εἰςτὸνκύκλονδιαχθῇτιςεὐθεῖατέμνουσατὸνκύκλον,ἃς also equal to two right-angles [Prop. 1.13]. Thus, DBF
ποιεῖγωνίαςπρὸςτῇἐφαπτομένῃ,ἴσαιἔσονταιταῖςἐντοῖς and DBE is equal to BAD and BCD, of which BAD
ἐναλλὰξτοῦκύκλουτμήμασιγωνίαις·ὅπερἔδειδεῖξαι. was shown (to be) equal to DBF . Thus, the remaining
(angle) DBE is equal to the angle DCB in the alternate
segment DCB of the circle.
Thus, if some straight-line touches a circle, and some
(other) straight-line is drawn across, from the point of
contact into the circle, cutting the circle (in two), then
those angles the (straight-line) makes with the tangent
will be equal to the angles in the alternate segments of
the circle. (Which is) the very thing it was required to
show.
Ð. Proposition 33
᾿Επὶτῆςδοθείσηςεὐθείαςγράψαιτμῆμακύκλουδεχόμε- To draw a segment of a circle, accepting an angle
νονγωνίανἴσηντῇδοθείσῃγωνίᾳεὐθυγράμμῳ. equal to a given rectilinear angle, on a given straight-line.
∆
Β
Ζ
Γ
Α
Ε
Η
∆
Ε
Γ
Α
Β
Ζ
Β
Γ
Θ
Ζ
Α
Ε
Η
∆
D
῎Εστω ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ, ἡ δὲ δοθεῖσα γωνία Let AB be the given straight-line, and C the given
εὐθύγραμμοςἡπρὸςτῷΓ·δεῖδὴἐπὶτῆςδοθείσηςεὐθείας rectilinear angle. So it is required to draw a segment
τῆςΑΒγράψαιτμῆμακύκλουδεχόμενονγωνίανἴσηντῇ of a circle, accepting an angle equal to C, on the given
πρὸςτῷΓ.
straight-line AB.
῾Η δὴπρὸς τῷ Γ [γωνία] ἤτοιὀξεῖάἐστιν ἢ ὀρθὴἢ So the [angle] C is surely either acute, a right-angle,
ἀμβλεῖα·ἔστωπρότερονὀξεῖα,καὶὡςἐπὶτῆςπρώτηςκα- or obtuse. First of all, let it be acute. And, as in the first
ταγραφῆςσυνεστάτωπρὸςτῇΑΒεὐθείᾳκαὶτῷΑσημείῳ diagram (from the left), let (angle) BAD, equal to angle
τῇπρὸςτῷΓγωνίᾳἴσηἡὑπὸΒΑΔ·ὀξεῖαἄραἐστὶκαὶἡ C, have been constructed on the straight-line AB, at the
ὑπὸΒΑΔ.ἤχθωτῇΔΑπρὸςὀρθὰςἡΑΕ,καὶτετμήσθω point A (on it) [Prop. 1.23]. Thus, BAD is also acute. Let
ἡΑΒδίχακατὰτὸΖ,καὶἤχθωἀπὸτοῦΖσημείουτῇΑΒ AE have been drawn, at right-angles to DA [Prop. 1.11].
B
F
C
A
E
G
D
E
C
A
B
F
B
C
H
F
A
E
G
D
101

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
πρὸςὀρθὰςἡΖΗ,καὶἐπεζεύχθωἡΗΒ.
And let AB have been cut in half at F [Prop. 1.10]. And
ΚαὶἐπεὶἴσηἐστὶνἡΑΖτῇΖΒ,κοινὴδὲἡΖΗ,δύοδὴ let FG have been drawn from point F , at right-angles to
αἱΑΖ,ΖΗδύοταῖςΒΖ,ΖΗἴσαιεἰσίν· καὶγωνίαἡὑπὸ AB [Prop. 1.11]. And let GB have been joined.
ΑΖΗ[γωνίᾳ]τῇὑπὸΒΖΗἴση·βάσιςἄραἡΑΗβάσειτῇ And since AF is equal to FB, and FG (is) common,
ΒΗἴσηἐστίν. ὁἄρακέντρῳμὲντῷΗδιαστήματιδὲτῷ the two (straight-lines) AF , FG are equal to the two
ΗΑκύκλοςγραφόμενοςἥξεικαὶδιὰτοῦΒ.γεγράφθωκαὶ (straight-lines) BF , FG (respectively). And angle AFG
ἔστωὁΑΒΕ,καὶἐπεζεύχθωἡΕΒ.ἐπεὶοὖνἀπ᾿ἄκραςτῆς (is) equal to [angle] BFG. Thus, the base AG is equal to
ΑΕδιαμέτρουἀπὸτοῦΑτῇΑΕπρὸςὀρθάςἐστινἡΑΔ, the base BG [Prop. 1.4]. Thus, the circle drawn with
ἡΑΔἄραἐφάπτεταιτοῦΑΒΕκύκλου·ἐπεὶοὖνκύκλου center G, and radius GA, will also go through B (as
τοῦΑΒΕἐφάπτεταίτιςεὐθεῖαἡΑΔ,καὶἀπὸτῆςκατὰτὸ well as A). Let it have been drawn, and let it be (de-
ΑἁφῆςεἰςτὸνΑΒΕκύκλονδιῆκταίτιςεὐθεῖαἡΑΒ,ἡ noted) ABE. And let EB have been joined. Therefore,
ἄραὑπὸΔΑΒγωνίαἴσηἐστὶτῇἐντῷἐναλλὰξτοῦκύκλου since AD is at the extremity of diameter AE, (namely,
τμήματιγωνίᾳτῇὑπὸΑΕΒ.ἀλλ᾿ἡὑπὸΔΑΒτῇπρὸςτῷΓ point) A, at right-angles to AE, the (straight-line) AD
ἐστινἴση·καὶἡπρὸςτῷΓἄραγωνίαἴσηἐστὶτῇὑπὸΑΕΒ. thus touches the circle ABE [Prop. 3.16 corr.]. There-
᾿ΕπὶτῆςδοθείσηςἄραεὐθείαςτῆςΑΒτμῆμακύκλου fore, since some straight-line AD touches the circle ABE,
γέγραπταιτὸΑΕΒδεχόμενονγωνίαντὴνὑπὸΑΕΒἴσην and some (other) straight-line AB has been drawn across
τῇδοθείσῃτῇπρὸςτῷΓ.
from the point of contact A into circle ABE, angle DAB
ἈλλὰδὴὀρθὴἔστωἡπρὸςτῷΓ·καὶδέονπάλινἔστω is thus equal to the angle AEB in the alternate segment
ἐπὶτῆςΑΒγράψαιτμῆμακύκλουδεχόμενονγωνίανἴσηντῇ of the circle [Prop. 3.32]. But, DAB is equal to C. Thus,
πρὸςτῷΓὀρθῇ[γωνίᾳ]. συνεστάτω[πάλιν]τῇπρὸςτῷΓ angle C is also equal to AEB.
ὀρθῇγωνίᾳἴσηἡὑπὸΒΑΔ,ὡςἔχειἐπὶτῆςδευτέραςκατα- Thus, a segment AEB of a circle, accepting the angle
γραφῆς,καὶτετμήσθωἡΑΒδίχακατὰτὸΖ,καὶκέντρῳτῷ AEB (which is) equal to the given (angle) C, has been
Ζ,διαστήματιδὲὁποτέρῳτῶνΖΑ,ΖΒ,κύκλοςγεγράφθω drawn on the given straight-line AB.
ὁΑΕΒ. And so let C be a right-angle. And let it again be
᾿ΕφάπτεταιἄραἡΑΔεὐθεῖατοῦΑΒΕκύκλουδιὰτὸ necessary to draw a segment of a circle on AB, accepting
ὀρθὴνεἶναιτὴνπρὸςτῷΑγωνίαν. καὶἴσηἐστὶνἡὑπὸ an angle equal to the right-[angle] C. Let the (angle)
ΒΑΔγωνίατῇἐντῷΑΕΒτμήματι·ὀρθὴγὰρκαὶαὐτὴἐν BAD [again] have been constructed, equal to the right-
ἡμικυκλίῳοὖσα. ἀλλὰκαὶἡὑπὸΒΑΔτῇπρὸςτῷΓἴση angle C [Prop. 1.23], as in the second diagram (from the
ἐστίν.καὶἡἐντῷΑΕΒἄραἴσηἐστὶτῇπρὸςτῷΓ. left). And let AB have been cut in half at F [Prop. 1.10].
ΓέγραπταιἄραπάλινἐπὶτῆςΑΒτμῆμακύκλουτὸΑΕΒ And let the circle AEB have been drawn with center F ,
δεχόμενονγωνίανἴσηντῇπρὸςτῷΓ.
and radius either FA or FB.
ἈλλὰδὴἡπρὸςτῷΓἀμβλεῖαἔστω· καὶσυνεστάτω Thus, the straight-line AD touches the circle ABE, on
αὐτῇἴσηπρὸςτῇΑΒεὐθείᾳκαὶτῷΑσημείῳἡὑπὸΒΑΔ, account of the angle at A being a right-angle [Prop. 3.16
ὡςἔχειἐπὶτῆςτρίτηςκαταγραφῆς,καὶτῇΑΔπρὸςὀρθὰς corr.]. And angle BAD is equal to the angle in segment
ἤχθωἡΑΕ,καὶτετμήσθωπάλινἡΑΒδίχακατὰτὸΖ,καὶ AEB. For (the latter angle), being in a semi-circle, is also
τῇΑΒπρὸςὀρθὰςἤχθωἡΖΗ,καὶἐπεζεύχθωἡΗΒ. a right-angle [Prop. 3.31]. But, BAD is also equal to C.
ΚαὶἐπεὶπάλινἴσηἐστὶνἡΑΖτῇΖΒ,καὶκοινὴἡΖΗ, Thus, the (angle) in (segment) AEB is also equal to C.
δύοδὴαἱΑΖ,ΖΗδύοταῖςΒΖ,ΖΗἴσαιεἰσίν·καὶγωνίαἡ Thus, a segment AEB of a circle, accepting an angle
ὑπὸΑΖΗγωνίᾳτῇὑπὸΒΖΗἴση·βάσιςἄραἡΑΗβάσει equal to C, has again been drawn on AB.
τῇΒΗἴσηἐστίν·ὁἄρακέντρῳμὲντῷΗδιαστήματιδὲτῷ And so let (angle) C be obtuse. And let (angle) BAD,
ΗΑκύκλοςγραφόμενοςἥξεικαὶδιὰτοῦΒ.ἐρχέσθωὡςὁ equal to (C), have been constructed on the straight-line
ΑΕΒ.καὶἐπεὶτῇΑΕδιαμέτρῳἀπ᾿ἄκραςπρὸςὀρθάςἐστιν AB, at the point A (on it) [Prop. 1.23], as in the third
ἡΑΔ,ἡΑΔἄραἐφάπτεταιτοῦΑΕΒκύκλου.καὶἀπὸτῆς diagram (from the left). And let AE have been drawn, at
κατὰτὸΑἐπαφῆςδιῆκταιἡΑΒ·ἡἄραὑπὸΒΑΔγωνία right-angles to AD [Prop. 1.11]. And let AB have again
ἴσηἐστὶτῇἐντῷἐναλλὰξτοῦκύκλουτμήματιτῷΑΘΒ been cut in half at F [Prop. 1.10]. And let FG have been
συνισταμένῃγωνίᾳ. ἀλλ᾿ἡὑπὸΒΑΔγωνίατῇπρὸςτῷΓ drawn, at right-angles to AB [Prop. 1.10]. And let GB
ἴσηἐστίν.καὶἡἐντῷΑΘΒἄρατμήματιγωνίαἴσηἐστὶτῇ have been joined.
πρὸςτῷΓ.
And again, since AF is equal to FB, and FG (is)
᾿ΕπὶτῆςἄραδοθείσηςεὐθείαςτῆςΑΒγέγραπταιτμῆμα common, the two (straight-lines) AF , FG are equal to
κύκλουτὸΑΘΒδεχόμενονγωνίανἴσηντῇπρὸςτῷΓ·ὅπερ the two (straight-lines) BF , FG (respectively). And an-
ἔδειποιῆσαι.
gle AFG (is) equal to angle BFG. Thus, the base AG is
102

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
equal to the base BG [Prop. 1.4]. Thus, a circle of center
G, and radius GA, being drawn, will also go through B
(as well as A). Let it go like AEB (in the third diagram
from the left). And since AD is at right-angles to the diameter
AE, at its extremity, AD thus touches circle AEB
[Prop. 3.16 corr.]. And AB has been drawn across (the
circle) from the point of contact A. Thus, angle BAD is
equal to the angle constructed in the alternate segment
AHB of the circle [Prop. 3.32]. But, angle BAD is equal
to C. Thus, the angle in segment AHB is also equal to
C.
Thus, a segment AHB of a circle, accepting an angle
equal to C, has been drawn on the given straight-line AB.
(Which is) the very thing it was required to do.
Ð. Proposition 34
Ἀπὸ τοῦδοθέντοςκύκλουτμῆμαἀφελεῖνδεχόμενον To cut off a segment, accepting an angle equal to a
γωνίανἴσηντῇδοθείσῃγωνίᾳεὐθυγράμμῳ.
given rectilinear angle, from a given circle.
Γ
C
Ζ
F
Β
B
∆
Ε Α
E
A
῎Εστω ὁδοθεὶςκύκλοςὁΑΒΓ,ἡδὲδοθεῖσαγωνία Let ABC be the given circle, and D the given rectilinεὐθύγραμμοςἡπρὸςτῷΔ·δεῖδὴἀπὸτοῦΑΒΓκύκλου
ear angle. So it is required to cut off a segment, accepting
τμῆμαἀφελεῖνδεχόμενονγωνίανἴσηντῇδοθείσῃγωνίᾳ an angle equal to the given rectilinear angle D, from the
εὐθυγράμμῳτῇπρὸςτῷΔ.
given circle ABC.
῎ΗχθωτοῦΑΒΓἐφαπτομένηἡΕΖκατὰτὸΒσημεῖον, Let EF have been drawn touching ABC at point B. †
καὶσυνεστάτωπρὸςτῇΖΒεὐθείᾳκαὶτῷπρὸςαὐτῇσημείῳ And let (angle) FBC, equal to angle D, have been conτῷΒτῇπρὸςτῷΔγωνίᾳἴσηἡὑπὸΖΒΓ.
structed on the straight-line FB, at the point B on it
᾿ΕπεὶοὖνκύκλουτοῦΑΒΓἐφάπτεταίτιςεὐθεῖαἡΕΖ, [Prop. 1.23].
καὶἀπὸτῆςκατὰτὸΒἐπαφῆςδιῆκταιἡΒΓ,ἡὑπὸΖΒΓἄρα Therefore, since some straight-line EF touches the
γωνίαἴσηἐστὶτῇἐντῷΒΑΓἐναλλὰξτμήματισυνισταμένῃ circle ABC, and BC has been drawn across (the circle)
γωνίᾳ. ἀλλ᾿ἡὑπὸΖΒΓτῇπρὸςτῷΔἐστινἴση·καὶἡἐν from the point of contact B, angle FBC is thus equal
τῷΒΑΓἄρατμήματιἴσηἐστὶτῇπρὸςτῷΔ[γωνίᾳ]. to the angle constructed in the alternate segment BAC
ἈπὸτοῦδοθέντοςἄρακύκλουτοῦΑΒΓτμῆμαἀφῄρηται [Prop. 1.32]. But, FBC is equal to D. Thus, the (angle)
τὸΒΑΓδεχόμενονγωνίανἴσηντῇδοθείσῃγωνίᾳεὐθυγράμ- in the segment BAC is also equal to [angle] D.
μῳτῇπρὸςτῷΔ·ὅπερἔδειποιῆσαι.
Thus, the segment BAC, accepting an angle equal to
the given rectilinear angle D, has been cut off from the
given circle ABC. (Which is) the very thing it was required
to do.
† Presumably, by finding the center of ABC [Prop. 3.1], drawing a straight-line between the center and point B, and then drawing EF through
D
103

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
point B, at right-angles to the aforementioned straight-line [Prop. 1.11].
Ð. Proposition 35
᾿Εὰνἐνκύκλῳδύοεὐθεῖαιτέμνωσινἀλλήλας,τὸὑπὸ If two straight-lines in a circle cut one another then
τῶντῆςμιᾶςτμημάτωνπεριεχόμενονὀρθογώνιονἴσονἐστὶ the rectangle contained by the pieces of one is equal to
τῷὑπὸτῶντῆςἑτέραςτμημάτωνπεριεχομένῳὀρθογωνίῳ. the rectangle contained by the pieces of the other.
Α
A
∆
D
Α
A
Β
Ε
∆
Η
Ζ
Θ
Ε
E
Γ
Β Γ
C
B C
᾿Εν γὰρ κύκλῳ τῷ ΑΒΓΔ δύο εὐθεῖαι αἱ ΑΓ, ΒΔ For let the two straight-lines AC and BD, in the circle
τεμνέτωσανἀλλήλαςκατὰτὸΕσημεῖον·λέγω,ὅτιτὸὑπὸ ABCD, cut one another at point E. I say that the rectτῶνΑΕ,ΕΓπεριεχόμενονὀρθογώνιονἴσονἐστὶτῷὑπὸ
angle contained by AE and EC is equal to the rectangle
τῶνΔΕ,ΕΒπεριεχομένῳὀρθογωνίῳ.
contained by DE and EB.
ΕἰμὲνοὖναἱΑΓ,ΒΔδιὰτοῦκέντρουεἰσὶνὥστετὸΕ In fact, if AC and BD are through the center (as in
κέντρονεἶναιτοῦΑΒΓΔκύκλου,φανερόν,ὅτιἴσωνοὐσῶν the first diagram from the left), so that E is the center of
τῶνΑΕ,ΕΓ,ΔΕ,ΕΒκαὶτὸὑπὸτῶνΑΕ,ΕΓπεριεχόμενον circle ABCD, then (it is) clear that, AE, EC, DE, and
ὀρθογώνιονἴσονἐστὶτῷὑπὸτῶνΔΕ,ΕΒπεριεχομένῳ EB being equal, the rectangle contained by AE and EC
ὀρθογωνίῳ.
is also equal to the rectangle contained by DE and EB.
Μὴ ἔστωσαν δὴ αἱ ΑΓ, ΔΒ διὰ τοῦ κέντρου, καὶ So let AC and DB not be though the center (as in
εἰλήφθωτὸκέντροντοῦΑΒΓΔ,καὶἔστωτὸΖ,καὶἀπὸ the second diagram from the left), and let the center of
τοῦΖἐπὶτὰςΑΓ,ΔΒεὐθείαςκάθετοιἤχθωσαναἱΖΗ, ABCD have been found [Prop. 3.1], and let it be (at) F .
ΖΘ,καὶἐπεζεύχθωσαναἱΖΒ,ΖΓ,ΖΕ.
And let FG and FH have been drawn from F , perpen-
ΚαὶἐπεὶεὐθεῖάτιςδιὰτοῦκέντρουἡΗΖεὐθεῖάντινα dicular to the straight-lines AC and DB (respectively)
μὴδιὰτοῦκέντρουτὴνΑΓπρὸςὀρθὰςτέμνει,καὶδίχα [Prop. 1.12]. And let FB, FC, and FE have been joined.
αὐτὴντέμνει·ἴσηἄραἡΑΗτῇΗΓ.ἐπεὶοὖνεὐθεῖαἡΑΓ And since some straight-line, GF , through the center,
τέτμηταιεἰςμὲνἴσακατὰτὸΗ,εἰςδὲἄνισακατὰτὸΕ,τὸ cuts at right-angles some (other) straight-line, AC, not
ἄραὑπὸτῶνΑΕ,ΕΓπεριεχόμενονὀρθογώνιονμετὰτοῦ through the center, then it also cuts it in half [Prop. 3.3].
ἀπὸτῆςΕΗτετραγώνουἴσονἐστὶτῷἀπὸτῆςΗΓ·[κοινὸν] Thus, AG (is) equal to GC. Therefore, since the straightπροσκείσθωτὸἀπὸτῆςΗΖ·τὸἄραὑπὸτῶνΑΕ,ΕΓμετὰ
line AC is cut equally at G, and unequally at E, the
τῶνἀπὸτῶνΗΕ,ΗΖἴσονἐστὶτοῖςἀπὸτῶνΓΗ,ΗΖ.ἀλλὰ rectangle contained by AE and EC plus the square on
τοῖςμὲνἀπὸτῶνΕΗ,ΗΖἴσονἐστὶτὸἀπὸτῆςΖΕ,τοὶς EG is thus equal to the (square) on GC [Prop. 2.5]. Let
δὲἀπὸτῶνΓΗ,ΗΖἴσονἐστὶτὸἀπὸτῆςΖΓ·τὸἄραὑπὸ the (square) on GF have been added [to both]. Thus,
τῶνΑΕ,ΕΓμετὰτοῦἀπὸτῆςΖΕἴσονἐστὶτῷἀπὸτῆς the (rectangle contained) by AE and EC plus the (sum
ΖΓ.ἴσηδὲἡΖΓτῇΖΒ·τὸἄραὑπὸτῶνΑΕ,ΕΓμετὰτοῦ of the squares) on GE and GF is equal to the (sum of
ἀπὸτῆςΕΖἴσονἐστὶτῷἀπὸτῆςΖΒ.διὰτὰαὐτὰδὴκαὶ the squares) on CG and GF . But, the (square) on FE
τὸὑπὸτῶνΔΕ,ΕΒμετὰτοῦἀπὸτῆςΖΕἰσονἐστὶτῷἀπὸ is equal to the (sum of the squares) on EG and GF
τῆςΖΒ.ἐδείχθηδὲκαὶτὸὑπὸτῶνΑΕ,ΕΓμετὰτοῦἀπὸ [Prop. 1.47], and the (square) on FC is equal to the (sum
τῆςΖΕἴσοντῷἀπὸτῆςΖΒ·τὸἄραὑπὸτῶνΑΕ,ΕΓμετὰ of the squares) on CG and GF [Prop. 1.47]. Thus, the
τοῦἀπὸτῆςΖΕἴσονἐστὶτῷὑπὸτῶνΔΕ,ΕΒμετὰτοῦ (rectangle contained) by AE and EC plus the (square)
ἀπὸτῆςΖΕ.κοινὸνἀφῇρήσθωτὸἀπὸτῆςΖΕ·λοιπὸνἄρα on FE is equal to the (square) on FC. And FC (is)
τὸὑπὸτῶνΑΕ,ΕΓπεριεχόμενονὀρθογώνιονἴσονἐστὶτῷ equal to FB. Thus, the (rectangle contained) by AE
ὑπὸτῶνΔΕ,ΕΒπεριεχομένῳὀρθογωνίῳ.
and EC plus the (square) on FE is equal to the (square)
᾿Εὰνἄραἐνκύκλῳεὐθεῖαιδύοτέμνωσινἀλλήλας,τὸ on FB. So, for the same (reasons), the (rectangle con-
ὑπὸτῶντῆςμιᾶςτμημάτωνπεριεχόμενονὀρθογώνιονἴσον tained) by DE and EB plus the (square) on FE is equal
B
E
D
G
F
H
104

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
ἐστὶτῷὑπὸτῶντῆςἑτέραςτμημάτωνπεριεχομένῳὀρθο- to the (square) on FB. And the (rectangle contained)
γωνίῳ·ὅπερἔδειδεῖξαι.
by AE and EC plus the (square) on FE was also shown
(to be) equal to the (square) on FB. Thus, the (rectangle
contained) by AE and EC plus the (square) on
FE is equal to the (rectangle contained) by DE and EB
plus the (square) on FE. Let the (square) on FE have
been taken from both. Thus, the remaining rectangle contained
by AE and EC is equal to the rectangle contained
by DE and EB.
Thus, if two straight-lines in a circle cut one another
then the rectangle contained by the pieces of one is equal
to the rectangle contained by the pieces of the other.
(Which is) the very thing it was required to show.
Ц. Proposition 36
᾿Εὰν κύκλουληφθῇ τι σημεῖονἐκτός, καὶἀπ᾿ αὐτοῦ If some point is taken outside a circle, and two
πρὸςτὸνκύκλονπροσπίπτωσιδύοεὐθεῖαι,καὶἡμὲναὐτῶν straight-lines radiate from it towards the circle, and (one)
τέμνῃτὸνκύκλον,ἡδὲἐφάπτηται,ἔσταιτὸὑπὸὅληςτῆς of them cuts the circle, and the (other) touches (it), then
τεμνούσηςκαὶτῆςἐκτὸςἀπολαμβανομένηςμεταξὺτοῦτε the (rectangle contained) by the whole (straight-line)
σημείουκαὶτῆςκυρτῆςπεριφερείαςἴσοντῷἀπὸτῆςἐφα- cutting (the circle), and the (part of it) cut off outside
πτομένηςτετραγώνῳ.
(the circle), between the point and the convex circumference,
will be equal to the square on the tangent (line).
Α
A
Β
Γ
Ζ
∆
Γ
Ε
Β
Ζ
Α
B
C
F
D
C
E
B
F
A
∆
D
ΚύκλουγὰρτοῦΑΒΓεἰλήφθωτισημεῖονἐκτὸςτὸΔ, For let some point D have been taken outside circle
καὶἀπὸτοῦΔπρὸςτὸνΑΒΓκύκλονπροσπιπτέτωσανδύο ABC, and let two straight-lines, DC[A] and DB, radiεὐθεῖαιαἱΔΓ[Α],ΔΒ·καὶἡμὲνΔΓΑτεμνέτωτὸνΑΒΓ
ate from D towards circle ABC. And let DCA cut circle
κύκλον,ἡδὲΒΔἐφαπτέσθω·λέγω,ὅτιτὸὑπὸτῶνΑΔ, ABC, and let BD touch (it). I say that the rectangle
ΔΓπεριεχόμενονὀρθογώνιονἴσονἐστὶτῷἀπὸτῆςΔΒ contained by AD and DC is equal to the square on DB.
τετραγώνῳ.
[D]CA is surely either through the center, or not. Let
῾Ηἄρα[Δ]ΓΑἤτοιδιὰτοῦκέντρουἐστὶνἢοὔ. ἔστω it first of all be through the center, and let F be the cenπρότερονδιὰτοῦκέντρου,καὶἔστωτὸΖκέντροντοῦΑΒΓ
ter of circle ABC, and let FB have been joined. Thus,
κύκλου,καὶἐπεζεύχθωἡΖΒ·ὀρθὴἄραἐστὶνἡὑπὸΖΒΔ. (angle) FBD is a right-angle [Prop. 3.18]. And since
καὶἐπεὶεὐθεῖαἡΑΓδίχατέτμηταικατὰτὸΖ,πρόσκειται straight-line AC is cut in half at F , let CD have been
δὲαὐτῇἡΓΔ,τὸἄραὑπὸτῶνΑΔ,ΔΓμετὰτοῦἀπὸτῆς added to it. Thus, the (rectangle contained) by AD and
ΖΓἴσονἐστὶτῷἀπὸτῆςΖΔ.ἴσηδὲἡΖΓτῇΖΒ·τὸἄρα DC plus the (square) on FC is equal to the (square) on
ὑπὸτῶνΑΔ,ΔΓμετὰτοῦἀπὸτῆςΖΒἴσονἐστὶτῷἀπὸ FD [Prop. 2.6]. And FC (is) equal to FB. Thus, the
τὴςΖΔ.τῷδὲἀπὸτῆςΖΔἴσαἐστὶτὰἀπὸτῶνΖΒ,ΒΔ· (rectangle contained) by AD and DC plus the (square)
τὸἄραὑπὸτῶνΑΔ,ΔΓμετὰτοῦἀπὸτῆςΖΒἴσονἐστὶ on FB is equal to the (square) on FD. And the (square)
τοῖςἀπὸτῶνΖΒ,ΒΔ.κοινὸνἀφῃρήσθωτὸἀπὸτῆςΖΒ· on FD is equal to the (sum of the squares) on FB and
λοιπὸνἄρατὸὑπὸτῶνΑΔ,ΔΓἴσονἐστὶτῷἀπὸτῆςΔΒ BD [Prop. 1.47]. Thus, the (rectangle contained) by AD
105

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
ἐφαπτομένης.
and DC plus the (square) on FB is equal to the (sum
ἈλλὰδὴἡΔΓΑμὴἔστωδιὰτοῦκέντρουτοῦΑΒΓ of the squares) on FB and BD. Let the (square) on
κύκλου,καὶεἰλήφθωτὸκέντροντὸΕ,καὶἀπὸτοῦΕἐπὶ FB have been subtracted from both. Thus, the remainτὴνΑΓκάθετοςἤχθωἡΕΖ,καὶἐπεζεύχθωσαναἱΕΒ,ΕΓ,
ing (rectangle contained) by AD and DC is equal to the
ΕΔ·ὀρθὴἄραἐστὶνἡὑπὸΕΒΔ.καὶἐπεὶεὐθεῖάτιςδιὰτοῦ (square) on the tangent DB.
κέντρουἡΕΖεὐθεῖάντιναμὴδιὰτοῦκέντρουτὴνΑΓπρὸς And so let DCA not be through the center of cir-
ὀρθὰςτέμνει,καὶδίχααὐτὴντέμνει·ἡΑΖἄρατῇΖΓἐστιν cle ABC, and let the center E have been found, and
ἴση.καὶἐπεὶεὐθεῖαἡΑΓτέτμηταιδίχακατὰτὸΖσημεῖον, let EF have been drawn from E, perpendicular to AC
πρόσκειταιδὲαὐτῇἡΓΔ,τὸἄραὑπὸτῶνΑΔ,ΔΓμετὰτοῦ [Prop. 1.12]. And let EB, EC, and ED have been joined.
ἀπὸτῆςΖΓἴσονἐστὶτῷἀπὸτῆςΖΔ.κοινὸνπροσκείσθω (Angle) EBD (is) thus a right-angle [Prop. 3.18]. And
τὸἀπὸτῆςΖΕ·τὸἄραὑπὸτῶνΑΔ,ΔΓμετὰτῶνἀπὸτῶν since some straight-line, EF , through the center, cuts
ΓΖ,ΖΕἴσονἐστὶτοῖςἀπὸτῶνΖΔ,ΖΕ.τοῖςδὲἀπὸτῶν some (other) straight-line, AC, not through the center,
ΓΖ,ΖΕἴσονἐστὶτὸἀπὸτῆςΕΓ·ὀρθὴγὰρ[ἐστιν]ἡὑπὸ at right-angles, it also cuts it in half [Prop. 3.3]. Thus,
ΕΖΓ[γωνία]·τοῖςδὲἀπὸτῶνΔΖ,ΖΕἴσονἐστὶτὸἀπὸτῆς AF is equal to FC. And since the straight-line AC is cut
ΕΔ·τὸἄραὑπὸτῶνΑΔ,ΔΓμετὰτοῦἀπὸτῆςΕΓἴσον in half at point F , let CD have been added to it. Thus, the
ἐστὶτῷἀπὸτῆςΕΔ.ἴσηδὲἡΕΓτῂΕΒ·τὸἄραὑπὸτῶν (rectangle contained) by AD and DC plus the (square)
ΑΔ,ΔΓμετὰτοῦἀπὸτῆςΕΒἴσονἐστὶτῷἄπὸτῆςΕΔ. on FC is equal to the (square) on FD [Prop. 2.6]. Let
τῷδὲἀπὸτῆςΕΔἴσαἐστὶτὰἀπὸτῶνΕΒ,ΒΔ·ὀρθὴγὰρ the (square) on FE have been added to both. Thus, the
ἡὑπὸΕΒΔγωνία·τὸἄραὑπὸτῶνΑΔ,ΔΓμετὰτοῦἀπὸ (rectangle contained) by AD and DC plus the (sum of
τῆςΕΒἴσονἐστὶτοῖςἀπὸτῶνΕΒ,ΒΔ.κοινὸνἀφῃρήσθω the squares) on CF and FE is equal to the (sum of the
τὸἀπὸτῆςΕΒ·λοιπὸνἄρατὸὑπὸτῶνΑΔ,ΔΓἴσονἐστὶ squares) on FD and FE. But the (square) on EC is equal
τῷἀπὸτῆςΔΒ.
to the (sum of the squares) on CF and FE. For [angle]
᾿Εὰνἄρακύκλουληφθῇτισημεῖονἐκτός,καὶἀπ᾿αὐτοῦ EFC [is] a right-angle [Prop. 1.47]. And the (square)
πρὸςτὸνκύκλονπροσπίπτωσιδύοεὐθεῖαι,καὶἡμὲναὐτῶν on ED is equal to the (sum of the squares) on DF and
τέμνῃτὸνκύκλον,ἡδὲἐφάπτηται,ἔσταιτὸὑπὸὅληςτῆς FE [Prop. 1.47]. Thus, the (rectangle contained) by AD
τεμνούσηςκαὶτῆςἐκτὸςἀπολαμβανομένηςμεταξὺτοῦτε and DC plus the (square) on EC is equal to the (square)
σημείουκαὶτῆςκυρτῆςπεριφερείαςἴσοντῷἀπὸτῆςἐφα- on ED. And EC (is) equal to EB. Thus, the (rectanπτομένηςτετραγώνῳ·ὅπερἔδειδεῖξαι.
gle contained) by AD and DC plus the (square) on EB
is equal to the (square) on ED. And the (sum of the
squares) on EB and BD is equal to the (square) on ED.
For EBD (is) a right-angle [Prop. 1.47]. Thus, the (rectangle
contained) by AD and DC plus the (square) on
EB is equal to the (sum of the squares) on EB and BD.
Let the (square) on EB have been subtracted from both.
Thus, the remaining (rectangle contained) by AD and
DC is equal to the (square) on BD.
Thus, if some point is taken outside a circle, and two
straight-lines radiate from it towards the circle, and (one)
of them cuts the circle, and (the other) touches (it), then
the (rectangle contained) by the whole (straight-line)
cutting (the circle), and the (part of it) cut off outside
(the circle), between the point and the convex circumference,
will be equal to the square on the tangent (line).
(Which is) the very thing it was required to show.
ÐÞ. Proposition 37
᾿Εὰν κύκλου ληφθῇ τι σημεῖον ἐκτός, ἀπὸ δὲ τοῦ If some point is taken outside a circle, and two
σημείου πρὸς τὸν κύκλον προσπίπτωσι δύο εὐθεῖαι, καὶ straight-lines radiate from the point towards the circle,
ἡμὲναὐτῶντέμνῃτὸνκύκλον,ἡδὲπροσπίπτῃ,ᾖδὲτὸ and one of them cuts the circle, and the (other) meets
ὑπὸ [τῆς] ὅληςτῆς τεμνούσηςκαὶ τῆς ἐκτὸς ἀπολαμβα- (it), and the (rectangle contained) by the whole (straight-
106

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
νομένηςμεταξὺτοῦτεσημείουκαὶτῆςκυρτῆςπεριφερείας line) cutting (the circle), and the (part of it) cut off out-
ἴσοντῷἀπὸτῆςπροσπιπτούσης,ἡπροσπίπτουσαἐφάψεται side (the circle), between the point and the convex cirτοῦκύκλου.
cumference, is equal to the (square) on the (straight-line)
meeting (the circle), then the (straight-line) meeting (the
circle) will touch the circle.
∆
Ε
D
E
Γ
Ζ
C
F
Β
Α
ΚύκλουγὰρτοῦΑΒΓεἰλήφθωτισημεῖονἐκτὸςτὸΔ, For let some point D have been taken outside circle
καὶἀπὸτοῦΔπρὸςτὸνΑΒΓκύκλονπροσπιπτέτωσανδύο ABC, and let two straight-lines, DCA and DB, radiate
εὐθεῖαιαἱΔΓΑ,ΔΒ,καὶἡμὲνΔΓΑτεμνέτωτὸνκύκλον,ἡ from D towards circle ABC, and let DCA cut the circle,
δὲΔΒπροσπιπτέτω,ἔστωδὲτὸὑπὸτῶνΑΔ,ΔΓἴσοντῷ and let DB meet (the circle). And let the (rectangle con-
ἀπὸτῆςΔΒ.λέγω,ὅτιἡΔΒἐφάπτεταιτοῦΑΒΓκύκλου. tained) by AD and DC be equal to the (square) on DB.
῎ΗχθωγὰρτοῦΑΒΓἐφαπτομένηἡΔΕ,καὶεἰλήφθωτὸ I say that DB touches circle ABC.
κέντροντοῦΑΒΓκύκλου,καὶἔστωτὸΖ,καὶἐπεζεύχθωσαν For let DE have been drawn touching ABC [Prop.
αἱΖΕ,ΖΒ,ΖΔ.ἡἄραὑπὸΖΕΔὀρθήἐστιν.καὶἐπεὶἡΔΕ 3.17], and let the center of the circle ABC have been
ἐφάπτεταιτοῦΑΒΓκύκλου,τέμνειδὲἡΔΓΑ,τὸἄραὑπὸ found, and let it be (at) F . And let FE, FB, and FD
τῶνΑΔ,ΔΓἴσονἐστὶτῷἀπὸτῆςΔΕ.ἦνδὲκαὶτὸὑπὸ have been joined. (Angle) FED is thus a right-angle
τῶνΑΔ,ΔΓἴσοντῷἀπὸτῆςΔΒ·τὸἄραἀπὸτῆςΔΕ [Prop. 3.18]. And since DE touches circle ABC, and
ἴσονἐστὶτῷἀπὸτῆςΔΒ·ἴσηἄραἡΔΕτῇΔΒ.ἐστὶδὲ DCA cuts (it), the (rectangle contained) by AD and DC
καὶἡΖΕτῇΖΒἴση·δύοδὴαἱΔΕ,ΕΖδύοταῖςΔΒ,ΒΖ is thus equal to the (square) on DE [Prop. 3.36]. And the
ἴσαιεἰσίν·καὶβάσιςαὐτῶνκοινὴἡΖΔ·γωνίαἄραἡὑπὸ (rectangle contained) by AD and DC was also equal to
ΔΕΖγωνίᾳτῇὑπὸΔΒΖἐστινἴση. ὀρθὴδὲἡὑπὸΔΕΖ· the (square) on DB. Thus, the (square) on DE is equal
ὀρθὴἄρακαὶἡὑπὸΔΒΖ.καίἐστινἡΖΒἐκβαλλομένη to the (square) on DB. Thus, DE (is) equal to DB. And
διάμετρος·ἡδὲτῇδιαμέτρῳτοῦκύκλουπρὸςὀρθὰςἀπ᾿ FE is also equal to FB. So the two (straight-lines) DE,
ἄκραςἀγομένηἐφάπτεταιτοῦκύκλου·ἡΔΒἄραἐφάπτεται EF are equal to the two (straight-lines) DB, BF (reτοῦΑΒΓκύκλου.ὁμοίωςδὴδειχθήσεται,κἂντὸκέντρον
spectively). And their base, FD, is common. Thus, angle
ἐπὶτῆςΑΓτυγχάνῃ.
DEF is equal to angle DBF [Prop. 1.8]. And DEF (is)
᾿Εὰνἄρακύκλουληφθῇτισημεῖονἐκτός,ἀπὸδὲτοῦ a right-angle. Thus, DBF (is) also a right-angle. And
σημείουπρὸςτὸνκύκλονπροσπίπτωσιδύοεὐθεῖαι,καὶἡ FB produced is a diameter, And a (straight-line) drawn
μὲναὐτῶντέμνῃτὸνκύκλον,ἡδὲπροσπίπτῃ,ᾖδὲτὸὑπὸ at right-angles to a diameter of a circle, at its extremity,
ὅληςτῆςτεμνούσηςκαὶτῆςἐκτὸςἀπολαμβανομένηςμεταξὺ touches the circle [Prop. 3.16 corr.]. Thus, DB touches
τοῦτεσημείουκαὶτῆςκυρτῆςπεριφερείαςἴσοντῷἀπὸ circle ABC. Similarly, (the same thing) can be shown,
τῆςπροσπιπτούσης,ἡπροσπίπτουσαἐφάψεταιτοῦκύκλου· even if the center happens to be on AC.
ὅπερἔδειδεῖξαι.
Thus, if some point is taken outside a circle, and two
straight-lines radiate from the point towards the circle,
and one of them cuts the circle, and the (other) meets
(it), and the (rectangle contained) by the whole (straightline)
cutting (the circle), and the (part of it) cut off outside
(the circle), between the point and the convex circumference,
is equal to the (square) on the (straight-line)
meeting (the circle), then the (straight-line) meeting (the
circle) will touch the circle. (Which is) the very thing it
B
A
107

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 3
was required to show.
108

ELEMENTS BOOK 4
Construction of Rectilinear Figures In and
Around Circles
109

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 4
ÎÇÖÓ.
αʹ.Σχῆμαεὐθύγραμμονεἰςσχῆμαεὐθύγραμμονἐγγράφ- 1. A rectilinear figure is said to be inscribed in
Definitions
εσθαιλέγεται,ὅτανἑκάστητῶντοῦἐγγραφομένουσχήματ- a(nother) rectilinear figure when the respective angles
οςγωνιῶνἑκάστηςπλευρᾶςτοῦ,εἰςὃἐγγράφεται,ἅπτηται. of the inscribed figure touch the respective sides of the
βʹ.Σχῆμαδὲὁμοίωςπερὶσχῆμαπεριγράφεσθαιλέγεται, (figure) in which it is inscribed.
ὅτανἑκάστηπλευρὰτοῦπεριγραφομένουἑκάστηςγωνίας 2. And, similarly, a (rectilinear) figure is said to be cirτοῦ,περὶὃπεριγράφεται,ἅπτηται.
cumscribed about a(nother rectilinear) figure when the
γʹ.Σχῆμαεὐθύγραμμονεἰςκύκλονἐγγράφεσθαιλέγεται, respective sides of the circumscribed (figure) touch the
ὅταν ἑκάστη γωνία τοῦ ἐγγραφομένου ἅπτηται τῆς τοῦ respective angles of the (figure) about which it is circumκύκλουπεριφερείας.
scribed.
δʹ. Σχῆμα δὲ εὐθύγραμμον περὶ κύκλον περιγράφε- 3. A rectilinear figure is said to be inscribed in a cirσθαι
λέγεται, ὅταν ἑκάστη πλευρὰ τοῦ περιγραφομένου cle when each angle of the inscribed (figure) touches the
ἐφάπτηταιτῆςτοῦκύκλουπεριφερείας.
circumference of the circle.
εʹ.Κύκλοςδὲεἰςσχῆμαὁμοίωςἐγγράφεσθαιλέγεται, 4. And a rectilinear figure is said to be circumscribed
ὅτανἡτοῦκύκλουπεριφέρειαἑκάστηςπλευρᾶςτοῦ,εἰςὃ about a circle when each side of the circumscribed (fig-
ἐγγράφεται,ἅπτηται.
ure) touches the circumference of the circle.
ϛʹ.Κύκλοςδὲπερὶσχῆμαπεριγράφεσθαιλέγεται,ὅταν 5. And, similarly, a circle is said to be inscribed in a
ἡτοῦκύκλουπεριφέρειαἑκάστηςγωνίαςτοῦ,περὶὃπε- (rectilinear) figure when the circumference of the circle
ριγράφεται,ἅπτηται.
touches each side of the (figure) in which it is inscribed.
ζʹ.Εὐθεῖαεἰςκύκλονἐναρμόζεσθαιλέγεται,ὅταντὰ 6. And a circle is said to be circumscribed about a
πέρατααὐτῆςἐπὶτῆςπεριφερείαςᾖτοῦκύκλου. rectilinear (figure) when the circumference of the circle
touches each angle of the (figure) about which it is circumscribed.
7. A straight-line is said to be inserted into a circle
when its extemities are on the circumference of the circle.
. Proposition 1
Εἰςτὸνδοθέντακύκλοντῇδοθείσῃεὐθείᾳμὴμείζονι To insert a straight-line equal to a given straight-line
οὔσῃτῆςτοῦκύκλουδιαμέτρουἴσηνεὐθεῖανἐναρμόσαι. into a circle, (the latter straight-line) not being greater
than the diameter of the circle.
∆
D
Α
A
Β
Ε
Γ
B
E
C
Ζ
F
῎ΕστωὁδοθεὶςκύκλοςὁΑΒΓ,ἡδὲδοθεῖσαεὐθεῖαμὴ Let ABC be the given circle, and D the given straightμείζωντῆςτοῦκύκλουδιαμέτρουἡΔ.δεῖδὴεἰςτὸνΑΒΓ
line (which is) not greater than the diameter of the cirκύκλοντῇΔεὐθείᾳἴσηνεὐθεῖανἐναρμόσαι.
cle. So it is required to insert a straight-line, equal to the
῎ΗχθωτοῦΑΒΓκύκλουδιάμετροςἡΒΓ.εἰμὲνοὖνἴση straight-line D, into the circle ABC.
ἐστὶνἡΒΓτῇΔ,γεγονὸςἂνεἴητὸἐπιταχθέν·ἐνήρμοσται Let a diameter BC of circle ABC have been drawn. †
110

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 4
γὰρεἰςτὸνΑΒΓκύκλοντῇΔεὐθείᾳἴσηἡΒΓ.εἰδὲμείζων Therefore, if BC is equal to D then that (which) was
ἐστὶνἡΒΓτῆςΔ,κείσθωτῇΔἴσηἡΓΕ,καὶκέντρῳ prescribed has taken place. For the (straight-line) BC,
τῷΓδιαστήματιδὲτῷΓΕκύκλοςγεγράφθωὁΕΑΖ,καὶ equal to the straight-line D, has been inserted into the
ἐπεζεύχθωἡΓΑ.
circle ABC. And if BC is greater than D then let CE be
᾿ΕπεὶοὖντοΓσημεῖονκέντρονἐστὶτοῦΕΑΖκύκλου, made equal to D [Prop. 1.3], and let the circle EAF have
ἴσηἐστὶνἡΓΑτῇΓΕ.ἀλλὰτῇΔἡΓΕἐστινἴση·καὶἡΔ been drawn with center C and radius CE. And let CA
ἄρατῇΓΑἐστινἴση.
have been joined.
ΕἰςἄρατὸνδοθέντακύκλοντὸνΑΒΓτῇδοθείσῃεὐθείᾳ Therefore, since the point C is the center of circle
τῇΔἴσηἐνήρμοσταιἡΓΑ·ὅπερἔδειποιῆσαι. EAF , CA is equal to CE. But, CE is equal to D. Thus,
D is also equal to CA.
Thus, CA, equal to the given straight-line D, has been
inserted into the given circle ABC. (Which is) the very
thing it was required to do.
† Presumably, by finding the center of the circle [Prop. 3.1], and then drawing a line through it.
. Proposition 2
Εἰςτὸνδοθέντακύκλοντῷδοθέντιτριγώνῳἰσογώνιον To inscribe a triangle, equiangular with a given trianτρίγωνονἐγγράψαι.
gle, in a given circle.
Β
Ε
B
E
Η
Γ
∆
Ζ
G
C
D
F
Α
A
Θ
H
῎ΕστωὁδοθεὶςκύκλοςὁΑΒΓ,τὸδὲδοθὲντριγωνον Let ABC be the given circle, and DEF the given triτὸΔΕΖ·δεῖδὴεἰςτὸνΑΒΓκύκλοντῷΔΕΖτριγώνῳ
angle. So it is required to inscribe a triangle, equiangular
ἰσογώνιοντρίγωνονἐγγράψαι.
with triangle DEF , in circle ABC.
῎ΗχθωτοῦΑΒΓκύκλουἐφαπτομένηἡΗΘκατὰτὸΑ, Let GH have been drawn touching circle ABC at A. †
καὶσυνεστάτωπρὸςτῇΑΘεὐθείᾳκαὶτῷπρὸςαὐτῇσημείῳ And let (angle) HAC, equal to angle DEF , have been
τῷΑτῇὑπὸΔΕΖγωνίᾳἴσηἡὑπὸΘΑΓ,πρὸςδὲτῇΑΗ constructed on the straight-line AH at the point A on it,
εὐθείᾳκαὶτῷπρὸςαὐτῇσημείῳτῷΑτῇὑπὸΔΖΕ[γωνίᾳ] and (angle) GAB, equal to [angle] DFE, on the straight-
ἴσηἡὑπὸΗΑΒ,καὶἐπεζεύχθωἡΒΓ.
line AG at the point A on it [Prop. 1.23]. And let BC
᾿ΕπεὶοὖνκύκλουτοῦΑΒΓἐφάπτεταίτιςεὐθεῖαἡΑΘ, have been joined.
καὶἀπὸτῆςκατὰτὸΑἐπαφῆςεἰςτὸνκύκλονδιῆκταιεὐθεῖα Therefore, since some straight-line AH touches the
ἡΑΓ,ἡἄραὑπὸΘΑΓἴσηἐστὶτῇἐντῷἐναλλὰξτοῦκύκλου circle ABC, and the straight-line AC has been drawn
τμήματιγωνίᾳτῇὑπὸΑΒΓ.ἀλλ᾿ἡὑπὸΘΑΓτῇὑπὸΔΕΖ across (the circle) from the point of contact A, (angle)
ἐστινἴση· καὶἡὑπὸΑΒΓἄραγωνίατῇὑπὸΔΕΖἐστιν HAC is thus equal to the angle ABC in the alternate
ἴση. διὰτὰαὐτὰδὴκαὶἡὑπὸΑΓΒτῇὑπὸΔΖΕἐστιν segment of the circle [Prop. 3.32]. But, HAC is equal to
ἴση·καὶλοιπὴἄραἡὑπὸΒΑΓλοιπῇτῇὑπὸΕΔΖἐστινἴση DEF . Thus, angle ABC is also equal to DEF . So, for the
[ἰσογώνιονἄραἐστὶτὸΑΒΓτρίγωνοντῷΔΕΖτριγώνῳ, same (reasons), ACB is also equal to DFE. Thus, the reκαὶἐγγέγραπταιεἰςτὸνΑΒΓκύκλον].
maining (angle) BAC is equal to the remaining (angle)
Εἰς τὸν δοθέντα ἄρα κύκλον τῷ δοθέντι τριγώνῳ EDF [Prop. 1.32]. [Thus, triangle ABC is equiangu-
ἰσογώνιοντρίγωνονἐγγέγραπται·ὅπερἔδειποιῆσαι. lar with triangle DEF , and has been inscribed in circle
111

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 4
ABC].
Thus, a triangle, equiangular with the given triangle,
has been inscribed in the given circle. (Which is) the very
thing it was required to do.
† See the footnote to Prop. 3.34.. Proposition 3
Περὶτὸνδοθέντακύκλοντῷδοθέντιτριγώνῳἰσογώνιον To circumscribe a triangle, equiangular with a given
τρίγωνονπεριγράψαι.
triangle, about a given circle.
Λ
Α
Μ
Κ
Γ
Θ
Β
Ζ
Ν
Ε
Η
∆
A
M
K
H
L C N G
῎ΕστωὁδοθεὶςκύκλοςὁΑΒΓ,τὸδὲδοθὲντρίγωνον Let ABC be the given circle, and DEF the given triτὸΔΕΖ·δεῖδὴπερὶτὸνΑΒΓκύκλοντῷΔΕΖτριγώνῳ
angle. So it is required to circumscribe a triangle, equian-
ἰσογώνιοντρίγωνονπεριγράψαι.
gular with triangle DEF , about circle ABC.
᾿ΕκβεβλήσθωἡΕΖἐφ᾿ἑκάτερατὰμέρηκατὰτὰΗ,Θ Let EF have been produced in each direction to
σημεῖα,καὶεἰλήφθωτοῦΑΒΓκύκλουκέντροντὸΚ,καὶ points G and H. And let the center K of circle ABC
διήχθω,ὡςἔτυχεν,εὐθεῖαἡΚΒ,καὶσυνεστάτωπρὸςτῇ have been found [Prop. 3.1]. And let the straight-line
ΚΒεὐθείᾳκαὶτῷπρὸςαὐτῇσημείῳτῷΚτῇμὲνὑπὸΔΕΗ KB have been drawn, at random, across (ABC). And
γωνίᾳἴσηἡὑπὸΒΚΑ,τῇδὲὑπὸΔΖΘἴσηἡὑπὸΒΚΓ,καὶ let (angle) BKA, equal to angle DEG, have been conδιὰτῶνΑ,Β,ΓσημείωνἤχθωσανἐφαπτόμεναιτοῦΑΒΓ
structed on the straight-line KB at the point K on it,
κύκλουαἱΛΑΜ,ΜΒΝ,ΝΓΛ.
and (angle) BKC, equal to DFH [Prop. 1.23]. And let
ΚαὶἐπεὶἐφάπτονταιτοῦΑΒΓκύκλουαἱΛΜ,ΜΝ,ΝΛ the (straight-lines) LAM, MBN, and NCL have been
κατὰτὰΑ,Β,Γσημεῖα,ἀπὸδὲτοῦΚκέντρουἐπὶτὰΑ,Β, drawn through the points A, B, and C (respectively),
ΓσημεῖαἐπεζευγμέναιεἰσὶναἱΚΑ,ΚΒ,ΚΓ,ὀρθαὶἄραεἰσὶν touching the circle ABC. †
αἱπρὸςτοῖςΑ,Β,Γσημείοιςγωνίαι.καὶἐπεὶτοῦΑΜΒΚ And since LM, MN, and NL touch circle ABC at
τετραπλεύρουαἱτέσσαρεςγωνίαιτέτρασινὀρθαῖςἴσαιεἰσίν, points A, B, and C (respectively), and KA, KB, and
ἐπειδήπερκαὶεἰςδύοτρίγωναδιαιρεῖταιτὸΑΜΒΚ,καίεἰσιν KC are joined from the center K to points A, B, and
ὀρθαὶαἱὑπὸΚΑΜ,ΚΒΜγωνίαι,λοιπαὶἄρααἱὑπὸΑΚΒ, C (respectively), the angles at points A, B, and C are
ΑΜΒδυσὶνὀρθαῖςἴσαιεἰσίν. εἰσὶδὲκαὶαἱὑπὸΔΕΗ, thus right-angles [Prop. 3.18]. And since the (sum of the)
ΔΕΖδυσὶνὀρθαῖςἴσαι·αἱἄραὑπὸΑΚΒ,ΑΜΒταῖςὑπὸ four angles of quadrilateral AMBK is equal to four right-
ΔΕΗ,ΔΕΖἴσαιεἰσίν,ὧνἡὑπὸΑΚΒτῇὑπὸΔΕΗἐστιν angles, inasmuch as AMBK (can) also (be) divided into
ἴση·λοιπὴἄραἡὑπὸΑΜΒλοιπῇτῇὑπὸΔΕΖἐστινἴση. two triangles [Prop. 1.32], and angles KAM and KBM
ὁμοίωςδὴδειχθήσεται,ὅτικαὶἡὑπὸΛΝΒτῇὑπὸΔΖΕ are (both) right-angles, the (sum of the) remaining (an-
ἐστινἴση·καὶλοιπὴἄραἡὑπὸΜΛΝ[λοιπῇ]τῇὑπὸΕΔΖ gles), AKB and AMB, is thus equal to two right-angles.
ἐστινἴση.ἰσογώνιονἄραἐστὶτὸΛΜΝτρίγωνοντῷΔΕΖ And DEG and DEF is also equal to two right-angles
τριγώνῳ·καὶπεριγέγραπταιπερὶτὸνΑΒΓκύκλον. [Prop. 1.13]. Thus, AKB and AMB is equal to DEG
Περὶ τὸν δοθέντα ἄρα κύκλον τῷ δοθέντι τριγώνῳ and DEF , of which AKB is equal to DEG. Thus, the re-
ἰσογώνιοντρίγωνονπεριγέγραπται·ὅπερἔδειποιῆσαι. mainder AMB is equal to the remainder DEF . So, similarly,
it can be shown that LNB is also equal to DFE.
Thus, the remaining (angle) MLN is also equal to the
B
F
E
D
112

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 4
[remaining] (angle) EDF [Prop. 1.32]. Thus, triangle
LMN is equiangular with triangle DEF . And it has been
drawn around circle ABC.
Thus, a triangle, equiangular with the given triangle,
has been circumscribed about the given circle. (Which is)
the very thing it was required to do.
† See the footnote to Prop. 3.34.. Proposition 4
Εἰςτὸδοθὲντρίγωνονκύκλονἐγγράψαι.
Α
To inscribe a circle in a given triangle.
A
Ε
∆
Η
E
D
G
Β
Ζ
Γ
῎ΕστωτὸδοθὲντρίγωνοντὸΑΒΓ·δεῖδὴεἰςτὸΑΒΓ Let ABC be the given triangle. So it is required to
τρίγωνονκύκλονἐγγράψαι.
inscribe a circle in triangle ABC.
ΤετμήσθωσαναἱὑπὸΑΒΓ,ΑΓΒγωνίαιδίχαταῖςΒΔ, Let the angles ABC and ACB have been cut in half by
ΓΔ εὐθείαις, καὶ συμβαλλέτωσαν ἀλλήλαις κατὰ τὸ Δ the straight-lines BD and CD (respectively) [Prop. 1.9],
σημεῖον, καὶ ἤχθωσαν ἀπὸ τοῦ Δ ἐπὶ τὰς ΑΒ, ΒΓ, ΓΑ and let them meet one another at point D, and let DE,
εὐθείαςκάθετοιαἱΔΕ,ΔΖ,ΔΗ.
DF , and DG have been drawn from point D, perpendic-
Καὶἐπεὶἴσηἐστὶν ἡὑπὸΑΒΔ γωνίατῇ ὑπὸΓΒΔ, ular to the straight-lines AB, BC, and CA (respectively)
ἐστὶδὲκαὶὀρθὴἡὑπὸΒΕΔὀρθῇτῇὑπὸΒΖΔἴση,δύο [Prop. 1.12].
δὴτρίγωνάἐστιτὰΕΒΔ,ΖΒΔτὰςδύογωνίαςταῖςδυσὶ And since angle ABD is equal to CBD, and the rightγωνίαιςἴσαςἔχοντακαὶμίανπλευρὰνμιᾷπλευρᾷἴσηντὴν
angle BED is also equal to the right-angle BFD, EBD
ὑποτείνουσανὑπὸμίαντῶνἴσωνγωνιῶνκοινὴναὐτῶντὴν and FBD are thus two triangles having two angles equal
ΒΔ·καὶτὰςλοιπὰςἄραπλευρὰςταῖςλοιπαῖςπλευραῖςἴσας to two angles, and one side equal to one side—the (one)
ἕξουσιν·ἴσηἄραἡΔΕτῇΔΖ.διὰτὰαὐτὰδὴκαὶἡΔΗ subtending one of the equal angles (which is) common to
τῇΔΖἐστινἴση. αἱτρεῖςἄραεὐθεῖαιαἱΔΕ,ΔΖ,ΔΗ the (triangles)—(namely), BD. Thus, they will also have
ἴσαιἀλλήλαιςεἰσίν·ὁἄρακέντρῷτῷΔκαὶδιαστήματιἑνὶ the remaining sides equal to the (corresponding) remainτῶνΕ,Ζ,Ηκύκλοςγραφόμενοςἥξεικαὶδιὰτῶνλοιπῶν
ing sides [Prop. 1.26]. Thus, DE (is) equal to DF . So,
σημείων καὶ ἐφάψεται τῶν ΑΒ, ΒΓ, ΓΑ εὐθειῶν διὰ τὸ for the same (reasons), DG is also equal to DF . Thus,
ὀρθὰςεἶναιτὰςπρὸςτοῖςΕ,Ζ,Ησημείοιςγωνίας. εἰγὰρ the three straight-lines DE, DF , and DG are equal to
τεμεῖαὐτάς,ἔσταιἡτῇδιαμέτρῳτοῦκύκλουπρὸςὀρθὰς one another. Thus, the circle drawn with center D, and
ἀπ᾿ἄκραςἀγομένηἐντὸςπίπτουσατοῦκύκλου·ὅπερἄτο- radius one of E, F , or G, † will also go through the reπονἐδείχθη·οὐκἄραὁκέντρῳτῷΔδιαστήματιδὲἑνὶτῶν
maining points, and will touch the straight-lines AB, BC,
Ε,Ζ,ΗγραφόμενοςκύκλοςτεμεῖτὰςΑΒ,ΒΓ,ΓΑεὐθείας· and CA, on account of the angles at E, F , and G being
ἐφάψεταιἄρααὐτῶν,καὶἔσταιὁκύκλοςἐγγεγραμμένοςεἰς right-angles. For if it cuts (one of) them then it will be
τὸΑΒΓτρίγωνον.ἐγγεγράφθωὡςὁΖΗΕ.
a (straight-line) drawn at right-angles to a diameter of
ΕἰςἄρατὸδοθὲντρίγωνοντὸΑΒΓκύκλοςἐγγέγραπται the circle, from its extremity, falling inside the circle. The
ὁΕΖΗ·ὅπερἔδειποιῆσαι.
very thing was shown (to be) absurd [Prop. 3.16]. Thus,
the circle drawn with center D, and radius one of E, F ,
B
F
C
113

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 4
or G, does not cut the straight-lines AB, BC, and CA.
Thus, it will touch them and will be the circle inscribed
in triangle ABC. Let it have been (so) inscribed, like
FGE (in the figure).
Thus, the circle EFG has been inscribed in the given
triangle ABC. (Which is) the very thing it was required
to do.
† Here, and in the following propositions, it is understood that the radius is actually one of DE, DF , or DG.
. Proposition
Περὶτὸδοθὲντρίγωνονκύκλονπεριγράψαι.
Α
Α
∆
Β
∆ Ε
∆
Ε
Β Γ
Ζ
Ζ Ζ
Β
Γ
Α
Ε
D
Γ
B
5
To circumscribe a circle about a given triangle.
A
A
A
D
B
D E
E
B
F
C
F
F
C
E
C
῎ΕστωτὸδοθὲντρίγωνοντὸΑΒΓ·δεῖδὲπερὶτὸδοθὲν Let ABC be the given triangle. So it is required to
τρίγωνοντὸΑΒΓκύκλονπεριγράψαι.
circumscribe a circle about the given triangle ABC.
ΤετμήσθωσαναἱΑΒ,ΑΓεὐθεῖαιδίχακατὰτὰΔ,Ε Let the straight-lines AB and AC have been cut in
σημεῖα,καὶἀπὸτῶνΔ,ΕσημείωνταῖςΑΒ,ΑΓπρὸςὀρθὰς half at points D and E (respectively) [Prop. 1.10]. And
ἤχθωσαναἱΔΖ,ΕΖ·συμπεσοῦνταιδὴἤτοιἐντὸςτοῦΑΒΓ let DF and EF have been drawn from points D and E,
τριγώνουἢἐπὶτῆςΒΓεὐθείαςἢἐκτὸςτῆςΒΓ. at right-angles to AB and AC (respectively) [Prop. 1.11].
ΣυμπιπτέτωσανπρότερονἐντὸςκατὰτὸΖ,καὶἐπεζεύχθ- So (DF and EF ) will surely either meet inside triangle
ωσαναἱΖΒ,ΖΓ,ΖΑ.καὶἐπεὶἴσηἐστὶνἡΑΔτῇΔΒ,κοινὴ ABC, on the straight-line BC, or beyond BC.
δὲκαὶπρὸςὀρθὰςἡΔΖ,βάσιςἄραἡΑΖβάσειτῇΖΒἐστιν Let them, first of all, meet inside (triangle ABC) at
ἴση. ὁμοίωςδὴδείξομεν,ὅτικαὶἡΓΖτῇΑΖἐστινἴση· (point) F , and let FB, FC, and FA have been joined.
ὥστεκαὶἡΖΒτῇΖΓἐστινἴση·αἱτρεῖςἄρααἱΖΑ,ΖΒ,ΖΓ And since AD is equal to DB, and DF is common and
ἴσαιἀλλήλαιςεἰσίν. ὁἄρακέντρῳτῷΖδιαστήματιδὲἑνὶ at right-angles, the base AF is thus equal to the base FB
τῶνΑ,Β,Γκύκλοςγραφόμενοςἥξεικαὶδιὰτῶνλοιπῶν [Prop. 1.4]. So, similarly, we can show that CF is also
σημείων,καὶἔσταιπεριγεγραμμένοςὁκύκλοςπερὶτὸΑΒΓ equal to AF . So that FB is also equal to FC. Thus, the
τρίγωνον.περιγεγράφθωὡςὁΑΒΓ.
three (straight-lines) FA, FB, and FC are equal to one
ἈλλὰδὴαἱΔΖ,ΕΖσυμπιπτέτωσανἐπὶτῆςΒΓεὐθείας another. Thus, the circle drawn with center F , and radius
κατὰ τὸ Ζ, ὡς ἔχει ἐπὶ τῆς δευτέρας καταγραφῆς, καὶ one of A, B, or C, will also go through the remaining
ἐπεζεύχθωἡΑΖ.ὁμοίωςδὴδείξομεν, ὅτιτὸΖσημεῖον points. And the circle will have been circumscribed about
κέντρονἐστὶτοῦπερὶτὸΑΒΓτρίγωνονπεριγραφομένου triangle ABC. Let it have been (so) circumscribed, like
κύκλου.
ABC (in the first diagram from the left).
ἈλλὰδὴαἱΔΖ,ΕΖσυμπιπτέτωσανἐκτὸςτοῦΑΒΓ And so, let DF and EF meet on the straight-line
τριγώνουκατὰτὸΖπάλιν,ὡςἔχειἐπὶτῆςτρίτηςκατα- BC at (point) F , like in the second diagram (from the
γραφῆς,καίἐπεζεύχθωσαναἱΑΖ,ΒΖ,ΓΖ.καὶἐπεὶπάλιν left). And let AF have been joined. So, similarly, we can
ἴσηἐστὶνἡΑΔτῇΔΒ,κοινὴδὲκαὶπρὸςὀρθὰςἡΔΖ,βάσις show that point F is the center of the circle circumscribed
ἄραἡΑΖβάσειτῇΒΖἐστινἴση. ὁμοίωςδὴδείξομεν,ὅτι about triangle ABC.
καὶἡΓΖτῇΑΖἐστινἴση·ὥστεκαὶἡΒΖτῇΖΓἐστινἴση· And so, let DF and EF meet outside triangle ABC,
ὁἄρα[πάλιν]κέντρῳτῷΖδιαστήματιδὲἑνὶτῶνΖΑ,ΖΒ, again at (point) F , like in the third diagram (from the
ΖΓκύκλοςγραφόμενοςἥξεικαὶδιὰτῶνλοιπῶνσημείων, left). And let AF , BF , and CF have been joined. And,
καὶἔσταιπεριγεγραμμένοςπερὶτὸΑΒΓτρίγωνον. again, since AD is equal to DB, and DF is common and
Περὶ τὸ δοθὲν ἄρα τρίγωνον κύκλος περιγέγραπται· at right-angles, the base AF is thus equal to the base BF
ὅπερἔδειποιῆσαι.
[Prop. 1.4]. So, similarly, we can show that CF is also
equal to AF . So that BF is also equal to FC. Thus,
114

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 4
¦. Proposition
Εἰςτὸνδοθέντακύκλοντετράγωνονἐγγράψαι.
Α
[again] the circle drawn with center F , and radius one
of FA, FB, and FC, will also go through the remaining
points. And it will have been circumscribed about triangle
ABC.
Thus, a circle has been circumscribed about the given
triangle. (Which is) the very thing it was required to do.
6
To inscribe a square in a given circle.
A
Β
Ε
∆
B
E
D
Γ
῎ΕστωἡδοθεὶςκύκλοςὁΑΒΓΔ·δεῖδὴεἰςτὸνΑΒΓΔ Let ABCD be the given circle. So it is required to
κύκλοντετράγωνονἐγγράψαι.
inscribe a square in circle ABCD.
῎ΗχθωσαντοῦΑΒΓΔκύκλουδύοδιάμετροιπρὸςὀρθὰς Let two diameters of circle ABCD, AC and BD, have
ἀλλήλαιςαἱΑΓ,ΒΔ,καὶἐπεζεύχθωσαναἱΑΒ,ΒΓ,ΓΔ, been drawn at right-angles to one another. † And let AB,
ΔΑ.
BC, CD, and DA have been joined.
ΚαὶἐπεὶἴσηἐστὶνἡΒΕτῇΕΔ·κέντρονγὰρτὸΕ·κοινὴ And since BE is equal to ED, for E (is) the center
δὲκαὶπρὸςὀρθὰςἡΕΑ,βάσιςἄραἡΑΒβάσειτῇΑΔἴση (of the circle), and EA is common and at right-angles,
ἐστίν. διὰτὰαὐτὰδὴκαὶἑκατέρατῶνΒΓ,ΓΔἑκατέρᾳ the base AB is thus equal to the base AD [Prop. 1.4].
τῶν ΑΒ,ΑΔἴσηἐστίν· ἰσόπλευρονἄραἐστὶτὸΑΒΓΔ So, for the same (reasons), each of BC and CD is equal
τετράπλευρον. λέγωδή,ὅτικαὶὀρθογώνιον. ἐπεὶγὰρἡ to each of AB and AD. Thus, the quadrilateral ABCD
ΒΔεὐθεῖαδιάμετρόςἐστιτοῦΑΒΓΔκύκλου,ἡμικύκλιον is equilateral. So I say that (it is) also right-angled. For
ἄραἐστὶτὸΒΑΔ·ὀρθὴἄραἡὑπὸΒΑΔγωνία. διὰτὰ since the straight-line BD is a diameter of circle ABCD,
αὐτὰδὴκαὶἑκάστητῶνὑπὸΑΒΓ,ΒΓΔ,ΓΔΑὀρθήἐστιν· BAD is thus a semi-circle. Thus, angle BAD (is) a right-
ὀρθογώνιονἄραἐστὶτὸΑΒΓΔτετράπλευρον. ἐδείχθηδὲ angle [Prop. 3.31]. So, for the same (reasons), (angles)
καὶἰσόπλευρον·τετράγωνονἄραἐστίν.καὶἐγγέγραπταιεἰς ABC, BCD, and CDA are also each right-angles. Thus,
τὸνΑΒΓΔκύκλον.
the quadrilateral ABCD is right-angled. And it was also
Εἰςἄρατὸνδοθέντακύκλοντετράγωνονἐγγέγραπται shown (to be) equilateral. Thus, it is a square [Def. 1.22].
τὸΑΒΓΔ·ὅπερἔδειποιῆσαι.
And it has been inscribed in circle ABCD.
Thus, the square ABCD has been inscribed in the
given circle. (Which is) the very thing it was required
to do.
† Presumably, by finding the center of the circle [Prop. 3.1], drawing a line through it, and then drawing a second line through it, at right-angles
C
to the first [Prop. 1.11].
Þ.
Proposition
Περὶτὸνδοθέντακύκλοντετράγωνονπεριγράψαι.
7
To circumscribe a square about a given circle.
115

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 4
῎ΕστωὁδοθεὶςκύκλοςὁΑΒΓΔ·δεῖδὴπερὶτὸνΑΒΓΔ Let ABCD be the given circle. So it is required to
κύκλοντετράγωνονπεριγράψαι.
circumscribe a square about circle ABCD.
Η Α Ζ
G A F
Β
Ε
∆
B
E
D
Θ Γ Κ
H C K
῎ΗχθωσαντοῦΑΒΓΔκύκλουδύοδιάμετροιπρὸςὀρθὰς Let two diameters of circle ABCD, AC and BD, have
ἀλλήλαιςαἱΑΓ,ΒΔ,καὶδιὰτῶνΑ,Β,Γ,Δσημείωνἤχθω- been drawn at right-angles to one another. † And let FG,
σανἐφαπτόμεναιτοῦΑΒΓΔκύκλουαἱΖΗ,ΗΘ,ΘΚ,ΚΖ. GH, HK, and KF have been drawn through points A,
᾿ΕπεὶοὖνἐφάπτεταιἡΖΗτοῦΑΒΓΔκύκλου,ἀπὸδὲ B, C, and D (respectively), touching circle ABCD. ‡
τοῦΕκέντρουἐπὶτὴνκατὰτὸΑἐπαφὴνἐπέζευκταιἡ Therefore, since FG touches circle ABCD, and EA
ΕΑ,αἱἄραπρὸςτῷΑγωνίαιὀρθαίεἰσιν. διὰτὰαὐτὰ has been joined from the center E to the point of contact
δὴκαὶαἱπρὸςτοῖςΒ,Γ,Δσημείοιςγωνίαιὀρθαίεἰσιν. A, the angles at A are thus right-angles [Prop. 3.18]. So,
καὶἐπεὶὀρθήἐστινἡὑπὸΑΕΒγωνία,ἐστὶδὲὀρθὴκαὶἡ for the same (reasons), the angles at points B, C, and
ὑπὸΕΒΗ,παράλληλοςἄραἐστὶνἡΗΘτῇΑΓ.διὰτὰαὐτὰ D are also right-angles. And since angle AEB is a rightδὴκαὶἡΑΓτῇΖΚἐστιπαράλληλος.
ὥστεκαὶἡΗΘτῇ angle, and EBG is also a right-angle, GH is thus parallel
ΖΚἐστιπαράλληλος.ὁμοίωςδὴδείξομεν,ὅτικαὶἑκατέρα to AC [Prop. 1.29]. So, for the same (reasons), AC is
τῶνΗΖ,ΘΚτῇΒΕΔἐστιπαράλληλος.παραλληλόγραμμα also parallel to FK. So that GH is also parallel to FK
ἄραἐστὶτὰΗΚ,ΗΓ,ΑΚ,ΖΒ,ΒΚ·ἴσηἄραἐστὶνἡμὲν [Prop. 1.30]. So, similarly, we can show that GF and
ΗΖτῇΘΚ,ἡδὲΗΘτῇΖΚ.καὶἐπεὶἴσηἐστὶνἡΑΓτῇ HK are each parallel to BED. Thus, GK, GC, AK, FB,
ΒΔ,ἀλλὰκαὶἡμὲνΑΓἑκατέρᾳτῶνΗΘ,ΖΚ,ἡδὲΒΔ and BK are (all) parallelograms. Thus, GF is equal to
ἑκατέρᾳτῶνΗΖ,ΘΚἐστινἴση[καὶἑκατέραἄρατῶνΗΘ, HK, and GH to FK [Prop. 1.34]. And since AC is equal
ΖΚἑκατέρᾳτῶνΗΖ,ΘΚἐστινἴση],ἰσόπλευρονἄραἐστὶτὸ to BD, but AC (is) also (equal) to each of GH and FK,
ΖΗΘΚτετράπλευρον. λέγωδή,ὅτικαὶὀρθογώνιον. ἐπεὶ and BD is equal to each of GF and HK [Prop. 1.34]
γὰρπαραλληλόγραμμόνἐστιτὸΗΒΕΑ,καίἐστινὀρθὴἡ [and each of GH and FK is thus equal to each of GF
ὑπὸΑΕΒ,ὀρθὴἄρακαὶἡὑπὸΑΗΒ.ὁμοίωςδὴδείξομεν, and HK], the quadrilateral FGHK is thus equilateral.
ὅτικαὶαἱπρὸςτοῖςΘ,Κ,Ζγωνίαιὀρθαίεἰσιν.ὀρθογώνιον So I say that (it is) also right-angled. For since GBEA
ἄραἐστὶτὸΖΗΘΚ.ἐδείχθηδὲκαὶἰσόπλευρον·τετράγωνον is a parallelogram, and AEB is a right-angle, AGB is
ἄραἐστίν.καὶπεριγέγραπταιπερὶτὸνΑΒΓΔκύκλον. thus also a right-angle [Prop. 1.34]. So, similarly, we can
Περὶτὸνδοθένταἄρακύκλοντετράγωνονπεριγέγραπται· show that the angles at H, K, and F are also right-angles.
ὅπερἔδειποιῆσαι.
Thus, FGHK is right-angled. And it was also shown (to
be) equilateral. Thus, it is a square [Def. 1.22]. And it
has been circumscribed about circle ABCD.
Thus, a square has been circumscribed about the
given circle. (Which is) the very thing it was required
to do.
† See the footnote to the previous proposition.
‡ See the footnote to Prop. 3.34.
116

ËÌÇÁÉÁÏƺ ELEMENTS BOOK 4
. Proposition 8
Εἰςτὸδοθὲντετράγωνονκύκλονἐγγράψαι.
To inscribe a circle in a given square.
῎ΕστωτὸδοθὲντετράγωνοντὸΑΒΓΔ.δεῖδὴεἰςτὸ Let the given square be ABCD. So it is required to
ΑΒΓΔτετράγωνονκύκλονἐγγράψαι.
inscribe a circle in square ABCD.
Α Ε ∆
A E D
Ζ
Η
Κ
F
G
K
Β
Θ
Γ
B H C
ΤετμήσθωἑκατέρατῶνΑΔ,ΑΒδίχακατὰτὰΕ,Ζ Let AD and AB each have been cut in half at points E
σημεῖα,καὶδιὰμὲντοῦΕὁποτέρᾳτῶνΑΒ,ΓΔπαράλληλος and F (respectively) [Prop. 1.10]. And let EH have been
ἤχθωὁΕΘ,διὰδὲτοῦΖὁποτέρᾳτῶνΑΔ,ΒΓπαράλληλος drawn through E, parallel to either of AB or CD, and let
ἤχθω ἡ ΖΚ· παραλληλόγραμμονἄρα ἐστὶν ἕκαστον τῶν FK have been drawn through F , parallel to either of AD
ΑΚ,ΚΒ,ΑΘ,ΘΔ,ΑΗ,ΗΓ,ΒΗ,ΗΔ,καὶαἱἀπεναντίον or BC [Prop. 1.31]. Thus, AK, KB, AH, HD, AG, GC,
αὐτῶνπλευραὶδηλονότιἴσαι[εἰσίν]. καὶἐπεὶἴσηἐστὶνἡ BG, and GD are each parallelograms, and their opposite
ΑΔτῇΑΒ,καίἐστιτῆςμὲνΑΔἡμίσειαἡΑΕ,τῆςδὲΑΒ sides [are] manifestly equal [Prop. 1.34]. And since AD
ἡμίσειαἡΑΖ,ἴσηἄρακαὶἡΑΕτῇΑΖ·ὥστεκαὶαἱἀπε- is equal to AB, and AE is half of AD, and AF half of
ναντίον·ἴσηἄρακαὶἡΖΗτῇΗΕ.ὁμοίωςδὴδείξομεν,ὅτι AB, AE (is) thus also equal to AF . So that the opposite
καὶἑκατέρατῶνΗΘ,ΗΚἑκατέρᾳτῶνΖΗ,ΗΕἐστινἴση· (sides are) also (equal). Thus, FG (is) also equal to GE.
αἱτέσσαρεςἄρααἱΗΕ,ΗΖ,ΗΘ,ΗΚἴσαιἀλλήλαις[εἰσίν]. So, similarly, we can also show that each of GH and GK
ὁἄρακέντρῳμὲντῷΗδιαστήματιδὲἑνὶτῶνΕ,Ζ,Θ,Κ is equal to each of FG and GE. Thus, the four (straightκύκλοςγραφόμενοςἥξεικαὶδιὰτῶνλοιπῶνσημείων·καὶ
lines) GE, GF , GH, and GK [are] equal to one another.
ἐφάψεταιτῶνΑΒ,ΒΓ,ΓΔ,ΔΑεὐθειῶνδιὰτὸὀρθὰςεἶναι Thus, the circle drawn with center G, and radius one of
τὰςπρὸςτοῖςΕ,Ζ,Θ,Κγωνίας·εἰγὰρτεμεῖὁκύκλοςτὰς E, F , H, or K, will also go through the remaining points.
ΑΒ,ΒΓ,ΓΔ,ΔΑ,ἡτῇδιαμέτρῳτοῦκύκλουπρὸςὀρθὰς And it will touch the straight-lines AB, BC, CD, and
ἀπ᾿ἄκραςἀγομένηἐντὸςπεσεῖταιτοῦκύκλου·ὅπερἄτοπον DA, on account of the angles at E, F , H, and K being
ἐδείχθη.οὐκἄραὁκέντρῳτῷΗδιαστήματιδὲἑνὶτῶνΕ, right-angles. For if the circle cuts AB, BC, CD, or DA,
Ζ,Θ,ΚκύκλοςγραφόμενοςτεμεῖτὰςΑΒ,ΒΓ,ΓΔ,ΔΑ then a (straight-line) drawn at right-angles to a diameter
εὐθείας.ἐφάψεταιἄρααὐτῶνκαὶἔσταιἐγγεγραμμένοςεἰς of the circle, from its extremity, will fall inside the circle.
τὸΑΒΓΔτετράγωνον. The very thing was shown (to be) absurd [Prop. 3.16].
Εἰς ἄρα τὸ δοθὲν τετράγωνον κύκλος ἐγγέγραπται· Thus, the circle drawn with center G, and radius one of
ὅπερἔδειποιῆσαι.
E, F , H, or K, does not cut the straight-lines AB, BC,
CD, or DA. Thus, it will touch them, and will have been
inscribed in the square ABCD.
Thus, a circle has been inscribed in the given square.
(Which is) the very thing it was required to do.
. Proposition 9
Περὶτὸδοθὲντετράγωνονκύκλονπεριγράψαι.
To circumscribe a circle about a given square.
῎ΕστωτὸδοθὲντετράγωνοντὸΑΒΓΔ·δεῖδὴπερὶτὸ Let ABCD be the given square. So it is required to
ΑΒΓΔτετράγωνονκύκλονπεριγράψαι.
circumscribe a circle about square ABCD.
117

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 4
᾿Επιζευχθεῖσαι γὰρ αἱ ΑΓ, ΒΔ τεμνέτωσαν ἀλλήλας AC and BD being joined, let them cut one another at
κατὰτὸΕ. E.
Α
A
Β
Ε
∆
B
E
D
Γ
ΚαὶἐπεὶἴσηἐστὶνἡΔΑτῇΑΒ,κοινὴδὲἡΑΓ,δύο And since DA is equal to AB, and AC (is) common,
δὴαἱΔΑ,ΑΓδυσὶταῖςΒΑ,ΑΓἴσαιεἰσίν· καὶβάσιςἡ the two (straight-lines) DA, AC are thus equal to the two
ΔΓβάσειτῇΒΓἴση·γωνίαἄραἡὑπὸΔΑΓγωνίᾳτῇὑπὸ (straight-lines) BA, AC. And the base DC (is) equal to
ΒΑΓἴσηἐστίν·ἡἄραὑπὸΔΑΒγωνίαδίχατέτμηταιὑπὸ the base BC. Thus, angle DAC is equal to angle BAC
τῆςΑΓ.ὁμοίωςδὴδείξομεν,ὅτικαὶἑκάστητῶνὑπὸΑΒΓ, [Prop. 1.8]. Thus, the angle DAB has been cut in half
ΒΓΔ,ΓΔΑδίχατέτμηταιὑπὸτῶνΑΓ,ΔΒεὐθειῶν. καὶ by AC. So, similarly, we can show that ABC, BCD, and
ἐπεὶἴσηἐστὶνἡὑπὸΔΑΒγωνίατῇὑπὸΑΒΓ,καίἐστι CDA have each been cut in half by the straight-lines AC
τῆςμὲνὑπὸΔΑΒἡμίσειαἡὑπὸΕΑΒ,τῆςδὲὑπὸΑΒΓ and DB. And since angle DAB is equal to ABC, and
ἡμίσειαἡὑπὸΕΒΑ,καὶἡὑπὸΕΑΒἄρατῇὑπὸΕΒΑἐστιν EAB is half of DAB, and EBA half of ABC, EAB is
ἴση·ὥστεκαὶπλευρὰἡΕΑτῇΕΒἐστινἴση. ὁμοίωςδὴ thus also equal to EBA. So that side EA is also equal
δείξομεν,ὅτικαὶἑκατέρατῶνΕΑ,ΕΒ[εὐθειῶν]ἑκατέρᾳ to EB [Prop. 1.6]. So, similarly, we can show that each
τῶνΕΓ,ΕΔἴσηἐστίν. αἱτέσσαρεςἄρααἱΕΑ,ΕΒ,ΕΓ, of the [straight-lines] EA and EB are also equal to each
ΕΔἴσαιἀλλήλαιςεἰσίν.ὁἄρακέντρῳτῷΕκαὶδιαστήματι of EC and ED. Thus, the four (straight-lines) EA, EB,
ἑνὶτῶνΑ,Β,Γ,Δκύκλοςγραφόμενοςἥξεικαὶδιὰτῶν EC, and ED are equal to one another. Thus, the circle
λοιπῶνσημείωνκαὶἔσταιπεριγεγραμμένοςπερὶτὸΑΒΓΔ drawn with center E, and radius one of A, B, C, or D,
τετράγωνον.περιγεγράφθωὡςὁΑΒΓΔ.
will also go through the remaining points, and will have
Περὶτὸδοθὲνἄρατετράγωνονκύκλοςπεριγέγραπται· been circumscribed about the square ABCD. Let it have
ὅπερἔδειποιῆσαι.
been (so) circumscribed, like ABCD (in the figure).
Thus, a circle has been circumscribed about the given
square. (Which is) the very thing it was required to do.
. Proposition 10
᾿Ισοσκελὲςτρίγωνονσυστήσασθαιἔχονἑκατέραντῶν To construct an isosceles triangle having each of the
πρὸςτῇβάσειγωνιῶνδιπλασίονατῆςλοιπῆς.
angles at the base double the remaining (angle).
᾿Εκκείσθω τις εὐθεῖα ἡ ΑΒ, καὶ τετμήσθω κατὰ τὸ Let some straight-line AB be taken, and let it have
Γ σημεῖον, ὥστε τὸ ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον been cut at point C so that the rectangle contained by
ὀρθογώνιον ἴσον εἶναι τῷ ἀπὸ τῆς ΓΑ τετραγώνῳ· καὶ AB and BC is equal to the square on CA [Prop. 2.11].
κέντρῳ τῷ Α καὶ διαστήματι τῷ ΑΒ κύκλος γεγράφθω And let the circle BDE have been drawn with center A,
ὁΒΔΕ,καὶἐνηρμόσθωεἰςτὸνΒΔΕκύκλοντῇΑΓεὐθείᾳ and radius AB. And let the straight-line BD, equal to
μὴμείζονιοὔσῃτῆςτοῦΒΔΕκύκλουδιαμέτρουἴσηεὐθεῖα the straight-line AC, being not greater than the diame-
ἡΒΔ·καὶἐπεζεύχθωσαναἱΑΔ,ΔΓ,καὶπεριγεγράφθω ter of circle BDE, have been inserted into circle BDE
περὶτὸΑΓΔτρίγωνονκύκλοςὁΑΓΔ.
[Prop. 4.1]. And let AD and DC have been joined. And
let the circle ACD have been circumscribed about triangle
ACD [Prop. 4.5].
C
118

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 4
Β
B
Γ
∆
C
D
Α
A
Ε
E
ΚαὶἐπεὶτὸὑπὸτῶνΑΒ,ΒΓἴσονἐστὶτῷἀπὸτῆς And since the (rectangle contained) by AB and BC
ΑΓ,ἴσηδὲἡΑΓτῇΒΔ,τὸἄραὑπὸτῶνΑΒ,ΒΓἴσον is equal to the (square) on AC, and AC (is) equal to
ἐστὶτῷἀπὸτῆςΒΔ.καὶἐπεὶκύκλουτοῦΑΓΔεἴληπταίτι BD, the (rectangle contained) by AB and BC is thus
σημεῖονἐκτὸςτὸΒ,καὶἀπὸτοῦΒπρὸςτὸνΑΓΔκύκλον equal to the (square) on BD. And since some point B
προσπεπτώκασιδύοεὐθεῖαιαἱΒΑ,ΒΔ,καὶἡμὲναὐτῶν has been taken outside of circle ACD, and two straightτέμνει,ἡδὲπροσπίπτει,καίἐστιτὸὑπὸτῶνΑΒ,ΒΓἴσον
lines BA and BD have radiated from B towards the cirτῷἀπὸτῆςΒΔ,ἡΒΔἄραἐφάπτεταιτοῦΑΓΔκύκλου.ἐπεὶ
cle ACD, and (one) of them cuts (the circle), and (the
οὖνἐφάπτεταιμὲνἡΒΔ,ἀπὸδὲτῆςκατὰτὸΔἐπαφῆς other) meets (the circle), and the (rectangle contained)
διῆκταιἡΔΓ,ἡἄραὑπὸΒΔΓγωνιάἴσηἐστὶτῇἐντῷ by AB and BC is equal to the (square) on BD, BD thus
ἐναλλὰξτοῦκύκλουτμήματιγωνίᾳτῇὑπὸΔΑΓ.ἐπεὶοὖν touches circle ACD [Prop. 3.37]. Therefore, since BD
ἴσηἐστὶνἡὑπὸΒΔΓτῇὑπὸΔΑΓ,κοινὴπροσκείσθωἡ touches (the circle), and DC has been drawn across (the
ὑπὸΓΔΑ·ὅληἄραἡὑπὸΒΔΑἴσηἐστὶδυσὶταῖςὑπὸΓΔΑ, circle) from the point of contact D, the angle BDC is
ΔΑΓ.ἀλλὰταῖςὑπὸΓΔΑ,ΔΑΓἴσηἐστὶνἡἐκτὸςἡὑπὸ thus equal to the angle DAC in the alternate segment of
ΒΓΔ·καὶἡὑπὸΒΔΑἄραἴσηἐστὶτῇὑπὸΒΓΔ.ἀλλὰἡ the circle [Prop. 3.32]. Therefore, since BDC is equal
ὑπὸΒΔΑτῇὑπὸΓΒΔἐστινἴση,ἐπεὶκαὶπλευρὰἡΑΔ to DAC, let CDA have been added to both. Thus, the
τῇΑΒἐστινἴση·ὥστεκαὶἡὑπὸΔΒΑτῇὑπὸΒΓΔἐστιν whole of BDA is equal to the two (angles) CDA and
ἴση. αἱτρεῖςἄρααἱὑπὸΒΔΑ,ΔΒΑ,ΒΓΑἴσαιἀλλήλαις DAC. But, the external (angle) BCD is equal to CDA
εἰσίν. καὶἐπεὶἴσηἐστὶνἡὑπὸΔΒΓγωνίατῇὑπὸΒΓΔ, and DAC [Prop. 1.32]. Thus, BDA is also equal to
ἴσηἐστὶκαὶπλευρὰἡΒΔπλευρᾷτῇΔΓ.ἀλλὰἡΒΔτῇ BCD. But, BDA is equal to CBD, since the side AD is
ΓΑὑπόκειταιἴση·καὶἡΓΑἄρατῇΓΔἐστινἴση·ὥστεκαὶ also equal to AB [Prop. 1.5]. So that DBA is also equal
γωνίαἡὑπὸΓΔΑγωνίᾳτῇὑπὸΔΑΓἐστινἴση· αἱἄρα to BCD. Thus, the three (angles) BDA, DBA, and BCD
ὑπὸΓΔΑ,ΔΑΓτῆςὑπὸΔΑΓεἰσιδιπλασίους. ἴσηδὲἡ are equal to one another. And since angle DBC is equal
ὑπὸΒΓΔταῖςὑπὸΓΔΑ,ΔΑΓ·καὶἡὑπὸΒΓΔἄρατῆςὑπὸ to BCD, side BD is also equal to side DC [Prop. 1.6].
ΓΑΔἐστιδιπλῆ.ἴσηδὲἡὑπὸΒΓΔἑκατέρᾳτῶνὑπὸΒΔΑ, But, BD was assumed (to be) equal to CA. Thus, CA
ΔΒΑ·καὶἑκατέραἄρατῶνὑπὸΒΔΑ,ΔΒΑτῆςὑπὸΔΑΒ is also equal to CD. So that angle CDA is also equal to
ἐστιδιπλῆ.
angle DAC [Prop. 1.5]. Thus, CDA and DAC is double
᾿Ισοσκελὲς ἄρα τρίγωνον συνέσταται τὸ ΑΒΔ ἔχον DAC. But BCD (is) equal to CDA and DAC. Thus,
ἑκατέραντῶνπρὸςτῇΔΒβάσειγωνιῶνδιπλασίονατῆς BCD is also double CAD. And BCD (is) equal to to
λοιπῆς·ὅπερἔδειποιῆσαι.
each of BDA and DBA. Thus, BDA and DBA are each
double DAB.
Thus, the isosceles triangle ABD has been constructed
having each of the angles at the base BD double
the remaining (angle). (Which is) the very thing it was
required to do.
. Proposition 11
Εἰςτὸνδοθέντακύκλονπεντάγωνονἰσόπλευρόντεκαὶ To inscribe an equilateral and equiangular pentagon
119

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 4
ἰσογώνιονἐγγράψαι.
Α
Ζ
in a given circle.
A
F
Β
Ε
B
E
Γ
∆
Η
Θ
῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔΕ· δεῖ δὴ εἰς τὸν Let ABCDE be the given circle. So it is required to
ΑΒΓΔΕκύκλονπεντάγωνονἰσόπλευρόντεκαὶἰσογώνιον inscribed an equilateral and equiangular pentagon in cir-
ἐγγράψαι.
cle ABCDE.
᾿Εκκείσθω τρίγωνον ἰσοσκελὲς τὸ ΖΗΘ διπλασίονα Let the the isosceles triangle FGH be set up hav-
ἔχονἑκατέραντῶνπρὸςτοῖςΗ,ΘγωνιῶντῆςπρὸςτῷΖ, ing each of the angles at G and H double the (angle)
καὶἐγγεγράφθωεἰςτὸνΑΒΓΔΕκύκλοντῷΖΗΘτριγώνῳ at F [Prop. 4.10]. And let triangle ACD, equiangular
ἰσογώνιοντρίγωνοντὸΑΓΔ,ὥστετῇμὲνπρὸςτῷΖγωνίᾳ to FGH, have been inscribed in circle ABCDE, such
ἴσηνεἶναιτὴνὑπὸΓΑΔ,ἑκατέρανδὲτῶνπρὸςτοῖςΗ,Θ that CAD is equal to the angle at F , and the (angles)
ἴσηνἑκατέρᾳτῶνὑπὸΑΓΔ,ΓΔΑ·καὶἑκατέραἄρατῶν at G and H (are) equal to ACD and CDA, respectively
ὑπὸΑΓΔ,ΓΔΑτῆςὑπὸΓΑΔἐστιδιπλῆ. τετμήσθωδὴ [Prop. 4.2]. Thus, ACD and CDA are each double
ἑκατέρατῶνὑπὸΑΓΔ,ΓΔΑδίχαὑπὸἑκατέραςτῶνΓΕ, CAD. So let ACD and CDA have been cut in half by
ΔΒεὐθειῶν,καὶἐπεζεύχθωσαναἱΑΒ,ΒΓ,ΔΕ,ΕΑ. the straight-lines CE and DB, respectively [Prop. 1.9].
᾿Επεὶ οὖν ἑκατέρα τῶν ὑπὸ ΑΓΔ, ΓΔΑ γωνιῶν δι- And let AB, BC, DE and EA have been joined.
πλασίωνἐστὶτῆςὑπὸΓΑΔ,καὶτετμημέναιεἰσὶδίχαὑπὸ Therefore, since angles ACD and CDA are each douτῶνΓΕ,ΔΒεὐθειῶν,αἱπέντεἄραγωνίαιαἱὑπὸΔΑΓ,
ble CAD, and are cut in half by the straight-lines CE and
ΑΓΕ, ΕΓΔ, ΓΔΒ,ΒΔΑ ἴσαιἀλλήλαιςεἰσίν. αἱδὲ ἴσαι DB, the five angles DAC, ACE, ECD, CDB, and BDA
γωνίαιἐπὶἴσωνπεριφερειῶνβεβήκασιν· αἱπέντεἄραπε- are thus equal to one another. And equal angles stand
ριφέρειαιαἱΑΒ,ΒΓ,ΓΔ,ΔΕ,ΕΑἴσαιἀλλήλαιςεἰσίν.ὑπὸ upon equal circumferences [Prop. 3.26]. Thus, the five
δὲτὰςἴσαςπεριφερείαςἴσαιεὐθεῖαιὑποτείνουσιν·αἱπέντε circumferences AB, BC, CD, DE, and EA are equal to
ἄραεὐθεῖαιαἱΑΒ,ΒΓ,ΓΔ,ΔΕ,ΕΑἴσαιἀλλήλαιςεἰσίν· one another [Prop. 3.29]. Thus, the pentagon ABCDE
ἰσόπλευρονἄραἐστὶτὸΑΒΓΔΕπεντάγωνον. λέγωδή, is equilateral. So I say that (it is) also equiangular. For
ὅτικαὶἰσογώνιον. ἐπεὶγὰρἡΑΒπεριφέρειατῇΔΕπε- since the circumference AB is equal to the circumferριφερείᾳἐστὶνἴση,κοινὴπροσκείσθωἡΒΓΔ·ὅληἄραἡ
ence DE, let BCD have been added to both. Thus, the
ΑΒΓΔπεριφέριαὅλῃτῇΕΔΓΒπεριφερείᾳἐστὶνἴση. καὶ whole circumference ABCD is equal to the whole cirβέβηκενἐπὶμὲντῆςΑΒΓΔπεριφερείαςγωνίαἡὑπὸΑΕΔ,
cumference EDCB. And the angle AED stands upon
ἐπὶδὲτῆςΕΔΓΒπεριφερείαςγωνίαἡὑπὸΒΑΕ·καὶἡὑπὸ circumference ABCD, and angle BAE upon circumfer-
ΒΑΕἄραγωνίατῇὑπὸΑΕΔἐστινἴση. διὰτὰαὐτὰδὴ ence EDCB. Thus, angle BAE is also equal to AED
καὶἑκάστητῶνὑπὸΑΒΓ,ΒΓΔ,ΓΔΕγωνιῶνἑκατέρᾳτῶν [Prop. 3.27]. So, for the same (reasons), each of the an-
ὑπὸΒΑΕ,ΑΕΔἐστινἴση·ἰσογώνιονἄραἐστὶτὸΑΒΓΔΕ gles ABC, BCD, and CDE is also equal to each of BAE
πεντάγωνον.ἐδείχθηδὲκαὶἰσόπλευρον.
and AED. Thus, pentagon ABCDE is equiangular. And
Εἰςἄρατὸνδοθέντακύκλονπεντάγωνονἰσόπλευρόντε it was also shown (to be) equilateral.
καὶἰσογώνιονἐγγέγραπται·ὅπερἔδειποιῆσαι.
Thus, an equilateral and equiangular pentagon has
been inscribed in the given circle. (Which is) the very
thing it was required to do.
. Proposition 12
Περὶτὸνδοθέντακύκλονπεντάγωνονἰσόπλευρόντε To circumscribe an equilateral and equiangular penκαὶἰσογώνιονπεριγράψαι.
tagon about a given circle.
C
D
G
H
120

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 4
Η
G
Α
Ε
A
E
Θ
Ζ
Μ
H
F
M
Β
∆
B
D
Κ Γ Λ
K C L
῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔΕ· δεῖ δὲ περὶ τὸν Let ABCDE be the given circle. So it is required
ΑΒΓΔΕκύκλονπεντάγωνονἰσόπλευρόντεκαὶἰσογώνιον to circumscribe an equilateral and equiangular pentagon
περιγράψαι.
about circle ABCDE.
Νενοήσθωτοῦἐγγεγραμμένουπενταγώνουτῶνγωνιῶν Let A, B, C, D, and E have been conceived as the anσημεῖατὰΑ,Β,Γ,Δ,Ε,ὥστεἴσαςεἶναιτὰςΑΒ,ΒΓ,
gular points of a pentagon having been inscribed (in cir-
ΓΔ, ΔΕ, ΕΑ περιφερείας· καὶ διὰ τῶν Α, Β, Γ, Δ, Ε cle ABCDE) [Prop. 3.11], such that the circumferences
ἤχθωσαντοῦκύκλουἐφαπτόμεναιαἱΗΘ,ΘΚ,ΚΛ,ΛΜ, AB, BC, CD, DE, and EA are equal. And let GH, HK,
ΜΗ,καὶεἰλήφθωτοῦΑΒΓΔΕκύκλουκέντροντὸΖ,καὶ KL, LM, and MG have been drawn through (points) A,
ἐπεζεύχθωσαναἱΖΒ,ΖΚ,ΖΓ,ΖΛ,ΖΔ.
B, C, D, and E (respectively), touching the circle. † And
ΚαὶἐπεὶἡμὲνΚΛεὐθεῖαἐφάπτεταιτοῦΑΒΓΔΕκατὰ let the center F of the circle ABCDE have been found
τὸΓ,ἀπὸδὲτοῦΖκέντρουἐπὶτὴνκατὰτὸΓἐπαφὴν [Prop. 3.1]. And let FB, FK, FC, FL, and FD have
ἐπέζευκταιἡΖΓ,ἡΖΓἄρακάθετόςἐστινἐπὶτὴνΚΛ·ὀρθὴ been joined.
ἄραἐστὶνἑκατέρατῶνπρὸςτῷΓγωνιῶν. διὰτὰαὐτὰδὴ And since the straight-line KL touches (circle) ABCDE
καὶαἱπρὸςτοῖςΒ,Δσημείοιςγωνίαιὀρθαίεἰσιν.καὶἐπεὶ at C, and FC has been joined from the center F to the
ὀρθήἐστινἡὑπὸΖΓΚγωνία,τὸἄραἀπὸτῆςΖΚἴσονἐστὶ point of contact C, FC is thus perpendicular to KL
τοῖςἀπὸτῶνΖΓ,ΓΚ.διὰτὰαὐτὰδὴκαὶτοῖςἀπὸτῶνΖΒ, [Prop. 3.18]. Thus, each of the angles at C is a right-
ΒΚἴσονἐστὶτὸἀπὸτῆςΖΚ·ὥστετὰἀπὸτῶνΖΓ,ΓΚ angle. So, for the same (reasons), the angles at B and
τοῖςἀπὸτῶνΖΒ,ΒΚἐστινἴσα,ὧντὸἀπὸτῆςΖΓτῷἀπὸ D are also right-angles. And since angle FCK is a rightτῆςΖΒἐστινἴσον·λοιπὸνἄρατὸἀπὸτῆςΓΚτῷἀπὸτῆς
angle, the (square) on FK is thus equal to the (sum of
ΒΚἐστινἴσον. ἴσηἄραἡΒΚτῇΓΚ.καὶἐπεὶἴσηἐστὶν the squares) on FC and CK [Prop. 1.47]. So, for the
ἡΖΒτῇΖΓ,καὶκοινὴἡΖΚ,δύοδὴαἱΒΖ,ΖΚδυσὶταῖς same (reasons), the (square) on FK is also equal to the
ΓΖ,ΖΚἴσαιεἰσίν·καὶβάσιςἡΒΚβάσειτῇΓΚ[ἐστιν]ἴση· (sum of the squares) on FB and BK. So that the (sum
γωνίαἄραἡμὲνὑπὸΒΖΚ[γωνίᾳ]τῇὑπὸΚΖΓἐστινἴση· of the squares) on FC and CK is equal to the (sum of
ἡδὲὑπὸΒΚΖτῇὑπὸΖΚΓ·διπλῆἄραἡμὲνὑπὸΒΖΓτῆς the squares) on FB and BK, of which the (square) on
ὑπὸΚΖΓ,ἡδὲὑπὸΒΚΓτῆςὑπὸΖΚΓ.διὰτὰαὐτὰδὴκαὶ FC is equal to the (square) on FB. Thus, the remain-
ἡμὲνὑπὸΓΖΔτῆςὑπὸΓΖΛἐστιδιπλῆ,ἡδὲὑπὸΔΛΓ ing (square) on CK is equal to the remaining (square)
τῆςὑπὸΖΛΓ.καὶἐπεὶἴσηἐστὶνἡΒΓπεριφέρειατῇΓΔ, on BK. Thus, BK (is) equal to CK. And since FB is
ἴσηἐστὶκαὶγωνίαἡὑπὸΒΖΓτῇὑπὸΓΖΔ.καίἐστινἡ equal to FC, and FK (is) common, the two (straightμὲνὑπὸΒΖΓτῆςὑπὸΚΖΓδιπλῆ,ἡδὲὑπὸΔΖΓτῆςὑπὸ
lines) BF , FK are equal to the two (straight-lines) CF ,
ΛΖΓ·ἴσηἄρακαὶἡὑπὸΚΖΓτῇὑπὸΛΖΓ·ἐστὶδὲκαὶἡ FK. And the base BK [is] equal to the base CK. Thus,
ὑπὸΖΓΚγωνίατῇὑπὸΖΓΛἴση. δύοδὴτρίγωνάἐστιτὰ angle BFK is equal to [angle] KFC [Prop. 1.8]. And
ΖΚΓ,ΖΛΓτὰςδύογωνίαςταῖςδυσὶγωνίαιςἴσαςἔχοντα BKF (is equal) to FKC [Prop. 1.8]. Thus, BFC (is)
καὶμίανπλευρὰνμιᾷπλευρᾷἴσηνκοινὴναὐτῶντὴνΖΓ· double KFC, and BKC (is double) FKC. So, for the
καὶτὰςλοιπὰςἄραπλευρὰςταῖςλοιπαῖςπλευραῖςἴσαςἕξει same (reasons), CFD is also double CFL, and DLC (is
καὶτὴνλοιπὴνγωνίαντῇλοιπῇγωνίᾳ·ἴσηἄραἡμὲνΚΓ also double) FLC. And since circumference BC is equal
εὐθεῖατῇΓΛ,ἡδὲὑπὸΖΚΓγωνίατῇὑπὸΖΛΓ.καὶἐπεὶἴση to CD, angle BFC is also equal to CFD [Prop. 3.27].
ἐστὶνἡΚΓτῇΓΛ,διπλῆἄραἡΚΛτῆςΚΓ.διὰτὰαὐταδὴ And BFC is double KFC, and DFC (is double) LFC.
δειχθήσεταικαὶἡΘΚτῆςΒΚδιπλῆ.καίἐστινἡΒΚτῇΚΓ Thus, KFC is also equal to LFC. And angle FCK is also
ἴση·καὶἡΘΚἄρατῇΚΛἐστινἴση.ὁμοίωςδὴδειχθήσεται equal to FCL. So, FKC and FLC are two triangles hav-
121

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 4
καὶἑκάστητῶνΘΗ,ΗΜ,ΜΛἑκατέρᾳτῶνΘΚ,ΚΛἴση· ing two angles equal to two angles, and one side equal
ἰσόπλευρονἄραἐστὶτὸΗΘΚΛΜπεντάγωνον. λέγωδή, to one side, (namely) their common (side) FC. Thus,
ὅτικαὶἰσογώνιον.ἐπεὶγὰρἴσηἐστὶνἡὑπὸΖΚΓγωνίατῇ they will also have the remaining sides equal to the (cor-
ὑπὸΖΛΓ,καὶἐδείχθητῆςμὲνὑπὸΖΚΓδιπλῆἡὑπὸΘΚΛ, responding) remaining sides, and the remaining angle to
τῆςδὲὑπὸΖΛΓδιπλῆἡὑπὸΚΛΜ,καὶἡὑπὸΘΚΛἄρα the remaining angle [Prop. 1.26]. Thus, the straight-line
τῇὑπὸΚΛΜἐστινἴση.ὁμοίωςδὴδειχθήσεταικαὶἑκάστη KC (is) equal to CL, and the angle FKC to FLC. And
τῶνὑπὸΚΘΗ,ΘΗΜ,ΗΜΛἑκατέρᾳτῶνὑπὸΘΚΛ,ΚΛΜ since KC is equal to CL, KL (is) thus double KC. So,
ἴση·αἱπέντεἄραγωνίαιαἱὑπὸΗΘΚ,ΘΚΛ,ΚΛΜ,ΛΜΗ, for the same (reasons), it can be shown that HK (is) also
ΜΗΘἴσαιἀλλήλαιςεἰσίν.ἰσογώνιονἄραἐστὶτὸΗΘΚΛΜ double BK. And BK is equal to KC. Thus, HK is also
πεντάγωνον.ἐδείχθηδὲκαὶἰσόπλευρον,καὶπεριγέγραπται equal to KL. So, similarly, each of HG, GM, and ML
περὶτὸνΑΒΓΔΕκύκλον.
can also be shown (to be) equal to each of HK and KL.
[Περὶτὸνδοθένταἄρακύκλονπεντάγωνονἰσόπλευρόν Thus, pentagon GHKLM is equilateral. So I say that
τεκαὶἰσογώνιονπεριγέγραπται]·ὅπερἔδειποιῆσαι. (it is) also equiangular. For since angle FKC is equal to
FLC, and HKL was shown (to be) double FKC, and
KLM double FLC, HKL is thus also equal to KLM.
So, similarly, each of KHG, HGM, and GML can also
be shown (to be) equal to each of HKL and KLM. Thus,
the five angles GHK, HKL, KLM, LMG, and MGH
are equal to one another. Thus, the pentagon GHKLM
is equiangular. And it was also shown (to be) equilateral,
and has been circumscribed about circle ABCDE.
[Thus, an equilateral and equiangular pentagon has
been circumscribed about the given circle]. (Which is)
the very thing it was required to do.
† See the footnote to Prop. 3.34.. Proposition 13
Εἰςτὸδοθὲνπεντάγωνον,ὅἐστινἰσόπλευρόντεκαὶ To inscribe a circle in a given pentagon, which is equi-
ἰσογώνιον,κύκλονἐγγράψαι.
lateral and equiangular.
Α
A
Η
Μ
G
M
Β
Ζ
Ε
B
F
E
Θ
Λ
H
L
Γ Κ ∆
C K D
῎Εστωτὸδοθὲνπεντάγωνονἰσόπλευρόντεκαὶἰσογώνι- Let ABCDE be the given equilateral and equiangular
οντὸΑΒΓΔΕ·δεῖδὴεἰςτὸΑΒΓΔΕπεντάγωνονκύκλον pentagon. So it is required to inscribe a circle in pentagon
ἐγγράψαι.
ABCDE.
ΤετμήσθωγὰρἑκατέρατῶνὑπὸΒΓΔ,ΓΔΕγωνιῶν For let angles BCD and CDE have each been cut
δίχαὑπὸἑκατέραςτῶνΓΖ,ΔΖεὐθειῶν· καὶἀπὸτοῦΖ in half by each of the straight-lines CF and DF (reσημείου,καθ᾿ὃσυμβάλλουσινἀλλήλαιςαἱΓΖ,ΔΖεὐθεῖαι,
spectively) [Prop. 1.9]. And from the point F , at which
ἐπεζεύχθωσαναἱΖΒ,ΖΑ,ΖΕεὐθεῖαι. καὶἐπεὶἴσηἐστὶν the straight-lines CF and DF meet one another, let the
122

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 4
ἡΒΓτῇΓΔ,κοινὴδὲἡΓΖ,δύοδὴαἱΒΓ,ΓΖδυσὶταῖς straight-lines FB, FA, and FE have been joined. And
ΔΓ,ΓΖἴσαιεἰσίν· καὶγωνίαἡὑπὸΒΓΖγωνίᾳτῇὑπὸ since BC is equal to CD, and CF (is) common, the two
ΔΓΖ[ἐστιν]ἴση·βάσιςἄραἡΒΖβάσειτῇΔΖἐστινἴση, (straight-lines) BC, CF are equal to the two (straightκαὶτὸΒΓΖτρίγωνοντῷΔΓΖτριγώνῳἐστινἴσον,καὶαἱ
lines) DC, CF . And angle BCF [is] equal to angle
λοιπαὶγωνίαιταῖςλοιπαῖςγωνίαιςἴσαιἔσονται,ὑφ᾿ἃςαἱ DCF . Thus, the base BF is equal to the base DF , and
ἴσαιπλευραὶὑποτείνουσιν· ἴσηἄραἡὑπὸΓΒΖγωνίατῇ triangle BCF is equal to triangle DCF , and the remain-
ὑπὸΓΔΖ.καὶἐπεὶδιπλῆἐστινἡὑπὸΓΔΕτῆςὑπὸΓΔΖ, ing angles will be equal to the (corresponding) remain-
ἴσηδὲἡμὲνὑπὸΓΔΕτῇὑπὸΑΒΓ,ἡδὲὑπὸΓΔΖτῇὑπὸ ing angles which the equal sides subtend [Prop. 1.4].
ΓΒΖ,καὶἡὑπὸΓΒΑἄρατῆςὑπὸΓΒΖἐστιδιπλῆ· ἴση Thus, angle CBF (is) equal to CDF . And since CDE
ἄραἡὑπὸΑΒΖγωνίατῇὑπὸΖΒΓ·ἡἄραὑπὸΑΒΓγωνία is double CDF , and CDE (is) equal to ABC, and CDF
δίχατέτμηταιὑπὸτῆςΒΖεὐθείας.ὁμοίωςδὴδειχθήσεται, to CBF , CBA is thus also double CBF . Thus, angle
ὅτικαὶἑκατέρατῶνὑπὸΒΑΕ,ΑΕΔδίχατέτμηταιὑπὸ ABF is equal to FBC. Thus, angle ABC has been cut
ἑκατέραςτῶνΖΑ,ΖΕεὐθειῶν. ἤχθωσανδὴἀπὸτοῦΖ in half by the straight-line BF . So, similarly, it can be
σημείουἐπὶτὰςΑΒ,ΒΓ,ΓΔ,ΔΕ,ΕΑ εὐθείαςκάθετοι shown that BAE and AED have been cut in half by the
αἱΖΗ,ΖΘ,ΖΚ,ΖΛ,ΖΜ.καὶἐπεὶἴσηἐστὶνἡὑπὸΘΓΖ straight-lines FA and FE, respectively. So let FG, FH,
γωνίατῇὑπὸΚΓΖ,ἐστὶδὲκαὶὀρθὴἡὑπὸΖΘΓ[ὀρθῇ] FK, FL, and FM have been drawn from point F , perτῇὑπὸΖΚΓἴση,δύοδὴτρίγωνάἐστιτὰΖΘΓ,ΖΚΓτὰς
pendicular to the straight-lines AB, BC, CD, DE, and
δύογωνίαςδυσὶγωνίαιςἴσαςἔχοντακαὶμίανπλευρὰνμιᾷ EA (respectively) [Prop. 1.12]. And since angle HCF
πλευρᾷἴσηνκοινὴναὐτῶντὴνΖΓὑποτείνουσανὑπὸμίαν is equal to KCF , and the right-angle FHC is also equal
τῶνἴσωνγωνιῶν·καὶτὰςλοιπὰςἄραπλευρὰςταῖςλοιπαῖς to the [right-angle] FKC, FHC and FKC are two triπλευραῖςἴσαςἕξει·ἴσηἄραἡΖΘκάθετοςτῂΖΚκαθέτῳ.
angles having two angles equal to two angles, and one
ὁμοίωςδὴδειχθήσεται,ὅτικαὶἑκάστητῶνΖΛ,ΖΜ,ΖΗ side equal to one side, (namely) their common (side) FC,
ἑκατέρᾳτῶνΖΘ,ΖΚἴσηἐστίν·αἱπέντεἄραεὐθεῖαιαἱΖΗ, subtending one of the equal angles. Thus, they will also
ΖΘ,ΖΚ,ΖΛ,ΖΜἴσαιἀλλήλαιςεἰσίν. ὁἄρακέντρῳτῷΖ have the remaining sides equal to the (corresponding)
διαστήματιδὲἑνὶτῶνΗ,Θ,Κ,Λ,Μκύκλοςγραφόμενος remaining sides [Prop. 1.26]. Thus, the perpendicular
ἥξεικαὶδιὰτῶνλοιπῶνσημείωνκαὶἐφάψεταιτῶνΑΒ,ΒΓ, FH (is) equal to the perpendicular FK. So, similarly, it
ΓΔ,ΔΕ,ΕΑεὐθειῶνδιὰτὸὀρθὰςεἶναιτὰςπρὸςτοῖςΗ, can be shown that FL, FM, and FG are each equal to
Θ,Κ,Λ,Μσημείοιςγωνίας. εἰγὰροὐκἐφάψεταιαὐτῶν, each of FH and FK. Thus, the five straight-lines FG,
ἀλλὰτεμεῖαὐτάς,συμβήσεταιτὴντῇδιαμέτρῳτοῦκύκλου FH, FK, FL, and FM are equal to one another. Thus,
πρὸςὀρθὰςἀπ᾿ἄκραςἀγομένηνἐντὸςπίπτειντοῦκύκλου· the circle drawn with center F , and radius one of G, H,
ὅπερἄτοπονἐδείχθη.οὐκἄραὁκέντρῳτῷΖδιαστήματιδὲ K, L, or M, will also go through the remaining points,
ἑνὶτῶνΗ,Θ,Κ,Λ,Μσημείωνγραφόμενοςκύκλοςτεμεῖ and will touch the straight-lines AB, BC, CD, DE, and
τὰςΑΒ,ΒΓ,ΓΔ,ΔΕ,ΕΑεὐθείας·ἐφάψεταιἄρααὐτῶν. EA, on account of the angles at points G, H, K, L, and
γεγράφθωὡςὁΗΘΚΛΜ.
M being right-angles. For if it does not touch them, but
Εἰςἄρατὸδοθὲνπεντάγωνον,ὅἐστινἰσόπλευρόντε cuts them, it follows that a (straight-line) drawn at rightκαὶἰσογώνιον,κύκλοςἐγγέγραπται·ὅπερἔδειποιῆσαι.
angles to the diameter of the circle, from its extremity,
falls inside the circle. The very thing was shown (to be)
absurd [Prop. 3.16]. Thus, the circle drawn with center
F , and radius one of G, H, K, L, or M, does not cut
the straight-lines AB, BC, CD, DE, or EA. Thus, it will
touch them. Let it have been drawn, like GHKLM (in
the figure).
Thus, a circle has been inscribed in the given pentagon
which is equilateral and equiangular. (Which is)
the very thing it was required to do.
. Proposition 14
Περὶτὸδοθὲνπεντάγωνον,ὅἐστινἰσόπλευρόντεκαὶ To circumscribe a circle about a given pentagon which
ἰσογώνιον,κύκλονπεριγράψαι.
is equilateral and equiangular.
῎Εστωτὸδοθὲνπεντάγωνον,ὅἐστινἰσόπλευρόντεκαὶ Let ABCDE be the given pentagon which is equilat-
123

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 4
ἰσογώνιον,τὸΑΒΓΔΕ·δεῖδὴπερὶτὸΑΒΓΔΕπεντάγωνον eral and equiangular. So it is required to circumscribe a
κύκλονπεριγράψαι.
circle about the pentagon ABCDE.
Α
A
Β
Ζ
Ε
B
F
E
Γ
∆
Τετμήσθω δὴ ἑκατέρα τῶν ὑπὸ ΒΓΔ, ΓΔΕ γωνιῶν So let angles BCD and CDE have been cut in half by
δίχαὑπὸἑκατέραςτῶνΓΖ,ΔΖ,καὶἀπὸτοῦΖσημείου, the (straight-lines) CF and DF , respectively [Prop. 1.9].
καθ᾿ ὃσυμβάλλουσιναἱεὐθεῖαι, ἐπὶτὰΒ, Α,Εσημεῖα And let the straight-lines FB, FA, and FE have been
ἐπεζεύχθωσανεὐθεῖαιαἱΖΒ,ΖΑ,ΖΕ.ὁμοίωςδὴτῷπρὸ joined from point F , at which the straight-lines meet,
τούτουδειχθήσεται,ὅτικαὶἑκάστητῶνὑπὸΓΒΑ,ΒΑΕ, to the points B, A, and E (respectively). So, similarly,
ΑΕΔγωνιῶνδίχατέτμηταιὑπὸἑκάστηςτῶνΖΒ,ΖΑ,ΖΕ to the (proposition) before this (one), it can be shown
εὐθειῶν.καὶἐπεὶἴσηἐστὶνἡὑπὸΒΓΔγωνίατῇὑπὸΓΔΕ, that angles CBA, BAE, and AED have also been cut
καίἐστιτῆςμὲνὑπὸΒΓΔἡμίσειαἡὑπὸΖΓΔ,τῆςδὲὑπὸ in half by the straight-lines FB, FA, and FE, respec-
ΓΔΕἡμίσειαἡὑπὸΓΔΖ,καὶἡὑπὸΖΓΔἄρατῇὑπὸΖΔΓ tively. And since angle BCD is equal to CDE, and FCD
ἐστινἴση·ὥστεκαὶπλευρὰἡΖΓπλευρᾷτῇΖΔἐστινἴση. is half of BCD, and CDF half of CDE, FCD is thus
ὁμοίωςδὴδειχθήσεται,ὅτικαὶἑκάστητῶνΖΒ,ΖΑ,ΖΕ also equal to FDC. So that side FC is also equal to side
ἑκατέρᾳτῶν ΖΓ,ΖΔἐστιν ἴση· αἱπέντεἄραεὐθεῖαιαἱ FD [Prop. 1.6]. So, similarly, it can be shown that FB,
ΖΑ,ΖΒ,ΖΓ,ΖΔ,ΖΕἴσαιἀλλήλαιςεἰσίν. ὀἄρακέντρῳ FA, and FE are also each equal to each of FC and FD.
τῷΖκαὶδιαστήματιἑνὶτῶνΖΑ,ΖΒ,ΖΓ,ΖΔ,ΖΕκύκλος Thus, the five straight-lines FA, FB, FC, FD, and FE
γραφόμενοςἥξεικαὶδιὰτῶνλοιπῶνσημείωνκαὶἔσταιπε- are equal to one another. Thus, the circle drawn with
ριγεγραμμένος.περιγεγράφθωκαὶἔστωὁΑΒΓΔΕ. center F , and radius one of FA, FB, FC, FD, or FE,
Περὶἄρατὸδοθὲνπεντάγωνον,ὅἐστινἰσόπλευρόντε will also go through the remaining points, and will have
καὶἰσογώνιον,κύκλοςπεριγέγραπται·ὅπερἔδειποιῆσαι. been circumscribed. Let it have been (so) circumscribed,
and let it be ABCDE.
Thus, a circle has been circumscribed about the given
pentagon, which is equilateral and equiangular. (Which
is) the very thing it was required to do.
. Proposition 15
Εἰςτὸνδοθέντακύκλονἑξάγωνονἰσόπλευρόντεκαὶ To inscribe an equilateral and equiangular hexagon in
ἰσογώνιονἐγγράψαι.
a given circle.
῎Εστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔΕΖ· δεῖ δὴ εἰς τὸν Let ABCDEF be the given circle. So it is required to
ΑΒΓΔΕΖκύκλονἑξάγωνονἰσόπλευρόντεκαὶἰσογώνιον inscribe an equilateral and equiangular hexagon in circle
ἐγγράψαι. ABCDEF .
῎Ηχθω τοῦ ΑΒΓΔΕΖ κύκλου διάμετρος ἡ ΑΔ, καὶ Let the diameter AD of circle ABCDEF have been
εἰλήφθω τὸ κέντρον τοῦ κύκλουτὸ Η, καὶ κέντρῳ μὲν drawn, † and let the center G of the circle have been
τῷΔδιαστήματιδὲτῷΔΗκύκλοςγεγράφθωὁΕΗΓΘ, found [Prop. 3.1]. And let the circle EGCH have been
καὶἐπιζευχθεῖσαιαἱΕΗ,ΓΗδιήχθωσανἐπὶτὰΒ,Ζσημεῖα, drawn, with center D, and radius DG. And EG and CG
καὶἐπεζεύχθωσαναἱΑΒ,ΒΓ,ΓΔ,ΔΕ,ΕΖ,ΖΑ·λέγω,ὅτι being joined, let them have been drawn across (the cir-
C
D
124

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 4
τὸΑΒΓΔΕΖἑξάγωνονἰσόπλευρόντέἐστικαὶἰσογώνιον. cle) to points B and F (respectively). And let AB, BC,
CD, DE, EF , and FA have been joined. I say that the
hexagon ABCDEF is equilateral and equiangular.
Θ
H
∆
D
Γ
Η
Ε
C
G
E
Β
Ζ
B
F
Α
A
᾿Επεὶ γὰρ τὸ Η σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓΔΕΖ For since point G is the center of circle ABCDEF ,
κύκλου,ἴσηἐστὶνἡΗΕτῇΗΔ.πάλιν,ἐπεὶτὸΔσημεῖον GE is equal to GD. Again, since point D is the cenκέντρονἐστὶτοῦΗΓΘκύκλου,ἴσηἐστὶνἡΔΕτῇΔΗ.
ter of circle GCH, DE is equal to DG. But, GE was
ἀλλ᾿ἡΗΕτῇΗΔἐδείχθηἴση·καὶἡΗΕἄρατῇΕΔἴση shown (to be) equal to GD. Thus, GE is also equal to
ἐστίν·ἰσόπλευρονἄραἐστὶτὸΕΗΔτρίγωνον·καὶαἱτρεῖς ED. Thus, triangle EGD is equilateral. Thus, its three
ἄρααὐτοῦγωνίαιαἱὑπὸΕΗΔ,ΗΔΕ,ΔΕΗἴσαιἀλλήλαις angles EGD, GDE, and DEG are also equal to one anεἰσίν,ἐπειδήπερτῶνἰσοσκελῶντριγώνωναἱπρὸςτῇβάσει
other, inasmuch as the angles at the base of isosceles triγωνίαιἴσαιἀλλήλαιςεἰσίν·καίεἰσιναἱτρεῖςτοῦτριγώνου
angles are equal to one another [Prop. 1.5]. And the
γωνίαιδυσὶνὀρθαῖςἴσαι·ἡἄραὑπὸΕΗΔγωνίατρίτονἐστὶ three angles of the triangle are equal to two right-angles
δύοὀρθῶν. ὁμοίωςδὴδειχθήσεταικαὶἡὑπὸΔΗΓτρίτον [Prop. 1.32]. Thus, angle EGD is one third of two rightδύοὀρθῶν.καὶἐπεὶἡΓΗεὐθεῖαἐπὶτὴνΕΒσταθεῖσατὰς
angles. So, similarly, DGC can also be shown (to be)
ἐφεξῆςγωνίαςτὰςὑπὸΕΗΓ,ΓΗΒδυσὶνὀρθαῖςἴσαςποιεῖ, one third of two right-angles. And since the straight-line
καὶλοιπὴἄραἡὑπὸΓΗΒτρίτονἐστὶδύοὀρθῶν·αἱἄρα CG, standing on EB, makes adjacent angles EGC and
ὑπὸΕΗΔ,ΔΗΓ,ΓΗΒγωνίαιἴσαιἀλλήλαιςεἰσίν·ὥστεκαὶ CGB equal to two right-angles [Prop. 1.13], the remainαἱκατὰκορυφὴναὐταῖςαἱὑπὸΒΗΑ,ΑΗΖ,ΖΗΕἴσαιεἰσὶν
ing angle CGB is thus also one third of two right-angles.
[ταῖςὑπὸΕΗΔ,ΔΗΓ,ΓΗΒ].αἱἓξἄραγωνίαιαἱὑπὸΕΗΔ, Thus, angles EGD, DGC, and CGB are equal to one an-
ΔΗΓ,ΓΗΒ,ΒΗΑ,ΑΗΖ,ΖΗΕἴσαιἀλλήλαιςεἰσίν. αἱδὲ other. And hence the (angles) opposite to them BGA,
ἴσαιγωνίαιἐπὶἴσωνπεριφερειῶνβεβήκασιν·αἱἓξἄραπε- AGF , and FGE are also equal [to EGD, DGC, and
ριφέρειαιαἱΑΒ,ΒΓ,ΓΔ,ΔΕ,ΕΖ,ΖΑἴσαιἀλλήλαιςεἰσίν. CGB (respectively)] [Prop. 1.15]. Thus, the six angles
ὑπὸδὲτὰςἴσαςπεριφερείαςαἱἴσαιεὐθεῖαιὑποτείνουσιν· EGD, DGC, CGB, BGA, AGF , and FGE are equal
αἱἓξἄραεὐθεῖαιἴσαιἀλλήλαιςεἰσίν·ἰσόπλευρονἄραἐστὶ to one another. And equal angles stand on equal cirτοΑΒΓΔΕΖἑξάγωνον.
λέγωδή,ὅτικαὶἰσογώνιον. ἐπεὶ cumferences [Prop. 3.26]. Thus, the six circumferences
γὰρἴσηἐστὶνἡΖΑπεριφέρειατῇΕΔπεριφερείᾳ, κοινὴ AB, BC, CD, DE, EF , and FA are equal to one anπροσκείσθωἡΑΒΓΔπεριφέρεια·ὅληἄραἡΖΑΒΓΔὅλῃ
other. And equal circumferences are subtended by equal
τῇΕΔΓΒΑἐστινἴση· καὶβέβηκενἐπὶμὲντῆςΖΑΒΓΔ straight-lines [Prop. 3.29]. Thus, the six straight-lines
περιφερείαςἡὑπὸΖΕΔγωνία,ἐπὶδὲτῆςΕΔΓΒΑπερι- (AB, BC, CD, DE, EF , and FA) are equal to one
φερείαςἡὑπὸΑΖΕγωνία·ἴσηἄραἡὑπὸΑΖΕγωνίατῇ another. Thus, hexagon ABCDEF is equilateral. So,
ὑπὸΔΕΖ.ὁμοίωςδὴδειχθήσεται,ὅτικαὶαἱλοιπαὶγωνίαι I say that (it is) also equiangular. For since circumferτοῦΑΒΓΔΕΖἑξαγώνουκατὰμίανἴσαιεἰσὶνἑκατέρᾳτῶν
ence FA is equal to circumference ED, let circumference
ὑπὸΑΖΕ,ΖΕΔγωνιῶν·ἰσογώνιονἄραἐστὶτὸΑΒΓΔΕΖ ABCD have been added to both. Thus, the whole of
ἑξάγωνον.ἐδείχθηδὲκαὶἰσόπλευρον·καὶἐγγέγραπταιεἰς FABCD is equal to the whole of EDCBA. And angle
τὸνΑΒΓΔΕΖκύκλον.
FED stands on circumference FABCD, and angle AFE
Εἰςἄρατὸνδοθέντακύκλονἑξάγωνονἰσόπλευρόντε on circumference EDCBA. Thus, angle AFE is equal
125

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 4
καὶἰσογώνιονἐγγέγραπται·ὅπερἔδειποιῆσαι.
to DEF [Prop. 3.27]. Similarly, it can also be shown
that the remaining angles of hexagon ABCDEF are individually
equal to each of the angles AFE and FED.
Thus, hexagon ABCDEF is equiangular. And it was also
shown (to be) equilateral. And it has been inscribed in
circle ABCDE.
Thus, an equilateral and equiangular hexagon has
been inscribed in the given circle. (Which is) the very
thing it was required to do.
ÈÖ×Ñ.
᾿Εκδὴτούτουφανερόν,ὅτιἡτοῦἑξαγώνουπλευρὰἴση So, from this, (it is) manifest that a side of the
Corollary
ἐστὶτῇἐκτοῦκέντρουτοῦκύκλου.
hexagon is equal to the radius of the circle.
῾Ομοίωςδὲτοῖςἐπὶτοῦπενταγώνουἐὰνδιὰτῶνκατὰ And similarly to a pentagon, if we draw tangents
τὸνκύκλονδιαιρέσεωνἐφαπτομέναςτοῦκύκλουἀγάγωμεν, to the circle through the (sixfold) divisions of the (cirπεριγραφήσεται
περὶ τὸν κύκλον ἑξάγωνον ἰσόπλευρόν cumference of the) circle, an equilateral and equianguτε
καὶ ἰσογώνιον ἀκολούθως τοῖς ἐπὶ τοῦ πενταγώνου lar hexagon can be circumscribed about the circle, analoεἰρημένοις.καὶἔτιδιὰτῶνὁμοίωντοῖςἐπὶτοῦπενταγώνου
gously to the aforementioned pentagon. And, further, by
εἰρημένοιςεἰςτὸδοθὲνἑξάγωνονκύκλονἐγγράψομέντε (means) similar to the aforementioned pentagon, we can
καὶπεριγράψομεν·ὅπερἔδειποιῆσαι.
inscribe and circumscribe a circle in (and about) a given
hexagon. (Which is) the very thing it was required to do.
† See the footnote to Prop. 4.6.¦. Proposition 16
Εἰςτὸνδοθέντακύκλονπεντεκαιδεκάγωνονἰσόπλευρόν To inscribe an equilateral and equiangular fifteenτεκαὶἰσογώνιονἐγγράψαι.
sided figure in a given circle.
Α
A
Β
B
Ε
E
Γ
∆
C
D
῎ΕστωὁδοθεὶςκύκλοςὁΑΒΓΔ·δεῖδὴεἰςτὸνΑΒΓΔ Let ABCD be the given circle. So it is required to inκύκλονπεντεκαιδεκάγωνονἰσόπλευρόντε
καὶἰσογώνιον scribe an equilateral and equiangular fifteen-sided figure
ἐγγράψαι.
in circle ABCD.
᾿ΕγγεγράφθωεἰςτὸνΑΒΓΔκύκλοντριγώνουμὲνἰσο- Let the side AC of an equilateral triangle inscribed
πλεύρουτοῦεἰςαὐτὸνἐγγραφομένουπλευρὰἡΑΓ,πεν- in (the circle) [Prop. 4.2], and (the side) AB of an (inταγώνουδὲἰσοπλεύρουἡΑΒ·οἵωνἄραἐστὶνὁΑΒΓΔ
scribed) equilateral pentagon [Prop. 4.11], have been inκύκλοςἴσωντμήματωνδεκαπέντε,
τοιούτωνἡμὲνΑΒΓ scribed in circle ABCD. Thus, just as the circle ABCD
περιφέρειατρίτονοὖσατοῦκύκλουἔσταιπέντε,ἡδὲΑΒ is (made up) of fifteen equal pieces, the circumference
περιφέρειαπέμτονοὖσατοῦκύκλουἔσταιτριῶν·λοιπὴἄρα ABC, being a third of the circle, will be (made up) of five
126

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 4
ἡ ΒΓτῶν ἴσων δύο. τετμήσθω ἡΒΓδίχακατὰτὸΕ· such (pieces), and the circumference AB, being a fifth of
ἑκατέραἄρατῶνΒΕ,ΕΓπεριφερειῶνπεντεκαιδέκατόνἐστι the circle, will be (made up) of three. Thus, the remainτοῦΑΒΓΔκύκλου.
der BC (will be made up) of two equal (pieces). Let (cir-
᾿ΕὰνἄραἐπιζεύξαντεςτὰςΒΕ,ΕΓἴσαςαὐταῖςκατὰτὸ cumference) BC have been cut in half at E [Prop. 3.30].
συνεχὲςεὐθείαςἐναρμόσωμενεἰςτὸνΑΒΓΔ[Ε]κύκλον, Thus, each of the circumferences BE and EC is one fif-
ἔσταιεἰςαὐτὸνἐγγεγραμμένονπεντεκαιδεκάγωνονἰσόπλευ- teenth of the circle ABCDE.
ρόντεκαὶἰσογώνιον·ὅπερἔδειποιῆσαι.
Thus, if, joining BE and EC, we continuously in-
῾Ομοίως δὲ τοῖς ἐπὶ τοῦ πενταγώνου ἐὰν διὰ τῶν sert straight-lines equal to them into circle ABCD[E]
κατὰ τὸν κύκλον διαιρέσεων ἐφαπτομένας τοῦ κύκλου [Prop. 4.1], then an equilateral and equiangular fifteen-
ἀγάγωμεν, περιγραφήσεται περὶ τὸν κύκλον πεντεκαι- sided figure will have been inserted into (the circle).
δεκάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον. ἔτι δὲ διὰ (Which is) the very thing it was required to do.
τῶνὁμοίωντοῖςἐπὶτοῦπενταγώνουδείξεωνκαὶεἰςτὸ And similarly to the pentagon, if we draw tangents to
δοθὲνπεντεκαιδεκάγωνονκύκλονἐγγράψομέντεκαὶπε- the circle through the (fifteenfold) divisions of the (cirριγράψομεν·ὅπερἔδειποιῆσαι.
cumference of the) circle, we can circumscribe an equilateral
and equiangular fifteen-sided figure about the circle.
And, further, through similar proofs to the pentagon, we
can also inscribe and circumscribe a circle in (and about)
a given fifteen-sided figure. (Which is) the very thing it
was required to do.
127

128

ELEMENTS BOOK 5
Proportion †
† The theory of proportion set out in this book is generally attributed to Eudoxus of Cnidus. The novel feature of this theory is its ability to deal
with irrational magnitudes, which had hitherto been a major stumbling block for Greek mathematicians. Throughout the footnotes in this book,
α, β, γ, etc., denote general (possibly irrational) magnitudes, whereas m, n, l, etc., denote positive integers.
129

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
ÎÇÖÓ.
αʹ. Μέρος ἐστὶ μέγεθος μεγέθους τὸ ἔλασσον τοῦ 1. A magnitude is a part of a(nother) magnitude, the
Definitions
μείζονος,ὅτανκαταμετρῇτὸμεῖζον. lesser of the greater, when it measures the greater. †
βʹ.Πολλαπλάσιονδὲτὸμεῖζοντοῦἐλάττονος,ὅτανκα- 2. And the greater (magnitude is) a multiple of the
ταμετρῆταιὑπὸτοῦἐλάττονος.
lesser when it is measured by the lesser.
γʹ. Λόγος ἐστὶ δύο μεγεθῶν ὁμογενῶν ἡ κατὰ πη- 3. A ratio is a certain type of condition with respect to
λικότητάποιασχέσις. size of two magnitudes of the same kind. ‡
δʹ.Λόγονἔχεινπρὸςἄλληλαμεγέθηλέγεται,ἃδύναται 4. (Those) magnitudes are said to have a ratio with reπολλαπλασιαζόμεναἀλλήλωνὑπερέχειν.
spect to one another which, being multiplied, are capable
εʹ. ᾿Εν τῷ αὐτῷ λόγῳ μεγέθη λέγεται εἶναι πρῶτον of exceeding one another. §
πρὸς δεύτερον καὶ τρίτον πρὸς τέταρτον, ὅταν τὰ τοῦ 5. Magnitudes are said to be in the same ratio, the first
πρώτουκαίτρίτουἰσάκιςπολλαπλάσιατῶντοῦδευτέρου to the second, and the third to the fourth, when equal
καὶτετάρτουἰσάκιςπολλαπλασίωνκαθ᾿ὁποιονοῦνπολλα- multiples of the first and the third either both exceed, are
πλασιασμὸνἑκάτερονἑκατέρουἢἅμαὑπερέχῃἢἅμαἴσαᾖ both equal to, or are both less than, equal multiples of the
ἢἅμαἐλλείπῇληφθέντακατάλληλα.
second and the fourth, respectively, being taken in correϛʹ.Τὰδὲτὸν
αὐτὸνἔχονταλόγονμεγέθηἀνάλογον sponding order, according to any kind of multiplication
καλείσθω. whatever.
ζʹ. ῞Οταν δὲ τῶν ἰσάκις πολλαπλασίων τὸ μὲν τοῦ 6. And let magnitudes having the same ratio be called
πρώτου πολλαπλάσιον ὑπερέχῃ τοῦ τοῦ δευτέρου πολ- proportional. ∗
λαπλασίου, τὸ δὲ τοῦ τρίτου πολλαπλάσιον μὴ ὑπερέχῃ 7. And when for equal multiples (as in Def. 5), the
τοῦτοῦτετάρτουπολλαπλασίου,τότετὸπρῶτονπρὸςτὸ multiple of the first (magnitude) exceeds the multiple of
δεύτερονμείζοναλόγονἔχεινλέγεται,ἤπερτὸτρίτονπρὸς the second, and the multiple of the third (magnitude)
τὸτέταρτον.
does not exceed the multiple of the fourth, then the first
ηʹ.Ἀναλογίαδὲἐντρισὶνὅροιςἐλαχίστηἐστίν. (magnitude) is said to have a greater ratio to the second
θʹ.῞Οτανδὲτρίαμεγέθηἀνάλογονᾖ,τὸπρῶτονπρὸς than the third (magnitude has) to the fourth.
τὸ τρίτον διπλασίονα λόγον ἔχειν λέγεταιἤπερ πρὸς τὸ 8. And a proportion in three terms is the smallest
δεύτερον. (possible). $
ιʹ.῞Οτανδὲτέσσαραμεγέθηἀνάλογονᾖ, τὸπρῶτον 9. And when three magnitudes are proportional, the
πρὸς τὸ τέταρτον τριπλασίοναλόγονἔχειν λέγεταιἤπερ first is said to have to the third the squared ‖ ratio of that
πρὸςτὸδεύτερον,καὶἀεὶἑξῆςὁμοίως,ὡςἂνἡἀναλογία (it has) to the second. ††
ὑπάρχῃ.
10. And when four magnitudes are (continuously)
ιαʹ. ῾Ομόλογα μεγέθη λέγεται τὰ μὲν ἡγούμενα τοῖς proportional, the first is said to have to the fourth the
ἡγουμένοιςτὰδὲἑπόμενατοῖςἑπομένοις.
cubed ‡‡ ratio of that (it has) to the second. §§ And so on,
ιβʹ.᾿Εναλλὰξλόγοςἐστὶλῆψιςτοῦἡγουμένουπρὸςτὸ similarly, in successive order, whatever the (continuous)
ἡγούμενονκαὶτοῦἑπομένουπρὸςτὸἑπόμενον. proportion might be.
ιγʹ. Ἀνάπαλιν λόγος ἐστὶ λῆψις τοῦ ἑπομένου ὡς 11. These magnitudes are said to be corresponding
ἡγουμένουπρὸςτὸἡγούμενονὡςἑπόμενον.
(magnitudes): the leading to the leading (of two ratios),
ιδʹ.Σύνθεσιςλόγουἐστὶλῆψιςτοῦἡγουμένουμετὰτοῦ and the following to the following.
ἑπομένουὡςἑνὸςπρὸςαὐτὸτὸἑπόμενον.
12. An alternate ratio is a taking of the (ratio of the)
ιεʹ.Διαίρεσιςλόγουἐστὶλῆψιςτῆςὑπεροχῆς,ᾗὑπερέχει leading (magnitude) to the leading (of two equal ratios),
τὸἡγούμενοντοῦἑπομένου,πρὸςαὐτὸτὸἑπόμενον. and (setting it equal to) the (ratio of the) following (magιϛʹ.Ἀναστροφὴλόγουἐστὶλῆψιςτοῦἡγουμένουπρὸς
nitude) to the following.
τὴνὑπεροχήν,ᾗὑπερέχειτὸἡγούμενοντοῦἑπομένου. 13. An inverse ratio is a taking of the (ratio of the) folιζʹ.Δι᾿ἴσουλόγοςἐστὶπλειόνωνὄντωνμεγεθῶνκαὶ
lowing (magnitude) as the leading and the leading (mag-
ἄλλωναὐτοῖςἴσωντὸπλῆθοςσύνδυολαμβανομένωνκαὶ nitude) as the following. ∗∗
ἐντῷαὐτῷλόγῳ,ὅτανᾖὡςἐντοῖςπρώτοιςμεγέθεσιτὸ 14. A composition of a ratio is a taking of the (ratio of
πρῶτονπρὸςτὸἔσχατον,οὕτωςἐντοῖςδευτέροιςμεγέθεσι the) leading plus the following (magnitudes), as one, to
τὸπρῶτονπρὸςτὸἔσχατον· ἢἄλλως·λῆψιςτῶνἄκρων the following (magnitude) by itself. $$
130

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
καθ᾿ὑπεξαίρεσιντῶνμέσων.
15. A separation of a ratio is a taking of the (ratio
ιηʹ.Τεταραγμένηδὲἀναλογίαἐστίν,ὅταντριῶνὄντων of the) excess by which the leading (magnitude) exceeds
μεγεθῶνκαὶἄλλωναὐτοῖςἴσωντὸπλῆθοςγίνηταιὡςμὲν the following to the following (magnitude) by itself. ‖‖
ἐντοῖςπρώτοιςμεγέθεσινἡγούμενονπρὸςἐπόμενον,οὕτως 16. A conversion of a ratio is a taking of the (ratio
ἐντοῖςδευτέροιςμεγέθεσινἡγούμενονπρὸςἑπόμενον,ὡς of the) leading (magnitude) to the excess by which the
δὲἐντοῖςπρώτοιςμεγέθεσινἑπόμενονπρὸςἄλλοτι,οὕτως leading (magnitude) exceeds the following. †††
ἐντοῖςδευτέροιςἄλλοτιπρὸςἡγούμενον.
17. There being several magnitudes, and other (magnitudes)
of equal number to them, (which are) also in the
same ratio taken two by two, a ratio via equality (or ex
aequali) occurs when as the first is to the last in the first
(set of) magnitudes, so the first (is) to the last in the second
(set of) magnitudes. Or alternately, (it is) a taking of
the (ratio of the) outer (magnitudes) by the removal of
the inner (magnitudes). ‡‡‡
18. There being three magnitudes, and other (magnitudes)
of equal number to them, a perturbed proportion
occurs when as the leading is to the following in the first
(set of) magnitudes, so the leading (is) to the following
in the second (set of) magnitudes, and as the following
(is) to some other (i.e., the remaining magnitude) in the
first (set of) magnitudes, so some other (is) to the leading
in the second (set of) magnitudes. §§§
† In other words, α is said to be a part of β if β = m α.
‡ In modern notation, the ratio of two magnitudes, α and β, is denoted α : β.
§ In other words, α has a ratio with respect to β if m α > β and n β > α, for some m and n.
In other words, α : β :: γ : δ if and only if m α > n β whenever m γ > n δ, and m α = n β whenever m γ = n δ, and m α < n β whenever
m γ < n δ, for all m and n. This definition is the kernel of Eudoxus’ theory of proportion, and is valid even if α, β, etc., are irrational.
∗ Thus if α and β have the same ratio as γ and δ then they are proportional. In modern notation, α : β :: γ : δ.
$ In modern notation, a proportion in three terms—α, β, and γ—is written: α : β :: β : γ.
‖ Literally, “double”.
†† In other words, if α : β :: β : γ then α : γ :: α 2 : β 2 .
‡‡ Literally, “triple”.
§§ In other words, if α : β :: β : γ :: γ : δ then α : δ :: α 3 : β 3 .
In other words, if α : β :: γ : δ then the alternate ratio corresponds to α : γ :: β : δ.
∗∗ In other words, if α : β then the inverse ratio corresponds to β : α.
$$ In other words, if α : β then the composed ratio corresponds to α + β : β.
‖‖ In other words, if α : β then the separated ratio corresponds to α − β : β.
††† In other words, if α : β then the converted ratio corresponds to α : α − β.
‡‡‡ In other words, if α, β, γ are the first set of magnitudes, and δ, ǫ, ζ the second set, and α : β : γ :: δ : ǫ : ζ, then the ratio via equality (or ex
aequali) corresponds to α : γ :: δ : ζ.
§§§ In other words, if α, β, γ are the first set of magnitudes, and δ, ǫ, ζ the second set, and α : β :: δ : ǫ as well as β : γ :: ζ : δ, then the proportion
is said to be perturbed.
.
Proposition 1 †
᾿Εὰνᾖὁποσαοῦνμεγέθηὁποσωνοῦνμεγεθῶνἴσωντὸ If there are any number of magnitudes whatsoever
πλῆθοςἕκαστονἑκάστουἰσάκιςπολλαπλάσιον,ὁσαπλάσιόν (which are) equal multiples, respectively, of some (other)
ἐστινἓντῶνμεγεθῶνἑνός,τοσαυταπλάσιαἔσταικαὶτὰ magnitudes, of equal number (to them), then as many
131

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
πάντατῶνπάντων.
Α
Η Β Γ Θ ∆
times as one of the (first) magnitudes is (divisible) by
one (of the second), so many times will all (of the first
magnitudes) also (be divisible) by all (of the second).
A G B C H D
Ε
Ζ
῎Εστω ὁποσαοῦνμεγέθητὰΑΒ,ΓΔὁποσωνοῦνμε- Let there be any number of magnitudes whatsoever,
γεθῶντῶνΕ,Ζἴσωντὸπλῆθοςἕκαστονἑκάστουἰσάκις AB, CD, (which are) equal multiples, respectively, of
πολλαπλάσιον·λέγω,ὅτιὁσαπλάσιόνἐστιτὸΑΒτοῦΕ, some (other) magnitudes, E, F , of equal number (to
τοσαυταπλάσιαἔσταικαὶτὰΑΒ,ΓΔτῶνΕ,Ζ. them). I say that as many times as AB is (divisible) by E,
᾿ΕπεὶγὰρἰσάκιςἐστὶπολλαπλάσιοντὸΑΒτοῦΕκαὶ so many times will AB, CD also be (divisible) by E, F .
τὸΓΔτοῦΖ,ὅσαἄραἐστὶνἐντῷΑΒμεγέθηἴσατῷΕ, For since AB, CD are equal multiples of E, F , thus
τοσαῦτακαὶἐντῷΓΔἴσατῷΖ.διῃρήσθωτὸμὲνΑΒεἰςτὰ as many magnitudes as (there) are in AB equal to E, so
τῷΕμεγέθηἴσατὰΑΗ,ΗΒ,τὸδὲΓΔεἰςτὰτῷΖἴσατὰ many (are there) also in CD equal to F . Let AB have
ΓΘ,ΘΔ·ἔσταιδὴἴσοντὸπλῆθοςτῶνΑΗ,ΗΒτῷπλήθει been divided into magnitudes AG, GB, equal to E, and
τῶνΓΘ,ΘΔ.καὶἐπεὶἴσονἐστὶτὸμὲνΑΗτῷΕ,τὸδὲΓΘ CD into (magnitudes) CH, HD, equal to F . So, the
τῷΖ,ἴσονἄρατὸΑΗτῷΕ,καὶτὰΑΗ,ΓΘτοῖςΕ,Ζ.διὰ number of (divisions) AG, GB will be equal to the numτὰαὐτὰδὴἴσονἐστὶτὸΗΒτῷΕ,καὶτὰΗΒ,ΘΔτοῖςΕ,
ber of (divisions) CH, HD. And since AG is equal to E,
Ζ·ὅσαἄραἐστὶνἐντῷΑΒἴσατῷΕ,τοσαῦτακαὶἐντοῖς and CH to F , AG (is) thus equal to E, and AG, CH to E,
ΑΒ,ΓΔἴσατοῖςΕ,Ζ·ὁσαπλάσιονἄραἐστὶτὸΑΒτοῦΕ, F . So, for the same (reasons), GB is equal to E, and GB,
τοσαυταπλάσιαἔσταικαὶτὰΑΒ,ΓΔτῶνΕ,Ζ. HD to E, F . Thus, as many (magnitudes) as (there) are
᾿Εὰν ἄρα ᾖ ὁποσαοῦν μεγέθη ὁποσωνοῦν μεγεθῶν in AB equal to E, so many (are there) also in AB, CD
ἴσων τὸ πλῆθος ἕκαστον ἑκάστου ἰσάκις πολλαπλάσιον, equal to E, F . Thus, as many times as AB is (divisible)
ὁσαπλάσιόν ἐστιν ἓν τῶν μεγεθῶν ἑνός, τοσαυταπλάσια by E, so many times will AB, CD also be (divisible) by
ἔσταικαὶτὰπάντατῶνπάντων·ὅπερἔδειδεῖξαι. E, F .
Thus, if there are any number of magnitudes whatsoever
(which are) equal multiples, respectively, of some
(other) magnitudes, of equal number (to them), then as
many times as one of the (first) magnitudes is (divisible)
by one (of the second), so many times will all (of the
first magnitudes) also (be divisible) by all (of the second).
(Which is) the very thing it was required to show.
† In modern notation, this proposition reads m α + m β + · · · = m(α + β + · · ·).
. Proposition 2 †
᾿Εὰνπρῶτονδευτέρουἰσάκιςᾖπολλαπλάσιονκαὶτρίτον If a first (magnitude) and a third are equal multiples
τετάρτου,ᾖδὲκαὶπέμπτονδευτέρουἰσάκιςπολλαπλάσιον of a second and a fourth (respectively), and a fifth (magκαὶ
ἕκτον τετάρτου, καὶ συντεθὲν πρῶτον καὶ πέμπτον nitude) and a sixth (are) also equal multiples of the secδευτέρουἰσάκιςἔσταιπολλαπλάσιονκαὶτρίτονκαὶἕκτον
ond and fourth (respectively), then the first (magnitude)
τετάρτου.
and the fifth, being added together, and the third and the
ΠρῶτονγὰρτὸΑΒδευτέρουτοῦΓἰσάκιςἔστωπολ- sixth, (being added together), will also be equal multiples
λαπλάσιονκαὶτρίτοντὸΔΕτετάρτουτοῦΖ,ἔστωδὲκαὶ of the second (magnitude) and the fourth (respectively).
πέμπτοντὸΒΗδευτέρουτοῦΓἰσάκιςπολλαπλάσιονκαὶ For let a first (magnitude) AB and a third DE be
ἕκτον τὸ ΕΘ τετάρτου τοῦ Ζ· λέγω, ὅτι καὶ συντεθὲν equal multiples of a second C and a fourth F (respecπρῶτονκαὶπέμπτοντὸΑΗδευτέρουτοῦΓἰσάκιςἔσται
tively). And let a fifth (magnitude) BG and a sixth EH
πολλαπλάσιονκαὶτρίτονκαὶἕκτοντὸΔΘτετάρτουτοῦΖ. also be (other) equal multiples of the second C and the
fourth F (respectively). I say that the first (magnitude)
and the fifth, being added together, (to give) AG, and the
third (magnitude) and the sixth, (being added together,
E
F
132

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
Α Β Η
to give) DH, will also be equal multiples of the second
(magnitude) C and the fourth F (respectively).
A B G
Γ
∆ Ε Θ
C
D
E
H
Ζ
᾿ΕπεὶγὰρἰσάκιςἐστὶπολλαπλάσιοντὸΑΒτοῦΓκαὶτὸ For since AB and DE are equal multiples of C and F
ΔΕτοῦΖ,ὅσαἄραἐστὶνἐντῷΑΒἴσατῷΓ,τοσαῦτακαὶ (respectively), thus as many (magnitudes) as (there) are
ἐντῷΔΕἴσατῷΖ.διὰτὰαὐτὰδὴκαὶὅσαἐστὶνἐντῷΒΗ in AB equal to C, so many (are there) also in DE equal to
ἴσατῷΓ,τοσαῦτακαὶἐντῷΕΘἴσατῷΖ·ὅσαἄραἐστὶνἐν F . And so, for the same (reasons), as many (magnitudes)
ὅλῳτῷΑΗἴσατῷΓ,τοσαῦτακαὶἐνὅλῳτῷΔΘἴσατῷΖ· as (there) are in BG equal to C, so many (are there)
ὁσαπλάσιονἄραἐστὶτὸΑΗτοῦΓ,τοσαυταπλάσιονἔσται also in EH equal to F . Thus, as many (magnitudes) as
καὶτὸΔΘτοῦΖ.καὶσυντεθὲνἄραπρῶτονκαὶπέμπτοντὸ (there) are in the whole of AG equal to C, so many (are
ΑΗδευτέρουτοῦΓἰσάκιςἔσταιπολλαπλάσιονκαὶτρίτον there) also in the whole of DH equal to F . Thus, as many
καὶἕκτοντὸΔΘτετάρτουτοῦΖ.
times as AG is (divisible) by C, so many times will DH
᾿Εὰνἄραπρῶτονδευτέρουἰσάκιςᾖπολλαπλάσιονκαὶ also be divisible by F . Thus, the first (magnitude) and
τρίτοντετάρτου, ᾖδὲκαὶπέμπτονδευτέρουἰσάκιςπολ- the fifth, being added together, (to give) AG, and the
λαπλάσιονκαὶἕκτοντετάρτου,καὶσυντεθὲνπρῶτονκαὶ third (magnitude) and the sixth, (being added together,
πέμπτονδευτέρουἰσάκιςἔσταιπολλαπλάσιονκαὶτρίτονκαὶ to give) DH, will also be equal multiples of the second
ἕκτοντετάρτου·ὅπερἔδειδεῖξαι.
(magnitude) C and the fourth F (respectively).
Thus, if a first (magnitude) and a third are equal multiples
of a second and a fourth (respectively), and a fifth
(magnitude) and a sixth (are) also equal multiples of the
second and fourth (respectively), then the first (magnitude)
and the fifth, being added together, and the third
and sixth, (being added together), will also be equal multiples
of the second (magnitude) and the fourth (respectively).
(Which is) the very thing it was required to show.
† In modern notation, this propostion reads m α + n α = (m + n) α.
. Proposition 3 †
᾿Εὰνπρῶτονδευτέρουἰσάκιςᾖπολλαπλάσιονκαὶτρίτον If a first (magnitude) and a third are equal multiples
τετάρτου, ληφθῇ δὲ ἰσάκις πολλαπλάσια τοῦ τε πρώτου of a second and a fourth (respectively), and equal multiκαὶτρίτου,καὶδι᾿ἴσουτῶνληφθέντωνἑκάτερονἑκατέρου
ples are taken of the first and the third, then, via equality,
ἰσάκιςἔσταιπολλαπλάσιοντὸμὲντοῦδευτέρουτὸδὲτοῦ the (magnitudes) taken will also be equal multiples of the
τετάρτου.
second (magnitude) and the fourth, respectively.
ΠρῶτονγὰρτὸΑδευτέρουτοῦΒἰσάκιςἔστωπολ- For let a first (magnitude) A and a third C be equal
λαπλάσιονκαὶτρίτοντὸΓτετάρτουτοῦΔ,καὶεἰλήφθω multiples of a second B and a fourth D (respectively),
τῶνΑ,ΓἰσάκιςπολλαπλάσιατὰΕΖ,ΗΘ·λέγω,ὅτιἰσάκις and let the equal multiples EF and GH have been taken
ἐστὶπολλαπλάσιοντὸΕΖτοῦΒκαὶτὸΗΘτοῦΔ. of A and C (respectively). I say that EF and GH are
᾿ΕπεὶγὰρἰσάκιςἐστὶπολλαπλάσιοντὸΕΖτοῦΑκαὶ equal multiples of B and D (respectively).
τὸΗΘτοῦΓ,ὅσαἄραἐστὶνἐντῷΕΖἴσατῷΑ,τοσαῦτα For since EF and GH are equal multiples of A and
καὶἐντῷΗΘἴσατῷΓ.διῃρήσθωτὸμὲνΕΖεἰςτὰτῷΑ C (respectively), thus as many (magnitudes) as (there)
μεγέθηἴσατὰΕΚ,ΚΖ,τὸδὲΗΘεἰςτὰτῷΓἴσατὰΗΛ, are in EF equal to A, so many (are there) also in GH
F
133

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
ΛΘ·ἔσταιδὴἴσοντὸπλῆθοςτῶνΕΚ,ΚΖτῷπλήθειτῶν equal to C. Let EF have been divided into magnitudes
ΗΛ,ΛΘ.καὶἐπεὶἰσάκιςἐστὶπολλαπλάσιοντὸΑτοῦΒκαὶ EK, KF equal to A, and GH into (magnitudes) GL, LH
τὸΓτοῦΔ,ἴσονδὲτὸμὲνΕΚτῷΑ,τὸδὲΗΛτῷΓ, equal to C. So, the number of (magnitudes) EK, KF
ἰσάκιςἄραἐστὶπολλαπλάσιοντὸΕΚτοῦΒκαὶτὸΗΛτοῦ will be equal to the number of (magnitudes) GL, LH.
Δ.διὰτὰαὐτὰδὴἰσάκιςἐστὶπολλαπλάσιοντὸΚΖτοῦΒ And since A and C are equal multiples of B and D (reκαὶτὸΛΘτοῦΔ.ἐπεὶοὖνπρῶτοντὸΕΚδευτέρουτοῦΒ
spectively), and EK (is) equal to A, and GL to C, EK
ἴσάκιςἐστὶπολλαπλάσιονκαὶτρίτοντὸΗΛτετάρτουτοῦ and GL are thus equal multiples of B and D (respec-
Δ,ἔστιδὲκαὶπέμπτοντὸΚΖδευτέρουτοῦΒἰσάκιςπολ- tively). So, for the same (reasons), KF and LH are equal
λαπλάσιονκαὶἕκτοντὸΛΘτετάρτουτοῦΔ,καὶσυντεθὲν multiples of B and D (respectively). Therefore, since the
ἄραπρῶτονκαὶπέμπτοντὸΕΖδευτέρουτοῦΒἰσάκιςἐστὶ first (magnitude) EK and the third GL are equal mulπολλαπλάσιονκαὶτρίτονκαὶἕκτοντὸΗΘτετάρτουτοῦΔ.
tiples of the second B and the fourth D (respectively),
and the fifth (magnitude) KF and the sixth LH are also
equal multiples of the second B and the fourth D (respectively),
then the first (magnitude) and fifth, being
added together, (to give) EF , and the third (magnitude)
and sixth, (being added together, to give) GH, are thus
also equal multiples of the second (magnitude) B and the
fourth D (respectively) [Prop. 5.2].
Α
Β
Ε
Κ
Ζ
A
B
E K F
Γ
∆
Η Λ Θ
C
D
G
L
H
᾿Εὰν ἄρα πρῶτον δευτέρου ἰσάκις ᾖ πολλαπλάσιον Thus, if a first (magnitude) and a third are equal mulκαὶ
τρίτον τετάρτου, ληφθῇ δὲ τοῦ πρώτου καὶ τρίτου tiples of a second and a fourth (respectively), and equal
ἰσάκιςπολλαπλάσια,καὶδι᾿ἴσουτῶνληφθέντωνἑκάτερον multiples are taken of the first and the third, then, via
ἑκατέρουἰσάκιςἔσταιπολλαπλάσιοντὸμὲντοῦδευτέρου equality, the (magnitudes) taken will also be equal mulτὸδὲτοῦτετάρτου·ὅπερἔδειδεῖξαι.
tiples of the second (magnitude) and the fourth, respectively.
(Which is) the very thing it was required to show.
† In modern notation, this proposition reads m(n α) = (m n) α.
. Proposition 4 †
᾿Εὰνπρῶτονπρὸςδεύτεροντὸναὐτὸνἔχῃλόγονκαὶ If a first (magnitude) has the same ratio to a second
τρίτονπρὸςτέταρτον, καὶτὰἰσάκιςπολλαπλάσιατοῦτε that a third (has) to a fourth then equal multiples of the
πρώτουκαὶτρίτουπρὸςτὰἰσάκιςπολλαπλάσιατοῦδευτέρου first (magnitude) and the third will also have the same
καὶτετάρτουκαθ᾿ὁποιονοῦνπολλαπλασιασμὸντὸναὐτὸν ratio to equal multiples of the second and the fourth, be-
ἕξειλόγονληφθέντακατάλληλα.
ing taken in corresponding order, according to any kind
ΠρῶτονγὰρτὸΑπρὸςδεύτεροντὸΒτὸναὐτὸνἐχέτω of multiplication whatsoever.
λόγονκαὶτρίτοντὸΓπρὸςτέταρτοντὸΔ,καὶεἰλήφθω For let a first (magnitude) A have the same ratio to
τῶνμὲνΑ,ΓἰσάκιςπολλαπλάσιατὰΕ,Ζ,τῶνδὲΒ,Δ a second B that a third C (has) to a fourth D. And let
ἄλλα,ἃἔτυχεν,ἰσάκιςπολλαπλάσιατὰΗ,Θ·λέγω, ὅτι equal multiples E and F have been taken of A and C
ἐστὶνὡςτὸΕπρὸςτὸΗ,οὕτωςτὸΖπρὸςτὸΘ. (respectively), and other random equal multiples G and
134

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
Α
Β
Ε
Η
Κ
Μ
Γ
∆
Ζ
Θ
Λ
Ν
H of B and D (respectively). I say that as E (is) to G, so
F (is) to H.
ΕἰλήφθωγὰρτῶνμὲνΕ,ΖἰσάκιςπολλαπλάσιατὰΚ, For let equal multiples K and L have been taken of E
Λ,τῶνδὲΗ,Θἄλλα,ἃἔτυχεν,ἰσάκιςπολλαπλάσιατὰΜ, and F (respectively), and other random equal multiples
Ν. M and N of G and H (respectively).
[Καὶ]ἐπεὶἰσάκιςἐστὶπολλαπλάσιοντὸμὲνΕτοῦΑ,τὸ [And] since E and F are equal multiples of A and
δὲΖτοῦΓ,καὶεἴληπταιτῶνΕ,ΖἴσάκιςπολλαπλάσιατὰΚ, C (respectively), and the equal multiples K and L have
Λ,ἴσάκιςἄραἐστὶπολλαπλάσιοντὸΚτοῦΑκαὶτὸΛτοῦ been taken of E and F (respectively), K and L are thus
Γ.διὰτὰαὐτὰδὴἰσάκιςἐστὶπολλαπλάσιοντὸΜτοῦΒκαὶ equal multiples of A and C (respectively) [Prop. 5.3]. So,
τὸΝτοῦΔ.καὶἐπείἐστινὡςτὸΑπρὸςτὸΒ,οὕτωςτὸΓ for the same (reasons), M and N are equal multiples of
πρὸςτὸΔ,καὶεἴληπταιτῶνμὲνΑ,Γἰσάκιςπολλαπλάσια B and D (respectively). And since as A is to B, so C (is)
τὰΚ,Λ,τῶνδὲΒ,Δἄλλα,ἃἔτυχεν,ἰσάκιςπολλαπλάσια to D, and the equal multiples K and L have been taken
τὰΜ,Ν,εἰἄραὑπερέχειτὸΚτοῦΜ,ὑπερέχεικαὶτὸΛ of A and C (respectively), and the other random equal
τοῦΝ,καὶεἰἴσον,ἴσον,καὶεἰἔλαττον,ἔλαττον. καίἐστι multiples M and N of B and D (respectively), then if K
τὰμὲνΚ,ΛτῶνΕ,Ζἰσάκιςπολλαπλάσια,τὰδὲΜ,Ντῶν exceeds M then L also exceeds N, and if (K is) equal (to
Η,Θἄλλα,ἃἔτυχεν,ἰσάκιςπολλαπλάσια·ἔστινἄραὡςτὸ M then L is also) equal (to N), and if (K is) less (than M
ΕπρὸςτὸΗ,οὕτωςτὸΖπρὸςτὸΘ.
then L is also) less (than N) [Def. 5.5]. And K and L are
᾿Εὰνἄραπρῶτονπρὸςδεύτεροντὸναὐτὸνἔχῃλόγον equal multiples of E and F (respectively), and M and N
καὶτρίτονπρὸςτέταρτον,καὶτὰἰσάκιςπολλαπλάσιατοῦτε other random equal multiples of G and H (respectively).
πρώτουκαὶτρίτουπρὸςτὰἰσάκιςπολλαπλάσιατοῦδευτέρου Thus, as E (is) to G, so F (is) to H [Def. 5.5].
καὶτετάρτουτὸναὐτὸνἕξειλόγονκαθ᾿ὁποιονοῦνπολλα- Thus, if a first (magnitude) has the same ratio to a
πλασιασμὸνληφθέντακατάλληλα·ὅπερἔδειδεῖξαι. second that a third (has) to a fourth then equal multiples
of the first (magnitude) and the third will also have
the same ratio to equal multiples of the second and the
fourth, being taken in corresponding order, according to
any kind of multiplication whatsoever. (Which is) the
very thing it was required to show.
† In modern notation, this proposition reads that if α : β :: γ : δ then m α : n β :: m γ : n δ, for all m and n.
A
B
E
G
K
M
C
D
F
H
L
N
135

ËÌÇÁÉÁÏƺ ELEMENTS BOOK 5
. Proposition 5 †
᾿Εὰνμέγεθοςμεγέθουςἰσάκιςᾖπολλαπλάσιον,ὅπερ If a magnitude is the same multiple of a magnitude
ἀφαιρεθὲνἀφαιρεθέντος,καὶτὸλοιπὸντοῦλοιποῦἰσάκις that a (part) taken away (is) of a (part) taken away (re-
ἔσταιπολλαπλάσιον,ὁσαπλάσιόνἐστιτὸὅλοντοῦὅλου. spectively) then the remainder will also be the same multiple
of the remainder as that which the whole (is) of the
whole (respectively).
Α
Ε
Β
A
E
B
Η Γ Ζ ∆
G
C
F
D
ΜέγεθοςγὰρτὸΑΒμεγέθουςτοῦΓΔἰσάκιςἔστωπολ- For let the magnitude AB be the same multiple of the
λαπλάσιον,ὅπερἀφαιρεθὲντὸΑΕἀφαιρεθέντοςτοῦΓΖ· magnitude CD that the (part) taken away AE (is) of the
λέγω,ὅτικαὶλοιπὸντὸΕΒλοιποῦτοῦΖΔἰσάκιςἔσται (part) taken away CF (respectively). I say that the reπολλαπλάσιον,ὁσαπλάσιόνἐστινὅλοντὸΑΒὅλουτοῦΓΔ.
mainder EB will also be the same multiple of the remain-
῾ΟσαπλάσιονγάρἐστιτὸΑΕτοῦΓΖ,τοσαυταπλάσιον der FD as that which the whole AB (is) of the whole CD
γεγονέτωκαὶτὸΕΒτοῦΓΗ.
(respectively).
ΚαὶἐπεὶἰσάκιςἐστὶπολλαπλάσιοντὸΑΕτοῦΓΖκαὶτὸ For as many times as AE is (divisible) by CF , so many
ΕΒτοῦΗΓ,ἰσάκιςἄραἐστὶπολλαπλάσιοντὸΑΕτοῦΓΖ times let EB also have been made (divisible) by CG.
καὶτὸΑΒτοῦΗΖ.κεῖταιδὲἰσάκιςπολλαπλάσιοντὸΑΕ And since AE and EB are equal multiples of CF and
τοῦΓΖκαὶτὸΑΒτοῦΓΔ.ἰσάκιςἄραἐστὶπολλαπλάσιοντὸ GC (respectively), AE and AB are thus equal multiples
ΑΒἑκατέρουτῶνΗΖ,ΓΔ·ἴσονἄρατὸΗΖτῷΓΔ.κοινὸν of CF and GF (respectively) [Prop. 5.1]. And AE and
ἀφῃρήσθωτὸΓΖ·λοιπὸνἄρατὸΗΓλοιπῷτῷΖΔἴσον AB are assumed (to be) equal multiples of CF and CD
ἐστίν. καὶἐπεὶἰσάκιςἐστὶπολλαπλάσιοντὸΑΕτοῦΓΖ (respectively). Thus, AB is an equal multiple of each
καὶτὸΕΒτοῦΗΓ,ἴσονδὲτὸΗΓτῷΔΖ,ἰσάκιςἄραἐστὶ of GF and CD. Thus, GF (is) equal to CD. Let CF
πολλαπλάσιοντὸΑΕτοῦΓΖκαὶτὸΕΒτοῦΖΔ.ἰσάκιςδὲ have been subtracted from both. Thus, the remainder
ὑπόκειταιπολλαπλάσιοντὸΑΕτοῦΓΖκαὶτὸΑΒτοῦΓΔ· GC is equal to the remainder FD. And since AE and
ἰσάκιςἄραἐστὶπολλαπλάσιοντὸΕΒτοῦΖΔκαὶτὸΑΒ EB are equal multiples of CF and GC (respectively),
τοῦΓΔ.καὶλοιπὸνἄρατὸΕΒλοιποῦτοῦΖΔἰσάκιςἔσται and GC (is) equal to DF , AE and EB are thus equal
πολλαπλάσιον,ὁσαπλάσιόνἐστινὅλοντὸΑΒὅλουτοῦΓΔ. multiples of CF and FD (respectively). And AE and
᾿Εὰν ἄρα μέγεθος μεγέθους ἰσάκις ᾖ πολλαπλάσιον, AB are assumed (to be) equal multiples of CF and CD
ὅπερἀφαιρεθὲνἀφαιρεθέντος, καὶτὸλοιπὸντοῦλοιποῦ (respectively). Thus, EB and AB are equal multiples of
ἰσάκιςἔσταιπολλαπλάσιον,ὁσαπλάσιόνἐστικαὶτὸὅλον FD and CD (respectively). Thus, the remainder EB will
τοῦὅλου·ὅπερἔδειδεῖξαι.
also be the same multiple of the remainder FD as that
which the whole AB (is) of the whole CD (respectively).
Thus, if a magnitude is the same multiple of a magnitude
that a (part) taken away (is) of a (part) taken away
(respectively) then the remainder will also be the same
multiple of the remainder as that which the whole (is) of
the whole (respectively). (Which is) the very thing it was
required to show.
† In modern notation, this proposition reads m α − m β = m(α − β).
¦. Proposition 6 †
᾿Εὰνδύομεγέθηδύομεγεθῶνἰσάκιςᾖπολλαπλάσια, If two magnitudes are equal multiples of two (other)
καὶἀφαιρεθέντατινὰτῶναὐτῶνἰσάκιςᾖπολλαπλάσια,καὶ magnitudes, and some (parts) taken away (from the forτὰλοιπὰτοῖςαὐτοῖςἤτοιἴσαἐστὶνἢἰσάκιςαὐτῶνπολ-
mer magnitudes) are equal multiples of the latter (magλαπλάσια.
nitudes, respectively), then the remainders are also either
ΔύογὰρμεγέθητὰΑΒ,ΓΔδύομεγεθῶντῶνΕ,Ζ equal to the latter (magnitudes), or (are) equal multiples
136

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
ἰσάκιςἔστωπολλαπλάσια,καὶἀφαιρεθέντατὰΑΗ,ΓΘτῶν of them (respectively).
αὐτῶντῶνΕ,Ζἰσάκιςἔστωπολλαπλάσια·λέγω,ὅτικαὶ For let two magnitudes AB and CD be equal multiλοιπὰτὰΗΒ,ΘΔτοῖςΕ,Ζἤτοιἴσαἐστὶνἢἰσάκιςαὐτῶν
ples of two magnitudes E and F (respectively). And let
πολλαπλάσια.
the (parts) taken away (from the former) AG and CH be
equal multiples of E and F (respectively). I say that the
remainders GB and HD are also either equal to E and F
(respectively), or (are) equal multiples of them.
Α
Η Β
A
G
B
Ε
Κ
Γ
Θ ∆
E
K
C H D
Ζ
῎ΕστωγὰρπρότεροντὸΗΒτῷΕἴσον·λέγω,ὅτικαὶ For let GB be, first of all, equal to E. I say that HD is
τὸΘΔτῷΖἴσονἐστίν. also equal to F .
ΚείσθωγὰρτῷΖἴσοντὸΓΚ.ἐπεὶἰσάκιςἐστὶπολ- For let CK be made equal to F . Since AG and CH
λαπλάσιοντὸΑΗτοῦΕκαὶτὸΓΘτοῦΖ,ἴσονδὲτὸμὲνΗΒ are equal multiples of E and F (respectively), and GB
τῷΕ,τὸδὲΚΓτῷΖ,ἰσάκιςἄραἐστὶπολλαπλάσιοντὸΑΒ (is) equal to E, and KC to F , AB and KH are thus equal
τοῦΕκαὶτὸΚΘτοῦΖ.ἰσάκιςδὲὑπόκειταιπολλαπλάσιον multiples of E and F (respectively) [Prop. 5.2]. And AB
τὸΑΒτοῦΕκαὶτὸΓΔτοῦΖ·ἴσάκιςἄραἐστὶπολλαπλάσιον and CD are assumed (to be) equal multiples of E and F
τὸΚΘτοῦΖκαὶτὸΓΔτοῦΖ.ἐπεὶοὖνἑκάτεροντῶνΚΘ, (respectively). Thus, KH and CD are equal multiples of
ΓΔτοῦΖἰσάκιςἐστὶπολλαπλάσιον,ἴσονἄραἐστὶτὸΚΘ F and F (respectively). Therefore, KH and CD are each
τῷΓΔ.κοινὸνἀφῃρήσθωτὸΓΘ·λοιπὸνἄρατὸΚΓλοιπῷ equal multiples of F . Thus, KH is equal to CD. Let CH
τῷΘΔἴσονἐστίν. ἀλλὰτὸΖτῷΚΓἐστινἴσον·καὶτὸ have be taken away from both. Thus, the remainder KC
ΘΔἄρατῷΖἴσονἐστίν.ὥστεεἰτὸΗΒτῷΕἴσονἐστίν, is equal to the remainder HD. But, F is equal to KC.
καὶτὸΘΔἴσονἔσταιτῷΖ.
Thus, HD is also equal to F . Hence, if GB is equal to E
῾Ομοίωςδὴδείξομεν,ὅτι,κᾂνπολλαπλάσιονᾖτὸΗΒ then HD will also be equal to F .
τοῦΕ,τοσαυταπλάσιονἔσταικαὶτὸΘΔτοῦΖ.
So, similarly, we can show that even if GB is a multi-
᾿Εὰν ἄρα δύο μεγέθη δύο μεγεθῶν ἰσάκις ᾖ πολ- ple of E then HD will also be the same multiple of F .
λαπλάσια,καὶἀφαιρεθέντατινὰτῶναὐτῶνἰσάκιςᾖπολ- Thus, if two magnitudes are equal multiples of two
λαπλάσια,καὶτὰλοιπὰτοῖςαὐτοῖςἤτοιἴσαἐστὶνἢἰσάκις (other) magnitudes, and some (parts) taken away (from
αὐτῶνπολλαπλάσια·ὅπερἔδειδεῖξαι.
the former magnitudes) are equal multiples of the latter
(magnitudes, respectively), then the remainders are also
either equal to the latter (magnitudes), or (are) equal
multiples of them (respectively). (Which is) the very
thing it was required to show.
† In modern notation, this proposition reads m α − n α = (m − n) α.
Þ. Proposition 7
Τὰἴσαπρὸςτὸαὐτὸτὸναὐτὸνἔχειλόγονκαὶτὸαὐτὸ Equal (magnitudes) have the same ratio to the same
πρὸςτὰἴσα.
(magnitude), and the latter (magnitude has the same ra-
῎Εστω ἴσα μεγέθη τὰ Α, Β, ἄλλο δέ τι, ὃ ἔτυχεν, tio) to the equal (magnitudes).
μέγεθοςτὸΓ·λέγω,ὅτιἑκάτεροντῶνΑ,ΒπρὸςτὸΓ Let A and B be equal magnitudes, and C some other
τὸναὐτὸνἔχειλόγον,καὶτὸΓπρὸςἑκάτεροντῶνΑ,Β. random magnitude. I say that A and B each have the
F
137

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
Α
Β
Γ
∆
Ε
Ζ
same ratio to C, and (that) C (has the same ratio) to
each of A and B.
ΕἰλήφθωγὰρτῶνμὲνΑ,ΒἰσάκιςπολλαπλάσιατὰΔ, For let the equal multiples D and E have been taken
Ε,τοῦδὲΓἄλλο,ὃἔτυχεν,πολλαπλάσιοντὸΖ. of A and B (respectively), and the other random multiple
᾿ΕπεὶοὖνἰσάκιςἐστὶπολλαπλάσιοντὸΔτοῦΑκαὶτὸ F of C.
ΕτοῦΒ,ἴσονδὲτὸΑτῷΒ,ἴσονἄρακαὶτὸΔτῷΕ.ἄλλο Therefore, since D and E are equal multiples of A
δέ,ὅἔτυχεν,τὸΖ.ΕἰἄραὑπερέχειτὸΔτοῦΖ,ὑπερέχει and B (respectively), and A (is) equal to B, D (is) thus
καὶτὸΕτοῦΖ,καὶεἰἴσον,ἴσον,καὶεἰἔλαττον,ἔλαττον. also equal to E. And F (is) different, at random. Thus, if
καίἐστιτὰμὲνΔ,ΕτῶνΑ,Βἰσάκιςπολλαπλάσια,τὸδὲ D exceeds F then E also exceeds F , and if (D is) equal
ΖτοῦΓἄλλο,ὃἔτυχεν,πολλαπλάσιον·ἔστινἄραὡςτὸΑ (to F then E is also) equal (to F ), and if (D is) less
πρὸςτὸΓ,οὕτωςτὸΒπρὸςτὸΓ.
(than F then E is also) less (than F ). And D and E are
Λέγω[δή],ὅτικαὶτὸΓπρὸςἑκάτεροντῶνΑ,Βτὸν equal multiples of A and B (respectively), and F another
αὐτὸνἔχειλόγον.
random multiple of C. Thus, as A (is) to C, so B (is) to
Τῶνγὰραὐτῶνκατασκευασθέντωνὁμοίωςδείξομεν, C [Def. 5.5].
ὅτιἴσονἐστὶτὸΔτῷΕ·ἄλλοδέτιτὸΖ·εἰἄραὑπερέχει [So] I say that C † also has the same ratio to each of A
τὸΖτοῦΔ,ὑπερέχεικαὶτοῦΕ,καὶεἰἴσον,ἴσον,καὶεἰ and B.
ἔλαττον,ἔλαττον. καίἐστιτὸμὲνΖτοῦΓπολλαπλάσιον, For, similarly, we can show, by the same construction,
τὰδὲΔ,ΕτῶνΑ,Βἄλλα,ἃἔτυχεν,ἰσάκιςπολλαπλάσια· that D is equal to E. And F (has) some other (value).
ἔστινἄραὡςτὸΓπρὸςτὸΑ,οὕτωςτὸΓπρὸςτὸΒ. Thus, if F exceeds D then it also exceeds E, and if (F is)
Τὰἴσαἄραπρὸςτὸαὐτὸτὸναὐτὸνἔχειλόγονκαὶτὸ equal (to D then it is also) equal (to E), and if (F is) less
αὐτὸπρὸςτὰἴσα.
(than D then it is also) less (than E). And F is a multiple
of C, and D and E other random equal multiples of A
and B. Thus, as C (is) to A, so C (is) to B [Def. 5.5].
Thus, equal (magnitudes) have the same ratio to the
same (magnitude), and the latter (magnitude has the
same ratio) to the equal (magnitudes).
ÈÖ×Ñ. Corollary ‡
᾿Εκδὴτούτουφανερόν,ὅτιἐὰνμεγέθητινὰἀνάλογον So (it is) clear, from this, that if some magnitudes are
ᾖ,καὶἀνάπαλινἀνάλογονἔσται.ὅπερἔδειδεῖξαι. proportional then they will also be proportional inversely.
(Which is) the very thing it was required to show.
† The Greek text has “E”, which is obviously a mistake.
‡ In modern notation, this corollary reads that if α : β :: γ : δ then β : α :: δ : γ.
. Proposition 8
Τῶνἀνίσωνμεγεθῶντὸμεῖζονπρὸςτὸαὐτὸμείζονα For unequal magnitudes, the greater (magnitude) has
λόγονἔχειἤπερτὸἔλαττον.καὶτὸαὐτὸπρὸςτὸἔλαττον a greater ratio than the lesser to the same (magnitude).
μείζοναλόγονἔχειἤπερπρὸςτὸμεῖζον.
And the latter (magnitude) has a greater ratio to the
῎ΕστωἄνισαμεγέθητὰΑΒ,Γ,καὶἔστωμεῖζοντὸΑΒ, lesser (magnitude) than to the greater.
ἄλλοδέ, ὃἔτυχεν, τὸΔ·λέγω, ὅτιτὸΑΒπρὸςτὸΔ Let AB and C be unequal magnitudes, and let AB be
μείζοναλόγονἔχειἤπερτὸΓπρὸςτὸΔ,καὶτὸΔπρὸς the greater (of the two), and D another random magniτὸΓμείζοναλόγονἔχειἤπερπρὸςτὸΑΒ.
tude. I say that AB has a greater ratio to D than C (has)
to D, and (that) D has a greater ratio to C than (it has)
to AB.
A
B
C
D
E
F
138

ËÌÇÁÉÁÏƺ ELEMENTS
À Â À Â
ÃÄÅÆ Æ ÅÄÃ
BOOK 5
A E B A E B
C
C
F G H F G H
N
᾿ΕπεὶγὰρμεῖζόνἐστιτὸΑΒτοῦΓ,κείσθωτῷΓἴσον For since AB is greater than C, let BE be made equal
τὸΒΕ·τὸδὴἔλασσοντῶνΑΕ,ΕΒπολλαπλασιαζόμενον to C. So, the lesser of AE and EB, being multiplied, will
ἔσταιποτὲτοῦΔμεῖζον. ἔστωπρότεροντὸΑΕἔλαττον sometimes be greater than D [Def. 5.4]. First of all, let
τοῦΕΒ,καὶπεπολλαπλασιάσθωτὸΑΕ,καὶἔστωαὐτοῦ AE be less than EB, and let AE have been multiplied,
πολλαπλάσιοντὸΖΗμεῖζονὂντοῦΔ,καὶὁσαπλάσιόνἐστι and let FG be a multiple of it which (is) greater than
τὸΖΗτοῦΑΕ,τοσαυταπλάσιονγεγονέτωκαὶτὸμὲνΗΘ D. And as many times as FG is (divisible) by AE, so
τοῦΕΒτὸδὲΚτοῦΓ·καὶεἰλήφθωτοῦΔδιπλάσιονμὲν many times let GH also have become (divisible) by EB,
τὸΛ,τριπλάσιονδὲτὸΜ,καὶἑξῆςἑνὶπλεῖον,ἕωςἂντὸ and K by C. And let the double multiple L of D have
λαμβανόμενονπολλαπλάσιονμὲνγένηταιτοῦΔ,πρώτωςδὲ been taken, and the triple multiple M, and several more,
μεῖζοντοῦΚ.εἰλήφθω,καὶἔστωτὸΝτετραπλάσιονμὲν (each increasing) in order by one, until the (multiple)
τοῦΔ,πρώτωςδὲμεῖζοντοῦΚ.
taken becomes the first multiple of D (which is) greater
᾿ΕπεὶοὖντὸΚτοῦΝπρώτωςἐστὶνἔλαττον,τὸΚἄρα than K. Let it have been taken, and let it also be the
τοῦΜοὔκἐστινἔλαττον.καὶἐπεὶἰσάκιςἐστὶπολλαπλάσιον quadruple multiple N of D—the first (multiple) greater
τὸΖΗτοῦΑΕκαὶτὸΗΘτοῦΕΒ,ἰσάκιςἄραἐστὶπολ- than K.
λαπλάσιοντὸΖΗτοῦΑΕκαὶτὸΖΘτοῦΑΒ.ἰσάκιςδέ Therefore, since K is less than N first, K is thus not
ἐστιπολλαπλάσιοντὸΖΗτοῦΑΕκαὶτὸΚτοῦΓ·ἰσάκις less than M. And since FG and GH are equal multi-
ἄραἐστὶπολλαπλάσιοντὸΖΘτοῦΑΒκαὶτὸΚτοῦΓ.τὰ ples of AE and EB (respectively), FG and FH are thus
ΖΘ,ΚἄρατῶνΑΒ,Γἰσάκιςἐστὶπολλαπλάσια.πάλιν,ἐπεὶ equal multiples of AE and AB (respectively) [Prop. 5.1].
ἰσάκιςἐστὶπολλαπλάσιοντὸΗΘτοῦΕΒκαὶτὸΚτοῦΓ, And FG and K are equal multiples of AE and C (re-
ἴσονδὲτὸΕΒτῷΓ,ἴσονἄρακαὶτὸΗΘτῷΚ.τὸδὲΚ spectively). Thus, FH and K are equal multiples of AB
τοῦΜοὔκἐστινἔλαττον·οὐδ᾿ἄρατὸΗΘτοῦΜἔλαττόν and C (respectively). Thus, FH, K are equal multiples
ἐστιν. μεῖζονδὲτὸΖΗτοῦΔ·ὅλονἄρατὸΖΘσυναμ- of AB, C. Again, since GH and K are equal multiples
φοτέρωντῶνΔ,Μμεῖζόνἐστιν.ἀλλὰσυναμφότερατὰΔ, of EB and C, and EB (is) equal to C, GH (is) thus also
ΜτῷΝἐστινἴσα,ἐπειδήπερτὸΜτοῦΔτριπλάσιόνἐστιν, equal to K. And K is not less than M. Thus, GH not less
συναμφότεραδὲτὰΜ,ΔτοῦΔἐστιτετραπλάσια,ἔστιδὲ than M either. And FG (is) greater than D. Thus, the
καὶτὸΝτοῦΔτετραπλάσιον·συναμφότεραἄρατὰΜ,Δ whole of FH is greater than D and M (added) together.
τῷΝἴσαἐστίν. ἀλλὰτὸΖΘτῶνΜ,Δμεῖζόνἐστιν·τὸ But, D and M (added) together is equal to N, inasmuch
ΖΘἄρατοῦΝὑπερέχει·τὸδὲΚτοῦΝοὐχὑπερέχει.καί as M is three times D, and M and D (added) together is
ἐστιτὰμὲνΖΘ,ΚτῶνΑΒ,Γἰσάκιςπολλαπλάσια,τὸδὲΝ four times D, and N is also four times D. Thus, M and D
τοῦΔἄλλο,ὃἔτυχεν,πολλαπλάσιον·τὸΑΒἄραπρὸςτὸ (added) together is equal to N. But, FH is greater than
ΔμείζοναλόγονἔχειἤπερτὸΓπρὸςτὸΔ.
M and D. Thus, FH exceeds N. And K does not exceed
Λέγωδή,ὅτικαὶτὸΔπρὸςτὸΓμείζοναλόγονἔχει N. And FH, K are equal multiples of AB, C, and N
ἤπερτὸΔπρὸςτὸΑΒ.
another random multiple of D. Thus, AB has a greater
Τῶνγὰραὐτῶνκατασκευασθέντωνὁμοίωςδείξομεν, ratio to D than C (has) to D [Def. 5.7].
ὅτιτὸμὲνΝτοῦΚὑπερέχει,τὸδὲΝτοῦΖΘοὐχὑπερέχει. So, I say that D also has a greater ratio to C than D
καίἐστιτὸμὲνΝτοῦΔπολλαπλάσιον,τὰδὲΖΘ,Κτῶν (has) to AB.
ΑΒ,Γἄλλα,ἃἔτυχεν,ἰσάκιςπολλαπλάσια·τὸΔἄραπρὸς For, similarly, by the same construction, we can show
τὸΓμείζοναλόγονἔχειἤπερτὸΔπρὸςτὸΑΒ. that N exceeds K, and N does not exceed FH. And
ἈλλὰδὴτὸΑΕτοῦΕΒμεῖζονἔστω. τὸδὴἔλαττον N is a multiple of D, and FH, K other random equal
τὸΕΒπολλαπλασιαζόμενονἔσταιποτὲτοῦΔμεῖζον. πε- multiples of AB, C (respectively). Thus, D has a greater
K
D
L
M
K
D
L
M
N
139

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
πολλαπλασιάσθω,καὶἔστωτὸΗΘπολλαπλάσιονμὲντοῦ ratio to C than D (has) to AB [Def. 5.5].
ΕΒ,μεῖζονδὲτοῦΔ·καὶὁσαπλάσιόνἐστιτὸΗΘτοῦΕΒ, And so let AE be greater than EB. So, the lesser,
τοσαυταπλάσιονγεγονέτωκαὶτὸμὲνΖΗτοῦΑΕ,τὸδὲΚ EB, being multiplied, will sometimes be greater than D.
τοῦΓ.ὁμοίωςδὴδείξομεν,ὅτιτὰΖΘ,ΚτῶνΑΒ,Γἰσάκις Let it have been multiplied, and let GH be a multiple of
ἐστὶπολλαπλάσια·καὶεἰλήφθωὁμοίωςτὸΝπολλαπλάσιον EB (which is) greater than D. And as many times as
μὲντοῦΔ,πρώτωςδὲμεῖζοντοῦΖΗ·ὥστεπάλιντὸΖΗ GH is (divisible) by EB, so many times let FG also have
τοῦΜοὔκἐστινἔλασσον. μεῖζονδὲτὸΗΘτοῦΔ·ὅλον become (divisible) by AE, and K by C. So, similarly
ἄρατὸΖΘτῶνΔ,Μ,τουτέστιτοῦΝ,ὑπερέχει.τὸδὲΚ (to the above), we can show that FH and K are equal
τοῦΝοὐχὑπερέχει,ἐπειδήπερκαὶτὸΖΗμεῖζονὂντοῦ multiples of AB and C (respectively). And, similarly (to
ΗΘ,τουτέστιτοῦΚ,τοῦΝοὐχὑπερέχει. καὶὡσαύτως the above), let the multiple N of D, (which is) the first
κατακολουθοῦντεςτοῖςἐπάνωπεραίνομεντὴνἀπόδειξιν. (multiple) greater than FG, have been taken. So, FG
Τῶν ἄρα ἀνίσων μεγεθῶν τὸ μεῖζον πρὸς τὸ αὐτὸ is again not less than M. And GH (is) greater than D.
μείζοναλόγονἔχειἤπερτὸἔλαττον·καὶτὸαὐτὸπρὸςτὸ Thus, the whole of FH exceeds D and M, that is to say
ἔλαττονμείζοναλόγονἔχειἤπερπρὸςτὸμεῖζον·ὅπερἔδει N. And K does not exceed N, inasmuch as FG, which
δεῖξαι.
(is) greater than GH—that is to say, K—also does not
exceed N. And, following the above (arguments), we
(can) complete the proof in the same manner.
Thus, for unequal magnitudes, the greater (magnitude)
has a greater ratio than the lesser to the same (magnitude).
And the latter (magnitude) has a greater ratio to
the lesser (magnitude) than to the greater. (Which is) the
very thing it was required to show.
. Proposition 9
Τὰπρὸςτὸαὐτὸτὸναὐτὸνἔχονταλὸγονἴσαἀλλήλοις (Magnitudes) having the same ratio to the same
ἐστίν·καὶπρὸςἃτὸαὐτὸτὸναὐτὸνἕχειλόγον,ἐκεῖναἴσα (magnitude) are equal to one another. And those (mag-
ἐστίν.
nitudes) to which the same (magnitude) has the same
ratio are equal.
Α
Γ
Β
᾿ΕχέτωγὰρἑκάτεροντῶνΑ,ΒπρὸςτὸΓτὸναὐτὸν For let A and B each have the same ratio to C. I say
λόγον·λέγω,ὅτιἴσονἐστὶτὸΑτῷΒ. that A is equal to B.
Εἰγὰρμή,οὐκἂνἑκάτεροντῶνΑ,ΒπρὸςτὸΓτὸν For if not, A and B would not each have the same
αὐτὸνεἶχελόγον·ἔχειδέ·ἴσονἄραἐστὶτὸΑτῷΒ. ratio to C [Prop. 5.8]. But they do. Thus, A is equal to
᾿ΕχέτωδὴπάλιντὸΓπρὸςἑκάτεροντῶνΑ,Βτὸναὐτὸν B.
λόγον·λέγω,ὅτιἴσονἐστὶτὸΑτῷΒ.
So, again, let C have the same ratio to each of A and
Εἰγὰρμή,οὐκἂντὸΓπρὸςἑκάτεροντῶνΑ,Βτὸν B. I say that A is equal to B.
αὐτὸνεἶχελόγον·ἔχειδέ·ἴσονἄραἐστὶτὸΑτῷΒ. For if not, C would not have the same ratio to each of
Τὰ ἄρα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχοντα λόγον ἴσα A and B [Prop. 5.8]. But it does. Thus, A is equal to B.
ἀλλήλοιςἐστίν·καὶπρὸςἃτὸαὐτὸτὸναὐτὸνἔχειλόγον, Thus, (magnitudes) having the same ratio to the same
ἐκεῖναἴσαἐστίν·ὅπερἔδειδεῖξαι.
(magnitude) are equal to one another. And those (magnitudes)
to which the same (magnitude) has the same ratio
are equal. (Which is) the very thing it was required to
show.
. Proposition 10
Τῶνπρὸςτὸαὐτὸλόγονἐχόντωντὸμείζοναλόγον For (magnitudes) having a ratio to the same (mag-
ἔχονἐκεῖνομεῖζόνἐστιν·πρὸςὃδὲτὸαὐτὸμείζοναλόγον nitude), that (magnitude which) has the greater ratio is
A
C
B
140

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
ἔχει,ἐκεῖνοἔλαττόνἐστιν.
Α
Γ
Β
(the) greater. And that (magnitude) to which the latter
(magnitude) has a greater ratio is (the) lesser.
᾿ΕχέτωγὰρτὸΑπρὸςτὸΓμείζοναλόγονἤπερτὸΒ For let A have a greater ratio to C than B (has) to C.
πρὸςτὸΓ·λέγω,ὅτιμεῖζόνἐστιτὸΑτοῦΒ. I say that A is greater than B.
Εἰγὰρμή,ἤτοιἴσονἐστὶτὸΑτῷΒἢἔλασσον. ἴσον For if not, A is surely either equal to or less than B.
μὲνοὖνοὔκἐστὶτὸΑτῷΒ·ἑκάτερονγὰρἂντῶνΑ,Β In fact, A is not equal to B. For (then) A and B would
πρὸςτὸΓτὸναὐτὸνεἶχελόγον.οὐκἔχειδέ·οὐκἄραἴσον each have the same ratio to C [Prop. 5.7]. But they do
ἐστὶτὸΑτῷΒ.οὐδὲμὴνἔλασσόνἐστιτὸΑτοῦΒ·τὸΑ not. Thus, A is not equal to B. Neither, indeed, is A less
γὰρἂνπρὸςτὸΓἐλάσσοναλόγονεἶχενἤπερτὸΒπρὸςτὸ than B. For (then) A would have a lesser ratio to C than
Γ.οὐκἔχειδέ·οὐκἄραἔλασσόνἐστιτὸΑτοῦΒ.ἐδείχθη B (has) to C [Prop. 5.8]. But it does not. Thus, A is not
δὲοὐδὲἴσον·μεῖζονἄραἐστὶτὸΑτοῦΒ.
less than B. And it was shown not (to be) equal either.
᾿ΕχέτωδὴπάλιντὸΓπρὸςτὸΒμείζοναλόγονἤπερτὸ Thus, A is greater than B.
ΓπρὸςτὸΑ·λέγω,ὅτιἔλασσόνἐστιτὸΒτοῦΑ.
So, again, let C have a greater ratio to B than C (has)
Εἰγὰρμή,ἤτοιἴσονἐστὶνἢμεῖζον. ἴσονμὲνοὖνοὔκ to A. I say that B is less than A.
ἐστιτὸΒτῷΑ·τὸΓγὰρἂνπρὸςἑκάτεροντῶνΑ,Βτὸν For if not, (it is) surely either equal or greater. In fact,
αὐτὸνεἶχελόγον. οὐκἔχειδέ· οὐκἄραἴσονἐστὶτὸΑ B is not equal to A. For (then) C would have the same
τῷΒ.οὐδὲμὴνμεῖζόνἐστιτὸΒτοῦΑ·τὸΓγὰρἂνπρὸς ratio to each of A and B [Prop. 5.7]. But it does not.
τὸΒἐλάσσοναλόγονεἶχενἤπερπρὸςτὸΑ.οὐκἔχειδέ· Thus, A is not equal to B. Neither, indeed, is B greater
οὐκἄραμεῖζόνἐστιτὸΒτοῦΑ.ἐδείχθηδέ,ὅτιοὐδὲἴσον· than A. For (then) C would have a lesser ratio to B than
ἔλαττονἄραἐστὶτὸΒτοῦΑ.
(it has) to A [Prop. 5.8]. But it does not. Thus, B is not
Τῶνἄραπρὸςτὸαὐτὸλόγονἐχόντωντὸμείζοναλόγον greater than A. And it was shown that (it is) not equal
ἔχονμεῖζόνἐστιν·καὶπρὸςὃτὸαὐτὸμείζοναλόγονἔχει, (to A) either. Thus, B is less than A.
ἐκεῖνοἔλαττόνἐστιν·ὅπερἔδειδεῖξαι.
Thus, for (magnitudes) having a ratio to the same
(magnitude), that (magnitude which) has the greater
ratio is (the) greater. And that (magnitude) to which
the latter (magnitude) has a greater ratio is (the) lesser.
(Which is) the very thing it was required to show.
. Proposition 11 †
Οἱτῷαὐτῷλόγῳοἱαὐτοὶκαὶἀλλήλοιςεἰσὶνοἱαὐτοί. (Ratios which are) the same with the same ratio are
also the same with one another.
Α Γ Ε
A
C
E
Β ∆ Ζ
B
D
F
Η Θ Κ
G
H
K
Λ Μ Ν
L
M
N
῎ΕστωσανγὰρὡςμὲντὸΑπρὸςτὸΒ,οὕτωςτὸΓπρὸς For let it be that as A (is) to B, so C (is) to D, and as
τὸΔ,ὡςδὲτὸΓπρὸςτὸΔ,οὕτωςτὸΕπρὸςτὸΖ·λέγω, C (is) to D, so E (is) to F . I say that as A is to B, so E
ὅτιἐστὶνὡςτὸΑπρὸςτὸΒ,οὕτωςτὸΕπρὸςτὸΖ. (is) to F .
ΕἰλήφθωγὰρτῶνΑ,Γ,ΕἰσάκιςπολλαπλάσιατὰΗ,Θ, For let the equal multiples G, H, K have been taken
Κ,τῶνδὲΒ,Δ,Ζἄλλα,ἃἔτυχεν,ἰσάκιςπολλαπλάσιατὰ of A, C, E (respectively), and the other random equal
Λ,Μ,Ν.
multiples L, M, N of B, D, F (respectively).
ΚαὶἐπείἐστινὡςτὸΑπρὸςτὸΒ,οὕτωςτὸΓπρὸςτὸ And since as A is to B, so C (is) to D, and the equal
Δ,καὶεἴληπταιτῶνμὲνΑ,ΓἰσάκιςπολλαπλάσιατὰΗ,Θ, multiples G and H have been taken of A and C (respecτῶνδὲΒ,Δἄλλα,ἃἔτυχεν,ἰσάκιςπολλαπλάσιατὰΛ,Μ,
tively), and the other random equal multiples L and M
εἰἄραὑπερέχειτὸΗτοῦΛ,ὑπερέχεικαὶτὸΘτοῦΜ,καὶεἰ of B and D (respectively), thus if G exceeds L then H
ἴσονἐστίν,ἴσον,καὶεἰἐλλείπει,ἐλλείπει.πάλιν,ἐπείἐστιν also exceeds M, and if (G is) equal (to L then H is also)
A
C
B
141

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
ὡςτὸΓπρὸςτὸΔ,οὕτωςτὸΕπρὸςτὸΖ,καὶεἴληπται equal (to M), and if (G is) less (than L then H is also)
τῶνΓ,ΕἰσάκιςπολλαπλάσιατὰΘ,Κ,τῶνδὲΔ,Ζἄλλα, less (than M) [Def. 5.5]. Again, since as C is to D, so
ἃἔτυχεν,ἰσάκιςπολλαπλάσιατὰΜ,Ν,εἰἄραὑπερέχειτὸ E (is) to F , and the equal multiples H and K have been
ΘτοῦΜ,ὑπερέχεικαὶτὸΚτοῦΝ,καὶεἰἴσον,ἴσον,καὶεἰ taken of C and E (respectively), and the other random
ἔλλατον,ἔλαττον. ἀλλὰεἰὑπερεῖχετὸΘτοῦΜ,ὑπερεῖχε equal multiples M and N of D and F (respectively), thus
καὶτὸΗτοῦΛ,καὶεἰἴσον,ἴσον,καὶεἰἔλαττον,ἔλαττον· if H exceeds M then K also exceeds N, and if (H is)
ὥστεκαὶεἰὑπερέχειτὸΗτοῦΛ,ὑπερέχεικαὶτὸΚτοῦ equal (to M then K is also) equal (to N), and if (H is)
Ν,καὶεἰἴσον,ἴσον,καὶεἰἔλαττον,ἔλαττον. καίἐστιτὰ less (than M then K is also) less (than N) [Def. 5.5]. But
μὲνΗ,ΚτῶνΑ,Εἰσάκιςπολλαπλάσια,τὰδὲΛ,ΝτῶνΒ, (we saw that) if H was exceeding M then G was also ex-
Ζἄλλα,ἃἔτυχεν,ἰσάκιςπολλαπλάσια·ἔστινἄραὡςτὸΑ ceeding L, and if (H was) equal (to M then G was also)
πρὸςτὸΒ,οὕτωςτὸΕπρὸςτὸΖ.
equal (to L), and if (H was) less (than M then G was
Οἱἄρατῷαὐτῷλόγῳοἱαὐτοὶκαὶἀλλήλοιςεἰσὶνοἱ also) less (than L). And, hence, if G exceeds L then K
αὐτοί·ὅπερἔδειδεῖξαι.
also exceeds N, and if (G is) equal (to L then K is also)
equal (to N), and if (G is) less (than L then K is also)
less (than N). And G and K are equal multiples of A
and E (respectively), and L and N other random equal
multiples of B and F (respectively). Thus, as A is to B,
so E (is) to F [Def. 5.5].
Thus, (ratios which are) the same with the same ratio
are also the same with one another. (Which is) the very
thing it was required to show.
† In modern notation, this proposition reads that if α : β :: γ : δ and γ : δ :: ǫ : ζ then α : β :: ǫ : ζ.
. Proposition 12 †
᾿Εὰνᾖὁποσαοῦνμεγέθηἀνάλογον,ἔσταιὡςἓντῶν If there are any number of magnitudes whatsoever
ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ (which are) proportional then as one of the leading (mag-
ἡγούμεναπρὸςἅπαντατὰἑπόμενα.
nitudes is) to one of the following, so will all of the leading
(magnitudes) be to all of the following.
Α Γ Ε
A C E
Β ∆ Ζ
B D F
Η
Θ
Κ
Λ
Μ
Ν
῎ΕστωσανὁποσαοῦνμεγέθηἀνάλογοντὰΑ,Β,Γ,Δ, Let there be any number of magnitudes whatsoever,
Ε,Ζ,ὡςτὸΑπρὸςτὸΒ,οὕτωςτὸΓπρὸςτὸΔ,καὶτὸΕ A, B, C, D, E, F , (which are) proportional, (so that) as
πρὸςτοΖ·λέγω,ὅτιἐστὶνὡςτὸΑπρὸςτὸΒ,οὕτωςτὰ A (is) to B, so C (is) to D, and E to F . I say that as A is
Α,Γ,ΕπρὸςτὰΒ,Δ,Ζ. to B, so A, C, E (are) to B, D, F .
ΕἰλήφθωγὰρτῶνμὲνΑ,Γ,ΕἰσάκιςπολλαπλάσιατὰΗ, For let the equal multiples G, H, K have been taken
Θ,Κ,τῶνδὲΒ,Δ,Ζἄλλα,ἃἔτυχεν,ἰσάκιςπολλαπλάσια of A, C, E (respectively), and the other random equal
τὰΛ,Μ,Ν.
multiples L, M, N of B, D, F (respectively).
ΚαὶἐπείἐστινὡςτὸΑπρὸςτὸΒ,οὕτωςτὸΓπρὸςτὸ And since as A is to B, so C (is) to D, and E to F , and
Δ,καὶτὸΕπρὸςτὸΖ,καὶεἴληπταιτῶνμὲνΑ,Γ,Εἰσάκις the equal multiples G, H, K have been taken of A, C, E
πολλαπλάσιατὰΗ,Θ,ΚτῶνδὲΒ,Δ,Ζἄλλα,ἃἔτυχεν, (respectively), and the other random equal multiples L,
ἰσάκιςπολλαπλάσιατὰΛ,Μ,Ν,εἰἄραὑπερέχειτὸΗτοῦΛ, M, N of B, D, F (respectively), thus if G exceeds L then
ὑπερέχεικαὶτὸΘτοῦΜ,καὶτὸΚτοῦΝ,καὶεἰἴσον,ἴσον, H also exceeds M, and K (exceeds) N, and if (G is)
καὶεἰἔλαττον,ἔλαττον.ὥστεκαὶεἰὑπερέχειτὸΗτοῦΛ, equal (to L then H is also) equal (to M, and K to N),
G
H
K
L
M
N
142

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
ὑπερέχεικαὶτὰΗ,Θ,ΚτῶνΛ,Μ,Ν,καὶεἰἴσον,ἴσα,καὶ and if (G is) less (than L then H is also) less (than M,
εἰἔλαττον,ἔλαττονα. καίἐστιτὸμὲνΗκαὶτὰΗ,Θ,Κ and K than N) [Def. 5.5]. And, hence, if G exceeds L
τοῦΑκαὶτῶνΑ,Γ,Εἰσάκιςπολλαπλάσια,ἐπειδήπερἐὰν then G, H, K also exceed L, M, N, and if (G is) equal
ᾖὁποσαοῦνμεγέθηὁποσωνοῦνμεγεθῶνἴσωντὸπλῆθος (to L then G, H, K are also) equal (to L, M, N) and
ἕκαστονἑκάστουἰσάκιςπολλαπλάσιον,ὁσαπλάσιόνἐστιν if (G is) less (than L then G, H, K are also) less (than
ἓντῶνμεγεθῶνἑνός,τοσαυταπλάσιαἔσταικαὶτὰπάντα L, M, N). And G and G, H, K are equal multiples of
τῶνπάντων. διὰτὰαὐτὰδὴκαὶτὸΛκαὶτὰΛ,Μ,Ντοῦ A and A, C, E (respectively), inasmuch as if there are
ΒκαὶτῶνΒ,Δ,Ζἰσάκιςἐστὶπολλαπλάσια·ἔστινἄραὡς any number of magnitudes whatsoever (which are) equal
τὸΑπρὸςτὸΒ,οὕτωςτὰΑ,Γ,ΕπρὸςτὰΒ,Δ,Ζ. multiples, respectively, of some (other) magnitudes, of
᾿Εὰνἄραᾖὁποσαοῦνμεγέθηἀνάλογον,ἔσταιὡςἓν equal number (to them), then as many times as one of the
τῶνἡγουμένωνπρὸςἓντῶνἑπομένων,οὕτωςἅπαντατὰ (first) magnitudes is (divisible) by one (of the second),
ἡγούμεναπρὸςἅπαντατὰἑπόμενα·ὅπερἔδειδεῖξαι. so many times will all (of the first magnitudes) also (be
divisible) by all (of the second) [Prop. 5.1]. So, for the
same (reasons), L and L, M, N are also equal multiples
of B and B, D, F (respectively). Thus, as A is to B, so
A, C, E (are) to B, D, F (respectively).
Thus, if there are any number of magnitudes whatsoever
(which are) proportional then as one of the leading
(magnitudes is) to one of the following, so will all of the
leading (magnitudes) be to all of the following. (Which
is) the very thing it was required to show.
† In modern notation, this proposition reads that if α : α ′ :: β : β ′ :: γ : γ ′ etc. then α : α ′ :: (α + β + γ + · · ·) : (α ′ + β ′ + γ ′ + · · · ).
. Proposition 13 †
᾿Εὰνπρῶτονπρὸςδεύτεροντὸναὐτὸνἒχῃλόγονκαὶ If a first (magnitude) has the same ratio to a second
τρίτον πρὸς τέταρτον, τρίτον δὲ πρὸς τέταρτον μείζονα that a third (has) to a fourth, and the third (magnitude)
λόγονἔχῃἢπέμπτονπρὸςἕκτον,καὶπρῶτονπρὸςδεύτερον has a greater ratio to the fourth than a fifth (has) to a
μείζοναλόγονἕξειἢπέμπτονπρὸςἕκτον.
sixth, then the first (magnitude) will also have a greater
ratio to the second than the fifth (has) to the sixth.
Α Γ Ε
A
C
E
Β ∆ Ζ
B
D
F
Μ Η Θ
M
G
H
Ν Κ Λ
N
K
L
ΠρῶτονγὰρτὸΑπρὸςδεύτεροντὸΒτὸναὐτὸνἐχέτω For let a first (magnitude) A have the same ratio to a
λόγονκαὶτρίτοντὸΓπρὸςτέταρτοντὸΔ,τρίτονδὲτὸΓ second B that a third C (has) to a fourth D, and let the
πρὸςτέταρτοντὸΔμείζοναλόγονἐχέτωἢπέμπτοντὸΕ third (magnitude) C have a greater ratio to the fourth
πρὸςἕκτοντὸΖ.λέγω,ὅτικαὶπρῶτοντὸΑπρὸςδεύτερον D than a fifth E (has) to a sixth F . I say that the first
τὸΒμείζοναλόγονἕξειἤπερπέμπτοντὸΕπρὸςἕκτοντὸ (magnitude) A will also have a greater ratio to the second
Ζ. B than the fifth E (has) to the sixth F .
᾿ΕπεὶγὰρἔστιτινὰτῶνμὲνΓ,Εἰσάκιςπολλαπλάσια, For since there are some equal multiples of C and
τῶνδὲΔ,Ζἄλλα,ἃἔτυχεν,ἰσάκιςπολλαπλάσια,καὶτὸμὲν E, and other random equal multiples of D and F , (for
τοῦΓπολλαπλάσιοντοῦτοῦΔπολλαπλασίουὑπερέχει, which) the multiple of C exceeds the (multiple) of D,
τὸδὲτοῦΕπολλαπλάσιοντοῦτοῦΖπολλαπλασίουοὐχ and the multiple of E does not exceed the multiple of F
ὑπερέχει, εἰλήφθω, καὶ ἔστω τῶν μὲν Γ, Ε ἰσάκις πολ- [Def. 5.7], let them have been taken. And let G and H be
λαπλάσιατὰΗ,Θ,τῶνδὲΔ,Ζἄλλα,ἃἔτυχεν, ἰσάκις equal multiples of C and E (respectively), and K and L
πολλαπλάσιατὰΚ,Λ,ὥστετὸμὲνΗτοῦΚὑπερέχειν,τὸ other random equal multiples of D and F (respectively),
δὲΘτοῦΛμὴὑπερέχειν·καὶὁσαπλάσιονμένἐστιτὸΗ such that G exceeds K, but H does not exceed L. And as
τοῦΓ,τοσαυταπλάσιονἔστωκαὶτὸΜτοῦΑ,ὁσαπλάσιον many times as G is (divisible) by C, so many times let M
δὲτὸΚτοῦΔ,τοσαυταπλάσιονἔστωκαὶτὸΝτοῦΒ. be (divisible) by A. And as many times as K (is divisible)
143

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
ΚαὶἐπείἐστινὡςτὸΑπρὸςτὸΒ,οὕτωςτὸΓπρὸς by D, so many times let N be (divisible) by B.
τὸΔ,καὶεἴληπταιτῶνμὲνΑ,Γἰσάκιςπολλαπλάσιατὰ And since as A is to B, so C (is) to D, and the equal
Μ,Η,τῶνδὲΒ,Δἄλλα,ἃἔτυχεν,ἰσάκιςπολλαπλάσιατὰ multiples M and G have been taken of A and C (respec-
Ν,Κ,εἰἄραὑπερέχειτὸΜτοῦΝ,ὑπερέχεικαὶτὸΗτοῦ tively), and the other random equal multiples N and K
Κ,καὶεἰἴσον,ἴσον,καὶεἰἔλαττον,ἔλλατον. ὑπερέχειδὲ of B and D (respectively), thus if M exceeds N then G
τὸΗτοῦΚ·ὑπερέχειἄρακαὶτὸΜτοῦΝ.τὸδὲΘτοῦ exceeds K, and if (M is) equal (to N then G is also)
Λοὐχὑπερέχει· καίἐστιτὰμὲνΜ,ΘτῶνΑ,Εἰσάκις equal (to K), and if (M is) less (than N then G is also)
πολλαπλάσια,τὰδὲΝ,ΛτῶνΒ,Ζἄλλα,ἃἔτυχεν,ἰσάκις less (than K) [Def. 5.5]. And G exceeds K. Thus, M
πολλαπλάσια·τὸἄραΑπρὸςτὸΒμείζοναλόγονἔχειἤπερ also exceeds N. And H does not exceeds L. And M and
τὸΕπρὸςτὸΖ.
H are equal multiples of A and E (respectively), and N
᾿Εὰνἄραπρῶτονπρὸςδεύτεροντὸναὐτὸνἒχῃλόγον and L other random equal multiples of B and F (respecκαὶτρίτονπρὸςτέταρτον,τρίτονδὲπρὸςτέταρτονμείζονα
tively). Thus, A has a greater ratio to B than E (has) to
λόγονἔχῃἢπέμπτονπρὸςἕκτον,καὶπρῶτονπρὸςδεύτερον F [Def. 5.7].
μείζοναλόγονἕξειἢπέμπτονπρὸςἕκτον·ὅπερἔδειδεῖξαι. Thus, if a first (magnitude) has the same ratio to a
second that a third (has) to a fourth, and a third (magnitude)
has a greater ratio to a fourth than a fifth (has)
to a sixth, then the first (magnitude) will also have a
greater ratio to the second than the fifth (has) to the
sixth. (Which is) the very thing it was required to show.
† In modern notation, this proposition reads that if α : β :: γ : δ and γ : δ > ǫ : ζ then α : β > ǫ : ζ.
. Proposition 14 †
᾿Εὰνπρῶτονπρὸςδεύτεροντὸναὐτὸνἔχῃλόγονκαὶ If a first (magnitude) has the same ratio to a second
τρίτονπρὸςτέταρτον,τὸδὲπρῶτοντοῦτρίτουμεῖζονᾖ, that a third (has) to a fourth, and the first (magnitude)
καὶτὸδεύτεροντοῦτετάρτουμεῖζονἔσται,κἂνἴσον,ἴσον, is greater than the third, then the second will also be
κἂνἔλαττον,ἔλαττον.
greater than the fourth. And if (the first magnitude is)
equal (to the third then the second will also be) equal (to
the fourth). And if (the first magnitude is) less (than the
third then the second will also be) less (than the fourth).
Α
Β
Γ
∆
ΠρῶτονγὰρτὸΑπρὸςδεύτεροντὸΒαὐτὸνἐχέτω For let a first (magnitude) A have the same ratio to a
λόγονκαὶτρίτοντὸΓπρὸςτέταρτοντὸΔ,μεῖζονδὲἔστω second B that a third C (has) to a fourth D. And let A be
τὸΑτοῦΓ·λέγω,ὅτικαὶτὸΒτοῦΔμεῖζόνἐστιν. greater than C. I say that B is also greater than D.
᾿ΕπεὶγὰρτὸΑτοῦΓμεῖζόνἐστιν,ἄλλοδέ,ὃἔτυχεν, For since A is greater than C, and B (is) another ran-
[μέγεθος]τὸΒ,τὸΑἄραπρὸςτὸΒμείζοναλόγονἔχει dom [magnitude], A thus has a greater ratio to B than C
ἤπερτὸΓπρὸςτὸΒ.ὡςδὲτὸΑπρὸςτὸΒ,οὕτωςτὸ (has) to B [Prop. 5.8]. And as A (is) to B, so C (is) to
ΓπρὸςτὸΔ·καὶτὸΓἄραπρὸςτὸΔμείζοναλόγονἔχει D. Thus, C also has a greater ratio to D than C (has) to
ἤπερτὸΓπρὸςτὸΒ.πρὸςὃδὲτὸαὐτὸμείζοναλόγον B. And that (magnitude) to which the same (magnitude)
ἔχει,ἐκεῖνοἔλασσόνἐστιν·ἔλασσονἄρατὸΔτοῦΒ·ὥστε has a greater ratio is the lesser [Prop. 5.10]. Thus, D (is)
μεῖζόνἐστιτὸΒτοῦΔ. less than B. Hence, B is greater than D.
῾Ομοίωςδὴδεῖξομεν,ὅτικἂνἴσονᾖτὸΑτῷΓ,ἴσον So, similarly, we can show that even if A is equal to C
ἔσταικαὶτὸΒτῷΔ,κἄνἔλασσονᾖτὸΑτοῦΓ,ἔλασσον then B will also be equal to D, and even if A is less than
ἔσταικαὶτὸΒτοῦΔ. C then B will also be less than D.
᾿Εὰνἄραπρῶτονπρὸςδεύτεροντὸναὐτὸνἔχῃλόγον Thus, if a first (magnitude) has the same ratio to a
καὶτρίτονπρὸςτέταρτον,τὸδὲπρῶτοντοῦτρίτουμεῖζονᾖ, second that a third (has) to a fourth, and the first (magκαὶτὸδεύτεροντοῦτετάρτουμεῖζονἔσται,κἂνἴσον,ἴσον,
nitude) is greater than the third, then the second will also
κἂνἔλαττον,ἔλαττον·ὅπερἔδειδεῖξαι.
be greater than the fourth. And if (the first magnitude is)
A
B
C
D
144

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
† In modern notation, this proposition reads that if α : β :: γ : δ then α γ as β δ.
equal (to the third then the second will also be) equal (to
the fourth). And if (the first magnitude is) less (than the
third then the second will also be) less (than the fourth).
(Which is) the very thing it was required to show.
. Proposition 15 †
Τὰμέρητοῖςὡσαύτωςπολλαπλασίοιςτὸναὐτὸνἔχει Parts have the same ratio as similar multiples, taken
λόγονληφθέντακατάλληλα.
in corresponding order.
Α
Η
∆ Κ Λ
Θ
Ε
Β
Γ
Ζ
A G H
D K L
῎ΕστωγὰρἰσάκιςπολλαπλάσιοντὸΑΒτοῦΓκαὶτοΔΕ For let AB and DE be equal multiples of C and F
τοῦΖ·λέγω,ὅτιἐστὶνὡςτὸΓπρὸςτὸΖ,οὕτωςτὸΑΒ (respectively). I say that as C is to F , so AB (is) to DE.
πρὸςτὸΔΕ.
For since AB and DE are equal multiples of C and
᾿ΕπεὶγὰρἰσάκιςἐστὶπολλαπλάσιοντὸΑΒτοῦΓκαὶ F (respectively), thus as many magnitudes as there are
τὸΔΕτοῦΖ,ὅσαἄραἐστὶνἐντῷΑΒμεγέθηἴσατῷ in AB equal to C, so many (are there) also in DE equal
Γ,τοσαῦτακαὶἐντῷΔΕἴσατῷΖ.διῃρήσθωτὸμὲνΑΒ to F . Let AB have been divided into (magnitudes) AG,
εἰςτὰτῷΓἴσατὰΑΗ,ΗΘ,ΘΒ,τὸδὲΔΕεἰςτὰτῷΖ GH, HB, equal to C, and DE into (magnitudes) DK,
ἴσατὰΔΚ,ΚΛ,ΛΕ·ἔσταιδὴἴσοντὸπλῆθοςτῶνΑΗ, KL, LE, equal to F . So, the number of (magnitudes)
ΗΘ,ΘΒτῷπλήθειτῶνΔΚ,ΚΛ,ΛΕ.καὶἐπεὶἴσαἐστὶτὰ AG, GH, HB will equal the number of (magnitudes)
ΑΗ,ΗΘ,ΘΒἀλλήλοις,ἔστιδὲκαὶτὰΔΚ,ΚΛ,ΛΕἴσα DK, KL, LE. And since AG, GH, HB are equal to one
ἀλλήλοις,ἔστινἄραὡςτὸΑΗπρὸςτὸΔΚ,οὕτωςτὸΗΘ another, and DK, KL, LE are also equal to one another,
πρὸςτὸΚΛ,καὶτὸΘΒπρὸςτὸΛΕ.ἔσταιἄρακαὶὡςἓν thus as AG is to DK, so GH (is) to KL, and HB to LE
τῶνἡγουμένωνπρὸςἓντῶνἑπομένων,οὕτωςἅπαντατὰ [Prop. 5.7]. And, thus (for proportional magnitudes), as
ἡγουμέναπρὸςἅπαντατὰἑπόμενα·ἔστινἄραὡςτὸΑΗ one of the leading (magnitudes) will be to one of the folπρὸςτὸΔΚ,οὕτωςτὸΑΒπρὸςτὸΔΕ.ἴσονδὲτὸμὲνΑΗ
lowing, so all of the leading (magnitudes will be) to all of
τῷΓ,τὸδὲΔΚτῷΖ·ἔστινἄραὡςτὸΓπρὸςτὸΖοὕτως the following [Prop. 5.12]. Thus, as AG is to DK, so AB
τὸΑΒπρὸςτὸΔΕ.
(is) to DE. And AG is equal to C, and DK to F . Thus,
Τὰἄραμέρητοῖςὡσαύτωςπολλαπλασίοιςτὸναὐτὸν as C is to F , so AB (is) to DE.
ἔχειλόγονληφθέντακατάλληλα·ὅπερἔδειδεῖξαι.
Thus, parts have the same ratio as similar multiples,
taken in corresponding order. (Which is) the very thing
it was required to show.
† In modern notation, this proposition reads that α : β :: m α : m β.
¦. Proposition 16 †
᾿Εὰντέσσαραμεγέθηἀνάλογονᾖ,καὶἐναλλὰξἀνάλογον If four magnitudes are proportional then they will also
ἔσται.
be proportional alternately.
῎ΕστωτέσσαραμεγέθηἀνάλογοντὰΑ,Β,Γ,Δ,ὡςτὸ Let A, B, C and D be four proportional magnitudes,
ΑπρὸςτὸΒ,οὕτωςτὸΓπρὸςτὸΔ·λέγω,ὅτικαὶἐναλλὰξ (such that) as A (is) to B, so C (is) to D. I say that they
[ἀνάλογον]ἔσται,ὡςτὸΑπρὸςτὸΓ,οὕτωςτὸΒπρὸςτὸ will also be [proportional] alternately, (so that) as A (is)
Δ. to C, so B (is) to D.
ΕἰλήφθωγὰρτῶνμὲνΑ,ΒἰσάκιςπολλαπλάσιατὰΕ, For let the equal multiples E and F have been taken
Ζ,τῶνδὲΓ,Δἄλλα,ἃἔτυχεν,ἰσάκιςπολλαπλάσιατὰΗ, of A and B (respectively), and the other random equal
Θ. multiples G and H of C and D (respectively).
E
B
C
F
145

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
Α
Β
Ε
Ζ
Γ
∆
Η
Θ
ΚαὶἐπεὶἰσάκιςἐστὶπολλαπλάσιοντὸΕτοῦΑκαὶτὸΖ And since E and F are equal multiples of A and B
τοῦΒ,τὰδὲμέρητοῖςὡσαύτωςπολλαπλασίοιςτὸναὐτὸν (respectively), and parts have the same ratio as similar
ἔχειλόγον,ἔστινἄραὡςτὸΑπρὸςτὸΒ,οὕτωςτὸΕπρὸς multiples [Prop. 5.15], thus as A is to B, so E (is) to F .
τὸΖ.ὡςδὲτὸΑπρὸςτὸΒ,οὕτωςτὸΓπρὸςτὸΔ·καὶὡς But as A (is) to B, so C (is) to D. And, thus, as C (is)
ἄρατὸΓπρὸςτὸΔ,οὕτωςτὸΕπρὸςτὸΖ.πάλιν,ἐπεὶτὰ to D, so E (is) to F [Prop. 5.11]. Again, since G and H
Η,ΘτῶνΓ,Δἰσάκιςἐστὶπολλαπλάσια,ἔστινἄραὡςτὸΓ are equal multiples of C and D (respectively), thus as C
πρὸςτὸΔ,οὕτωςτὸΗπρὸςτὸΘ.ὡςδὲτὸΓπρὸςτὸΔ, is to D, so G (is) to H [Prop. 5.15]. But as C (is) to D,
[οὕτως]τὸΕπρὸςτὸΖ·καὶὡςἄρατὸΕπρὸςτὸΖ,οὕτως [so] E (is) to F . And, thus, as E (is) to F , so G (is) to
τὸΗπρὸςτὸΘ.ἐὰνδὲτέσσαραμεγέθηἀνάλογονᾖ,τὸδὲ H [Prop. 5.11]. And if four magnitudes are proportional,
πρῶτοντοῦτρίτουμεῖζονᾖ,καὶτὸδεύτεροντοῦτετάρτου and the first is greater than the third then the second will
μεῖζονἔσται,κἂνἴσον,ἴσον,κἄνἔλαττον,ἔλαττον.εἰἄρα also be greater than the fourth, and if (the first is) equal
ὑπερέχειτὸΕτοῦΗ,ὑπερέχεικαὶτὸΖτοῦΘ,καὶεἰἴσον, (to the third then the second will also be) equal (to the
ἴσον,καὶεἰἔλαττον,ἔλαττον. καίἐστιτὰμὲνΕ,Ζτῶν fourth), and if (the first is) less (than the third then the
Α,Βἰσάκιςπολλαπλάσια,τὰδὲΗ,ΘτῶνΓ,Δἄλλα,ἃ second will also be) less (than the fourth) [Prop. 5.14].
ἔτυχεν,ἰσάκιςπολλαπλάσια·ἔστινἄραὡςτὸΑπρὸςτὸΓ, Thus, if E exceeds G then F also exceeds H, and if (E is)
οὕτωςτὸΒπρὸςτὸΔ.
equal (to G then F is also) equal (to H), and if (E is) less
᾿Εὰν ἄρα τέσσαρα μεγέθη ἀνάλογον ᾖ, καὶ ἐναλλὰξ (than G then F is also) less (than H). And E and F are
ἀνάλογονἔσται·ὅπερἔδειδεῖξαι.
equal multiples of A and B (respectively), and G and H
other random equal multiples of C and D (respectively).
Thus, as A is to C, so B (is) to D [Def. 5.5].
Thus, if four magnitudes are proportional then they
will also be proportional alternately. (Which is) the very
thing it was required to show.
† In modern notation, this proposition reads that if α : β :: γ : δ then α : γ :: β : δ.
Þ. Proposition 17 †
᾿Εὰνσυγκείμεναμεγέθηἀνάλογονᾖ,καὶδιαιρεθέντα If composed magnitudes are proportional then they
ἀνάλογονἔσται.
will also be proportional (when) separarted.
Α Ε Β Γ Ζ ∆
A E B C F D
A
B
E
F
C
D
G
H
Η Θ Κ Ξ
G
H
K
O
Λ
Μ Ν Π
῎ΕστωσυγκείμεναμεγέθηἀνάλογοντὰΑΒ,ΒΕ,ΓΔ, Let AB, BE, CD, and DF be composed magnitudes
ΔΖ,ὡςτὸΑΒπρὸςτὸΒΕ,οὕτωςτὸΓΔπρὸςτὸΔΖ· (which are) proportional, (so that) as AB (is) to BE, so
λέγω,ὅτικαὶδιαιρεθένταἀνάλογονἔσται,ὡςτὸΑΕπρὸς CD (is) to DF . I say that they will also be proportional
τὸΕΒ,οὕτωςτὸΓΖπρὸςτὸΔΖ.
(when) separated, (so that) as AE (is) to EB, so CF (is)
ΕἰλήφθωγὰρτῶνμὲνΑΕ,ΕΒ,ΓΖ,ΖΔἰσάκιςπολ- to DF .
λαπλάσιατὰΗΘ,ΘΚ,ΛΜ,ΜΝ,τῶνδὲΕΒ,ΖΔἄλλα,ἃ For let the equal multiples GH, HK, LM, and MN
ἔτυχεν,ἰσάκιςπολλαπλάσιατὰΚΞ,ΝΠ.
have been taken of AE, EB, CF , and FD (respectively),
ΚαὶἐπεὶἰσάκιςἐστὶπολλαπλάσιοντὸΗΘτοῦΑΕκαὶ and the other random equal multiples KO and NP of
τὸΘΚτοῦΕΒ,ἰσάκιςἄραἐστὶπολλαπλάσιοντὸΗΘτοῦ EB and FD (respectively).
L
M
N
P
146

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
ΑΕκαὶτὸΗΚτοῦΑΒ.ἰσάκιςδέἐστιπολλαπλάσιοντὸΗΘ And since GH and HK are equal multiples of AE and
τοῦΑΕκαὶτὸΛΜτοῦΓΖ·ἰσάκιςἄραἐστὶπολλαπλάσιον EB (respectively), GH and GK are thus equal multiples
τὸΗΚτοῦΑΒκαὶτὸΛΜτοῦΓΖ.πάλιν,ἐπεὶἰσάκιςἐστὶ of AE and AB (respectively) [Prop. 5.1]. But GH and
πολλαπλάσιοντὸΛΜτοῦΓΖκαὶτὸΜΝτοῦΖΔ,ἰσάκιςἄρα LM are equal multiples of AE and CF (respectively).
ἐστὶπολλαπλάσιοντὸΛΜτοῦΓΖκαὶτὸΛΝτοῦΓΔ.ἰσάκις Thus, GK and LM are equal multiples of AB and CF
δὲἦνπολλαπλάσιοντὸΛΜτοῦΓΖκαὶτὸΗΚτοῦΑΒ· (respectively). Again, since LM and MN are equal mul-
ἰσάκιςἄραἐστὶπολλαπλάσιοντὸΗΚτοῦΑΒκαὶτὸΛΝτοῦ tiples of CF and FD (respectively), LM and LN are thus
ΓΔ.τὰΗΚ,ΛΝἄρατῶνΑΒ,ΓΔἰσάκιςἐστὶπολλαπλάσια. equal multiples of CF and CD (respectively) [Prop. 5.1].
πάλιν,ἐπεὶἰσάκιςἐστὶπολλαπλασίοντὸΘΚτοῦΕΒκαὶτὸ And LM and GK were equal multiples of CF and AB
ΜΝτοῦΖΔ,ἔστιδὲκαὶτὸΚΞτοῦΕΒἰσάκιςπολλαπλάσιον (respectively). Thus, GK and LN are equal multiples
καὶτὸΝΠτοῦΖΔ,καὶσυντεθὲντὸΘΞτοῦΕΒἰσάκιςἐστὶ of AB and CD (respectively). Thus, GK, LN are equal
πολλαπλάσιονκαὶτὸΜΠτοῦΖΔ.καὶἐπείἐστινὡςτὸΑΒ multiples of AB, CD. Again, since HK and MN are
πρὸςτὸΒΕ,οὕτωςτὸΓΔπρὸςτὸΔΖ,καὶεἴληπταιτῶν equal multiples of EB and FD (respectively), and KO
μὲνΑΒ,ΓΔἰσάκιςπολλαπλάσιατὰΗΚ,ΛΝ,τῶνδὲΕΒ, and NP are also equal multiples of EB and FD (respec-
ΖΔἰσάκιςπολλαπλάσιατὰΘΞ,ΜΠ,εἰἄραὑπερέχειτὸ tively), then, added together, HO and MP are also equal
ΗΚτοῦΘΞ,ὑπερέχεικαὶτὸΛΝτοῦΜΠ,καὶεἰἴσον,ἴσον, multiples of EB and FD (respectively) [Prop. 5.2]. And
καὶεἰἔλαττον,ἔλαττον. ὑπερεχέτωδὴτὸΗΚτοῦΘΞ, since as AB (is) to BE, so CD (is) to DF , and the equal
καὶκοινοῦἀφαιρεθέντοςτοῦΘΚὑπερέχειἄρακαὶτὸΗΘ multiples GK, LN have been taken of AB, CD, and the
τοῦΚΞ.ἀλλαεἰὑπερεῖχετὸΗΚτοῦΘΞὑπερεῖχεκαὶτὸ equal multiples HO, MP of EB, FD, thus if GK exceeds
ΛΝτοῦΜΠ·ὑπερέχειἄρακαὶτὸΛΝτοῦΜΠ,καὶκοινοῦ HO then LN also exceeds MP , and if (GK is) equal (to
ἀφαιρεθέντοςτοῦΜΝὑπερέχεικαὶτὸΛΜτοῦΝΠ·ὥστε HO then LN is also) equal (to MP ), and if (GK is) less
εἰὑπερέχειτὸΗΘτοῦΚΞ,ὑπερέχεικαὶτὸΛΜτοῦΝΠ. (than HO then LN is also) less (than MP ) [Def. 5.5].
ὁμοίωςδὴδεῖξομεν,ὅτικἂνἴσονᾖτὸΗΘτῷΚΞ,ἴσον So let GK exceed HO, and thus, HK being taken away
ἔσταικαὶτὸΛΜτῷΝΠ,κἂνἔλαττον,ἔλαττον.καίἐστιτὰ from both, GH exceeds KO. But (we saw that) if GK
μὲνΗΘ,ΛΜτῶνΑΕ,ΓΖἰσάκιςπολλαπλάσια,τὰδὲΚΞ, was exceeding HO then LN was also exceeding MP .
ΝΠτῶνΕΒ,ΖΔἄλλα,ἃἔτυχεν,ἰσάκιςπολλαπλάσια·ἔστιν Thus, LN also exceeds MP , and, MN being taken away
ἄραὡςτὸΑΕπρὸςτὸΕΒ,οὕτωςτὸΓΖπρὸςτὸΖΔ. from both, LM also exceeds NP . Hence, if GH exceeds
᾿Εὰν ἄρα συγκείμενα μεγέθη ἀνάλογον ᾖ, καὶ διαι- KO then LM also exceeds NP . So, similarly, we can
ρεθένταἀνάλογονἔσται·ὅπερἔδειδεῖξαι.
show that even if GH is equal to KO then LM will also
be equal to NP , and even if (GH is) less (than KO then
LM will also be) less (than NP ). And GH, LM are equal
multiples of AE, CF , and KO, NP other random equal
multiples of EB, FD. Thus, as AE is to EB, so CF (is)
to FD [Def. 5.5].
Thus, if composed magnitudes are proportional then
they will also be proportional (when) separarted. (Which
is) the very thing it was required to show.
† In modern notation, this proposition reads that if α + β : β :: γ + δ : δ then α : β :: γ : δ.
. Proposition 18 †
᾿Εὰν διῃρημένα μεγέθη ἀνάλογον ᾖ, καὶ συντεθέντα If separated magnitudes are proportional then they
ἀνάλογονἔσται.
will also be proportional (when) composed.
Α
Ε
Β
A
E
B
Γ
Ζ Η
∆
C
F
G
D
῎ΕστωδιῃρημέναμεγέθηἀνάλογοντὰΑΕ,ΕΒ,ΓΖ,ΖΔ, Let AE, EB, CF , and FD be separated magnitudes
ὡςτὸΑΕπρὸςτὸΕΒ,οὕτωςτὸΓΖπρὸςτὸΖΔ·λέγω, (which are) proportional, (so that) as AE (is) to EB, so
ὅτικαὶσυντεθένταἀνάλογονἔσται,ὡςτὸΑΒπρὸςτὸΒΕ, CF (is) to FD. I say that they will also be proportional
147

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
οὕτωςτὸΓΔπρὸςτὸΖΔ.
(when) composed, (so that) as AB (is) to BE, so CD (is)
ΕἰγὰρμήἐστὶνὡςτὸΑΒπρὸςτὸΒΕ,οὕτωςτὸΓΔ to FD.
πρὸςτὸΔΖ,ἔσταιὡςτὸΑΒπρὸςτὸΒΕ,οὕτωςτὸΓΔ For if (it is) not (the case that) as AB is to BE, so
ἤτοιπρὸςἔλασσόντιτοῦΔΖἢπρὸςμεῖζον.
CD (is) to FD, then it will surely be (the case that) as
῎ΕστωπρότερονπρὸςἔλασσοντὸΔΗ.καὶἐπείἐστινὡς AB (is) to BE, so CD is either to some (magnitude) less
τὸΑΒπρὸςτὸΒΕ,οὕτωςτὸΓΔπρὸςτὸΔΗ,συγκείμενα than DF , or (some magnitude) greater (than DF ). ‡
μεγέθη ἀνάλογόνἐστιν· ὥστε καὶ διαιρεθένταἀνάλογον Let it, first of all, be to (some magnitude) less (than
ἔσται.ἔστινἄραὡςτὸΑΕπρὸςτὸΕΒ,οὕτωςτὸΓΗπρὸς DF ), (namely) DG. And since composed magnitudes
τὸΗΔ.ὑπόκειταιδὲκαὶὡςτὸΑΕπρὸςτὸΕΒ,οὕτωςτὸ are proportional, (so that) as AB is to BE, so CD (is) to
ΓΖπρὸςτὸΖΔ.καὶὡςἄρατὸΓΗπρὸςτὸΗΔ,οὕτωςτὸ DG, they will thus also be proportional (when) separated
ΓΖπρὸςτὸΖΔ.μεῖζονδὲτὸπρῶτοντὸΓΗτοῦτρίτουτοῦ [Prop. 5.17]. Thus, as AE is to EB, so CG (is) to GD.
ΓΖ·μεῖζονἄρακαὶτὸδεύτεροντὸΗΔτοῦτετάρτουτοῦ But it was also assumed that as AE (is) to EB, so CF
ΖΔ.ἀλλὰκαὶἔλαττον·ὅπερἐστὶνἀδύνατον·οὐκἄραἐστὶν (is) to FD. Thus, (it is) also (the case that) as CG (is)
ὡςτὸΑΒπρὸςτὸΒΕ,οὕτωςτὸΓΔπρὸςἔλασσοντοῦ to GD, so CF (is) to FD [Prop. 5.11]. And the first
ΖΔ.ὁμοίωςδὴδείξομεν,ὅτιοὐδὲπρὸςμεῖζον·πρὸςαὐτὸ (magnitude) CG (is) greater than the third CF . Thus,
ἄρα.
the second (magnitude) GD (is) also greater than the
᾿Εὰνἄραδιῃρημέναμεγέθηἀνάλογονᾖ,καὶσυντεθέντα fourth FD [Prop. 5.14]. But (it is) also less. The very
ἀνάλογονἔσται·ὅπερἔδειδεῖξαι.
thing is impossible. Thus, (it is) not (the case that) as AB
is to BE, so CD (is) to less than FD. Similarly, we can
show that neither (is it the case) to greater (than FD).
Thus, (it is the case) to the same (as FD).
Thus, if separated magnitudes are proportional then
they will also be proportional (when) composed. (Which
is) the very thing it was required to show.
† In modern notation, this proposition reads that if α : β :: γ : δ then α + β : β :: γ + δ : δ.
‡ Here, Euclid assumes, without proof, that a fourth magnitude proportional to three given magnitudes can always be found.
. Proposition 19 †
᾿Εὰνᾖὡςὅλονπρὸςὅλον,οὕτωςἀφαιρεθὲνπρὸςἀφαι- If as the whole is to the whole so the (part) taken
ρεθέν,καὶτὸλοιπὸνπρὸςτὸλοιπὸνἔσταιὡςὅλονπρὸς away is to the (part) taken away then the remainder to
ὅλον.
the remainder will also be as the whole (is) to the whole.
Α Ε Β
A E
B
Γ
Ζ ∆ C F
D
῎ΕστωγὰρὡςὅλοντὸΑΒπρὸςὅλοντὸΓΔ,οὕτως For let the whole AB be to the whole CD as the (part)
ἀφαιρεθὲν τὸ ΑΕ πρὸς ἀφειρεθὲν τὸ ΓΖ· λέγω, ὅτι καὶ taken away AE (is) to the (part) taken away CF . I say
λοιπὸντὸΕΒπρὸςλοιπὸντὸΖΔἔσταιὡςὅλοντὸΑΒ that the remainder EB to the remainder FD will also be
πρὸςὅλοντὸΓΔ.
as the whole AB (is) to the whole CD.
᾿ΕπεὶγάρἐστινὡςτὸΑΒπρὸςτὸΓΔ,οὕτωςτὸΑΕ For since as AB is to CD, so AE (is) to CF , (it is)
πρὸςτὸΓΖ,καὶἐναλλὰξὡςτὸΒΑπρὸςτὸΑΕ,οὕτως also (the case), alternately, (that) as BA (is) to AE, so
τὸΔΓπρὸςτὸΓΖ.καὶἐπεὶσυγκείμεναμεγέθηἀνάλογόν DC (is) to CF [Prop. 5.16]. And since composed magni-
ἐστιν,καὶδιαιρεθένταἀνάλογονἔσται,ὡςτὸΒΕπρὸςτὸ tudes are proportional then they will also be proportional
ΕΑ,οὕτωςτὸΔΖπρὸςτὸΓΖ·καὶἐναλλάξ,ὡςτὸΒΕπρὸς (when) separated, (so that) as BE (is) to EA, so DF (is)
τὸΔΖ,οὕτωςτὸΕΑπρὸςτὸΖΓ.ὡςδὲτὸΑΕπρὸςτὸΓΖ, to CF [Prop. 5.17]. Also, alternately, as BE (is) to DF ,
οὕτωςὑπόκειταιὅλοντὸΑΒπρὸςὅλοντὸΓΔ.καὶλοιπὸν so EA (is) to FC [Prop. 5.16]. And it was assumed that
ἄρατὸΕΒπρὸςλοιπὸντὸΖΔἔσταιὡςὅλοντὸΑΒπρὸς as AE (is) to CF , so the whole AB (is) to the whole CD.
ὅλοντὸΓΔ.
And, thus, as the remainder EB (is) to the remainder
᾿Εὰνἄραᾖὡςὅλονπρὸςὅλον,οὕτωςἀφαιρεθὲνπρὸς FD, so the whole AB will be to the whole CD.
148

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
ἀφαιρεθέν, καὶτὸλοιπὸνπρὸςτὸλοιπὸνἔσταιὡςὅλον Thus, if as the whole is to the whole so the (part)
πρὸςὅλον[ὅπερἔδειδεῖξαι].
taken away is to the (part) taken away then the remain-
[ΚαὶἐπεὶἐδείχθηὡςτὸΑΒπρὸςτὸΓΔ,οὕτωςτὸΕΒ der to the remainder will also be as the whole (is) to
πρὸςτὸΖΔ,καὶἐναλλὰξὡςτὸΑΒπρὸςτὸΒΕοὕτωςτὸ the whole. [(Which is) the very thing it was required to
ΓΔπρὸςτὸΖΔ,συγκείμεναἄραμεγέθηἀνάλογόνἐστιν· show.]
ἐδείχθηδὲὡςτὸΒΑπρὸςτὸΑΕ,οὕτωςτὸΔΓπρὸςτὸ [And since it was shown (that) as AB (is) to CD, so
ΓΖ·καίἐστινἀναστρέψαντι].
EB (is) to FD, (it is) also (the case), alternately, (that)
as AB (is) to BE, so CD (is) to FD. Thus, composed
magnitudes are proportional. And it was shown (that)
as BA (is) to AE, so DC (is) to CF . And (the latter) is
converted (from the former).]
ÈÖ×Ñ. Corollary ‡
᾿Εκ δὴ τούτου φανερόν, ὅτι ἐὰν συγκείμενα μεγέθη So (it is) clear, from this, that if composed magni-
ἀνάλογονᾖ,καὶἀναστρέψαντιἀνάλογονἔσται·ὅπερἔδει tudes are proportional then they will also be proportional
δεῖξαι.
(when) converted. (Which is) the very thing it was required
to show.
† In modern notation, this proposition reads that if α : β :: γ : δ then α : β :: α − γ : β − δ.
‡ In modern notation, this corollary reads that if α : β :: γ : δ then α : α − β :: γ : γ − δ.
. Proposition 20 †
᾿Εὰν ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖςἴσα τὸ πλῆθος, If there are three magnitudes, and others of equal
σύνδυολαμβανόμενακαὶἐντῷαὐτῷλόγω,δι᾿ἴσουδὲτὸ number to them, (being) also in the same ratio taken two
πρῶτοντοῦτρίτουμεῖζονᾖ, καὶτὸτέταρτοντοῦἕκτου by two, and (if), via equality, the first is greater than the
μεῖζονἔσται,κἂνἴσον,ἴσον,κἂνἔλαττον,ἔλαττον. third then the fourth will also be greater than the sixth.
And if (the first is) equal (to the third then the fourth
will also be) equal (to the sixth). And if (the first is) less
(than the third then the fourth will also be) less (than the
sixth).
Α ∆
A
D
Β Ε
B
E
Γ Ζ
C
F
῎ΕστωτρίαμεγέθητὰΑ,Β,Γ,καὶἄλλααὐτοῖςἴσατὸ Let A, B, and C be three magnitudes, and D, E, F
πλῆθοςτὰΔ,Ε,Ζ,σύνδυολαμβανόμεναἐντῷαὐτῷλόγῳ, other (magnitudes) of equal number to them, (being) in
ὡςμὲντὸΑπρὸςτὸΒ,οὕτωςτὸΔπρὸςτὸΕ,ὡςδὲτὸΒ the same ratio taken two by two, (so that) as A (is) to B,
πρὸςτὸΓ,οὕτωςτὸΕπρὸςτὸΖ,δι᾿ἴσουδὲμεῖζονἔστω so D (is) to E, and as B (is) to C, so E (is) to F . And let
τὸΑτοῦΓ·λέγω,ὅτικαὶτὸΔτοῦΖμεῖζονἔσται,κἂν A be greater than C, via equality. I say that D will also
ἴσον,ἴσον,κἂνἔλαττον,ἔλαττον.
be greater than F . And if (A is) equal (to C then D will
᾿ΕπεὶγὰρμεῖζόνἐστιτὸΑτοῦΓ,ἄλλοδέτιτὸΒ,τὸδὲ also be) equal (to F ). And if (A is) less (than C then D
μεῖζονπρὸςτὸαὐτὸμείζοναλόγονἔχειἤπερτὸἔλαττον, will also be) less (than F ).
τὸΑἄραπρὸςτὸΒμείζοναλόγονἔχειἤπερτὸΓπρὸςτὸ For since A is greater than C, and B some other (mag-
Β.ἀλλ᾿ὡςμὲντὸΑπρὸςτὸΒ[οὕτως]τὸΔπρὸςτὸΕ,ὡς nitude), and the greater (magnitude) has a greater ratio
δὲτὸΓπρὸςτὸΒ,ἀνάπαλινοὕτωςτὸΖπρὸςτὸΕ·καὶτὸ than the lesser to the same (magnitude) [Prop. 5.8], A
ΔἄραπρὸςτὸΕμείζοναλόγονἔχειἤπερτὸΖπρὸςτὸΕ. thus has a greater ratio to B than C (has) to B. But as A
τῶνδὲπρὸςτὸαὐτὸλόγονἐχόντωντὸμείζοναλόγονἔχον (is) to B, [so] D (is) to E. And, inversely, as C (is) to B,
μεῖζόνἐστιν. μεῖζονἄρατὸΔτοῦΖ.ὁμοίωςδὴδείξομεν, so F (is) to E [Prop. 5.7 corr.]. Thus, D also has a greater
ὅτικἂνἴσονᾖτὸΑτῷΓ,ἴσονἔσταικαὶτὸΔτῷΖ,κἂν ratio to E than F (has) to E [Prop. 5.13]. And for (mag-
149

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
ἔλαττον,ἔλαττον.
nitudes) having a ratio to the same (magnitude), that
᾿Εὰνἄραᾖτρίαμεγέθηκαὶἄλλααὐτοῖςἴσατὸπλῆθος, having the greater ratio is greater [Prop. 5.10]. Thus,
σύνδυολαμβανόμενακαὶἐντῷαὐτῷλόγω,δι᾿ἴσουδὲτὸ D (is) greater than F . Similarly, we can show that even if
πρῶτοντοῦτρίτουμεῖζονᾖ, καὶτὸτέταρτοντοῦἕκτου A is equal to C then D will also be equal to F , and even
μεῖζονἔσται,κἂνἴσον,ἴσον,κἂνἔλαττον,ἔλαττον·ὅπερ if (A is) less (than C then D will also be) less (than F ).
ἔδειδεῖξαι.
Thus, if there are three magnitudes, and others of
equal number to them, (being) also in the same ratio
taken two by two, and (if), via equality, the first is greater
than the third, then the fourth will also be greater than
the sixth. And if (the first is) equal (to the third then the
fourth will also be) equal (to the sixth). And (if the first
is) less (than the third then the fourth will also be) less
(than the sixth). (Which is) the very thing it was required
to show.
† In modern notation, this proposition reads that if α : β :: δ : ǫ and β : γ :: ǫ : ζ then α γ as δ ζ.
. Proposition 21 †
᾿Εὰν ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος If there are three magnitudes, and others of equal
σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, ᾖ δὲ τετα- number to them, (being) also in the same ratio taken two
ραγμένη αὐτῶν ἡ ἀναλογία, δι᾿ ἴσου δὲ τὸ πρῶτον τοῦ by two, and (if) their proportion (is) perturbed, and (if),
τρίτουμεῖζονᾖ,καὶτὸτέταρτοντοῦἕκτουμεῖζονἔσται, via equality, the first is greater than the third then the
κἂνἴσον,ἴσον,κἂνἔλαττον,ἔλαττον.
fourth will also be greater than the sixth. And if (the first
is) equal (to the third then the fourth will also be) equal
(to the sixth). And if (the first is) less (than the third then
the fourth will also be) less (than the sixth).
Α ∆
A
D
Β Ε
B
E
Γ Ζ
C
F
῎ΕστωτρίαμεγέθητὰΑ,Β,Γκαὶἄλλααὐτοῖςἴσατὸ Let A, B, and C be three magnitudes, and D, E, F
πλῆθοςτὰΔ,Ε,Ζ,σύνδυολαμβανόμενακαὶἐντῷαὐτῷ other (magnitudes) of equal number to them, (being) in
λόγῳ,ἔστωδὲτεταραγμένηαὐτῶνἡἀναλογία,ὡςμὲντὸ the same ratio taken two by two. And let their proportion
ΑπρὸςτὸΒ,οὕτωςτὸΕπρὸςτὸΖ,ὡςδὲτὸΒπρὸςτὸΓ, be perturbed, (so that) as A (is) to B, so E (is) to F , and
οὕτωςτὸΔπρὸςτὸΕ,δι᾿ἴσουδὲτὸΑτοῦΓμεῖζονἔστω· as B (is) to C, so D (is) to E. And let A be greater than
λέγω,ὅτικαὶτὸΔτοῦΖμεῖζονἔσται,κἂνἴσον,ἴσον,κἂν C, via equality. I say that D will also be greater than F .
ἒλαττον,ἒλαττον.
And if (A is) equal (to C then D will also be) equal (to
᾿ΕπεὶγὰρμεῖζόνἐστιτὸΑτοῦΓ,ἄλλοδέτιτὸΒ,τὸ F ). And if (A is) less (than C then D will also be) less
ΑἄραπρὸςτὸΒμείζοναλόγονἔχειἤπερτὸΓπρὸςτὸΒ. (than F ).
ἀλλ᾿ὡςμὲντὸΑπρὸςτὸΒ,οὕτωςτὸΕπρὸςτὸΖ,ὡς For since A is greater than C, and B some other (magδὲτὸΓπρὸςτὸΒ,ἀνάπαλινοὕτωςτὸΕπρὸςτὸΔ.καὶ
nitude), A thus has a greater ratio to B than C (has) to
τὸΕἄραπρὸςτὸΖμείζοναλόγονἔχειἤπερτὸΕπρὸςτὸ B [Prop. 5.8]. But as A (is) to B, so E (is) to F . And,
Δ.πρὸςὃδὲτὸαὐτὸμείζοναλόγονἔχει,ἐκεῖνοἔλασσόν inversely, as C (is) to B, so E (is) to D [Prop. 5.7 corr.].
ἐστιν·ἔλασσονἄραἐστὶτὸΖτοῦΔ·μεῖζονἄραἐστὶτὸΔ Thus, E also has a greater ratio to F than E (has) to D
τοῦΖ.ὁμοίωςδὴδείξομεν,ὅτικἂνἴσονᾖτὸΑτῷΓ,ἴσον [Prop. 5.13]. And that (magnitude) to which the same
ἔσταικαὶτὸΔτῷΖ,κἂνἔλαττον,ἔλαττον.
(magnitude) has a greater ratio is (the) lesser (magni-
᾿Εὰνἄραᾖτρίαμεγέθηκαὶἄλλααὐτοῖςἴσατὸπλῆθος, tude) [Prop. 5.10]. Thus, F is less than D. Thus, D is
σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, ᾖ δὲ τετα- greater than F . Similarly, we can show that even if A is
ραγμένηαὐτῶνἡἀναλογία,δι᾿ἴσουδὲτὸπρῶτοντοῦτρίτου equal to C then D will also be equal to F , and even if (A
μεῖζονᾖ,καὶτὸτέταρτοντοῦἕκτουμεῖζονἔσται,κἂνἴσον, is) less (than C then D will also be) less (than F ).
150

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
ἴσον,κἂνἔλαττον,ἔλαττον·ὅπερἔδειδεῖξαι.
Thus, if there are three magnitudes, and others of
equal number to them, (being) also in the same ratio
taken two by two, and (if) their proportion (is) perturbed,
and (if), via equality, the first is greater than the
third then the fourth will also be greater than the sixth.
And if (the first is) equal (to the third then the fourth
will also be) equal (to the sixth). And if (the first is) less
(than the third then the fourth will also be) less (than the
sixth). (Which is) the very thing it was required to show.
† In modern notation, this proposition reads that if α : β :: ǫ : ζ and β : γ :: δ : ǫ then α γ as δ ζ.
. Proposition 22 †
᾿Εὰνᾖὁποσαοῦνμεγέθηκαὶἄλλααὐτοῖςἴσατὸπλῆθος, If there are any number of magnitudes whatsoever,
σύνδυολαμβανόμενακαὶἐντῷαὐτῷλόγῳ,καὶδι᾿ἴσουἐν and (some) other (magnitudes) of equal number to them,
τῷαὐτῷλόγῳἔσται.
(which are) also in the same ratio taken two by two, then
they will also be in the same ratio via equality.
Α Β
Γ
A
B
C
∆ Ε
Ζ
D
E
F
Η Κ Μ
G
K
M
Θ Λ
Ν
H
L
N
῎ΕστωὁποσαοῦνμεγέθητὰΑ,Β,Γκαὶἄλλααὐτοῖςἴσα Let there be any number of magnitudes whatsoever,
τὸπλῆθοςτὰΔ,Ε,Ζ,σύνδυολαμβανόμεναἐντῷαὐτῷ A, B, C, and (some) other (magnitudes), D, E, F , of
λόγῳ,ὡςμὲντὸΑπρὸςτὸΒ,οὕτωςτὸΔπρὸςτὸΕ,ὡς equal number to them, (which are) in the same ratio
δὲτὸΒπρὸςτὸΓ,οὕτωςτὸΕπρὸςτὸΖ·λέγω,ὅτικαὶ taken two by two, (so that) as A (is) to B, so D (is) to
δι᾿ἴσουἐντῷαὐτῳλόγῳἔσται.
E, and as B (is) to C, so E (is) to F . I say that they will
ΕἰλήφθωγὰρτῶνμὲνΑ,ΔἰσάκιςπολλαπλάσιατὰΗ, also be in the same ratio via equality. (That is, as A is to
Θ,τῶνδὲΒ,Εἄλλα,ἃἔτυχεν,ἰσάκιςπολλαπλάσιατὰΚ, C, so D is to F .)
Λ,καὶἔτιτῶνΓ,Ζἄλλα,ἃἔτυχεν,ἰσάκιςπολλαπλάσιατὰ For let the equal multiples G and H have been taken
Μ,Ν.
of A and D (respectively), and the other random equal
ΚαὶἐπείἐστινὡςτὸΑπρὸςτὸΒ,οὕτωςτὸΔπρὸςτὸ multiples K and L of B and E (respectively), and the
Ε,καὶεἴληπταιτῶνμὲνΑ,ΔἰσάκιςπολλαπλάσιατὰΗ,Θ, yet other random equal multiples M and N of C and F
τῶνδὲΒ,Εἄλλα,ἃἔτυχεν,ἰσάκιςπολλαπλάσιατὰΚ,Λ, (respectively).
ἔστινἄραὡςτὸΗπρὸςτὸΚ,οὕτωςτὸΘπρὸςτὸΛ.δὶα And since as A is to B, so D (is) to E, and the equal
τὰαὐτὰδὴκαὶὡςτὸΚπρὸςτὸΜ,οὕτωςτὸΛπρὸςτὸ multiples G and H have been taken of A and D (respec-
Ν.ἐπεὶοὖντρίαμεγέθηἐστὶτὰΗ,Κ,Μ,καὶἄλλααὐτοῖς tively), and the other random equal multiples K and L
ἴσατὸπλῆθοςτὰΘ,Λ,Ν,σύνδυολαμβανόμενακαὶἐντῷ of B and E (respectively), thus as G is to K, so H (is) to
αὐτῷλόγῳ,δι᾿ἴσουἄρα,εἰὑπερέχειτὸΗτοῦΜ,ὑπερέχει L [Prop. 5.4]. And, so, for the same (reasons), as K (is)
καὶτὸΘτοῦΝ,καὶεἰἴσον,ἴσον,καὶεἰἔλαττον,ἔλαττον. to M, so L (is) to N. Therefore, since G, K, and M are
καίἐστιτὰμὲνΗ,ΘτῶνΑ,Δἰσάκιςπολλαπλάσια,τὰδὲ three magnitudes, and H, L, and N other (magnitudes)
Μ,ΝτῶνΓ,Ζἄλλα,ἃἔτυχεν,ἰσάκιςπολλαπλάσια.ἔστιν of equal number to them, (which are) also in the same
ἄραὡςτὸΑπρὸςτὸΓ,οὕτωςτὸΔπρὸςτὸΖ. ratio taken two by two, thus, via equality, if G exceeds M
᾿Εὰνἄραᾖὁποσαοῦνμεγέθηκαὶἄλλααὐτοῖςἴσατὸ then H also exceeds N, and if (G is) equal (to M then H
πλῆθος,σύνδυολαμβανόμεναἐντῷαὐτῷλόγῳ,καὶδι᾿ἴσου is also) equal (to N), and if (G is) less (than M then H is
ἐντῷαὐτῷλόγῳἔσται·ὅπερἔδειδεῖξαι.
also) less (than N) [Prop. 5.20]. And G and H are equal
multiples of A and D (respectively), and M and N other
random equal multiples of C and F (respectively). Thus,
as A is to C, so D (is) to F [Def. 5.5].
Thus, if there are any number of magnitudes whatsoever,
and (some) other (magnitudes) of equal number
to them, (which are) also in the same ratio taken two by
151

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
two, then they will also be in the same ratio via equality.
(Which is) the very thing it was required to show.
† In modern notation, this proposition reads that if α : β :: ǫ : ζ and β : γ :: ζ : η and γ : δ :: η : θ then α : δ :: ǫ : θ.
. Proposition 23 †
᾿Εὰν ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος If there are three magnitudes, and others of equal
σύνδυολαμβανόμεναἐντῷαὐτῷλόγῳ,ᾖδὲτεταραγμένη number to them, (being) in the same ratio taken two by
αὐτῶνἡἀναλογία,καὶδι᾿ἴσουἐντῷαὐτῷλόγῳἔσται. two, and (if) their proportion is perturbed, then they will
also be in the same ratio via equality.
Α Β Γ
A
B
∆ Ε Ζ
D
E
Η Θ Λ
G
H
Κ Μ Ν
K
M
῎ΕστωτρίαμεγέθητὰΑ,Β,Γκαὶἄλλααὐτοῖςἴσατὸ Let A, B, and C be three magnitudes, and D, E and F
πλῆθοςσύνδυολαμβανόμεναἐντῷαὐτῷλόγῳτὰΔ,Ε,Ζ, other (magnitudes) of equal number to them, (being) in
ἔστωδὲτεταραγμένηαὐτῶνἡἀναλογία,ὡςμὲντὸΑπρὸς the same ratio taken two by two. And let their proportion
τὸΒ,οὕτωςτὸΕπρὸςτὸΖ,ὡςδὲτὸΒπρὸςτὸΓ,οὕτως be perturbed, (so that) as A (is) to B, so E (is) to F , and
τὸΔπρὸςτὸΕ·λέγω,ὅτιἐστὶνὡςτὸΑπρὸςτὸΓ,οὕτως as B (is) to C, so D (is) to E. I say that as A is to C, so
τὸΔπρὸςτὸΖ. D (is) to F .
ΕἰλήφθωτῶνμὲνΑ,Β,ΔἰσάκιςπολλαπλάσιατὰΗ,Θ, Let the equal multiples G, H, and K have been taken
Κ,τῶνδὲΓ,Ε,Ζἄλλα,ἃἔτυχεν,ἰσάκιςπολλαπλάσιατὰ of A, B, and D (respectively), and the other random
Λ,Μ,Ν.
equal multiples L, M, and N of C, E, and F (respec-
ΚαὶἐπεὶἰσάκιςἐστὶπολλαπλάσιατὰΗ,ΘτῶνΑ,Β,τὰ tively).
δὲμέρητοὶςὡσαύτωςπολλαπλασίοιςτὸναὐτὸνἔχειλόγον, And since G and H are equal multiples of A and B
ἔστινἄραὡςτὸΑπρὸςτὸΒ,οὕτωςτὸΗπρὸςτὸΘ.διὰ (respectively), and parts have the same ratio as similar
τὰαὐτὰδὴκαὶὡςτὸΕπρὸςτὸΖ,οὕτωςτὸΜπρὸςτὸΝ· multiples [Prop. 5.15], thus as A (is) to B, so G (is) to
καίἐστινὡςτὸΑπρὸςτὸΒ,οὕτωςτὸΕπρὸςτὸΖ·καὶὡς H. And, so, for the same (reasons), as E (is) to F , so M
ἄρατὸΗπρὸςτὸΘ,οὕτωςτὸΜπρὸςτὸΝ.καὶἐπείἐστιν (is) to N. And as A is to B, so E (is) to F . And, thus, as
ὡςτὸΒπρὸςτὸΓ,οὕτωςτὸΔπρὸςτὸΕ,καὶἐναλλὰξ G (is) to H, so M (is) to N [Prop. 5.11]. And since as B
ὡςτὸΒπρὸςτὸΔ,οὕτωςτὸΓπρὸςτὸΕ.καὶἐπεὶτὰ is to C, so D (is) to E, also, alternately, as B (is) to D, so
Θ,ΚτῶνΒ,Δἰσάκιςἐστὶπολλαπλάσια,τὰδὲμέρητοῖς C (is) to E [Prop. 5.16]. And since H and K are equal
ἰσάκιςπολλαπλασίοιςτὸναὐτὸνἔχειλόγον,ἔστινἄραὡς multiples of B and D (respectively), and parts have the
τὸΒπρὸςτὸΔ,οὕτωςτὸΘπρὸςτὸΚ.ἀλλ᾿ὡςτὸΒπρὸς same ratio as similar multiples [Prop. 5.15], thus as B is
τὸΔ,οὕτωςτὸΓπρὸςτὸΕ·καὶὡςἄρατὸΘπρὸςτὸΚ, to D, so H (is) to K. But, as B (is) to D, so C (is) to
οὕτωςτὸΓπρὸςτὸΕ.πάλιν,ἐπεὶτὰΛ,ΜτῶνΓ,Εἰσάκις E. And, thus, as H (is) to K, so C (is) to E [Prop. 5.11].
ἐστιπολλαπλάσια,ἔστινἄραὡςτὸΓπρὸςτὸΕ,οὕτωςτὸ Again, since L and M are equal multiples of C and E (re-
ΛπρὸςτὸΜ.ἀλλ᾿ὡςτὸΓπρὸςτὸΕ,οὕτωςτὸΘπρὸς spectively), thus as C is to E, so L (is) to M [Prop. 5.15].
τὸΚ·καὶὡςἄρατὸΘπρὸςτὸΚ,οὕτωςτὸΛπρὸςτὸΜ, But, as C (is) to E, so H (is) to K. And, thus, as H (is)
καὶἐναλλὰξὡςτὸΘπρὸςτὸΛ,τὸΚπρὸςτὸΜ.ἐδείχθη to K, so L (is) to M [Prop. 5.11]. Also, alternately, as H
δὲκαὶὡςτὸΗπρὸςτὸΘ,οὕτωςτὸΜπρὸςτὸΝ.ἐπεὶ (is) to L, so K (is) to M [Prop. 5.16]. And it was also
οὖντρίαμεγέθηἐστὶτὰΗ,Θ,Λ,καὶἄλλααὐτοιςἴσατὸ shown (that) as G (is) to H, so M (is) to N. Therefore,
πλῆθοςτὰΚ,Μ,Νσύνδυολαμβανόμεναἐντῷαὐτῷλόγῳ, since G, H, and L are three magnitudes, and K, M, and
καίἐστιναὐτῶντεταραγμένηἡἀναλογία,δι᾿ἴσουἄρα,εἰ N other (magnitudes) of equal number to them, (being)
ὑπερέχειτὸΗτοῦΛ,ὑπερέχεικαὶτὸΚτοῦΝ,καὶεἰἴσον, in the same ratio taken two by two, and their proportion
ἴσον,καὶεἰἔλαττον,ἔλαττον.καίἐστιτὰμὲνΗ,ΚτῶνΑ, is perturbed, thus, via equality, if G exceeds L then K
Δἰσάκιςπολλαπλάσια,τὰδὲΛ,ΝτῶνΓ,Ζ.ἔστινἄραὡς also exceeds N, and if (G is) equal (to L then K is also)
τὸΑπρὸςτὸΓ,οὕτωςτὸΔπρὸςτὸΖ.
equal (to N), and if (G is) less (than L then K is also)
᾿Εὰνἄραᾖτρίαμεγέθηκαὶἄλλααὐτοῖςἴσατὸπλῆθος less (than N) [Prop. 5.21]. And G and K are equal mulσύνδυολαμβανόμεναἐντῷαὐτῷλόγῳ,ᾖδὲτεταραγμένη
tiples of A and D (respectively), and L and N of C and
C
F
L
N
152

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
αὐτῶνἡἀναλογία,καὶδι᾿ἴσουἐντῷαὐτῷλόγῳἔσται·ὅπερ F (respectively). Thus, as A (is) to C, so D (is) to F
ἔδειδεῖξαι. [Def. 5.5].
Thus, if there are three magnitudes, and others of
equal number to them, (being) in the same ratio taken
two by two, and (if) their proportion is perturbed, then
they will also be in the same ratio via equality. (Which is)
the very thing it was required to show.
† In modern notation, this proposition reads that if α : β :: ǫ : ζ and β : γ :: δ : ǫ then α : γ :: δ : ζ.
. Proposition 24 †
᾿Εὰνπρῶτονπρὸςδεύτεροντὸναὐτὸνἔχῃλόγονκαὶ If a first (magnitude) has to a second the same ratio
τρίτονπρὸςτέταρτον,ἔχῃδὲκαὶπέμπτονπρὸςδεύτερον that third (has) to a fourth, and a fifth (magnitude) also
τὸναὐτὸνλόγονκαὶἕκτονπρὸςτέταρτον,καὶσυντεθὲν has to the second the same ratio that a sixth (has) to the
πρῶτονκαὶπέμπτονπρὸςδεύτεροντὸναὐτὸνἕξειλόγον fourth, then the first (magnitude) and the fifth, added
καὶτρίτονκαὶἕκτονπρὸςτέταρτον.
together, will also have the same ratio to the second that
the third (magnitude) and sixth (added together, have)
to the fourth.
Α
Γ
∆
Ζ
Ε
Β
Θ
Η
ΠρῶτονγὰρτὸΑΒπρὸςδεύρεροντὸΓτὸναὐτὸνἐχέτω For let a first (magnitude) AB have the same ratio to
λόγονκαὶτρίτοντὸΔΕπρὸςτέταρτοντὸΖ,ἐχέτωδὲκαὶ a second C that a third DE (has) to a fourth F . And let
πέμπτοντὸΒΗπρὸςδεύτεροντὸΓτὸναὐτὸνλόγονκαὶ a fifth (magnitude) BG also have the same ratio to the
ἕκτοντὸΕΘπρὸςτέταρτοντὸΖ·λέγω,ὅτικαὶσυντεθὲν second C that a sixth EH (has) to the fourth F . I say
πρῶτονκαὶπέμπτοντὸΑΗπρὸςδεύτεροντὸΓτὸναὐτὸν that the first (magnitude) and the fifth, added together,
ἕξειλόγον,καὶτρίτονκαὶἕκτοντὸΔΘπρὸςτέταρτοντὸ AG, will also have the same ratio to the second C that the
Ζ. third (magnitude) and the sixth, (added together), DH,
᾿ΕπεὶγάρἐστινὡςτὸΒΗπρὸςτὸΓ,οὕτωςτὸΕΘπρὸς (has) to the fourth F .
τὸΖ,ἀνάπαλινἄραὡςτὸΓπρὸςτὸΒΗ,οὕτωςτὸΖπρὸς For since as BG is to C, so EH (is) to F , thus, inτὸΕΘ.ἐπεὶοὖνἐστινὡςτὸΑΒπρὸςτὸΓ,οὕτωςτὸΔΕ
versely, as C (is) to BG, so F (is) to EH [Prop. 5.7 corr.].
πρὸςτὸΖ,ὡςδὲτὸΓπρὸςτὸΒΗ,οὕτωςτὸΖπρὸςτὸ Therefore, since as AB is to C, so DE (is) to F , and as C
ΕΘ,δι᾿ἴσουἄραἐστὶνὡςτὸΑΒπρὸςτὸΒΗ,οὕτωςτὸΔΕ (is) to BG, so F (is) to EH, thus, via equality, as AB is to
πρὸςτὸΕΘ.καὶἐπεὶδιῃρημέναμεγέθηἀνάλογόνἐστιν,καὶ BG, so DE (is) to EH [Prop. 5.22]. And since separated
συντεθένταἀνάλογονἔσται·ἔστινἄραὡςτὸΑΗπρὸςτὸ magnitudes are proportional then they will also be pro-
ΗΒ,οὕτωςτὸΔΘπρὸςτὸΘΕ.ἔστιδὲκαὶὡςτὸΒΗπρὸς portional (when) composed [Prop. 5.18]. Thus, as AG is
τὸΓ,οὕτωςτὸΕΘπρὸςτὸΖ·δι᾿ἴσουἄραἐστὶνὡςτὸΑΗ to GB, so DH (is) to HE. And, also, as BG is to C, so
πρὸςτὸΓ,οὕτωςτὸΔΘπρὸςτὸΖ.
EH (is) to F . Thus, via equality, as AG is to C, so DH
᾿Εὰνἄραπρῶτονπρὸςδεύτεροντὸναὐτὸνἔχῃλόγον (is) to F [Prop. 5.22].
καὶτρίτονπρὸςτέταρτον,ἔχῃδὲκαὶπέμπτονπρὸςδεύτερον Thus, if a first (magnitude) has to a second the same
τὸναὐτὸνλόγονκαὶἕκτονπρὸςτέταρτον,καὶσυντεθὲν ratio that a third (has) to a fourth, and a fifth (magniπρῶτονκαὶπέμπτονπρὸςδεύτεροντὸναὐτὸνἕξειλόγον
tude) also has to the second the same ratio that a sixth
καὶτρίτονκαὶἕκτονπρὸςτέταρτον·ὅπερἔδειδεὶξαι. (has) to the fourth, then the first (magnitude) and the
fifth, added together, will also have the same ratio to the
second that the third (magnitude) and the sixth (added
A
C
D
F
E
B
H
G
153

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 5
† In modern notation, this proposition reads that if α : β :: γ : δ and ǫ : β :: ζ : δ then α + ǫ : β :: γ + ζ : δ.
together, have) to the fourth. (Which is) the very thing it
was required to show.
. Proposition 25 †
᾿Εὰντέσσαραμεγέθηἀνάλογονᾖ,τὸμέγιστον[αὐτῶν] If four magnitudes are proportional then the (sum of
καὶτὸἐλάχιστονδύοτῶνλοιπῶνμείζονάἐστιν. the) largest and the smallest [of them] is greater than the
(sum of the) remaining two (magnitudes).
Α
Ε
Γ
Ζ
Θ
Η
∆
Β
῎ΕστωτέσσαραμεγέθηἀνάλογοντὰΑΒ,ΓΔ,Ε,Ζ,ὡς Let AB, CD, E, and F be four proportional magniτὸΑΒπρὸςτὸΓΔ,οὕτωςτὸΕπρὸςτὸΖ,ἔστωδὲμέγιστον
tudes, (such that) as AB (is) to CD, so E (is) to F . And
μὲναὐτῶντὸΑΒ,ἐλάχιστονδὲτὸΖ·λέγω,ὅτιτὰΑΒ,Ζ let AB be the greatest of them, and F the least. I say that
τῶνΓΔ,Εμείζονάἐστιν. AB and F is greater than CD and E.
ΚείσθωγὰρτῷμὲνΕἴσοντὸΑΗ,τῷδὲΖἴσοντὸΓΘ. For let AG be made equal to E, and CH equal to F .
᾿Επεὶ[οὖν]ἐστινὡςτὸΑΒπρὸςτὸΓΔ,οὕτωςτὸΕ [In fact,] since as AB is to CD, so E (is) to F , and
πρὸςτὸΖ,ἴσονδὲτὸμὲνΕτῷΑΗ,τὸδὲΖτῷΓΘ,ἔστιν E (is) equal to AG, and F to CH, thus as AB is to CD,
ἄραὡςτὸΑΒπρὸςτὸΓΔ,οὕτωςτὸΑΗπρὸςτὸΓΘ. so AG (is) to CH. And since the whole AB is to the
καὶἐπείἐστινὡςὅλοντὸΑΒπρὸςὅλοντὸΓΔ,οὕτως whole CD as the (part) taken away AG (is) to the (part)
ἀφαιρεθὲντὸΑΗπρὸςἀφαιρεθὲντὸΓΘ,καὶλοιπὸνἄρα taken away CH, thus the remainder GB will also be to
τὸΗΒπρὸςλοιπὸντὸΘΔἔσταιὡςὅλοντὸΑΒπρὸςὅλον the remainder HD as the whole AB (is) to the whole CD
τὸΓΔ.μεῖζονδὲτὸΑΒτοῦΓΔ·μεῖζονἄρακαὶτὸΗΒτοῦ [Prop. 5.19]. And AB (is) greater than CD. Thus, GB
ΘΔ.καὶἐπεὶἴσονἐστὶτὸμὲνΑΗτῷΕ,τὸδὲΓΘτῷΖ, (is) also greater than HD. And since AG is equal to E,
τὰἄραΑΗ,ΖἴσαἐστὶτοῖςΓΘ,Ε.καὶ[ἐπεὶ]ἐὰν[ἀνίσοις and CH to F , thus AG and F is equal to CH and E.
ἴσαπροστεθῇ,τὰὅλαἄνισάἐστιν,ἐὰνἄρα]τῶνΗΒ,ΘΔ And [since] if [equal (magnitudes) are added to unequal
ἀνίσωνὄντωνκαὶμείζονοςτοῦΗΒτῷμὲνΗΒπροστεθῇ (magnitudes) then the wholes are unequal, thus if] AG
τὰΑΗ,Ζ,τῷδὲΘΔπροστεθῇτὰΓΘ,Ε,συνάγεταιτὰ and F are added to GB, and CH and E to HD—GB
ΑΒ,ΖμείζονατῶνΓΔ,Ε.
and HD being unequal, and GB greater—it is inferred
᾿Εὰν ἄρα τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ μέγιστον that AB and F (is) greater than CD and E.
αὐτῶνκαὶτὸἐλάχιστονδύοτῶνλοιπῶνμείζονάἐστιν.ὅπερ Thus, if four magnitudes are proportional then the
ἔδειδεῖξαι.
(sum of the) largest and the smallest of them is greater
than the (sum of the) remaining two (magnitudes).
(Which is) the very thing it was required to show.
† In modern notation, this proposition reads that if α : β :: γ : δ, and α is the greatest and δ the least, then α + δ > β + γ.
A
E
C
F
H
G
D
B
154

ELEMENTS BOOK 6
Similar Figures
155

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
ÎÇÖÓ.
῞Ομοιασχήματα εὐθύγραμμά ἐστιν, ὅσα τάς τε 1. Similar rectilinear figures are those (which) have
Definitions
αʹ.
γωνίαςἴσαςἔχεικατὰμίανκαὶτὰςπερὶτὰςἴσαςγωνίας (their) angles separately equal and the (corresponding)
πλευρὰςἀνάλογον.
sides about the equal angles proportional.
βʹ.῎Ακρονκαὶμέσονλόγονεὐθεῖατετμῆσθαιλέγεται, 2. A straight-line is said to have been cut in extreme
ὅτανᾖὡςἡὅληπρὸςτὸμεῖζοντμῆμα,οὕτωςτὸμεῖζον and mean ratio when as the whole is to the greater segπρὸςτὸἔλαττὸν.
ment so the greater (segment is) to the lesser.
γʹ.῞Υψοςἐστὶπάντοςσχήματοςἡἀπὸτῆςκορυφῆςἐπὶ 3. The height of any figure is the (straight-line) drawn
τὴνβάσινκάθετοςἀγομένη.
from the vertex perpendicular to the base.
. Proposition 1 †
Τὰτρίγωνακαὶτὰπαραλληλόγραμματὰὑπὸτὸαὐτὸ Triangles and parallelograms which are of the same
ὕψοςὄνταπρὸςἄλληλάἐστινὡςαἱβάσεις.
height are to one another as their bases.
Ε
Α
Ζ
E
A
F
Θ
Η Β Γ
∆
Κ
Λ
H G B C D K L
῎ΕστωτρίγωναμὲντὰΑΒΓ,ΑΓΔ,παραλληλόγραμμα Let ABC and ACD be triangles, and EC and CF parδὲτὰΕΓ,ΓΖὑπὸτὸαὐτὸὕψοςτὸΑΓ·λέγω,ὅτιἐστὶνὡς
allelograms, of the same height AC. I say that as base BC
ἡΒΓβάσιςπρὸςτὴνΓΔβάσις,οὕτωςτὸΑΒΓτρίγωνον is to base CD, so triangle ABC (is) to triangle ACD, and
πρὸςτὸΑΓΔτρίγωνον,καὶτὸΕΓπαραλληλόγραμμονπρὸς parallelogram EC to parallelogram CF .
τὸΓΖπαραλληλόγραμμον.
For let the (straight-line) BD have been produced in
᾿ΕκβεβλήσθωγὰρἡΒΔἐφ᾿ἑκάτερατὰμέρηἐπὶτὰΘ,Λ each direction to points H and L, and let [any number]
σημεῖα,καὶκείσθωσαντῇμὲνΒΓβάσειἴσαι[ὁσαιδηποτοῦν] (of straight-lines) BG and GH be made equal to base
αἱΒΗ,ΗΘ,τῇδὲΓΔβάσειἴσαιὁσαιδηποτοῦναἱΔΚ,ΚΛ, BC, and any number (of straight-lines) DK and KL
καὶἐπεζεύχθωσαναἱΑΗ,ΑΘ,ΑΚ,ΑΛ.
equal to base CD. And let AG, AH, AK, and AL have
ΚαὶἐπεὶἴσαιεἰσὶναἱΓΒ,ΒΗ,ΗΘἀλλήλαις,ἴσαἐστὶκαὶ been joined.
τὰΑΘΗ,ΑΗΒ,ΑΒΓτρίγωναἀλλήλοις. ὁσαπλασίωνἄρα And since CB, BG, and GH are equal to one another,
ἐστὶνἡΘΓβάσιςτῆςΒΓβάσεως, τοσαυταπλάσιόνἐστι triangles AHG, AGB, and ABC are also equal to one
καὶτὸΑΘΓτρίγωνοντοῦΑΒΓτριγώνου. διὰτὰαὐτὰ another [Prop. 1.38]. Thus, as many times as base HC
δὴὁσαπλασίωνἐστὶνἡΛΓβάσιςτῆςΓΔβάσεως,τοσαυ- is (divisible by) base BC, so many times is triangle AHC
ταπλάσιόνἐστικαὶτὸΑΛΓτρίγωνοντοῦΑΓΔτριγώνου· also (divisible) by triangle ABC. So, for the same (reaκαὶεἰἴσηἐστὶνἡΘΓβάσιςτῇΓΛβάσει,ἴσονἐστὶκαὶτὸ
sons), as many times as base LC is (divisible) by base
ΑΘΓτρίγωνοντῳΑΓΛτριγώνῳ,καὶεἰὑπερέχειἡΘΓβάσις CD, so many times is triangle ALC also (divisible) by
τῆςΓΛβάσεως,ὑπερέχεικαὶτὸΑΘΓτρίγωνοντοῦΑΓΛ triangle ACD. And if base HC is equal to base CL then
τριγώνου,καὶεἰἐλάσσων,ἔλασσον. τεσσάρωνδὴὄντων triangle AHC is also equal to triangle ACL [Prop. 1.38].
μεγεθῶνδύομὲνβάσεωντῶνΒΓ,ΓΔ,δύοδὲτριγώνων And if base HC exceeds base CL then triangle AHC
τῶνΑΒΓ,ΑΓΔεἴληπταιἰσάκιςπολλαπλάσιατῆςμὲνΒΓ also exceeds triangle ACL. ‡ And if (HC is) less (than
βάσεωςκαὶτοῦΑΒΓτριγώνουἥτεΘΓβάσιςκαὶτὸΑΘΓ CL then AHC is also) less (than ACL). So, their being
τρίγωνον,τῆςδὲΓΔβάσεωςκαὶτοῦΑΔΓτριγώνουἄλλα, four magnitudes, two bases, BC and CD, and two trian-
156

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
ἃἔτυχεν,ἰσάκιςπολλαπλάσιαἥτεΛΓβάσιςκαὶτὸΑΛΓ gles, ABC and ACD, equal multiples have been taken of
τρίγωνον·καὶδέδεικται,ὅτι,εἰὑπερέχειἡΘΓβάσιςτῆςΓΛ base BC and triangle ABC—(namely), base HC and triβάσεως,ὑπερέχεικαὶτὸΑΘΓτρίγωνοντοῦΑΛΓτριγώνου,
angle AHC—and other random equal multiples of base
καίεἰἴση,ἴσον,καὶεἰἔλασσων,ἔλασσον·ἔστινἄραὡςἡ CD and triangle ADC—(namely), base LC and triangle
ΒΓβάσιςπρὸςτὴνΓΔβάσιν,οὕτωςτὸΑΒΓτρίγωνονπρὸς ALC. And it has been shown that if base HC exceeds
τὸΑΓΔτρίγωνον.
base CL then triangle AHC also exceeds triangle ALC,
ΚαὶἐπεὶτοῦμὲνΑΒΓτριγώνουδιπλάσιόνἐστιτὸΕΓ and if (HC is) equal (to CL then AHC is also) equal (to
παραλληλόγραμμον,τοῦδὲΑΓΔτριγώνουδιπλάσιόνἐστι ALC), and if (HC is) less (than CL then AHC is also)
τὸΖΓπαραλληλόγραμμον,τὰδὲμέρητοῖςὡσαύτωςπολ- less (than ALC). Thus, as base BC is to base CD, so
λαπλασίοιςτὸναὐτὸνἔχειλόγον,ἔστινἄραὡςτὸΑΒΓ triangle ABC (is) to triangle ACD [Def. 5.5]. And since
τρίγωνον πρὸς τὸ ΑΓΔ τρίγωνον, οὕτως τὸ ΕΓ παραλ- parallelogram EC is double triangle ABC, and paralleloληλόγραμμον
πρὸς τὸ ΖΓ παραλληλόγραμμον. ἐπεὶ οὖν gram FC is double triangle ACD [Prop. 1.34], and parts
ἐδείχθη,ὡςμὲνἡΒΓβάσιςπρὸςτὴνΓΔ,οὕτωςτὸΑΒΓ have the same ratio as similar multiples [Prop. 5.15], thus
τρίγωνονπρὸςτὸΑΓΔτρίγωνον,ὡςδὲτὸΑΒΓτρίγωνον as triangle ABC is to triangle ACD, so parallelogram EC
πρὸςτὸΑΓΔτρίγωνον,οὕτωςτὸΕΓπαραλληλόγραμμον (is) to parallelogram FC. In fact, since it was shown that
πρὸςτὸΓΖπαραλληλόγραμμον,καὶὡςἄραἡΒΓβάσιςπρὸς as base BC (is) to CD, so triangle ABC (is) to triangle
τὴνΓΔβάσιν,οὕτωςτὸΕΓπαραλληλόγραμμονπρὸςτὸΖΓ ACD, and as triangle ABC (is) to triangle ACD, so parπαραλληλόγραμμον.
allelogram EC (is) to parallelogram CF , thus, also, as
Τὰἄρατρίγωνακαὶτὰπαραλληλόγραμματὰὑπὸτὸ base BC (is) to base CD, so parallelogram EC (is) to
αὐτὸὕψοςὄνταπρὸςἄλληλάἐστινὡςαἱβάσεις·ὅπερἔδει parallelogram FC [Prop. 5.11].
δεῖξαι.
Thus, triangles and parallelograms which are of the
same height are to one another as their bases. (Which is)
the very thing it was required to show.
† As is easily demonstrated, this proposition holds even when the triangles, or parallelograms, do not share a common side, and/or are not
right-angled.
‡ This is a straight-forward generalization of Prop. 1.38.
. Proposition 2
᾿Εὰντριγώνουπαρὰμίαντῶνπλευρῶνἀχθῇτιςεὐθεῖα, If some straight-line is drawn parallel to one of the
ἀνάλογοντεμεῖτὰςτοῦτριγώνουπλευράς·καὶἐὰναἱτοῦ sides of a triangle then it will cut the (other) sides of the
τριγώνουπλευραὶἀνάλογοντμηθῶσιν,ἡἐπὶτὰςτομὰςἐπι- triangle proportionally. And if (two of) the sides of a triζευγνυμένη
εὐθεῖαπαρὰ τὴν λοιπὴνἔσταιτοῦτριγώνου angle are cut proportionally then the straight-line joining
πλευράν.
the cutting (points) will be parallel to the remaining side
of the triangle.
Α
A
∆
Ε
D
E
Β
Γ
ΤριγώνουγὰρτοῦΑΒΓπαράλληλοςμιᾷτῶνπλευρῶν For let DE have been drawn parallel to one of the
τῇΒΓἤχθωἡΔΕ·λέγω,ὅτιἐστὶνὡςἡΒΔπρὸςτὴνΔΑ, sides BC of triangle ABC. I say that as BD is to DA, so
οὕτωςἡΓΕπρὸςτὴνΕΑ.
CE (is) to EA.
B
C
157

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
᾿ΕπεζεύχθωσανγὰραἱΒΕ,ΓΔ.
For let BE and CD have been joined.
῎ΙσονἄραἐστὶτὸΒΔΕ τρίγωνοντῷ ΓΔΕτριγώνῳ· Thus, triangle BDE is equal to triangle CDE. For
ἐπὶγὰρτῆςαὐτῆςβάσεώςἐστιτῆςΔΕκαὶἐνταῖςαὐταῖς they are on the same base DE and between the same
παραλλήλοιςταῖςΔΕ,ΒΓ·ἄλλοδέτιτὸΑΔΕτρίγωνον.τὰ parallels DE and BC [Prop. 1.38]. And ADE is some
δὲἴσαπρὸςτὸαὐτὸτὸναὐτὸνἔχειλόγον·ἔστινἄραὡςτὸ other triangle. And equal (magnitudes) have the same ra-
ΒΔΕτρίγωνονπρὸςτὸΑΔΕ[τρίγωνον],οὕτωςτὸΓΔΕ tio to the same (magnitude) [Prop. 5.7]. Thus, as triangle
τρίγωνονπρὸςτὸΑΔΕτρίγωνον. αλλ᾿ὡςμὲντὸΒΔΕ BDE is to [triangle] ADE, so triangle CDE (is) to trianτρίγωνονπρὸςτὸΑΔΕ,οὕτωςἡΒΔπρὸςτὴνΔΑ·ὑπὸ
gle ADE. But, as triangle BDE (is) to triangle ADE, so
γὰρτὸαὐτὸὕψοςὄντατὴνἀπὸτοῦΕἐπὶτὴνΑΒκάθετον (is) BD to DA. For, having the same height—(namely),
ἀγομένηνπρὸςἄλληλάεἰσινὡςαἱβάσεις. διὰτὰαὐτὰδὴ the (straight-line) drawn from E perpendicular to AB—
ὡςτὸΓΔΕτρίγωνονπρὸςτὸΑΔΕ,οὕτωςἡΓΕπρὸςτὴν they are to one another as their bases [Prop. 6.1]. So, for
ΕΑ·καὶὡςἄραἡΒΔπρὸςτὴνΔΑ,οὕτωςἡΓΕπρὸςτὴν the same (reasons), as triangle CDE (is) to ADE, so CE
ΕΑ.
(is) to EA. And, thus, as BD (is) to DA, so CE (is) to
Ἀλλὰ δὴ αἱ τοῦ ΑΒΓ τριγώνου πλευραὶ αἱ ΑΒ, ΑΓ EA [Prop. 5.11].
ἀνάλογοντετμήσθωσαν, ὡςἡΒΔπρὸςτὴνΔΑ,οὕτως And so, let the sides AB and AC of triangle ABC
ἡ ΓΕ πρὸς τὴν ΕΑ, καὶ ἐπεζεύχθω ἡ ΔΕ· λέγω, ὅτι have been cut proportionally (such that) as BD (is) to
παράλληλόςἐστινἡΔΕτῇΒΓ.
DA, so CE (is) to EA. And let DE have been joined. I
Τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπείἐστιν ὡς ἡ say that DE is parallel to BC.
ΒΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΑ, ἀλλ᾿ ὡς For, by the same construction, since as BD is to DA,
μὲνἡΒΔπρὸςτὴνΔΑ,οὕτωςτὸΒΔΕτρίγωνονπρὸς so CE (is) to EA, but as BD (is) to DA, so triangle BDE
τὸΑΔΕτρίγωνον,ὡςδὲἡΓΕπρὸςτὴνΕΑ,οὕτωςτὸ (is) to triangle ADE, and as CE (is) to EA, so triangle
ΓΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον, καὶ ὡς ἄρα τὸ CDE (is) to triangle ADE [Prop. 6.1], thus, also, as tri-
ΒΔΕτρίγωνονπρὸςτὸΑΔΕτρίγωνον, οὕτωςτὸΓΔΕ angle BDE (is) to triangle ADE, so triangle CDE (is)
τρίγωνονπρὸςτὸΑΔΕτρίγωνον.ἑκάτερονἄρατῶνΒΔΕ, to triangle ADE [Prop. 5.11]. Thus, triangles BDE and
ΓΔΕτριγώνωνπρὸςτὸΑΔΕτὸναὐτὸνἔχειλόγον. ἴσον CDE each have the same ratio to ADE. Thus, triangle
ἄραἐστὶτὸΒΔΕτρίγωνοντῷΓΔΕτριγώνῳ·καίεἰσινἐπὶ BDE is equal to triangle CDE [Prop. 5.9]. And they are
τῆςαὐτῆςβάσεωςτῆςΔΕ.τὰδὲἴσατρίγωνακαὶἐπὶτῆς on the same base DE. And equal triangles, which are
αὐτῆςβάσεωςὄντακαὶἐνταῖςαὐταῖςπαραλλήλοιςἐστίν. also on the same base, are also between the same paralπαράλληλοςἄραἐστὶνἡΔΕτῇΒΓ.
lels [Prop. 1.39]. Thus, DE is parallel to BC.
᾿Εὰν ἄρατριγώνουπαρὰμίαντῶνπλευρῶνἀχθῇτις Thus, if some straight-line is drawn parallel to one of
εὐθεῖα,ἀνάλογοντεμεῖτὰςτοῦτριγώνουπλευράς·καὶἐὰν the sides of a triangle, then it will cut the (other) sides
αἱ τοῦ τριγώνου πλευραὶἀνάλογον τμηθῶσιν, ἡ ἐπὶ τὰς of the triangle proportionally. And if (two of) the sides
τομὰς ἐπιζευγνυμένη εὐθεῖα παρὰ τὴν λοιπὴν ἔσται τοῦ of a triangle are cut proportionally, then the straight-line
τριγώνουπλευράν·ὅπερἔδειδεῖξαι.
joining the cutting (points) will be parallel to the remaining
side of the triangle. (Which is) the very thing it was
required to show.
. Proposition 3
᾿Εὰντριγώνουἡγωνίαδίχατμηθῇ,ἡδὲτέμνουσατὴν If an angle of a triangle is cut in half, and the straightγωνίανεὐθεῖατέμνῃκαὶτὴνβάσιν,τὰτῆςβάσεωςτμήματα
line cutting the angle also cuts the base, then the segτὸναὐτὸνἕξειλόγονταῖςλοιπαῖςτοῦτριγώνουπλευραῖς·
ments of the base will have the same ratio as the remainκαὶἐὰντὰτῆςβάσεωςτμήματατὸναὐτὸνἔχῃλόγονταῖς
ing sides of the triangle. And if the segments of the base
λοιπαῖςτοῦτριγώνουπλευραῖς,ἡἀπὸτῆςκορυφῆςἐπὶτὴν have the same ratio as the remaining sides of the trianτομὴνἐπιζευγνυμένηεὐθεῖαδίχατεμεῖτὴντοῦτριγώνου
gle, then the straight-line joining the vertex to the cutting
γωνίαν.
(point) will cut the angle of the triangle in half.
῎Εστω τρίγωνοντὸΑΒΓ,καὶτετμήσθωἡὑπὸΒΑΓ Let ABC be a triangle. And let the angle BAC have
γωνίαδίχαὑπὸτῆςΑΔεὐθείας·λέγω,ὅτιἐστὶνὡςἡΒΔ been cut in half by the straight-line AD. I say that as BD
πρὸςτὴνΓΔ,οὕτωςἡΒΑπρὸςτὴνΑΓ.
is to CD, so BA (is) to AC.
῎ΗχθωγὰρδιὰτοῦΓτῇΔΑπαράλληλοςἡΓΕ,καὶ For let CE have been drawn through (point) C parδιαχθεῖσαἡΒΑσυμπιπτέτωαὐτῇκατὰτὸΕ.
allel to DA. And, BA being drawn through, let it meet
158

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
Ε
(CE) at (point) E. †
E
Α
A
Β ∆ Γ
B D C
ΚαὶἐπεὶεἰςπαραλλήλουςτὰςΑΔ,ΕΓεὐθεῖαἐνέπεσεν And since the straight-line AC falls across the parallel
ἡΑΓ,ἡἄραὑπὸΑΓΕγωνίαἴσηἐστὶτῇὑπὸΓΑΔ.ἀλλ᾿ (straight-lines) AD and EC, angle ACE is thus equal to
ἡὑπὸΓΑΔτῇὑπὸΒΑΔὑπόκειταιἴση·καὶἡὑπὸΒΑΔ CAD [Prop. 1.29]. But, (angle) CAD is assumed (to
ἄρατῇὑπὸΑΓΕἐστινἴση. πάλιν,ἐπεὶεἰςπαραλλήλους be) equal to BAD. Thus, (angle) BAD is also equal to
τὰςΑΔ,ΕΓεὐθεῖαἐνέπεσενἡΒΑΕ,ἡἐκτὸςγωνίαἡὑπὸ ACE. Again, since the straight-line BAE falls across the
ΒΑΔἴσηἐστὶτῇἐντὸςτῇὑπὸΑΕΓ.ἐδείχθηδὲκαὶἡὑπὸ parallel (straight-lines) AD and EC, the external angle
ΑΓΕτῇὑπὸΒΑΔἴση·καὶἡὑπὸΑΓΕἄραγωνίατῇὑπὸ BAD is equal to the internal (angle) AEC [Prop. 1.29].
ΑΕΓἐστινἴση·ὥστεκαὶπλευρὰἡΑΕπλευρᾷτῇΑΓἐστιν And (angle) ACE was also shown (to be) equal to BAD.
ἴση. καὶἐπεὶτριγώνουτοῦΒΓΕπαρὰμίαντῶνπλευρῶν Thus, angle ACE is also equal to AEC. And, hence, side
τὴνΕΓἦκταιἡΑΔ,ἀνάλογονἄραἐστὶνὡςἡΒΔπρὸςτὴν AE is equal to side AC [Prop. 1.6]. And since AD has
ΔΓ,οὕτωςἡΒΑπρὸςτὴνΑΕ.ἴσηδὲἡΑΕτῇΑΓ·ὡςἄρα been drawn parallel to one of the sides EC of triangle
ἡΒΔπρὸςτὴνΔΓ,οὕτωςἡΒΑπρὸςτὴνΑΓ. BCE, thus, proportionally, as BD is to DC, so BA (is)
ἈλλὰδὴἔστωὡςἡΒΔπρὸςτὴνΔΓ,οὕτωςἡΒΑπρὸς to AE [Prop. 6.2]. And AE (is) equal to AC. Thus, as
τὴνΑΓ,καὶἐπεζεύχθωἡΑΔ·λέγω,ὅτιδίχατέτμηταιἡ BD (is) to DC, so BA (is) to AC.
ὑπὸΒΑΓγωνίαὑπὸτῆςΑΔεὐθείας.
And so, let BD be to DC, as BA (is) to AC. And let
Τῶνγὰραὐτῶνκατασκευασθέντων,ἐπείἐστινὡςἡΒΔ AD have been joined. I say that angle BAC has been cut
πρὸςτὴνΔΓ,οὕτωςἡΒΑπρὸςτὴνΑΓ,ἀλλὰκαὶὡςἡΒΔ in half by the straight-line AD.
πρὸςτὴνΔΓ,οὕτωςἐστὶνἡΒΑπρὸςτὴνΑΕ·τριγώνου For, by the same construction, since as BD is to DC,
γὰρτοῦΒΓΕπαρὰμίαντὴνΕΓἦκταιἡΑΔ·καὶὡςἄραἡ so BA (is) to AC, then also as BD (is) to DC, so BA is
ΒΑπρὸςτὴνΑΓ,οὕτωςἡΒΑπρὸςτὴνΑΕ.ἴσηἄραἡΑΓ to AE. For AD has been drawn parallel to one (of the
τῇΑΕ·ὥστεκαὶγωνίαἡὑπὸΑΕΓτῇὑπὸΑΓΕἐστινἴση. sides) EC of triangle BCE [Prop. 6.2]. Thus, also, as
ἀλλ᾿ἡμὲνὑπὸΑΕΓτῇἐκτὸςτῇὑπὸΒΑΔ[ἐστιν]ἴση,ἡ BA (is) to AC, so BA (is) to AE [Prop. 5.11]. Thus, AC
δὲὑπὸΑΓΕτῇἐναλλὰξτῇὑπὸΓΑΔἐστινἴση·καὶἡὑπὸ (is) equal to AE [Prop. 5.9]. And, hence, angle AEC
ΒΑΔἄρατῇὑπὸΓΑΔἐστινἴση. ἡἄραὑπὸΒΑΓγωνία is equal to ACE [Prop. 1.5]. But, AEC [is] equal to the
δίχατέτμηταιὑπὸτῆςΑΔεὐθείας.
external (angle) BAD, and ACE is equal to the alternate
᾿Εὰνἄρατριγώνουἡγωνίαδίχατμηθῇ,ἡδὲτέμνουσα (angle) CAD [Prop. 1.29]. Thus, (angle) BAD is also
τὴν γωνίαν εὐθεῖατέμνῃ καὶ τὴν βάσιν, τὰ τῆς βάσεως equal to CAD. Thus, angle BAC has been cut in half by
τμήματατὸναὐτὸνἕξειλόγονταῖςλοιπαῖςτοῦτριγώνου the straight-line AD.
πλευραῖς·καὶἐὰντὰτῆςβάσεωςτμήματατὸναὐτὸνἔχῃ Thus, if an angle of a triangle is cut in half, and the
λόγονταῖςλοιπαῖςτοῦτριγώνουπλευραῖς,ἡἀπὸτῆςκο- straight-line cutting the angle also cuts the base, then the
ρυφῆςἐπὶτὴντομὴνἐπιζευγνυμένηεὐθεῖαδίχατέμνειτὴν segments of the base will have the same ratio as the reτοῦτριγώνουγωνίαν·ὅπερἔδειδεῖξαι.
maining sides of the triangle. And if the segments of the
base have the same ratio as the remaining sides of the
triangle, then the straight-line joining the vertex to the
cutting (point) will cut the angle of the triangle in half.
(Which is) the very thing it was required to show.
† The fact that the two straight-lines meet follows because the sum of ACE and CAE is less than two right-angles, as can easily be demonstrated.
159

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
See Post. 5.
.
Proposition 4
Τῶνἰσογωνίωντριγώνωνἀνάλογόνεἰσιναἱπλευραὶαἱ In equiangular triangles the sides about the equal anπερὶτὰςἴσαςγωνίαςκαὶὁμόλογοιαἱὑπὸτὰςἴσαςγωνίας
gles are proportional, and those (sides) subtending equal
ὑποτείνουσαι.
angles correspond.
Ζ
F
Α
A
∆
D
Β Γ Ε
῎ΕστωἰσογώνιατρίγωνατὰΑΒΓ,ΔΓΕἴσηνἔχοντατὴν Let ABC and DCE be equiangular triangles, having
μὲνὑπὸΑΒΓγωνίαντῇὑπὸΔΓΕ,τὴνδὲὑπὸΒΑΓτῇὑπὸ angle ABC equal to DCE, and (angle) BAC to CDE,
ΓΔΕκαὶἔτιτὴνὑπὸΑΓΒτῇὑπὸΓΕΔ·λέγω,ὅτιτῶνΑΒΓ, and, further, (angle) ACB to CED. I say that in trian-
ΔΓΕτριγώνωνἀνάλογόνεἰσιναἱπλευραὶαἱπερὶτὰςἴσας gles ABC and DCE the sides about the equal angles are
γωνίαςκαὶὁμόλογοιαἱὑπὸτὰςἴσαςγωνίαςὑποτείνουσαι. proportional, and those (sides) subtending equal angles
Κείσθωγὰρἐπ᾿εὐθείαςἡΒΓτῇΓΕ.καὶἐπεὶαἱὑπὸ correspond.
ΑΒΓ, ΑΓΒ γωνίαι δύο ὀρθῶν ἐλάττονές εἰσιν, ἴση δὲ Let BC be placed straight-on to CE. And since
ἡ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΕΓ, αἱ ἄρα ὑπὸ ΑΒΓ, ΔΕΓ δύο angles ABC and ACB are less than two right-angles
ὀρθῶνἐλάττονέςεἰσιν·αἱΒΑ,ΕΔἄραἐκβαλλόμεναισυμ- [Prop 1.17], and ACB (is) equal to DEC, thus ABC
πεσοῦνται.ἐκβεβλήσθωσανκαὶσυμπιπτέτωσανκατὰτὸΖ. and DEC are less than two right-angles. Thus, BA and
Καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΔΓΕ γωνία τῇ ὑπὸ ΑΒΓ, ED, being produced, will meet [C.N. 5]. Let them have
παράλληλόςἐστινἡΒΖτῇΓΔ.πάλιν,ἐπεὶἴσηἐστὶνἡὑπὸ been produced, and let them meet at (point) F .
ΑΓΒτῇὑπὸΔΕΓ,παράλληλόςἐστινἡΑΓτῇΖΕ.παραλ- And since angle DCE is equal to ABC, BF is parallel
ληλόγραμμονἄραἐστὶτὸΖΑΓΔ·ἴσηἄραἡμὲνΖΑτῇΔΓ, to CD [Prop. 1.28]. Again, since (angle) ACB is equal to
ἡδὲΑΓτῇΖΔ.καὶἐπεὶτριγώνουτοῦΖΒΕπαρὰμίαντὴν DEC, AC is parallel to FE [Prop. 1.28]. Thus, FACD
ΖΕἦκταιἡΑΓ,ἐστινἄραὡςἡΒΑπρὸςτὴνΑΖ,οὕτωςἡ is a parallelogram. Thus, FA is equal to DC, and AC to
ΒΓπρὸςτὴνΓΕ.ἴσηδὲἡΑΖτῇΓΔ·ὡςἄραἡΒΑπρὸςτὴν FD [Prop. 1.34]. And since AC has been drawn parallel
ΓΔ,οὕτωςἡΒΓπρὸςτὴνΓΕ,καὶἐναλλὰξὡςἡΑΒπρὸς to one (of the sides) FE of triangle FBE, thus as BA
τὴνΒΓ,οὕτωςἡΔΓπρὸςτὴνΓΕ.πάλιν,ἐπεὶπαράλληλός is to AF , so BC (is) to CE [Prop. 6.2]. And AF (is)
ἐστινἡΓΔτῇΒΖ,ἔστινἄραὡςἡΒΓπρὸςτὴνΓΕ,οὕτως equal to CD. Thus, as BA (is) to CD, so BC (is) to CE,
ἡΖΔπρὸςτὴνΔΕ.ἴσηδὲἡΖΔτῇΑΓ·ὡςἄραἡΒΓπρὸς and, alternately, as AB (is) to BC, so DC (is) to CE
τὴνΓΕ,οὕτωςἡΑΓπρὸςτὴνΔΕ,καὶἐναλλὰξὡςἡΒΓ [Prop. 5.16]. Again, since CD is parallel to BF , thus as
πρὸςτὴνΓΑ,οὕτωςἡΓΕπρὸςτὴνΕΔ.ἐπεὶοὖνἐδείχθη BC (is) to CE, so FD (is) to DE [Prop. 6.2]. And FD
ὡςμὲνἡΑΒπρὸςτὴνΒΓ,οὕτωςἡΔΓπρὸςτὴνΓΕ,ὡς (is) equal to AC. Thus, as BC is to CE, so AC (is) to
δὲἡΒΓπρὸςτὴνΓΑ,οὕτωςἡΓΕπρὸςτὴνΕΔ,δι᾿ἴσου DE, and, alternately, as BC (is) to CA, so CE (is) to
ἄραὡςἡΒΑπρὸςτὴνΑΓ,οὕτωςἡΓΔπρὸςτὴνΔΕ. ED [Prop. 6.2]. Therefore, since it was shown that as
Τῶνἄραἰσογωνίωντριγώνωνἀνάλογόνεἰσιναἱπλευραὶ AB (is) to BC, so DC (is) to CE, and as BC (is) to CA,
αἱπερὶτὰςἴσαςγωνίαςκαὶὁμόλογοιαἱὑπὸτὰςἴσαςγωνίας so CE (is) to ED, thus, via equality, as BA (is) to AC, so
ὑποτείνουσαι·ὅπερἔδειδεῖξαι. CD (is) to DE [Prop. 5.22].
Thus, in equiangular triangles the sides about the
equal angles are proportional, and those (sides) subtend-
B
C
E
160

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
ing equal angles correspond. (Which is) the very thing it
was required to show.
. Proposition 5
᾿Εὰνδύοτρίγωνατὰςπλευρὰςἀνάλογονἔχῃ,ἰσογώνια If two triangles have proportional sides then the trian-
ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, ὑφ᾿ ἃς αἱ gles will be equiangular, and will have the angles which
ὁμόλογοιπλευραὶὑποτείνουσιν.
corresponding sides subtend equal.
Α
∆
A
D
Ε
Ζ
E
F
Β
Γ
Η
῎ΕστωδύοτρίγωνατὰΑΒΓ,ΔΕΖτὰςπλευρὰςἀνάλογον Let ABC and DEF be two triangles having propor-
ἔχοντα,ὡςμὲντὴνΑΒπρὸςτὴνΒΓ,οὕτωςτὴνΔΕπρὸς tional sides, (so that) as AB (is) to BC, so DE (is) to
τὴνΕΖ,ὡςδὲτὴνΒΓπρὸςτὴνΓΑ,οὕτωςτὴνΕΖπρὸς EF , and as BC (is) to CA, so EF (is) to FD, and, furτὴνΖΔ,καὶἔτιὡςτὴνΒΑπρὸςτὴνΑΓ,οὕτωςτὴνΕΔ
ther, as BA (is) to AC, so ED (is) to DF . I say that
πρὸςτὴνΔΖ.λέγω,ὅτιἰσογώνιόνἐστιτὸΑΒΓτρίγωνον triangle ABC is equiangular to triangle DEF , and (that
τῷΔΕΖτριγώνῳκαὶἴσαςἕξουσιτὰςγωνίας,ὑφ᾿ἃςαἱ the triangles) will have the angles which corresponding
ὁμόλογοιπλευραὶὑποτείνουσιν,τὴνμὲνὑπὸΑΒΓτῇὑπὸ sides subtend equal. (That is), (angle) ABC (equal) to
ΔΕΖ,τὴνδὲὑπὸΒΓΑτῇὑπὸΕΖΔκαὶἔτιτὴνὑπὸΒΑΓ DEF , BCA to EFD, and, further, BAC to EDF .
τῇὑπὸΕΔΖ.
For let (angle) FEG, equal to angle ABC, and (an-
ΣυνεστάτωγὰρπρὸςτῇΕΖεὐθείᾳκαὶτοῖςπρὸςαὐτῇ gle) EFG, equal to ACB, have been constructed on the
σημείοιςτοῖςΕ,ΖτῇμὲνὑπὸΑΒΓγωνίᾳἴσηἡὑπὸΖΕΗ, straight-line EF at the points E and F on it (respectively)
τῇδὲὑπὸΑΓΒἴσηἡὑπὸΕΖΗ·λοιπὴἄραἡπρὸςτῷΑ [Prop. 1.23]. Thus, the remaining (angle) at A is equal
λοιπῇτῇπρὸςτῷΗἐστινἴση. to the remaining (angle) at G [Prop. 1.32].
῎ΙσογώνιονἄραἐστὶτὸΑΒΓτρίγωνοντῷΕΗΖ[τριγών- Thus, triangle ABC is equiangular to [triangle] EGF .
ῳ].τῶνἄραΑΒΓ,ΕΗΖτριγώνωνἀνάλογόνεἰσιναἱπλευραὶ Thus, for triangles ABC and EGF , the sides about the
αἱπερὶτὰςἴσαςγωνίαςκαὶὁμόλογοιαἱὑπὸτὰςἴσαςγωνίας equal angles are proportional, and (those) sides subtend-
ὑποτείνουσαι· ἔστινἄραὡςἡΑΒπρὸςτὴνΒΓ,[οὕτως] ing equal angles correspond [Prop. 6.4]. Thus, as AB
ἡΗΕπρὸςτὴνΕΖ.ἀλλ᾿ὡςἡΑΒπρὸςτὴνΒΓ,οὕτως is to BC, [so] GE (is) to EF . But, as AB (is) to BC,
ὑπόκειταιἡΔΕπρὸςτὴνΕΖ·ὡςἄραἡΔΕπρὸςτὴνΕΖ, so, it was assumed, (is) DE to EF . Thus, as DE (is) to
οὕτωςἡΗΕπρὸςτὴνΕΖ.ἑκατέραἄρατῶνΔΕ,ΗΕπρὸς EF , so GE (is) to EF [Prop. 5.11]. Thus, DE and GE
τὴνΕΖτὸναὐτὸνἔχειλόγον·ἴσηἄραἐστὶνἡΔΕτῇΗΕ. each have the same ratio to EF . Thus, DE is equal to
διὰτὰαὐτὰδὴκαὶἡΔΖτῇΗΖἐστινἴση.ἐπεὶοὖνἴσηἐστὶν GE [Prop. 5.9]. So, for the same (reasons), DF is also
ἡΔΕτῇΕΗ,κοινὴδὲἡΕΖ,δύοδὴαἱΔΕ,ΕΖδυσὶταῖςΗΕ, equal to GF . Therefore, since DE is equal to EG, and
ΕΖἴσαιεἰσίν·καὶβάσιςἡΔΖβάσειτῇΖΗ[ἐστιν]ἴση·γωνία EF (is) common, the two (sides) DE, EF are equal to
ἄραἡὑπὸΔΕΖγωνίᾳτῇὑπὸΗΕΖἐστινἴση,καὶτὸΔΕΖ the two (sides) GE, EF (respectively). And base DF
τρίγωνοντῷΗΕΖτριγώνῳἴσον,καὶαἱλοιπαὶγωνίαιταῖς [is] equal to base FG. Thus, angle DEF is equal to
λοιπαῖςγωνίαιςἴσαι,ὑφ᾿ἃςαἱἴσαιπλευραὶὑποτείνουσιν. angle GEF [Prop. 1.8], and triangle DEF (is) equal
ἴσηἄραἐστὶκαὶἡμὲνὑπὸΔΖΕγωνίατῇὑπὸΗΖΕ,ἡδὲ to triangle GEF , and the remaining angles (are) equal
ὑπὸΕΔΖτῇὑπὸΕΗΖ.καὶἐπεὶἡμὲνὑπὸΖΕΔτῇὑπὸ to the remaining angles which the equal sides subtend
ΗΕΖἐστινἴση,ἀλλ᾿ἡὑπὸΗΕΖτῇὑπὸΑΒΓ,καὶἡὑπὸ [Prop. 1.4]. Thus, angle DFE is also equal to GFE, and
B
C
G
161

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
ΑΒΓἄραγωνίατῇὑπὸΔΕΖἐστινἴση.διὰτὰαὐτὰδὴκαὶ (angle) EDF to EGF . And since (angle) FED is equal
ἡὑπὸΑΓΒτῇὑπὸΔΖΕἐστινἴση,καὶἔτιἡπρὸςτῷΑτῇ to GEF , and (angle) GEF to ABC, angle ABC is thus
πρὸςτῷΔ·ἰσογώνιονἄραἐστὶτὸΑΒΓτρίγωνοντῷΔΕΖ also equal to DEF . So, for the same (reasons), (angle)
τριγώνῳ.
ACB is also equal to DFE, and, further, the (angle) at A
᾿Εὰν ἄρα δύο τρίγωνα τὰς πλευρὰς ἀνάλογον ἔχῃ, to the (angle) at D. Thus, triangle ABC is equiangular
ἰσογώνιαἔσταιτὰτρίγωνακαὶἴσαςἕξειτὰςγωνίας,ὑφ᾿ to triangle DEF .
ἃςαἱὁμόλογοιπλευραὶὑποτείνουσιν·ὅπερἔδειδεῖξαι.
Thus, if two triangles have proportional sides then the
triangles will be equiangular, and will have the angles
which corresponding sides subtend equal. (Which is) the
very thing it was required to show.
¦. Proposition 6
᾿Εὰνδύοτρίγωναμίανγωνίανμιᾷγωνίᾳἴσηνἔχῃ,περὶ If two triangles have one angle equal to one angle,
δὲτὰςἴσαςγωνίαςτὰςπλευρὰςἀνάλογον,ἰσογώνιαἔσται and the sides about the equal angles proportional, then
τὰτρίγωνακαὶἴσαςἕξειτὰςγωνίας,ὑφ᾿ἃςαἱὁμόλογοι the triangles will be equiangular, and will have the angles
πλευραὶὑποτείνουσιν.
which corresponding sides subtend equal.
Α
∆
A
D
Η
G
Ε
Ζ
E
F
Β
Γ
῎ΕστωδύοτρίγωνατὰΑΒΓ,ΔΕΖμίανγωνίαντὴνὑπὸ Let ABC and DEF be two triangles having one angle,
ΒΑΓμιᾷγωνίᾳτῇὑπὸΕΔΖἴσηνἔχοντα,περὶδὲτὰςἴσας BAC, equal to one angle, EDF (respectively), and the
γωνίαςτὰςπλευρὰςἀνάλογον,ὡςτὴνΒΑπρὸςτὴνΑΓ, sides about the equal angles proportional, (so that) as BA
οὕτωςτὴνΕΔπρὸςτὴνΔΖ·λέγω,ὅτιἰσογώνιόνἐστιτὸ (is) to AC, so ED (is) to DF . I say that triangle ABC is
ΑΒΓτρίγωνοντῷΔΕΖτριγώνῳκαὶἴσηνἕξειτὴνὑπὸΑΒΓ equiangular to triangle DEF , and will have angle ABC
γωνίαντῇὑπὸΔΕΖ,τὴνδὲὑπὸΑΓΒτῇὑπὸΔΖΕ. equal to DEF , and (angle) ACB to DFE.
ΣυνεστάτωγὰρπρὸςτῇΔΖεὐθείᾳκαὶτοῖςπρὸςαὐτῇ For let (angle) FDG, equal to each of BAC and
σημείοιςτοῖςΔ,ΖὁποτέρᾳμὲντῶνὑπὸΒΑΓ,ΕΔΖἴση EDF , and (angle) DFG, equal to ACB, have been con-
ἡὑπὸΖΔΗ,τῇδὲὑπὸΑΓΒἴσηἡὑπὸΔΖΗ·λοιπὴἄραἡ structed on the straight-line AF at the points D and F on
πρὸςτῷΒγωνίαλοιπῇτῇπρὸςτῷΗἴσηἐστίν. it (respectively) [Prop. 1.23]. Thus, the remaining angle
᾿Ισογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΗΖ at B is equal to the remaining angle at G [Prop. 1.32].
τριγώνῳ.ἀνάλογονἄραἐστὶνὡςἡΒΑπρὸςτὴνΑΓ,οὕτως Thus, triangle ABC is equiangular to triangle DGF .
ἡΗΔπρὸςτὴνΔΖ.ὑπόκειταιδὲκαὶὡςἡΒΑπρὸςτὴνΑΓ, Thus, proportionally, as BA (is) to AC, so GD (is) to
οὕτωςἡΕΔπρὸςτὴνΔΖ·καὶὡςἄραἡΕΔπρὸςτὴνΔΖ, DF [Prop. 6.4]. And it was also assumed that as BA
οὕτωςἡΗΔπρὸςτὴνΔΖ.ἴσηἄραἡΕΔτῇΔΗ·καὶκοινὴ is) to AC, so ED (is) to DF . And, thus, as ED (is)
ἡΔΖ·δύοδὴαἱΕΔ,ΔΖδυσὶταῖςΗΔ,ΔΖἴσαςεἰσίν·καὶ to DF , so GD (is) to DF [Prop. 5.11]. Thus, ED (is)
γωνίαἡὑπὸΕΔΖγωνίᾳτῇὑπὸΗΔΖ[ἐστιν]ἴση·βάσις equal to DG [Prop. 5.9]. And DF (is) common. So, the
ἄραἡΕΖβάσειτῇΗΖἐστινἴση,καὶτὸΔΕΖτρίγωνοντῷ two (sides) ED, DF are equal to the two (sides) GD,
ΗΔΖτριγώνῳἴσονἐστίν,καὶαἱλοιπαὶγωνίαιταῖςλοιπαῖς DF (respectively). And angle EDF [is] equal to angle
γωνίαιςἴσαςἔσονται,ὐφ᾿ἃςἴσαςπλευραὶὑποτείνουσιν.ἴση GDF . Thus, base EF is equal to base GF , and triangle
ἄραἐστὶνἡμὲνὑπὸΔΖΗτῇὑποΔΖΕ,ἡδὲὑπὸΔΗΖ DEF is equal to triangle GDF , and the remaining angles
B
C
162

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
τῇὑπὸΔΕΖ.ἀλλ᾿ἡὑπὸΔΖΗτῇὑπὸΑΓΒἐστινἴση·καὶ will be equal to the remaining angles which the equal
ἡὑπὸΑΓΒἄρατῇὑπὸΔΖΕἐστινἴση. ὑπόκειταιδὲκαὶ sides subtend [Prop. 1.4]. Thus, (angle) DFG is equal
ἡὑπὸΒΑΓτῇὑπὸΕΔΖἴση·καὶλοιπὴἄραἡπρὸςτῷΒ to DFE, and (angle) DGF to DEF . But, (angle) DFG
λοιπῇτῇπρὸςτῷΕἴσηἐστίν·ἰσογώνιονἄραἐστὶτὸΑΒΓ is equal to ACB. Thus, (angle) ACB is also equal to
τρίγωνοντῷΔΕΖτριγώνῳ.
DFE. And (angle) BAC was also assumed (to be) equal
᾿Εὰνἄραδύοτρίγωναμίανγωνίανμιᾷγωνίᾳἴσηνἔχῃ, to EDF . Thus, the remaining (angle) at B is equal to the
περὶδὲτὰςἴσαςγωνίαςτὰςπλευρὰςἀνάλογον,ἰσογώνια remaining (angle) at E [Prop. 1.32]. Thus, triangle ABC
ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, ὑφ᾿ ἃς αἱ is equiangular to triangle DEF .
ὁμόλογοιπλευραὶὑποτείνουσιν·ὅπερἔδειδεῖξαι.
Thus, if two triangles have one angle equal to one
angle, and the sides about the equal angles proportional,
then the triangles will be equiangular, and will have the
angles which corresponding sides subtend equal. (Which
is) the very thing it was required to show.
Þ. Proposition 7
᾿Εὰν δύο τρίγωνα μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχῃ, If two triangles have one angle equal to one angle,
περὶ δὲ ἄλλας γωνίας τὰς πλευρὰς ἀνάλογον, τῶν δὲ and the sides about other angles proportional, and the
λοιπῶνἑκατέρανἅμαἤτοιἐλάσσοναἢμὴἐλάσσοναὀρθῆς, remaining angles either both less than, or both not less
ἰσογώνιαἔσταιτὰτρίγωνακαὶἴσαςἕξειτὰςγωνίας,περὶ than, right-angles, then the triangles will be equiangular,
ἃςἀνάλογόνεἰσιναἱπλευραί.
and will have the angles about which the sides are proportional
equal.
Α
A
∆
D
Β
Ε
B
E
Η
Ζ
G
F
Γ
C
῎Εστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ μίαν γωνίαν μιᾷ Let ABC and DEF be two triangles having one anγωνίᾳ
ἴσην ἔχοντα τὴν ὑπὸ ΒΑΓ τῇ ὑπὸΕΔΖ, περὶ δὲ gle, BAC, equal to one angle, EDF (respectively), and
ἄλλαςγωνίαςτὰςὑπὸΑΒΓ,ΔΕΖτὰςπλευρὰςἀνάλογον, the sides about (some) other angles, ABC and DEF (re-
ὡςτὴνΑΒπρὸςτὴνΒΓ,οὕτωςτὴνΔΕπρὸςτὴνΕΖ, spectively), proportional, (so that) as AB (is) to BC, so
τῶν δὲ λοιπῶν τῶν πρὸς τοῖς Γ, Ζ πρότερον ἑκατέραν DE (is) to EF , and the remaining (angles) at C and F ,
ἅμαἐλάσσοναὀρθῆς· λέγω,ὅτιἰσογώνιόνἐστιτὸΑΒΓ first of all, both less than right-angles. I say that triangle
τρίγωνοντῷΔΕΖτριγώνῳ,καὶἴσηἔσταιἡὑπὸΑΒΓγωνία ABC is equiangular to triangle DEF , and (that) angle
τῇὑπὸΔΕΖ,καὶλοιπὴδηλονότιἡπρὸςτῷΓλοιπῇτῇπρὸς ABC will be equal to DEF , and (that) the remaining
τῷΖἴση.
(angle) at C (will be) manifestly equal to the remaining
ΕἰγὰρἄνισόςἐστινἡὑπὸΑΒΓγωνίατῇὑπὸΔΕΖ, (angle) at F .
μίααὐτῶνμείζωνἐστίν. ἔστωμείζωνἡὑπὸΑΒΓ.καὶσυ- For if angle ABC is not equal to (angle) DEF then
νεστάτωπρὸςτῇΑΒεὐθείᾳκαὶτῷπρὸςαὐτῇσημείῳτῷ one of them is greater. Let ABC be greater. And let (an-
ΒτῇὑπὸΔΕΖγωνίᾳἴσηἡὑπὸΑΒΗ.
gle) ABG, equal to (angle) DEF , have been constructed
ΚαὶἐπεὶἴσηἐστὶνἡμὲνΑγωνίατῇΔ,ἡδὲὑπὸΑΒΗ on the straight-line AB at the point B on it [Prop. 1.23].
τῇὑπὸΔΕΖ,λοιπὴἄραἡὑπὸΑΗΒλοιπῇτῇὑπὸΔΖΕ And since angle A is equal to (angle) D, and (angle)
ἐστινἴση. ἰσογώνιονἄραἐστὶτὸΑΒΗτρίγωνοντῷΔΕΖ ABG to DEF , the remaining (angle) AGB is thus equal
163

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
τριγώνῳ. ἔστινἄραὡςἡΑΒπρὸςτὴνΒΗ,οὕτωςἡΔΕ to the remaining (angle) DFE [Prop. 1.32]. Thus, trianπρὸςτὴνΕΖ.ὡςδὲἡΔΕπρὸςτὴνΕΖ,[οὕτως]ὑπόκειταιἡ
gle ABG is equiangular to triangle DEF . Thus, as AB is
ΑΒπρὸςτὴνΒΓ·ἡΑΒἄραπρὸςἑκατέραντῶνΒΓ,ΒΗτὸν to BG, so DE (is) to EF [Prop. 6.4]. And as DE (is) to
αὐτὸνἔχειλόγον·ἴσηἄραἡΒΓτῇΒΗ.ὥστεκαὶγωνίαἡ EF , [so] it was assumed (is) AB to BC. Thus, AB has
πρὸςτῷΓγωνίᾳτῇὑπὸΒΗΓἐστινἴση.ἐλάττωνδὲὀρθῆς the same ratio to each of BC and BG [Prop. 5.11]. Thus,
ὑπόκειταιἡπρὸςτῷΓ·ἐλάττωνἄραἐστὶνὀρθῆςκαὶὑπὸ BC (is) equal to BG [Prop. 5.9]. And, hence, the angle
ΒΗΓ·ὥστεἡἐφεξῆςαὐτῇγωνίαἡὑπὸΑΗΒμείζωνἐστὶν at C is equal to angle BGC [Prop. 1.5]. And the angle
ὀρθῆς.καὶἐδείχθηἴσηοὖσατῇπρὸςτῷΖ·καὶἡπρὸςτῷΖ at C was assumed (to be) less than a right-angle. Thus,
ἄραμείζωνἐστὶνὀρθῆς.ὑπόκειταιδὲἐλάσσωνὀρθῆς·ὅπερ (angle) BGC is also less than a right-angle. Hence, the
ἐστὶνἄτοπον. οὐκἄραἄνισόςἐστινἡὑπὸΑΒΓγωνίατῇ adjacent angle to it, AGB, is greater than a right-angle
ὑπὸΔΕΖ·ἴσηἄρα.ἔστιδὲκαὶἡπρὸςτῷΑἴσητῇπρὸςτῷ [Prop. 1.13]. And (AGB) was shown to be equal to the
Δ·καὶλοιπὴἄραἡπρὸςτῷΓλοιπῇτῇπρὸςτῷΖἴσηἐστίν. (angle) at F . Thus, the (angle) at F is also greater than a
ἰσογώνιονἄραἐστὶτὸΑΒΓτρίγωνοντῷΔΕΖτριγώνῳ. right-angle. But it was assumed (to be) less than a right-
ἈλλὰδὴπάλινὑποκείσθωἑκατέρατῶνπρὸςτοῖςΓ,Ζμὴ angle. The very thing is absurd. Thus, angle ABC is not
ἐλάσσωνὀρθῆς·λέγωπάλιν,ὅτικαὶοὕτωςἐστὶνἰσογώνιον unequal to (angle) DEF . Thus, (it is) equal. And the
τὸΑΒΓτρίγωνοντῷΔΕΖτριγώνῳ.
(angle) at A is also equal to the (angle) at D. And thus
Τῶνγὰραὐτῶνκατασκευασθέντωνὁμοίωςδείξομεν, the remaining (angle) at C is equal to the remaining (an-
ὅτιἴσηἐστὶνἡΒΓτῇΒΗ·ὥστεκαὶγωνίαἡπρὸςτῷΓτῇ gle) at F [Prop. 1.32]. Thus, triangle ABC is equiangular
ὑπὸΒΗΓἴσηἐστίν. οὐκἐλάττωνδὲὀρθῆςἡπρὸςτῷΓ· to triangle DEF .
οὐκἐλάττωνἄραὀρθῆςοὐδὲἡὑπὸΒΗΓ.τριγώνουδὴτοῦ But, again, let each of the (angles) at C and F be
ΒΗΓαἱδύογωνίαιδύοὀρθῶνοὔκεἰσινἐλάττονες·ὅπερ assumed (to be) not less than a right-angle. I say, again,
ἐστὶνἀδύνατον. οὐκἄραπάλινἄνισόςἐστινἡὑπὸΑΒΓ that triangle ABC is equiangular to triangle DEF in this
γωνίατῇὑπὸΔΕΖ·ἴσηἄρα. ἔστιδὲκαὶἡπρὸςτῷΑτῇ case also.
πρὸςτῷΔἴση·λοιπὴἄραἡπρὸςτῷΓλοιπῇτῇπρὸςτῷΖ For, with the same construction, we can similarly
ἴσηἐστίν. ἰσογώνιονἄραἐστὶτὸΑΒΓτρίγωνοντῷΔΕΖ show that BC is equal to BG. Hence, also, the angle
τριγώνῳ.
at C is equal to (angle) BGC. And the (angle) at C (is)
᾿Εὰνἄραδύοτρίγωναμίανγωνίανμιᾷγωνίᾳἴσηνἔχῃ, not less than a right-angle. Thus, BGC (is) not less than
περὶδὲἄλλαςγωνίαςτὰςπλευρὰςἀνάλογον,τῶνδὲλοιπῶν a right-angle either. So, in triangle BGC the (sum of)
ἑκατέραν ἅμα ἐλάττονα ἢ μὴ ἐλάττονα ὀρθῆς, ἰσογώνια two angles is not less than two right-angles. The very
ἔσταιτὰτρίγωνακαὶἴσαςἕξειτὰςγωνίας,περὶἃςἀνάλογόν thing is impossible [Prop. 1.17]. Thus, again, angle ABC
εἰσιναἱπλευραί·ὅπερἔδειδεῖξαι.
is not unequal to DEF . Thus, (it is) equal. And the (angle)
at A is also equal to the (angle) at D. Thus, the
remaining (angle) at C is equal to the remaining (angle)
at F [Prop. 1.32]. Thus, triangle ABC is equiangular to
triangle DEF .
Thus, if two triangles have one angle equal to one
angle, and the sides about other angles proportional, and
the remaining angles both less than, or both not less than,
right-angles, then the triangles will be equiangular, and
will have the angles about which the sides (are) proportional
equal. (Which is) the very thing it was required to
show.
. Proposition 8
᾿Εὰνἐνὀρθογωνίῳτριγώνῳἀπότῆςὀρθῆςγωνίαςἐπὶ If, in a right-angled triangle, a (straight-line) is drawn
τὴνβάσινκάθετοςἀχθῇ,τὰπρὸςτῇκαθέτῳτρίγωναὅμοιά from the right-angle perpendicular to the base then the
ἐστιτῷτεὅλῳκαὶἀλλήλοις.
triangles around the perpendicular are similar to the
῎ΕστωτρίγωνονὀρθογώνιοντὸΑΒΓὀρθὴνἔχοντὴν whole (triangle), and to one another.
ὑπὸΒΑΓγωνίαν,καὶἤχθωἀπὸτοῦΑἐπὶτὴνΒΓκάθετος Let ABC be a right-angled triangle having the angle
ἡΑΔ·λέγω,ὅτιὅμοιόνἐστινἑκάτεροντῶνΑΒΔ,ΑΔΓ BAC a right-angle, and let AD have been drawn from
164

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
τριγώνωνὅλῳτῷΑΒΓκαὶἔτιἀλλήλοις.
Α
A, perpendicular to BC [Prop. 1.12]. I say that triangles
ABD and ADC are each similar to the whole (triangle)
ABC and, further, to one another.
A
Β
∆
Γ
᾿Επεὶ γὰρἴσηἐστὶν ἡὑπὸΒΑΓτῇὑπὸΑΔΒ· ὀρθὴ For since (angle) BAC is equal to ADB—for each
γὰρἑκατέρα·καὶκοινὴτῶνδύοτριγώνωντοῦτεΑΒΓκαὶ (are) right-angles—and the (angle) at B (is) common
τοῦΑΒΔἡπρὸςτῷΒ,λοιπὴἄραἡὑπὸΑΓΒλοιπῇτῇ to the two triangles ABC and ABD, the remaining (an-
ὑποΒΑΔἐστινἴση·ἰσογώνιονἄραἐστὶτὸΑΒΓτρίγωνον gle) ACB is thus equal to the remaining (angle) BAD
τῷΑΒΔτριγώνῳ. ἔστινἄραὡςἡΒΓὑποτείνουσατὴν [Prop. 1.32]. Thus, triangle ABC is equiangular to tri-
ὀρθὴντοῦΑΒΓτριγώνουπρὸςτὴνΒΑὑποτείνουσαντὴν angle ABD. Thus, as BC, subtending the right-angle in
ὀρθὴντοῦΑΒΔτριγώνου,οὕτωςαὐτὴἡΑΒὑποτείνουσα triangle ABC, is to BA, subtending the right-angle in triτὴνπρὸςτῷΓγωνίαντοῦΑΒΓτριγώνουπρὸςτὴνΒΔ
angle ABD, so the same AB, subtending the angle at C
ὑποτείνουσαντὴνἴσηντὴνὑπὸΒΑΔτοῦΑΒΔτριγώνου, in triangle ABC, (is) to BD, subtending the equal (anκαὶἔτιἡΑΓπρὸςτὴνΑΔὑποτείνουσαντὴνπρὸςτῷΒ
gle) BAD in triangle ABD, and, further, (so is) AC to
γωνίανκοινὴντῶνδύοτριγώνων. τὸΑΒΓἄρατρίγωνον AD, (both) subtending the angle at B common to the
τῷΑΒΔτριγώνῳἰσογώνιόντέἐστικαὶτὰςπερὶτὰςἴσας two triangles [Prop. 6.4]. Thus, triangle ABC is equianγωνίαςπλευρὰςἀνάλογονἔχει.ὅμοιονἄμα[ἐστὶ]τὸΑΒΓ
gular to triangle ABD, and has the sides about the equal
τρίγωνοντῷΑΒΔτριγώνῳ. ὁμοίωςδὴδείξομεν,ὅτικαὶ angles proportional. Thus, triangle ABC [is] similar to
τῷΑΔΓτριγώνῳὅμοιόνἐστιτὸΑΒΓτρίγωνον·ἑκάτερον triangle ABD [Def. 6.1]. So, similarly, we can show that
ἄρατῶνΑΒΔ,ΑΔΓ[τριγώνων]ὅμοιόνἐστινὅλῳτῷΑΒΓ. triangle ABC is also similar to triangle ADC. Thus, [tri-
Λέγωδή,ὅτικαὶἀλλήλοιςἐστὶνὅμοιατὰΑΒΔ,ΑΔΓ angles] ABD and ADC are each similar to the whole
τρίγωνα.
(triangle) ABC.
᾿ΕπεὶγὰρὀρθὴἡὑπὸΒΔΑὀρθῇτῇὑπὸΑΔΓἐστιν So I say that triangles ABD and ADC are also similar
ἴση,ἀλλὰμὴνκαὶἡὑπὸΒΑΔτῇπρὸςτῷΓἐδείχθηἴση, to one another.
καὶλοιπὴἄραἡπρὸςτῷΒλοιπῇτῇὑπὸΔΑΓἐστινἴση· For since the right-angle BDA is equal to the right-
ἰσογώνιονἄραἐστὶτὸΑΒΔτρίγωνοντῷΑΔΓτριγώνῳ. angle ADC, and, indeed, (angle) BAD was also shown
ἔστινἄραὡςἡΒΔτοῦΑΒΔτριγώνουὑποτείνουσατὴν (to be) equal to the (angle) at C, thus the remaining (an-
ὑπὸΒΑΔπρὸςτὴνΔΑτοῦΑΔΓτριγώνουὑποτείνουσαν gle) at B is also equal to the remaining (angle) DAC
τὴνπρὸςτῷΓἴσηντῇὑπὸΒΑΔ,οὕτωςαὐτὴἡΑΔτοῦ [Prop. 1.32]. Thus, triangle ABD is equiangular to trian-
ΑΒΔτριγώνουὑποτείνουσατὴνπρὸςτῷΒγωνίανπρὸς gle ADC. Thus, as BD, subtending (angle) BAD in triτὴνΔΓὑποτείνουσαντὴνὑπὸΔΑΓτοῦΑΔΓτριγώνου
angle ABD, is to DA, subtending the (angle) at C in tri-
ἴσηντῇπρὸςτῷΒ,καὶἔτιἡΒΑπρὸςτὴνΑΓὑποτείνουσαι angle ADC, (which is) equal to (angle) BAD, so (is) the
τὰςὀρθάς· ὅμοιονἄραἐστὶτὸΑΒΔτρίγωνοντῷΑΔΓ same AD, subtending the angle at B in triangle ABD, to
τριγώνῳ.
DC, subtending (angle) DAC in triangle ADC, (which
᾿Εὰνἄραἐνὀρθογωνίῳτριγώνῳἀπὸτῆςὀρθῆςγωνίας is) equal to the (angle) at B, and, further, (so is) BA to
ἐπὶτὴνβάσινκάθετοςἀχθῇ,τὰπρὸςτῇκαθέτῳτρίγωνα AC, (each) subtending right-angles [Prop. 6.4]. Thus,
ὅμοιάἐστιτῷτεὅλῳκαὶἀλλήλοις[ὅπερἔδειδεῖξαι]. triangle ABD is similar to triangle ADC [Def. 6.1].
Thus, if, in a right-angled triangle, a (straight-line)
is drawn from the right-angle perpendicular to the base
B
D
C
165

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
then the triangles around the perpendicular are similar
to the whole (triangle), and to one another. [(Which is)
the very thing it was required to show.]
ÈÖ×Ñ.
᾿Εκδὴτούτουφανερόν,ὅτιἐὰνἐνὀρθογωνίῳτριγώνῳ So (it is) clear, from this, that if, in a right-angled tri-
Corollary
ἀπὸ τῆς ὀρθῆς γωνάις ἐπὶ τὴν βάσις κάθετος ἀχθῇ, ἡ angle, a (straight-line) is drawn from the right-angle per-
ἀχθεῖσατῶντῆςβάσεωςτμημάτωνμέσηἀνάλογόνἐστιν· pendicular to the base then the (straight-line so) drawn
ὅπερἔδειδεῖξαι.
is in mean proportion to the pieces of the base. † (Which
is) the very thing it was required to show.
† In other words, the perpendicular is the geometric mean of the pieces.
. Proposition
Τῆςδοθείσηςεὐθείαςτὸπροσταχθὲνμέροςἀφελεῖν.
9
To cut off a prescribed part from a given straight-line.
∆
Ε
Γ
D
E
C
Α
Ζ
Β
῎ΕστωἡδοθεῖσαεὐθεῖαἡΑΒ·δεῖδὴτῆςΑΒτὸπρο- Let AB be the given straight-line. So it is required to
σταχθὲνμέροςἀφελεῖν.
cut off a prescribed part from AB.
᾿Επιτετάχθωδὴτὸτρίτον. [καὶ]διήθχωτιςἀπὸτοῦΑ So let a third (part) have been prescribed. [And] let
εὐθεῖαἡΑΓγωνίανπεριέχουσαμετὰτῆςΑΒτυχοῦσαν·καὶ some straight-line AC have been drawn from (point) A,
εἰλήφθωτυχὸνσημεῖονἐπὶτῆςΑΓτὸΔ,καὶκείσθωσαν encompassing a random angle with AB. And let a ranτῇΑΔἴσαιαἱΔΕ,ΕΓ.καὶἐπεζεύχθωἡΒΓ,καὶδιὰτοῦΔ
dom point D have been taken on AC. And let DE and
παράλληλοςαὐτῇἤχθωἡΔΖ.
EC be made equal to AD [Prop. 1.3]. And let BC have
᾿ΕπεὶοὖντριγώνουτοῦΑΒΓπαρὰμίαντῶνπλευρῶν been joined. And let DF have been drawn through D
τὴνΒΓἦκταιἡΖΔ,ἀνάλογονἄραἐστὶνὡςἡΓΔπρὸςτὴν parallel to it [Prop. 1.31].
ΔΑ,οὕτωςἡΒΖπρὸςτὴνΖΑ.διπλῆδὲἡΓΔτῆςΔΑ· Therefore, since FD has been drawn parallel to one
διπλῆἄρακαὶἡΒΖτῆςΖΑ·τριπλῆἄραἡΒΑτῆςΑΖ. of the sides, BC, of triangle ABC, then, proportionally,
ΤῆςἄραδοθείσηςεὐθείαςτῆςΑΒτὸἐπιταχθὲντρίτον as CD is to DA, so BF (is) to FA [Prop. 6.2]. And CD
μέροςἀφῄρηταιτὸΑΖ·ὅπερἔδειποιῆσαι.
(is) double DA. Thus, BF (is) also double FA. Thus,
BA (is) triple AF .
Thus, the prescribed third part, AF , has been cut off
from the given straight-line, AB. (Which is) the very
thing it was required to do.
. Proposition 10
Τὴνδοθεῖσανεὐθεῖανἄτμητοντῇδοθείσῃτετμημένῃ To cut a given uncut straight-line similarly to a given
ὁμοίωςτεμεῖν.
cut (straight-line).
A
F
B
166

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
Γ
C
Ε
E
∆
Θ
Κ
D
H
K
Α Ζ Η Β
A F G B
῎ΕστωἡμὲνδοθεῖσαεὐθεῖαἅτμητοςἡΑΒ,ἡδὲτε- Let AB be the given uncut straight-line, and AC a
τμημένηἡΑΓκατὰτὰΔ,Εσημεῖα,καὶκείσθωσανὥστε (straight-line) cut at points D and E, and let (AC) be
γωνίαντυχοῦσανπεριέχειν,καὶἐπεζεύχθωἡΓΒ,καὶδιὰ laid down so as to encompass a random angle (with AB).
τῶνΔ,ΕτῇΒΓπαράλληλοιἤχθωσαναἱΔΖ,ΕΗ,διὰδὲ And let CB have been joined. And let DF and EG have
τοῦΔτῇΑΒπαράλληλοςἤχθωἡΔΘΚ.
been drawn through (points) D and E (respectively),
ΠαραλληλόγραμμονἄραἐστὶνἑκάτεροντῶνΖΘ,ΘΒ· parallel to BC, and let DHK have been drawn through
ἴσηἄραἡμὲνΔΘτῇΖΗ,ἡδὲΘΚτῇΗΒ.καὶἐπεὶτριγώνου (point) D, parallel to AB [Prop. 1.31].
τοῦΔΚΓπαρὰμίαντῶνπλευρῶντὴνΚΓεὐθεῖαἦκταιἡ Thus, FH and HB are each parallelograms. Thus,
ΘΕ,ἀνάλογονἄραἐστὶνὡςἡΓΕπρὸςτὴνΕΔ,οὕτωςἡ DH (is) equal to FG, and HK to GB [Prop. 1.34]. And
ΚΘπρὸςτὴνΘΔ.ἴσηδὲἡμὲνΚΘτῇΒΗ,ἡδὲΘΔτῇ since the straight-line HE has been drawn parallel to one
ΗΖ.ἔστινἄραὡςἡΓΕπρὸςτὴνΕΔ,οὕτωςἡΒΗπρὸςτὴν of the sides, KC, of triangle DKC, thus, proportionally,
ΗΖ.πάλιν,ἐπεὶτριγώνουτοῦΑΗΕπαρὰμίαντῶνπλευρῶν as CE is to ED, so KH (is) to HD [Prop. 6.2]. And
τὴνΗΕἦκταιἡΖΔ,ἀνάλογονἄραἐστὶνὡςἡΕΔπρὸςτὴν KH (is) equal to BG, and HD to GF . Thus, as CE is
ΔΑ,οὕτωςἡΗΖπρὸςτὴνΖΑ.ἐδείχθηδὲκαὶὡςἡΓΕ to ED, so BG (is) to GF . Again, since FD has been
πρὸςτὴνΕΔ,οὕτωςἡΒΗπρὸςτὴνΗΖ·ἔστινἄραὡςμὲν drawn parallel to one of the sides, GE, of triangle AGE,
ἡΓΕπρὸςτὴνΕΔ,οὕτωςἡΒΗπρὸςτὴνΗΖ,ὡςδὲἡΕΔ thus, proportionally, as ED is to DA, so GF (is) to FA
πρὸςτὴνΔΑ,οὕτωςἡΗΖπρὸςτὴνΖΑ.
[Prop. 6.2]. And it was also shown that as CE (is) to
῾ΗἄραδοθεῖσαεὐθεῖαἄτμητοςἡΑΒτῇδοθείσῃεὐθείᾳ ED, so BG (is) to GF . Thus, as CE is to ED, so BG (is)
τετμημένῃτῇΑΓὁμοίωςτέτμηται·ὅπερἔδειποιῆσαι· to GF , and as ED (is) to DA, so GF (is) to FA.
Thus, the given uncut straight-line, AB, has been cut
similarly to the given cut straight-line, AC. (Which is)
the very thing it was required to do.
. Proposition 11
Δύοδοθεισῶνεὐθειῶντρίτηνἀνάλογονπροσευρεῖν. To find a third (straight-line) proportional to two
῎Εστωσαν αἱ δοθεῖσαι [δύο εὐθεῖαι] αἱ ΒΑ, ΑΓ καὶ given straight-lines.
κείσθωσανγωνίανπεριέχουσαιτυχοῦσαν.δεῖδὴτῶνΒΑ, Let BA and AC be the [two] given [straight-lines],
ΑΓτρίτηνἀνάλογονπροσευρεῖν. ἐκβεβλήσθωσανγὰρἐπὶ and let them be laid down encompassing a random angle.
τὰΔ,Εσημεῖα,καὶκείσθωτῇΑΓἴσηἡΒΔ,καὶἐπεζεύχθω So it is required to find a third (straight-line) proportional
ἡΒΓ,καὶδιὰτοῦΔπαράλληλοςαὐτῇἤχθωἡΔΕ. to BA and AC. For let (BA and AC) have been produced
᾿ΕπεὶοὖντριγώνουτοῦΑΔΕπαρὰμίαντῶνπλευρῶν to points D and E (respectively), and let BD be made
τὴνΔΕἦκταιἡΒΓ,ἀνάλογόνἐστινὡςἡΑΒπρὸςτὴν equal to AC [Prop. 1.3]. And let BC have been joined.
ΒΔ,οὕτωςἡΑΓπρὸςτὴνΓΕ.ἴσηδὲἡΒΔτῇΑΓ.ἔστιν And let DE have been drawn through (point) D parallel
ἄραὡςἡΑΒπρὸςτὴνΑΓ,οὕτωςἡΑΓπρὸςτὴνΓΕ. to it [Prop. 1.31].
Therefore, since BC has been drawn parallel to one
of the sides DE of triangle ADE, proportionally, as AB is
to BD, so AC (is) to CE [Prop. 6.2]. And BD (is) equal
167

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
Α
to AC. Thus, as AB is to AC, so AC (is) to CE.
A
Β
B
Γ
C
∆
D
Ε
ΔύοἄραδοθεισῶνεὐθειῶντῶνΑΒ,ΑΓτρίτηἀνάλογον Thus, a third (straight-line), CE, has been found
αὐταῖςπροσεύρηταιἡΓΕ·ὅπερἔδειποιῆσαι.
(which is) proportional to the two given straight-lines,
AB and AC. (Which is) the very thing it was required to
do.
. Proposition 12
Τριῶν δοθεισῶν εὐθειῶν τετάρτην ἀνάλογον προ- To find a fourth (straight-line) proportional to three
σευρεῖν.
given straight-lines.
Α
Β
Γ
Ε
A
B
C
E
E
Η
G
∆
Θ
Ζ
῎ΕστωσαναἱδοθεῖσαιτρεῖςεὐθεῖαιαἱΑ,Β,Γ·δεῖδὴ Let A, B, and C be the three given straight-lines. So
τῶνΑ,Β,Γτετράτηνἀνάλογονπροσευρεῖν.
it is required to find a fourth (straight-line) proportional
᾿ΕκκείσθωσανδύοεὐθεῖαιαἱΔΕ,ΔΖγωνίανπεριέχους- to A, B, and C.
αι[τυχοῦσαν]τὴνὑπὸΕΔΖ·καὶκείσθωτῇμὲνΑἴσηἡΔΗ, Let the two straight-lines DE and DF be set out enτῇδὲΒἴσηἡΗΕ,καὶἔτιτῇΓἴσηἡΔΘ·καὶἐπιζευχθείσης
compassing the [random] angle EDF . And let DG be
τῆςΗΘπαράλληλοςαὐτῇἤχθωδιὰτοῦΕἡΕΖ. made equal to A, and GE to B, and, further, DH to C
᾿ΕπεὶοὖντριγώνουτοῦΔΕΖπαρὰμίαντὴνΕΖἦκταιἡ [Prop. 1.3]. And GH being joined, let EF have been
ΗΘ,ἔστινἄραὡςἡΔΗπρὸςτὴνΗΕ,οὕτωςἡΔΘπρὸς drawn through (point) E parallel to it [Prop. 1.31].
τὴνΘΖ.ἴσηδὲἡμὲνΔΗτῇΑ,ἡδὲΗΕτῇΒ,ἡδὲΔΘτῇ Therefore, since GH has been drawn parallel to one
Γ·ἔστινἄραὡςἡΑπρὸςτὴνΒ,οὕτωςἡΓπρὸςτὴνΘΖ. of the sides EF of triangle DEF , thus as DG is to GE,
Τριῶν ἄρα δοθεισῶν εὐθειῶν τῶν Α, Β, Γ τετάρτη so DH (is) to HF [Prop. 6.2]. And DG (is) equal to A,
ἀνάλογονπροσεύρηταιἡΘΖ·ὅπερἔδειποιῆσαι. and GE to B, and DH to C. Thus, as A is to B, so C (is)
D
H
F
168

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
. Proposition
Δύοδοθεισῶνεὐθειῶνμέσηνἀνάλογονπροσευρεῖν.
∆
to HF .
Thus, a fourth (straight-line), HF , has been found
(which is) proportional to the three given straight-lines,
A, B, and C. (Which is) the very thing it was required to
do.
13
To find the (straight-line) in mean proportion to two
given straight-lines. †
D
Α
Β
Γ
῎ΕστωσαναἱδοθεῖσαιδύοεὐθεῖαιαἱΑΒ,ΒΓ·δεῖδὴτῶν Let AB and BC be the two given straight-lines. So it
ΑΒ,ΒΓμέσηνἀνάλογονπροσευρεῖν.
is required to find the (straight-line) in mean proportion
Κείσθωσαν ἐπ᾿ εὐθείας, καὶ γεγράφθω ἐπὶ τῆς ΑΓ to AB and BC.
ἡμικύκλιοντὸΑΔΓ,καὶἤχθωἀπὸτοῦΒσημείουτῇΑΓ Let (AB and BC) be laid down straight-on (with reεὐθείᾳπρὸςὀρθὰςἡΒΑ,καὶἐπεζεύχθωσαναἱΑΔ,ΔΓ.
spect to one another), and let the semi-circle ADC have
᾿ΕπεὶἐνἡμικυκλίῳγωνίαἐστὶνἡὑπὸΑΔΓ,ὀρθήἐστιν. been drawn on AC [Prop. 1.10]. And let BD have been
καὶἐπεὶἐνὀρθογωνίῳτριγώνῳτῷΑΔΓἀπὸτῆςὀρθῆς drawn from (point) B, at right-angles to AC [Prop. 1.11].
γωνίαςἐπὶτὴνβάσινκάθετοςἦκταιἡΔΒ,ἡΔΒἄρατῶν And let AD and DC have been joined.
τῆςβάσεωςτμημάτωντῶνΑΒ,ΒΓμέσηἀνάλογόνἐστιν. And since ADC is an angle in a semi-circle, it is a
ΔύοἄραδοθεισῶνεὐθειῶντῶνΑΒ,ΒΓμέσηἀνάλογον right-angle [Prop. 3.31]. And since, in the right-angled
προσεύρηταιἡΔΒ·ὅπερἔδειποιῆσαι.
triangle ADC, the (straight-line) DB has been drawn
from the right-angle perpendicular to the base, DB is
thus the mean proportional to the pieces of the base, AB
and BC [Prop. 6.8 corr.].
Thus, DB has been found (which is) in mean proportion
to the two given straight-lines, AB and BC. (Which
is) the very thing it was required to do.
† In other words, to find the geometric mean of two given straight-lines.
. Proposition 14
Τῶνἴσωντεκαὶἴσογωνίωνπαραλληλογράμμωνἀντι- In equal and equiangular parallelograms the sides
πεπόνθασιναἱπλευραὶαἱπερὶτὰςἴσαςγωνίας·καὶὧνἰσο- about the equal angles are reciprocally proportional.
γωνίων παραλληλογράμμωνἀντιπεπόνθασιναἱπλευραὶαἱ And those equiangular parallelograms in which the sides
περὶτὰςἴσαςγωνίας,ἴσαἐστὶνἐκεῖνα.
about the equal angles are reciprocally proportional are
῎ΕστωἴσατεκαὶἰσογώνιαπαραλληλόγραμματὰΑΒ, equal.
ΒΓἴσαςἔχοντατὰςπρὸςτῷΒγωνίας,καὶκείσθωσανἐπ᾿ Let AB and BC be equal and equiangular paralleloεὐθείαςαἱΔΒ,ΒΕ·ἐπ᾿εὐθείαςἄραεἰσὶκαὶαἱΖΒ,ΒΗ.
grams having the angles at B equal. And let DB and BE
λέγω,ὅτιτῶνΑΒ,ΒΓἀντιπεπόνθασιναἱπλευραὶαἱπερὶ be laid down straight-on (with respect to one another).
τὰςἴσαςγωνίας,τουτέστιν,ὅτιἐστὶνὡςἡΔΒπρὸςτὴν Thus, FB and BG are also straight-on (with respect to
ΒΕ,οὕτωςἡΗΒπρὸςτὴνΒΖ.
one another) [Prop. 1.14]. I say that the sides of AB and
A
B
C
169

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
Ε
Γ
BC about the equal angles are reciprocally proportional,
that is to say, that as DB is to BE, so GB (is) to BF .
E
C
Ζ
Β
Η
F
B
G
Α
∆
ΣυμπεπληρώσθωγὰρτὸΖΕπαραλληλόγραμμον. ἐπεὶ For let the parallelogram FE have been completed.
οὖνἴσονἐστὶτὸΑΒπαραλληλόγραμμοντῷΒΓπαραλλη- Therefore, since parallelogram AB is equal to paralleloλογράμμῳ,ἄλλοδέτιτὸΖΕ,ἔστινἄραὡςτὸΑΒπρὸςτὸ
gram BC, and FE (is) some other (parallelogram), thus
ΖΕ,οὕτωςτὸΒΓπρὸςτὸΖΕ.ἀλλ᾿ὡςμὲντὸΑΒπρὸς as (parallelogram) AB is to FE, so (parallelogram) BC
τὸΖΕ,οὕτωςἡΔΒπρὸςτὴνΒΕ,ὡςδὲτὸΒΓπρὸςτὸ (is) to FE [Prop. 5.7]. But, as (parallelogram) AB (is) to
ΖΕ,οὕτωςἡΗΒπρὸςτὴνΒΖ·καὶὡςἄραἡΔΒπρὸςτὴν FE, so DB (is) to BE, and as (parallelogram) BC (is) to
ΒΕ,οὕτωςἡΗΒπρὸςτὴνΒΖ.τῶνἄραΑΒ,ΒΓπαραλ- FE, so GB (is) to BF [Prop. 6.1]. Thus, also, as DB (is)
ληλογράμμωνἀντιπεπόνθασιναἱπλευραὶαἱπερὶτὰςἴσας to BE, so GB (is) to BF . Thus, in parallelograms AB
γωνίας.
and BC the sides about the equal angles are reciprocally
ἈλλὰδὴἔστωὡςἡΔΒπρὸςτὴνΒΕ,οὕτωςἡΗΒπρὸς proportional.
τὴνΒΖ·λέγω,ὅτιἴσονἐστὶτὸΑΒπαραλληλόγραμμοντῷ And so, let DB be to BE, as GB (is) to BF . I say that
ΒΓπαραλληλογράμμῳ.
parallelogram AB is equal to parallelogram BC.
᾿ΕπεὶγάρἐστινὡςἡΔΒπρὸςτὴνΒΕ,οὕτωςἡΗΒ For since as DB is to BE, so GB (is) to BF , but as
πρὸςτὴνΒΖ,ἀλλ᾿ὡςμὲνἡΔΒπρὸςτὴνΒΕ,οὕτωςτὸ DB (is) to BE, so parallelogram AB (is) to parallelo-
ΑΒπαραλληλόγραμμονπρὸςτὸΖΕπαραλληλόγραμμον,ὡς gram FE, and as GB (is) to BF , so parallelogram BC
δὲἡΗΒπρὸςτὴνΒΖ,οὕτωςτὸΒΓπαραλληλόγραμμον (is) to parallelogram FE [Prop. 6.1], thus, also, as (parπρὸςτὸΖΕπαραλληλόγραμμον,καὶὡςἄρατὸΑΒπρὸς
allelogram) AB (is) to FE, so (parallelogram) BC (is)
τὸΖΕ,οὕτωςτὸΒΓπρὸςτὸΖΕ·ἴσονἄραἐστὶτὸΑΒ to FE [Prop. 5.11]. Thus, parallelogram AB is equal to
παραλληλόγραμμοντῷΒΓπαραλληλογράμμῳ. parallelogram BC [Prop. 5.9].
Τῶν ἄρα ἴσων τε καὶ ἰσογωνίων παραλληλογράμμων Thus, in equal and equiangular parallelograms the
ἀντιπεπόνθασιναἱπλευραὶαἱπερὶτὰςἴσαςγωνίας·καὶὧν sides about the equal angles are reciprocally propor-
ἰσογωνίωνπαραλληλογράμμωνἀντιπεπόνθασιναἱπλευραὶ tional. And those equiangular parallelograms in which
αἱπερὶτὰςἴσαςγωνίας,ἴσαἐστὶνἐκεῖνα·ὅπερἔδειδεῖξαι. the sides about the equal angles are reciprocally proportional
are equal. (Which is) the very thing it was required
to show.
. Proposition 15
Τῶνἴσωνκαὶμίανμιᾷἴσηνἐχόντωνγωνίαντριγώνων In equal triangles also having one angle equal to one
ἀντιπεπόνθασιναἱπλευραὶαἱπερὶτὰςἴσαςγωνίας·καὶὧν (angle) the sides about the equal angles are reciprocally
μίανμιᾷἴσηνἐχόντωνγωνίαντριγώνωνἀντιπεπόνθασιναἱ proportional. And those triangles having one angle equal
πλευραὶαἱπερὶτὰςἴσαςγωνίας,ἴσαἐστὶνἐκεῖνα. to one angle for which the sides about the equal angles
῎ΕστωἴσατρίγωνατὰΑΒΓ,ΑΔΕμίανμιᾷἴσηνἔχοντα (are) reciprocally proportional are equal.
γωνίαντὴνὑπὸΒΑΓτῇὑπὸΔΑΕ·λέγω,ὅτιτῶνΑΒΓ, Let ABC and ADE be equal triangles having one an-
ΑΔΕτριγώνωνἀντιπεπόνθασιναἱπλευραὶαἱπερὶτὰςἴσας gle equal to one (angle), (namely) BAC (equal) to DAE.
γωνίας,τουτέστιν,ὅτιἐστὶνὡςἡΓΑπρὸςτὴνΑΔ,οὕτως I say that, in triangles ABC and ADE, the sides about the
A
D
170

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
ἡΕΑπρὸςτὴνΑΒ.
Β
Γ
equal angles are reciprocally proportional, that is to say,
that as CA is to AD, so EA (is) to AB.
B
C
Α
A
∆
Ε
Κείσθωγὰρὥστεἐπ᾿εὐθείαςεἶναιτὴνΓΑτῇΑΔ·ἐπ᾿ For let CA be laid down so as to be straight-on (with
εὐθείαςἄραἐστὶκαὶἡΕΑτῇΑΒ.καὶἐπεζεύχθωἡΒΔ. respect) to AD. Thus, EA is also straight-on (with re-
᾿ΕπεὶοὖνἴσονἐστὶτὸΑΒΓτρίγωνοντῷΑΔΕτριγώνῳ, spect) to AB [Prop. 1.14]. And let BD have been joined.
ἄλλοδέτιτὸΒΑΔ,ἔστινἄραὡςτὸΓΑΒτρίγωνονπρὸς Therefore, since triangle ABC is equal to triangle
τὸΒΑΔτρίγωνον,οὕτωςτὸΕΑΔτρίγωνονπρὸςτὸΒΑΔ ADE, and BAD (is) some other (triangle), thus as triτρίγωνον.
ἀλλ᾿ὡςμὲντὸΓΑΒπρὸςτὸΒΑΔ,οὕτωςἡ angle CAB is to triangle BAD, so triangle EAD (is) to
ΓΑπρὸςτὴνΑΔ,ὡςδὲτὸΕΑΔπρὸςτὸΒΑΔ,οὕτωςἡ triangle BAD [Prop. 5.7]. But, as (triangle) CAB (is)
ΕΑπρὸςτὴνΑΒ.καὶὡςἄραἡΓΑπρὸςτὴνΑΔ,οὕτως to BAD, so CA (is) to AD, and as (triangle) EAD (is)
ἡΕΑπρὸςτὴνΑΒ.τῶνΑΒΓ,ΑΔΕἄρατριγώνωνἀντι- to BAD, so EA (is) to AB [Prop. 6.1]. And thus, as CA
πεπόνθασιναἱπλευραὶαἱπερὶτὰςἴσαςγωνίας.
(is) to AD, so EA (is) to AB. Thus, in triangles ABC and
ἈλλὰδὴἀντιπεπονθέτωσαναἱπλευραὶτῶνΑΒΓ,ΑΔΕ ADE the sides about the equal angles (are) reciprocally
τριγώνων,καὶἔστωὡςἡΓΑπρὸςτὴνΑΔ,οὕτωςἡΕΑ proportional.
πρὸςτὴνΑΒ·λέγω,ὅτιἴσονἐστὶτὸΑΒΓτρίγωνοντῷ And so, let the sides of triangles ABC and ADE be
ΑΔΕτριγώνῳ.
reciprocally proportional, and (thus) let CA be to AD,
᾿ΕπιζευχθείσηςγὰρπάλιντῆςΒΔ,ἐπείἐστινὡςἡΓΑ as EA (is) to AB. I say that triangle ABC is equal to
πρὸς τὴν ΑΔ, οὕτως ἡ ΕΑ πρὸς τὴν ΑΒ, ἀλλ᾿ ὡς μὲν triangle ADE.
ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ For, BD again being joined, since as CA is to AD, so
ΒΑΔτρίγωνον,ὡςδὲἡΕΑπρὸςτὴνΑΒ,οὕτωςτὸΕΑΔ EA (is) to AB, but as CA (is) to AD, so triangle ABC
τρίγωνονπρὸςτὸΒΑΔτρίγωνον,ὡςἄρατὸΑΒΓτρίγωνον (is) to triangle BAD, and as EA (is) to AB, so triangle
πρὸςτὸΒΑΔτρίγωνον, οὕτωςτὸΕΑΔτρίγωνονπρὸς EAD (is) to triangle BAD [Prop. 6.1], thus as triangle
τὸΒΑΔτρίγωνον. ἑκάτερονἄρατῶνΑΒΓ,ΕΑΔπρὸς ABC (is) to triangle BAD, so triangle EAD (is) to triτὸΒΑΔτὸναὐτὸνἔχειλόγον.
ἴσωνἄραἐστὶτὸΑΒΓ angle BAD. Thus, (triangles) ABC and EAD each have
[τρίγωνον]τῷΕΑΔτριγώνῳ.
the same ratio to BAD. Thus, [triangle] ABC is equal to
Τῶν ἄρα ἴσων καὶ μίαν μιᾷ ἴσην ἐχόντων γωνίαν triangle EAD [Prop. 5.9].
τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας Thus, in equal triangles also having one angle equal to
γωνίας· καὶὧςμίανμιᾷἴσηνἐχόντωνγωνίαντριγώνων one (angle) the sides about the equal angles (are) recip-
ἀντιπεπόνθασιναἱπλευραὶαἱπερὶτὰςἴσαςγωνίας,ἐκεῖνα rocally proportional. And those triangles having one an-
ἴσαἐστὶν·ὅπερἔδειδεῖξαι.
gle equal to one angle for which the sides about the equal
angles (are) reciprocally proportional are equal. (Which
is) the very thing it was required to show.
¦. Proposition 16
᾿Εὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν If four straight-lines are proportional then the rect-
ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν angle contained by the (two) outermost is equal to the
μέσων περιεχομένῳ ὀρθογωνίῳ· κἂν τὸ ὑπὸτῶν ἄκρων rectangle contained by the middle (two). And if the rect-
D
E
171

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
περιεχόμενονὀρθογώνιονἴσονᾖτῷὑπὸτῶνμέσωνπεριε- angle contained by the (two) outermost is equal to the
χομένῳὀρθογωνίῳ,αἱτέσσαρεςεὐθεῖαιἀνάλογονἔσονται. rectangle contained by the middle (two) then the four
straight-lines will be proportional.
Θ
H
Η
G
Α
Β
Γ
∆
A
B
C
D
Ε
Ζ
E
F
῎ΕστωσαντέσσαρεςεὐθεῖαιἀνάλογοναἱΑΒ,ΓΔ,Ε,Ζ, Let AB, CD, E, and F be four proportional straight-
ὡςἡΑΒπρὸςτὴνΓΔ,οὕτωςἡΕπρὸςτὴνΖ·λέγω,ὅτι lines, (such that) as AB (is) to CD, so E (is) to F . I say
τὸὑπὸτῶνΑΒ,Ζπεριεχόμενονὀρθογώνιονἴσονἐστὶτῷ that the rectangle contained by AB and F is equal to the
ὑπὸτῶνΓΔ,Επεριεχομένῳὀρθογωνίῳ. rectangle contained by CD and E.
῎Ηχθωσαν[γὰρ]ἀπὸτῶνΑ,ΓσημείωνταῖςΑΒ,ΓΔ [For] let AG and CH have been drawn from points
εὐθείαιςπρὸςὀρθὰςαἱΑΗ,ΓΘ,καὶκείσθωτῇμὲνΖἴση A and C at right-angles to the straight-lines AB and CD
ἡΑΗ,τῇδὲΕἴσηἡΓΘ.καὶσυμπεπληρώσθωτὰΒΗ,ΔΘ (respectively) [Prop. 1.11]. And let AG be made equal to
παραλληλόγραμμα.
F , and CH to E [Prop. 1.3]. And let the parallelograms
ΚαὶἐπείἐστινὡςἡΑΒπρὸςτὴνΓΔ,οὕτωςἡΕπρὸς BG and DH have been completed.
τὴνΖ,ἴσηδὲἡμὲνΕτῇΓΘ,ἡδὲΖτῇΑΗ,ἔστινἄραὡς And since as AB is to CD, so E (is) to F , and E
ἡΑΒπρὸςτὴνΓΔ,οὕτωςἡΓΘπρὸςτὴνΑΗ.τῶνΒΗ, (is) equal CH, and F to AG, thus as AB is to CD, so
ΔΘἄραπαραλληλογράμμωνἀντιπεπόνθασιναἱπλευραὶαἱ CH (is) to AG. Thus, in the parallelograms BG and DH
περὶτὰςἴσαςγωνίας.ὧνδὲἰσογωνίωνπαραλληλογράμμων the sides about the equal angles are reciprocally propor-
ἀντιπεπόνθασιναἱπλευραίαἱπερὶτὰςἴσαςγωνάις,ἴσαἐστὶν tional. And those equiangular parallelograms in which
ἐκεῖνα· ἴσονἄραἐστὶτὸΒΗπαραλληλόγραμμοντῷΔΘ the sides about the equal angles are reciprocally proporπαραλληλογράμμῳ.καίἐστιτὸμὲνΒΗτὸὑπὸτῶνΑΒ,Ζ·
tional are equal [Prop. 6.14]. Thus, parallelogram BG
ἴσηγὰρἡΑΗτῇΖ·τὸδὲΔΘτὸὑπὸτῶνΓΔ,Ε·ἴσηγὰρἡ is equal to parallelogram DH. And BG is the (rectangle
ΕτῇΓΘ·τὸἄραὑπὸτῶνΑΒ,Ζπεριεχόμενονὀρθογώνιον contained) by AB and F . For AG (is) equal to F . And
ἴσονἐστὶτῷὑπὸτῶνΓΔ,Επεριεχομένῳὀρθογώνιῳ. DH (is) the (rectangle contained) by CD and E. For E
ἈλλὰδὴτὸὑπὸτῶνΑΒ,Ζπεριεχόμενονὀρθογώνιον (is) equal to CH. Thus, the rectangle contained by AB
ἴσον ἔστω τῷ ὑπὸτῶν ΓΔ, Ε περιεχομένῳ ὀρθογωνίῳ. and F is equal to the rectangle contained by CD and E.
λέγω,ὅτιαἱτέσσαρεςεὐθεῖαιἀνάλογονἔσονται,ὡςἡΑΒ And so, let the rectangle contained by AB and F be
πρὸςτὴνΓΔ,οὕτωςἡΕπρὸςτὴνΖ.
equal to the rectangle contained by CD and E. I say that
Τῶνγὰραὐτῶνκατασκευασθέντων,ἐπεὶτὸὑπὸτῶν the four straight-lines will be proportional, (so that) as
ΑΒ,ΖἴσονἐστὶτῷὑπὸτῶνΓΔ,Ε,καίἐστιτὸμὲνὑπὸ AB (is) to CD, so E (is) to F .
τῶνΑΒ,ΖτὸΒΗ·ἴσηγάρἐστινἡΑΗτῇΖ·τὸδὲὑπὸτῶν For, with the same construction, since the (rectangle
ΓΔ,ΕτὸΔΘ·ἴσηγὰρἡΓΘτῇΕ·τὸἄραΒΗἴσονἐστὶ contained) by AB and F is equal to the (rectangle conτῷΔΘ.καίἐστινἰσογώνια.
τῶνδὲἴσωνκαὶἰσογωνίων tained) by CD and E. And BG is the (rectangle conπαραλληλογράμμωνἀντιπεπόνθασιναἱπλευραὶαἱπερὶτὰς
tained) by AB and F . For AG is equal to F . And DH
ἴσαςγωνίας.ἔστινἄραὡςἡΑΒπρὸςτὴνΓΔ,οὕτωςἡΓΘ (is) the (rectangle contained) by CD and E. For CH
πρὸςτὴνΑΗ.ἴσηδὲἡμὲνΓΘτῇΕ,ἡδὲΑΗτῇΖ·ἔστιν (is) equal to E. BG is thus equal to DH. And they
ἄραὡςἡΑΒπρὸςτὴνΓΔ,οὕτωςἡΕπρὸςτὴνΖ. are equiangular. And in equal and equiangular parallel-
᾿Εὰνἄρατέσσαρεςεὐθεῖαιἀνάλογονὦσιν,τὸὑπὸτῶν ograms the sides about the equal angles are reciprocally
ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν proportional [Prop. 6.14]. Thus, as AB is to CD, so CH
μέσων περιεχομένῳ ὀρθογωνίῳ· κἂν τὸ ὑπὸτῶν ἄκρων (is) to AG. And CH (is) equal to E, and AG to F . Thus,
περιεχόμενονὀρθογώνιονἴσονᾖτῷὑπὸτῶνμέσωνπεριε- as AB is to CD, so E (is) to F .
χομένῳὀρθογωνίῳ,αἱτέσσαρεςεὐθεῖαιἀνάλογονἔσονται· Thus, if four straight-lines are proportional then the
ὅπερἔδειδεῖξαι.
rectangle contained by the (two) outermost is equal to
the rectangle contained by the middle (two). And if the
rectangle contained by the (two) outermost is equal to
172

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
the rectangle contained by the middle (two) then the four
straight-lines will be proportional. (Which is) the very
thing it was required to show.
Þ. Proposition 17
᾿Εὰντρεῖςεὐθεῖαιἀνάλογονὦσιν,τὸὑπὸτῶνἄκρων If three straight-lines are proportional then the rectπεριεχόμενονὀρθογώνιονἴσονἐστὶτῷἀπὸτῆςμέσηςτε-
angle contained by the (two) outermost is equal to the
τραγώνῳ·κἂντὸὑπὸτῶνἄκρωνπεριεχόμενονὀρθογώνιον square on the middle (one). And if the rectangle con-
ἴσον ᾖ τῷ ἀπὸ τῆς μέσης τετραγώνῳ, αἱ τρεῖς εὐθεῖαι tained by the (two) outermost is equal to the square on
ἀνάλογονἔσονται.
the middle (one) then the three straight-lines will be proportional.
Α
Β
Γ
∆
῎ΕστωσαντρεῖςεὐθεῖαιἀνάλογοναἱΑ,Β,Γ,ὡςἡΑ Let A, B and C be three proportional straight-lines,
πρὸςτὴνΒ,οὕτωςἡΒπρὸςτὴνΓ·λέγω,ὅτιτὸὑπὸτῶν (such that) as A (is) to B, so B (is) to C. I say that the
Α,ΓπεριεχόμενονὀρθογώνιονἴσονἐστὶτῷἀπὸτῆςΒ rectangle contained by A and C is equal to the square on
τετραγώνῳ. B.
ΚείσθωτῇΒἴσηἡΔ. Let D be made equal to B [Prop. 1.3].
ΚαὶἐπείἐστινὡςἡΑπρὸςτὴνΒ,οὕτωςἡΒπρὸςτὴν And since as A is to B, so B (is) to C, and B (is)
Γ,ἴσηδὲἡΒτῇΔ,ἔστινἄραὡςἡΑπρὸςτὴνΒ,ἡΔπρὸς equal to D, thus as A is to B, (so) D (is) to C. And if
τὴνΓ.ἐὰνδὲτέσσαρεςεὐθεῖαιἀνάλογονὦσιν,τὸὑπὸτῶν four straight-lines are proportional then the [rectangle]
ἄκρωνπεριεχόμενον[ὀρθογώνιον]ἴσονἐστὶτῷὑπὸτῶν contained by the (two) outermost is equal to the rectanμέσωνπεριεχομένῳὀρθογωνίῳ.τὸἄραὑπὸτῶνΑ,Γἴσον
gle contained by the middle (two) [Prop. 6.16]. Thus,
ἐστὶτῷὑπὸτῶνΒ,Δ.ἀλλὰτὸὑπὸτῶνΒ,ΔτὸἀπὸτῆςΒ the (rectangle contained) by A and C is equal to the
ἐστιν·ἴσηγὰρἡΒτῇΔ·τὸἄραὑπὸτῶνΑ,Γπεριεχόμενον (rectangle contained) by B and D. But, the (rectangle
ὀρθογώνιονἴσονἐστὶτῷἀπὸτῆςΒτετραγώνῳ. contained) by B and D is the (square) on B. For B (is)
ἈλλὰδὴτὸὑπὸτῶνΑ,ΓἴσονἔστωτῷἀπὸτῆςΒ· equal to D. Thus, the rectangle contained by A and C is
λέγω,ὅτιἐστὶνὡςἡΑπρὸςτὴνΒ,οὕτωςἡΒπρὸςτὴνΓ. equal to the square on B.
Τῶνγὰραὐτῶνκατασκευασθέντων,ἐπεὶτὸὑπὸτῶνΑ, And so, let the (rectangle contained) by A and C be
ΓἴσονἐστὶτῷἀπὸτῆςΒ,ἀλλὰτὸἀπὸτῆςΒτὸὑπὸτῶν equal to the (square) on B. I say that as A is to B, so B
Β,Δἐστιν·ἴσηγὰρἡΒτῇΔ·τὸἄραὑπὸτῶνΑ,Γἴσον (is) to C.
ἐστὶτῷὑπὸτῶνΒ,Δ.ἐὰνδὲτὸὑπὸτῶνἄκρωνἴσονᾖτῷ For, with the same construction, since the (rectangle
ὑπὸτῶνμέσων,αἱτέσσαρεςεὐθεῖαιἀνάλογόνεἰσιν.ἔστιν contained) by A and C is equal to the (square) on B.
ἄραὡςἡΑπρὸςτὴνΒ,οὕτωςἡΔπρὸςτὴνΓ.ἴσηδὲἡΒ But, the (square) on B is the (rectangle contained) by B
τῇΔ·ὡςἄραἡΑπρὸςτὴνΒ,οὕτωςἡΒπρὸςτὴνΓ. and D. For B (is) equal to D. The (rectangle contained)
᾿Εὰνἄρατρεῖςεὐθεῖαιἀνάλογονὦσιν,τὸὑπὸτῶνἄκρων by A and C is thus equal to the (rectangle contained) by
περιεχόμενονὀρθογώνιονἴσονἐστὶτῷἀπὸτῆςμέσηςτε- B and D. And if the (rectangle contained) by the (two)
τραγώνῳ·κἂντὸὑπὸτῶνἄκρωνπεριεχόμενονὀρθογώνιον outermost is equal to the (rectangle contained) by the
ἴσον ᾖ τῷ ἀπὸ τῆς μέσης τετραγώνῳ, αἱ τρεῖς εὐθεῖαι middle (two) then the four straight-lines are proportional
ἀνάλογονἔσονται·ὅπερἔδειδεῖξαι.
[Prop. 6.16]. Thus, as A is to B, so D (is) to C. And B
(is) equal to D. Thus, as A (is) to B, so B (is) to C.
Thus, if three straight-lines are proportional then the
rectangle contained by the (two) outermost is equal to
the square on the middle (one). And if the rectangle contained
by the (two) outermost is equal to the square on
the middle (one) then the three straight-lines will be proportional.
(Which is) the very thing it was required to
A
B
C
D
173

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
show.
. Proposition 18
Ἀπὸ τῆς δοθείσης εὐθείας τῷ δοθέντι εὐθυγράμμῳ To describe a rectilinear figure similar, and simi-
ὅμοιόντεκαὶὁμοίωςκείμενονεὐθύγραμμονἀναγράψαι. larly laid down, to a given rectilinear figure on a given
straight-line.
Ζ
Ε
Η
Θ
F
E
G
H
Γ
∆
Α
Β
῎Εστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν Let AB be the given straight-line, and CE the given
εὐθύγραμμοντὸΓΕ·δεῖδὴἀπὸτῆςΑΒεὐθείαςτῷΓΕ rectilinear figure. So it is required to describe a rectilinear
εὐθυγράμμῳὅμοιόντεκαὶὁμοίωςκείμενονεὐθύγραμμον figure similar, and similarly laid down, to the rectilinear
ἀναγράψαι.
figure CE on the straight-line AB.
᾿ΕπεζεύχθωἡΔΖ,καὶσυνεστάτωπρὸςτῇΑΒεὐθείᾳ Let DF have been joined, and let GAB, equal to the
καὶτοῖςπρὸςαὐτῇσημείοιςτοῖςΑ,ΒτῇμὲνπρὸςτῷΓ angle at C, and ABG, equal to (angle) CDF , have been
γωνίᾳἴσηἡὑπὸΗΑΒ,τῇδὲὑπὸΓΔΖἴσηἡὑπὸΑΒΗ.λοιπὴ constructed on the straight-line AB at the points A and
ἄραἡὑπὸΓΖΔτῇὑπὸΑΗΒἐστινἴση·ἰσογώνιονἄραἐστὶ B on it (respectively) [Prop. 1.23]. Thus, the remainτὸΖΓΔτρίγωνοντῷΗΑΒτριγώνῳ.
ἀνάλογονἄραἐστὶν ing (angle) CFD is equal to AGB [Prop. 1.32]. Thus,
ὡςἡΖΔπρὸςτὴνΗΒ,οὕτωςἡΖΓπρὸςτὴνΗΑ,καὶἡ triangle FCD is equiangular to triangle GAB. Thus,
ΓΔπρὸςτὴνΑΒ.πάλινσυνεστάτωπρὸςτῇΒΗεὐθείᾳκαὶ proportionally, as FD is to GB, so FC (is) to GA, and
τοῖςπρὸςαὐτῇσημείοιςτοῖςΒ,ΗτῇμὲνὑπὸΔΖΕγωνίᾳ CD to AB [Prop. 6.4]. Again, let BGH, equal to an-
ἴσηἡὑπὸΒΗΘ,τῇδὲὑπὸΖΔΕἴσηἡὑπὸΗΒΘ.λοιπὴἄρα gle DFE, and GBH equal to (angle) FDE, have been
ἡπρὸςτῷΕλοιπῇτῇπρὸςτῷΘἐστινἴση·ἰσογώνιονἄρα constructed on the straight-line BG at the points G and
ἐστὶτὸΖΔΕτρίγωνοντῷΗΘΒτριγώνῳ·ἀνάλογονἄρα B on it (respectively) [Prop. 1.23]. Thus, the remain-
ἐστὶνὡςἡΖΔπρὸςτὴνΗΒ,οὕτωςἡΖΕπρὸςτὴνΗΘκαὶ ing (angle) at E is equal to the remaining (angle) at H
ἡΕΔπρὸςτὴνΘΒ.ἐδείχθηδὲκαὶὡςἡΖΔπρὸςτὴνΗΒ, [Prop. 1.32]. Thus, triangle FDE is equiangular to triοὕτωςἡΖΓπρὸςτὴνΗΑκαὶἡΓΔπρὸςτὴνΑΒ·καὶὡς
angle GHB. Thus, proportionally, as FD is to GB, so
ἄραἡΖΓπρὸςτὴνΑΗ,οὕτωςἥτεΓΔπρὸςτὴνΑΒκαὶἡ FE (is) to GH, and ED to HB [Prop. 6.4]. And it was
ΖΕπρὸςτὴνΗΘκαὶἔτιἡΕΑπρὸςτὴνΘΒ.καὶἐπεὶἴση also shown (that) as FD (is) to GB, so FC (is) to GA,
ἐστὶνἡμὲνὑπὸΓΖΔγωνίατῇὑπὸΑΗΒ,ἡδὲὑπὸΔΖΕτῇ and CD to AB. Thus, also, as FC (is) to AG, so CD (is)
ὑπὸΒΗΘ,ὅληἄραἡὑπὸΓΖΕὅλῃτῇὑπὸΑΗΘἐστινἴση. to AB, and FE to GH, and, further, ED to HB. And
διὰτὰαὐτὰδὴκαὶἡὑπὸΓΔΕτῇὑπὸΑΒΘἐστινἴση.ἔστι since angle CFD is equal to AGB, and DFE to BGH,
δὲκαὶἡμὲνπρὸςτῷΓτῇπρὸςτῷΑἴση,ἡδὲπρὸςτῷΕ thus the whole (angle) CFE is equal to the whole (anτῇπρὸςτῷΘ.ἰσογώνιονἄραἐστὶτὸΑΘτῷΓΕ·καὶτὰς
gle) AGH. So, for the same (reasons), (angle) CDE is
περὶτὰςἴσαςγωνίαςαὐτῶνπλευρὰςἀνάλογονἔχει·ὅμοιον also equal to ABH. And the (angle) at C is also equal
ἄραἐστὶτὸΑΘεὐθύγραμμοντῷΓΕεὐθυγράμμῳ. to the (angle) at A, and the (angle) at E to the (angle)
Ἀπὸ τῆς δοθείσης ἄρα εὐθείας τῆς ΑΒ τῷ δοθέντι at H. Thus, (figure) AH is equiangular to CE. And (the
εὐθυγράμμῳτῷΓΕὅμοιόντεκαὶὁμοίωςκείμενονεὐθύγρα- two figures) have the sides about their equal angles proμμονἀναγέγραπταιτὸΑΘ·ὅπερἔδειποιῆσαι.
portional. Thus, the rectilinear figure AH is similar to the
rectilinear figure CE [Def. 6.1].
Thus, the rectilinear figure AH, similar, and similarly
laid down, to the given rectilinear figure CE has been
constructed on the given straight-line AB. (Which is) the
C
D
A
B
174

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
very thing it was required to do.
. Proposition 19
Τὰὅμοιατρίγωναπρὸςἄλληλαἐνδιπλασίονιλόγῳἐστὶ Similar triangles are to one another in the squared †
τῶνὁμολόγωνπλευρῶν.
ratio of (their) corresponding sides.
Α
A
∆
D
Β Η Γ
Ε
Ζ
῎ΕστωὅμοιατρίγωνατὰΑΒΓ,ΔΕΖἴσηνἔχοντατὴν Let ABC and DEF be similar triangles having the
πρὸςτῷΒγωνίαντῇπρὸςτῷΕ,ὡςδὲτὴνΑΒπρὸςτὴνΒΓ, angle at B equal to the (angle) at E, and AB to BC, as
οὕτωςτὴνΔΕπρὸςτὴνΕΖ,ὥστεὁμόλογονεἶναιτὴνΒΓ DE (is) to EF , such that BC corresponds to EF . I say
τῇΕΖ·λέγω,ὅτιτὸΑΒΓτρίγωνονπρὸςτὸΔΕΖτρίγωνον that triangle ABC has a squared ratio to triangle DEF
διπλασίοναλόγονἔχειἤπερἡΒΓπρὸςτὴνΕΖ. with respect to (that side) BC (has) to EF .
ΕἰλήφθωγὰρτῶνΒΓ,ΕΖτρίτηἀνάλογονἡΒΗ,ὥστε For let a third (straight-line), BG, have been taken
εἶναιὡςτὴνΒΓπρὸςτὴνΕΖ,οὕτωςτὴνΕΖπρὸςτὴνΒΗ· (which is) proportional to BC and EF , so that as BC
καὶἐπεζεύχθωἡΑΗ.
(is) to EF , so EF (is) to BG [Prop. 6.11]. And let AG
᾿ΕπεὶοὖνἐστινὡςἡΑΒπρὸςτὴνΒΓ,οὕτωςἡΔΕπρὸς have been joined.
τὴνΕΖ,ἐναλλὰξἄραἐστὶνὡςἡΑΒπρὸςτὴνΔΕ,οὕτωςἡ Therefore, since as AB is to BC, so DE (is) to EF ,
ΒΓπρὸςτὴνΕΖ.ἀλλ᾿ὡςἡΒΓπρὸςΕΖ,οὕτωςἐστινἡΕΖ thus, alternately, as AB is to DE, so BC (is) to EF
πρὸςΒΗ.καὶὡςἄραἡΑΒπρὸςΔΕ,οὕτωςἡΕΖπρὸςΒΗ· [Prop. 5.16]. But, as BC (is) to EF , so EF is to BG.
τῶνΑΒΗ,ΔΕΖἄρατριγώνωνἀντιπεπόνθασιναἱπλευραὶαἱ And, thus, as AB (is) to DE, so EF (is) to BG. Thus,
περὶτὰςἴσαςγωνάις.ὧνδὲμίανμιᾷἴσηνἐχόντωνγωνίαν for triangles ABG and DEF , the sides about the equal
τριγώνωνἀντιπεπόνθασιναἱπλευραὶαἱπερὶτὰςἴσαςγωνάις, angles are reciprocally proportional. And those triangles
ἴσαἐστὶνἐκεῖνα.ἴσονἄραἐστὶτὸΑΒΗτρίγωνοντῷΔΕΖ having one (angle) equal to one (angle) for which the
τριγώνῳ. καὶἐπείἐστινὡςἡΒΓπρὸςτὴνΕΖ,οὕτωςἡ sides about the equal angles are reciprocally proportional
ΕΖπρὸςτὴνΒΗ,ἐὰνδὲτρεῖςεὐθεῖαιἀνάλογονὦσιν,ἡ are equal [Prop. 6.15]. Thus, triangle ABG is equal to
πρώτηπρὸςτὴντρίτηνδιπλασίοναλόγονἔχειἤπερπρὸς triangle DEF . And since as BC (is) to EF , so EF (is)
τὴνδευτέραν,ἡΒΓἄραπρὸςτὴνΒΗδιπλασίοναλόγον to BG, and if three straight-lines are proportional then
ἔχειἤπερἡΓΒπρὸςτὴνΕΖ.ὡςδὲἡΓΒπρὸςτὴνΒΗ, the first has a squared ratio to the third with respect to
οὕτωςτὸΑΒΓτρίγωνονπρὸςτὸΑΒΗτρίγωνον·καὶτὸ the second [Def. 5.9], BC thus has a squared ratio to BG
ΑΒΓἄρατρίγωνονπρὸςτὸΑΒΗδιπλασίοναλόγονἔχει with respect to (that) CB (has) to EF . And as CB (is)
ἤπερἡΒΓπρὸςτὴνΕΖ.ἴσονδὲτὸΑΒΗτρίγωνοντῷ to BG, so triangle ABC (is) to triangle ABG [Prop. 6.1].
ΔΕΖτριγώνῳ. καὶτὸΑΒΓἄρατρίγωνονπρὸςτὸΔΕΖ Thus, triangle ABC also has a squared ratio to (triangle)
τρίγωνονδιπλασίοναλόγονἔχειἤπερἡΒΓπρὸςτὴνΕΖ. ABG with respect to (that side) BC (has) to EF . And
Τὰἄραὅμοιατρίγωναπρὸςἄλληλαἐνδιπλασίονιλόγῳ triangle ABG (is) equal to triangle DEF . Thus, trian-
ἐστὶτῶνὁμολόγωνπλευρῶν.[ὅπερἔδειδεῖξαι.]
gle ABC also has a squared ratio to triangle DEF with
respect to (that side) BC (has) to EF .
Thus, similar triangles are to one another in the
squared ratio of (their) corresponding sides. [(Which
is) the very thing it was required to show].
ÈÖ×Ñ.
Corollary
᾿Εκδὴτούτουφανερόν,ὅτι,ἐὰντρεῖςεὐθεῖαιἀνάλογον So it is clear, from this, that if three straight-lines are
ὦσιν, ἔστινὡςἡπρώτηπρὸςτὴντρίτην, οὕτωςτὸἀπὸ proportional, then as the first is to the third, so the figure
B
G
C
E
F
175

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
τῆςπρώτηςεἶδοςπρὸςτὸἀπὸτῆςδευτέραςτὸὅμοιονκαὶ (described) on the first (is) to the similar, and similarly
ὁμοίωςἀναγραφόμενον.ὅπερἔδειδεῖξαι.
described, (figure) on the second. (Which is) the very
thing it was required to show.
† Literally, “double”.
.
Proposition 20
Τὰὅμοιαπολύγωναεἴςτεὅμοιατρίγωναδιαιρεῖταικαὶ Similar polygons can be divided into equal numbers
εἰςἴσατὸπλῆθοςκαὶὁμόλογατοῖςὅλοις,καὶτὸπολύγωνον of similar triangles corresponding (in proportion) to the
πρὸςτὸπολύγωνονδιπλασίοναλόγονἔχειἤπερἡὁμόλογος wholes, and one polygon has to the (other) polygon a
πλευρὰπρὸςτὴνὁμόλογονπλευράν.
squared ratio with respect to (that) a corresponding side
(has) to a corresponding side.
Α
A
Ζ
F
Β
Μ
Ε
Η
Ν
Λ
B
M
E
G
N
L
Γ
∆
Θ
Κ
῎ΕστωὅμοιαπολύγωνατὰΑΒΓΔΕ,ΖΗΘΚΛ,ὁμόλογος Let ABCDE and FGHKL be similar polygons, and
δὲ ἔστω ἡ ΑΒ τῇ ΖΗ· λέγω, ὅτι τὰ ΑΒΓΔΕ, ΖΗΘΚΛ let AB correspond to FG. I say that polygons ABCDE
πολύγωναεἴςτεὅμοιατρίγωναδιαιρεῖταικαὶεἰςἴσατὸ and FGHKL can be divided into equal numbers of simiπλῆθοςκαὶὁμόλογατοῖςὅλοις,καὶτὸΑΒΓΔΕπολύγωνον
lar triangles corresponding (in proportion) to the wholes,
πρὸςτὸΖΗΘΚΛπολύγωνονδιπλασίοναλόγονἔχειἤπερἡ and (that) polygon ABCDE has a squared ratio to poly-
ΑΒπρὸςτὴνΖΗ.
gon FGHKL with respect to that AB (has) to FG.
᾿ΕπεζεύχθωσαναἱΒΕ,ΕΓ,ΗΛ,ΛΘ.
Let BE, EC, GL, and LH have been joined.
Καὶ ἐπεὶ ὅμοιόν ἐστι τὸ ΑΒΓΔΕ πολύγωνον τῷ And since polygon ABCDE is similar to polygon
ΖΗΘΚΛπολυγώνῳ,ἴσηἐστὶνἡὑπὸΒΑΕγωνίατῇὑπὸ FGHKL, angle BAE is equal to angle GFL, and as BA
ΗΖΛ.καίἐστινὡςἡΒΑπρὸςΑΕ,οὕτωςἡΗΖπρὸςΖΛ. is to AE, so GF (is) to FL [Def. 6.1]. Therefore, since
ἐπεὶοὖνδύοτρίγωνάἐστιτὰΑΒΕ,ΖΗΛμίανγωνίανμιᾷ ABE and FGL are two triangles having one angle equal
γωνίᾳἴσηνἔχοντα,περὶδὲτὰςἴσαςγωνίαςτὰςπλευρὰς to one angle and the sides about the equal angles propor-
ἀνάλογον,ἰσογώνιονἄραἐστὶτὸΑΒΕτρίγωνοντῷΖΗΛ tional, triangle ABE is thus equiangular to triangle FGL
τριγώνῳ·ὥστεκαὶὅμοιον·ἴσηἄραἐστὶνἡὑπὸΑΒΕγωνία [Prop. 6.6]. Hence, (they are) also similar [Prop. 6.4,
τῇ ὑπὸ ΖΗΛ. ἔστι δὲ καὶ ὅλη ἡ ὑπὸ ΑΒΓ ὅλῃ τῇ ὑπὸ Def. 6.1]. Thus, angle ABE is equal to (angle) FGL.
ΖΗΘἴσηδιὰτὴνὁμοιότητατῶνπολυγώνων·λοιπὴἄραἡ And the whole (angle) ABC is equal to the whole (angle)
ὑπὸΕΒΓγωνίατῇὑπὸΛΗΘἐστινἴση. καὶἐπεὶδιὰτὴν FGH, on account of the similarity of the polygons. Thus,
ὁμοιότητατῶνΑΒΕ,ΖΗΛτριγώνωνἐστὶνὡςἡΕΒπρὸς the remaining angle EBC is equal to LGH. And since,
ΒΑ,οὕτωςἡΛΗπρὸςΗΖ,ἀλλὰμὴνκαὶδιὰτὴνὁμοιότητα on account of the similarity of triangles ABE and FGL,
τῶνπολυγώνωνἐστὶνὡςἡΑΒπρὸςΒΓ,οὕτωςἡΖΗπρὸς as EB is to BA, so LG (is) to GF , but also, on account of
ΗΘ,δι᾿ἴσουἄραἐστὶνὡςἡΕΒπρὸςΒΓ,οὕτωςἡΛΗπρὸς the similarity of the polygons, as AB is to BC, so FG (is)
ΗΘ,καὶπερὶτὰςἴσαςγωνάιςτὰςὑπὸΕΒΓ,ΛΗΘαἱπλευραὶ to GH, thus, via equality, as EB is to BC, so LG (is) to
ἀνάλογόνεἰσιν·ἰσογώνιονἄραἐστὶτὸΕΒΓτρίγωνοντῷ GH [Prop. 5.22], and the sides about the equal angles,
ΛΗΘτριγώνῳ· ὥστεκαὶὅμοιόνἐστιτὸΕΒΓτρίγωνον EBC and LGH, are proportional. Thus, triangle EBC is
τῷΛΗΘτριγώνω. διὰτὰαὐτὰδὴκαὶτὸΕΓΔτρίγωνον equiangular to triangle LGH [Prop. 6.6]. Hence, triangle
ὅμοιόνἐστιτῷΛΘΚτριγώνῳ.τὰἄραὅμοιαπολύγωνατὰ EBC is also similar to triangle LGH [Prop. 6.4, Def. 6.1].
ΑΒΓΔΕ,ΖΗΘΚΛεἴςτεὅμοιατρίγωναδιῄρηταικαὶεἰςἴσα So, for the same (reasons), triangle ECD is also similar
C
D
H
K
176

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
τὸπλῆθος.
to triangle LHK. Thus, the similar polygons ABCDE
Λέγω, ὅτι καὶ ὁμόλογα τοῖς ὅλοις, τουτέστιν ὥστε and FGHKL have been divided into equal numbers of
ἀνάλογονεἶναιτὰτρίγωνα,καὶἡγούμεναμὲνεἶναιτὰΑΒΕ, similar triangles.
ΕΒΓ,ΕΓΔ,ἑπόμεναδὲαὐτῶντὰΖΗΛ,ΛΗΘ,ΛΘΚ,καὶ I also say that (the triangles) correspond (in propor-
ὅτιτὸΑΒΓΔΕπολύγωνονπρὸςτὸΖΗΘΚΛπολύγωνον tion) to the wholes. That is to say, the triangles are
διπλασίοναλόγονἔχειἤπερἡὁμόλογοςπλευρὰπρὸςτὴν proportional: ABE, EBC, and ECD are the leading
ὁμόλογονπλευράν,τουτέστινἡΑΒπρὸςτὴνΖΗ. (magnitudes), and their (associated) following (magni-
᾿Επεζεύχθωσαν γὰρ αἱ ΑΓ, ΖΘ. καὶ ἐπεὶ διὰ τὴν tudes are) FGL, LGH, and LHK (respectively). (I) also
ὁμοιότητατῶνπολυγώνωνἴσηἐστὶνἡὑπὸΑΒΓγωνίατῇ (say) that polygon ABCDE has a squared ratio to poly-
ὑπὸΖΗΘ,καίἐστινὡςἡΑΒπρὸςΒΓ,οὕτωςἡΖΗπρὸς gon FGHKL with respect to (that) a corresponding side
ΗΘ,ἰσογώνιόνἐστιτὸΑΒΓτρίγωνοντῷΖΗΘτριγώνῳ· (has) to a corresponding side—that is to say, (side) AB
ἴσηἄραἐστὶνἡμὲνὑπὸΒΑΓγωνίατῇὑπὸΗΖΘ,ἡδὲὑπὸ to FG.
ΒΓΑτῇὑπὸΗΘΖ.καὶἐπεὶἴσηἐστὶνἡὑπὸΒΑΜγωνία For let AC and FH have been joined. And since angle
τῇὑπὸΗΖΝ,ἔστιδὲκαὶἡὑπὸΑΒΜτῇὑπὸΖΗΝἴση, ABC is equal to FGH, and as AB is to BC, so FG (is) to
καὶλοιπὴἄραἡὑπὸΑΜΒλοιπῇτῇὑπὸΖΝΗἴσηἐστίν· GH, on account of the similarity of the polygons, triangle
ἰσογώνιονἄραἐστὶτὸΑΒΜτρίγωνοντῷΖΗΝτριγώνῳ. ABC is equiangular to triangle FGH [Prop. 6.6]. Thus,
ὁμοίωςδὴδεῖξομεν,ὅτικαὶτὸΒΜΓτρίγωνονἰσογώνιόν angle BAC is equal to GFH, and (angle) BCA to GHF .
ἐστιτῷΗΝΘτριγώνῳ.ἀνάλογονἄραἐστίν,ὡςμὲνἡΑΜ And since angle BAM is equal to GFN, and (angle)
πρὸςΜΒ,οὕτωςἡΖΝπρὸςΝΗ,ὡςδὲἡΒΜπρὸςΜΓ, ABM is also equal to FGN (see earlier), the remaining
οὕτωςἡΗΝπρὸςΝΘ·ὥστεκαὶδι᾿ἴσου,ὡςἡΑΜπρὸς (angle) AMB is thus also equal to the remaining (angle)
ΜΓ,οὕτωςἡΖΝπρὸςΝΘ.ἀλλ᾿ὡςἡΑΜπρὸςΜΓ,οὕτως FNG [Prop. 1.32]. Thus, triangle ABM is equiangular
τὸΑΒΜ[τρίγωνον]πρὸςτὸΜΒΓ,καὶτὸΑΜΕπρὸςτὸ to triangle FGN. So, similarly, we can show that triangle
ΕΜΓ·πρὸςἄλληλαγάρεἰσινὡςαἱβάσεις. καὶὡςἄρα BMC is also equiangular to triangle GNH. Thus, pro-
ἓντῶνἡγουμένωνπρὸςἓντῶνἑπόμενων,οὕτωςἅπαντα portionally, as AM is to MB, so FN (is) to NG, and as
τὰἡγούμεναπρὸςἅπαντατὰἑπόμενα· ὡςἄρατὸΑΜΒ BM (is) to MC, so GN (is) to NH [Prop. 6.4]. Hence,
τρίγωνονπρὸςτὸΒΜΓ,οὕτωςτὸΑΒΕπρὸςτὸΓΒΕ.αλλ᾿ also, via equality, as AM (is) to MC, so FN (is) to NH
ὡςτὸΑΜΒπρὸςτὸΒΜΓ,οὕτωςἡΑΜπρὸςΜΓ·καὶ [Prop. 5.22]. But, as AM (is) to MC, so [triangle] ABM
ὡςἄραἡΑΜπρὸςΜΓ,οὕτωςτὸΑΒΕτρίγωνονπρὸςτὸ is to MBC, and AME to EMC. For they are to one an-
ΕΒΓτρίγωνον. διὰτὰαὐτὰδὴκαὶὡςἡΖΝπρὸςΝΘ, other as their bases [Prop. 6.1]. And as one of the leading
οὕτωςτὸΖΗΛτρίγωνονπρὸςτὸΗΛΘτρίγωνον.καίἐστιν (magnitudes) is to one of the following (magnitudes), so
ὡςἡΑΜπρὸςΜΓ,οὕτωςἡΖΝπρὸςΝΘ·καὶὡςἄρα (the sum of) all the leading (magnitudes) is to (the sum
τὸΑΒΕτρίγωνονπρὸςτὸΒΕΓτρίγωνον,οὕτωςτὸΖΗΛ of) all the following (magnitudes) [Prop. 5.12]. Thus, as
τρίγωνονπρὸςτὸΗΛΘτρίγωνον,καὶἐναλλὰξὡςτὸΑΒΕ triangle AMB (is) to BMC, so (triangle) ABE (is) to
τρίγωνονπρὸςτὸΖΗΛτρίγωνον,οὕτωςτὸΒΕΓτρίγωνον CBE. But, as (triangle) AMB (is) to BMC, so AM (is)
πρὸςτὸΗΛΘτρίγωνον.ὁμοίωςδὴδείξομενἐπιζευχθεισῶν to MC. Thus, also, as AM (is) to MC, so triangle ABE
τῶνΒΔ,ΗΚ,ὅτικαὶὡςτὸΒΕΓτρίγωνονπρὸςτὸΛΗΘ (is) to triangle EBC. And so, for the same (reasons), as
τρίγωνον,οὕτωςτὸΕΓΔτρίγωνονπρὸςτὸΛΘΚτρίγωνον. FN (is) to NH, so triangle FGL (is) to triangle GLH.
καὶἐπείἐστινὡςτὸΑΒΕτρίγωνονπρὸςτὸΖΗΛτρίγωνον, And as AM is to MC, so FN (is) to NH. Thus, also, as
οὕτωςτὸΕΒΓπρὸςτὸΛΗΘ,καὶἔτιτὸΕΓΔπρὸςτὸΛΘΚ, triangle ABE (is) to triangle BEC, so triangle FGL (is)
καὶὡςἄραἓντῶνἡγουμένωνπρὸςἓντῶνἑπομένων,οὕτως to triangle GLH, and, alternately, as triangle ABE (is)
ἅπαντατὰἡγούμεναπρὸςἅπαντατὰἑπόμενα· ἔστινἄρα to triangle FGL, so triangle BEC (is) to triangle GLH
ὡςτὸΑΒΕτρίγωνονπρὸςτὸΖΗΛτρίγωνον,οὕτωςτὸ [Prop. 5.16]. So, similarly, we can also show, by joining
ΑΒΓΔΕπολύγωνονπρὸςτὸΖΗΘΚΛπολύγωνον.ἀλλὰτὸ BD and GK, that as triangle BEC (is) to triangle LGH,
ΑΒΕτρίγωνονπρὸςτὸΖΗΛτρίγωνονδιπλασίοναλόγον so triangle ECD (is) to triangle LHK. And since as tri-
ἔχειἤπερἡΑΒὁμόλογοςπλευρὰπρὸςτὴνΖΗὁμόλογον angle ABE is to triangle FGL, so (triangle) EBC (is)
πλευράν· τὰγὰρὅμοιατρίγωναἐνδιπλασίονιλόγῳἐστὶ to LGH, and, further, (triangle) ECD to LHK, and also
τῶνὁμολόγωνπλευρῶν. καὶτὸΑΒΓΔΕἄραπολύγωνον as one of the leading (magnitudes is) to one of the folπρὸςτὸΖΗΘΚΛπολύγωνονδιπλασίοναλόγονἔχειἤπερἡ
lowing, so (the sum of) all the leading (magnitudes is) to
ΑΒὁμόλογοςπλευρὰπρὸςτὴνΖΗὁμόλογονπλευράν. (the sum of) all the following [Prop. 5.12], thus as trian-
Τὰἄραὅμοιαπολύγωναεἴςτεὅμοιατρίγωναδιαιρεῖται gle ABE is to triangle FGL, so polygon ABCDE (is) to
καὶ εἰς ἴσα τὸ πλῆθος καὶ ὁμόλογα τοῖς ὅλοις, καὶ τὸ polygon FGHKL. But, triangle ABE has a squared ratio
177

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
πολύγωνονπρὸςτὸπολύγωνονδιπλασίοναλόγονἔχειἤπερ to triangle FGL with respect to (that) the corresponding
ἡὁμόλογοςπλευρὰπρὸςτὴνὁμόλογονπλευράν[ὅπερἔδει side AB (has) to the corresponding side FG. For, similar
δεῖξαι].
triangles are in the squared ratio of corresponding sides
[Prop. 6.14]. Thus, polygon ABCDE also has a squared
ratio to polygon FGHKL with respect to (that) the corresponding
side AB (has) to the corresponding side FG.
Thus, similar polygons can be divided into equal numbers
of similar triangles corresponding (in proportion) to
the wholes, and one polygon has to the (other) polygon a
squared ratio with respect to (that) a corresponding side
(has) to a corresponding side. [(Which is) the very thing
it was required to show].
ÈÖ×Ñ.
Corollary
῾Ωσαύτωςδὲκαὶἐπὶτῶν[ὁμοίων]τετραπλεύρωνδειχθή- And, in the same manner, it can also be shown for
σεται,ὅτιἐνδιπλασίονιλόγῳεἰσὶτῶνὁμολόγωνπλευρῶν. [similar] quadrilaterals that they are in the squared ratio
ἐδείχθηδὲκαὶἐπὶτῶντριγώνων· ὥστεκαὶκαθόλουτὰ of (their) corresponding sides. And it was also shown for
ὅμοια εὐθύγραμμα σχήματα πρὸς ἄλληλα ἐν διπλασίονι triangles. Hence, in general, similar rectilinear figures are
λόγῳεἰσὶτῶνὁμολόγωνπλευρῶν.ὅπερἔδειδεῖξαι. also to one another in the squared ratio of (their) corresponding
sides. (Which is) the very thing it was required
to show.
. Proposition 21
Τὰ τῷ αὐτῷ εὐθυγράμμῳ ὅμοια καὶ ἀλλήλοις ἐστὶν (Rectilinear figures) similar to the same rectilinear fig-
ὅμοια.
ure are also similar to one another.
Α
Β
A
B
Γ
C
῎Εστω γὰρ ἑκάτερον τῶν Α, Β εὐθυγράμμων τῷ Γ Let each of the rectilinear figures A and B be similar
ὅμοιον·λέγω,ὅτικαὶτὸΑτῷΒἐστινὅμοιον.
to (the rectilinear figure) C. I say that A is also similar to
᾿ΕπεὶγὰρὅμοιόνἐστιτὸΑτῷΓ,ἰσογώνιόντέἐστιν B.
αὐτῷκαὶτὰςπερὶτὰςἴσαςγωνίαςπλευρὰςἀνάλογονἔχει. For since A is similar to C, (A) is equiangular to (C),
πάλιν, ἐπεὶ ὅμοιόν ἐστι τὸ Β τῷ Γ, ἰσογώνιόν τέ ἐστιν and has the sides about the equal angles proportional
αὐτῷκαὶτὰςπερὶτὰςἴσαςγωνίαςπλευρὰςἀνάλογονἔχει. [Def. 6.1]. Again, since B is similar to C, (B) is equian-
ἑκάτερονἄρατῶνΑ,ΒτῷΓἰσογώνιόντέἐστικαὶτὰς gular to (C), and has the sides about the equal angles
περὶτὰςἴσαςγωνίαςπλευρὰςἀνάλογονἔχει[ὥστεκαὶτὸ proportional [Def. 6.1]. Thus, A and B are each equian-
ΑτῷΒἰσογώνιόντέἐστικαὶτὰςπερὶτὰςἴσαςγωνίας gular to C, and have the sides about the equal angles
178

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
πλευρὰςἀνάλογονἔχει].ὅμοιονἄραἐστὶτὸΑτῷΒ·ὅπερ proportional [hence, A is also equiangular to B, and has
ἔδειδεῖξαι.
the sides about the equal angles proportional]. Thus, A
is similar to B [Def. 6.1]. (Which is) the very thing it was
required to show.
. Proposition 22
᾿Εὰντέσσαρεςεὐθεῖαιἀνάλογονὦσιν,καὶτὰἀπ᾿αὐτῶν If four straight-lines are proportional then similar, and
εὐθύγραμμαὅμοιάτεκαὶὁμοίωςἀναγεγραμμέναἀνάλογον similarly described, rectilinear figures (drawn) on them
ἔσται·κἂντὰἀπ᾿αὐτῶνεὐθύγραμμαὅμοιάτεκαὶὁμοίως will also be proportional. And if similar, and similarly
ἀναγεγραμμέναἀνάλογονᾖ,καὶαὐτὰιαἱεὐθεῖαιἀνάλογον described, rectilinear figures (drawn) on them are pro-
ἔσονται.
portional then the straight-lines themselves will also be
proportional.
Κ
Λ
K
L
Α Β Γ ∆
Μ Ν
A B C D
M
N
Ξ
Ε
Ζ
Ο
Η Θ
Σ
Π Ρ
῎ΕστωσαντέσσαρεςεὐθεῖαιἀνάλογοναἱΑΒ,ΓΔ,ΕΖ, Let AB, CD, EF , and GH be four proportional
ΗΘ,ὡςἡΑΒπρὸςτὴνΓΔ,οὕτωςἡΕΖπρὸςτὴνΗΘ,καὶ straight-lines, (such that) as AB (is) to CD, so EF (is)
ἀναγεγράφθωσανἀπὸμὲντῶνΑΒ,ΓΔὅμοιάτεκαὶὁμοίως to GH. And let the similar, and similarly laid out, recκείμεναεὐθύγραμματὰΚΑΒ,ΛΓΔ,ἀπὸδὲτῶνΕΖ,ΗΘ
tilinear figures KAB and LCD have been described on
ὅμοιάτεκαὶὁμοίωςκείμεναεὐθύγραμματὰΜΖ,ΝΘ·λέγω, AB and CD (respectively), and the similar, and similarly
ὅτιἐστὶνὡςτὸΚΑΒπρὸςτὸΛΓΔ,οὕτωςτὸΜΖπρὸςτὸ laid out, rectilinear figures MF and NH on EF and GH
ΝΘ.
(respectively). I say that as KAB is to LCD, so MF (is)
ΕἰλήφθωγὰρτῶνμὲνΑΒ,ΓΔτρίτηἀνάλογονἡΞ,τῶν to NH.
δὲΕΖ,ΗΘτρίτηἀνάλογονἡΟ.καὶἐπείἐστινὡςμὲνἡΑΒ For let a third (straight-line) O have been taken
πρὸςτὴνΓΔ,οὕτωςἡΕΖπρὸςτὴνΗΘ,ὡςδὲἡΓΔπρὸς (which is) proportional to AB and CD, and a third
τὴνΞ,οὕτωςἡΗΘπρὸςτὴνΟ,δι᾿ἴσουἄραἐστὶνὡςἡ (straight-line) P proportional to EF and GH [Prop. 6.11].
ΑΒπρὸςτὴνΞ,οὕτωςἡΕΖπρὸςτὴνΟ.ἀλλ᾿ὡςμὲνἡ And since as AB is to CD, so EF (is) to GH, and as CD
ΑΒπρὸςτὴνΞ,οὕτως[καὶ]τὸΚΑΒπρὸςτὸΛΓΔ,ὡςδὲ (is) to O, so GH (is) to P , thus, via equality, as AB is to
ἡΕΖπρὸςτὴνΟ,οὕτωςτὸΜΖπρὸςτὸΝΘ·καὶὡςἄρα O, so EF (is) to P [Prop. 5.22]. But, as AB (is) to O, so
τὸΚΑΒπρὸςτὸΛΓΔ,οὕτωςτὸΜΖπρὸςτὸΝΘ. [also] KAB (is) to LCD, and as EF (is) to P , so MF
ἈλλὰδὴἔστωὡςτὸΚΑΒπρὸςτὸΛΓΔ,οὕτωςτὸΜΖ (is) to NH [Prop. 5.19 corr.]. And, thus, as KAB (is) to
πρὸςτὸΝΘ·λέγω,ὅτιἐστὶκαὶὡςἡΑΒπρὸςτὴνΓΔ, LCD, so MF (is) to NH.
οὕτωςἡΕΖπρὸςτὴνΗΘ.εἰγὰρμήἐστιν,ὡςἡΑΒπρὸς And so let KAB be to LCD, as MF (is) to NH. I say
τὴνΓΔ,οὕτωςἡΕΖπρὸςτὴνΗΘ,ἔστωὡςἡΑΒπρὸςτὴν also that as AB is to CD, so EF (is) to GH. For if as AB
ΓΔ,οὕτωςἡΕΖπρὸςτὴνΠΡ,καὶἀναγεγράφθωἀπὸτῆς is to CD, so EF (is) not to GH, let AB be to CD, as EF
O
E
F
P
G
S
Q
H
R
179

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
ΠΡὁποτέρῳτῶνΜΖ,ΝΘὅμοιόντεκαὶὁμοίωςκείμενον (is) to QR [Prop. 6.12]. And let the rectilinear figure SR,
εὐθύγραμμοντὸΣΡ.
similar, and similarly laid down, to either of MF or NH,
᾿ΕπεὶοὖνἐστινὡςἡΑΒπρὸςτὴνΓΔ,οὕτωςἡΕΖπρὸς have been described on QR [Props. 6.18, 6.21].
τὴνΠΡ,καὶἀναγέγραπταιἀπὸμὲντῶνΑΒ,ΓΔὅμοιάτε Therefore, since as AB is to CD, so EF (is) to QR,
καὶὁμοίωςκείμενατὰΚΑΒ,ΛΓΔ,ἀπὸδὲτῶνΕΖ,ΠΡ and the similar, and similarly laid out, (rectilinear fig-
ὅμοιάτεκαὶὁμοίωςκείμενατὰΜΖ,ΣΡ,ἔστινἄραὡςτὸ ures) KAB and LCD have been described on AB and
ΚΑΒπρὸςτὸΛΓΔ,οὕτωςτὸΜΖπρὸςτὸΣΡ.ὑπόκειται CD (respectively), and the similar, and similarly laid out,
δὲκαὶὡςτὸΚΑΒπρὸςτὸΛΓΔ,οὕτωςτὸΜΖπρὸςτὸ (rectilinear figures) MF and SR on EF and QR (re-
ΝΘ·καὶὡςἄρατὸΜΖπρὸςτὸΣΡ,οὕτωςτὸΜΖπρὸς sespectively), thus as KAB is to LCD, so MF (is) to
τὸΝΘ.τὸΜΖἄραπρὸςἑκάτεροντῶνΝΘ,ΣΡτὸναὐτὸν SR (see above). And it was also assumed that as KAB
ἔχειλόγον·ἴσονἄραἐστὶτὸΝΘτῷΣΡ.ἔστιδὲαὐτῷκαὶ (is) to LCD, so MF (is) to NH. Thus, also, as MF (is)
ὅμοιονκαὶὁμοίωςκείμενον·ἴσηἄραἡΗΘτῇΠΡ.καὶἐπεί to SR, so MF (is) to NH [Prop. 5.11]. Thus, MF has
ἐστινὡςἡΑΒπρὸςτὴνΓΔ,οὕτωςἡΕΖπρὸςτὴνΠΡ,ἴση the same ratio to each of NH and SR. Thus, NH is equal
δὲἡΠΡτῇΗΘ,ἔστινἄραὡςἡΑΒπρὸςτὴνΓΔ,οὕτωςἡ to SR [Prop. 5.9]. And it is also similar, and similarly laid
ΕΖπρὸςτὴνΗΘ.
out, to it. Thus, GH (is) equal to QR. † And since AB is
᾿Εὰνἄρατέσσαρεςεὐθεῖαιἀνάλογονὦσιν,καὶτὰἀπ᾿ to CD, as EF (is) to QR, and QR (is) equal to GH, thus
αὐτῶν εὐθύγραμμα ὅμοιά τε καὶ ὁμοίως ἀναγεγραμμένα as AB is to CD, so EF (is) to GH.
ἀνάλογονἔσται· κἂντὰἀπ᾿αὐτῶνεὐθύγραμμαὅμοιάτε Thus, if four straight-lines are proportional, then simκαὶὁμοίωςἀναγεγραμμέναἀνάλογονᾖ,καὶαὐτὰιαἱεὐθεῖαι
ilar, and similarly described, rectilinear figures (drawn)
ἀνάλογονἔσονται·ὅπερἔδειδεῖξαι.
on them will also be proportional. And if similar, and
similarly described, rectilinear figures (drawn) on them
are proportional then the straight-lines themselves will
also be proportional. (Which is) the very thing it was
required to show.
† Here, Euclid assumes, without proof, that if two similar figures are equal then any pair of corresponding sides is also equal.
. Proposition 23
Τὰἰσογώνιαπαραλληλόγραμμαπρὸςἄλληλαλόγονἔχει Equiangular parallelograms have to one another the
τὸνσυγκείμενονἐκτῶνπλευρῶν.
ratio compounded † out of (the ratios of) their sides.
῎Εστω ἰσογώνια παραλληλόγραμμα τὰ ΑΓ, ΓΖ ἴσην Let AC and CF be equiangular parallelograms having
ἔχοντατὴνὑπὸΒΓΔγωνίαντῇὑπὸΕΓΗ·λέγω,ὅτιτὸΑΓ angle BCD equal to ECG. I say that parallelogram AC
παραλληλόγραμμονπρὸςτὸΓΖπαραλληλόγραμμονλόγον has to parallelogram CF the ratio compounded out of
ἔχειτὸνσυγκείμενονἐκτῶνπλευρῶν.
(the ratios of) their sides.
Κείσθωγὰρὥστεἐπ᾿εὐθείαςεἶναιτὴνΒΓτῇΓΗ·ἐπ᾿ For let BC be laid down so as to be straight-on to
εὐθείαςἄραἐστὶκαὶἡΔΓτῇΓΕ.καὶσυμπεπληρώσθωτὸ CG. Thus, DC is also straight-on to CE [Prop. 1.14].
ΔΗπαραλληλόγραμμον,καὶἐκκείσθωτιςεὐθεῖαἡΚ,καὶ And let the parallelogram DG have been completed. And
γεγονέτωὡςμὲνἡΒΓπρὸςτὴνΓΗ,οὕτωςἡΚπρὸςτὴν let some straight-line K have been laid down. And let it
Λ,ὡςδὲἡΔΓπρὸςτὴνΓΕ,οὕτωςἡΛπρὸςτὴνΜ. be contrived that as BC (is) to CG, so K (is) to L, and
ΟἱἄραλόγοιτῆςτεΚπρὸςτὴνΛκαὶτῆςΛπρὸς as DC (is) to CE, so L (is) to M [Prop. 6.12].
τὴνΜοἱαὐτοίεἰσιτοῖςλόγοιςτῶνπλευρῶν,τῆςτεΒΓ Thus, the ratios of K to L and of L to M are the same
πρὸςτὴνΓΗκαὶτῆςΔΓπρὸςτὴνΓΕ.ἀλλ᾿ὁτῆςΚπρὸς as the ratios of the sides, (namely), BC to CG and DC
ΜλόγοςσύγκειταιἔκτετοῦτῆςΚπρὸςΛλόγουκαὶ to CE (respectively). But, the ratio of K to M is comτοῦτῆςΛπρὸςΜ·ὥστεκαὶἡΚπρὸςτὴνΜλόγονἔχει
pounded out of the ratio of K to L and (the ratio) of L
τὸνσυγκείμενονἐκτῶνπλευρῶν. καὶἐπείἐστινὡςἡΒΓ to M. Hence, K also has to M the ratio compounded
πρὸςτὴνΓΗ,οὕτωςτὸΑΓπαραλληλόγραμμονπρὸςτὸ out of (the ratios of) the sides (of the parallelograms).
ΓΘ,ἀλλ᾿ὡςἡΒΓπρὸςτὴνΓΗ,οὕτωςἡΚπρὸςτὴνΛ, And since as BC is to CG, so parallelogram AC (is) to
καὶὡςἄραἡΚπρὸςτὴνΛ,οὕτωςτὸΑΓπρὸςτὸΓΘ. CH [Prop. 6.1], but as BC (is) to CG, so K (is) to L,
πάλιν,ἐπείἐστινὡςἡΔΓπρὸςτὴνΓΕ,οὕτωςτὸΓΘπα- thus, also, as K (is) to L, so (parallelogram) AC (is) to
ραλληλόγραμμονπρὸςτὸΓΖ,ἀλλ᾿ὡςἡΔΓπρὸςτὴνΓΕ, CH. Again, since as DC (is) to CE, so parallelogram
180

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
οὕτωςἡΛπρὸςτὴνΜ,καὶὡςἄραἡΛπρὸςτὴνΜ,οὕτως CH (is) to CF [Prop. 6.1], but as DC (is) to CE, so L
τὸΓΘπαραλληλόγραμμονπρὸςτὸΓΖπαραλληλόγραμμον. (is) to M, thus, also, as L (is) to M, so parallelogram
ἐπεὶοὖνἐδείχθη,ὡςμὲνἡΚπρὸςτὴνΛ,οὕτωςτὸΑΓ CH (is) to parallelogram CF . Therefore, since it was
παραλληλόγραμμονπρὸςτὸΓΘπαραλληλόγραμμον,ὡςδὲ shown that as K (is) to L, so parallelogram AC (is) to
ἡΛπρὸςτὴνΜ,οὕτωςτὸΓΘπαραλληλόγραμμονπρὸςτὸ parallelogram CH, and as L (is) to M, so parallelogram
ΓΖπαραλληλόγραμμον,δι᾿ἴσουἄραἐστὶνὡςἡΚπρὸςτὴν CH (is) to parallelogram CF , thus, via equality, as K is
Μ,οὕτωςτὸΑΓπρὸςτὸΓΖπαραλληλόγραμμον. ἡδὲΚ to M, so (parallelogram) AC (is) to parallelogram CF
πρὸςτὴνΜλόγονἔχειτὸνσυγκείμενονἐκτῶνπλευρῶν· [Prop. 5.22]. And K has to M the ratio compounded out
καὶτὸΑΓἄραπρὸςτὸΓΖλόγονἔχειτὸνσυγκείμενονἐκ of (the ratios of) the sides (of the parallelograms). Thus,
τῶνπλευρῶν.
(parallelogram) AC also has to (parallelogram) CF the
ratio compounded out of (the ratio of) their sides.
Α
∆
Θ
A
D
H
Κ
Λ
Μ
Β
Ε
Γ
Ζ
Η
Τὰἄραἰσογώνιαπαραλληλόγραμμαπρὸςἄλληλαλόγον Thus, equiangular parallelograms have to one another
ἔχειτὸνσυγκείμενονἐκτῶνπλευρῶν·ὅπερἔδειδεῖξαι. the ratio compounded out of (the ratio of) their sides.
(Which is) the very thing it was required to show.
† In modern terminology, if two ratios are “compounded” then they are multiplied together.
. Proposition 24
Παντὸςπαραλληλογράμμουτὰπερὶτὴνδιάμετρονπα- In any parallelogram the parallelograms about the diραλληλόγραμμαὅμοιάἐστιτῷτεὅλῳκαὶἀλλήλοις.
agonal are similar to the whole, and to one another.
῎Εστω παραλληλόγραμμον τὸ ΑΒΓΔ, διάμετρος δὲ Let ABCD be a parallelogram, and AC its diagonal.
αὐτοῦἡΑΓ,περὶδὲτὴνΑΓπαραλληλόγραμμαἔστωτὰΕΗ, And let EG and HK be parallelograms about AC. I say
ΘΚ·λέγω,ὅτιἑκάτεροντῶνΕΗ,ΘΚπαραλληλογράμμων that the parallelograms EG and HK are each similar to
ὅμοιόνἐστιὅλῳτῷΑΒΓΔκαὶἀλλήλοις.
the whole (parallelogram) ABCD, and to one another.
᾿ΕπεὶγὰρτριγώνουτοῦΑΒΓπαρὰμίαντῶνπλευρῶν For since EF has been drawn parallel to one of the
τὴνΒΓἦκταιἡΕΖ,ἀνάλογόνἐστινὡςἡΒΕπρὸςτὴν sides BC of triangle ABC, proportionally, as BE is to
ΕΑ,οὕτωςἡΓΖπρὸςτὴνΖΑ.πάλιν,ἐπεὶτριγώνουτοῦ EA, so CF (is) to FA [Prop. 6.2]. Again, since FG has
ΑΓΔπαρὰμίαντὴνΓΔἦκταιἡΖΗ,ἀνάλογόνἐστινὡςἡ been drawn parallel to one (of the sides) CD of trian-
ΓΖπρὸςτὴνΖΑ,οὕτωςἡΔΗπρὸςτὴνΗΑ.ἀλλ᾿ὡςἡ gle ACD, proportionally, as CF is to FA, so DG (is) to
ΓΖπρὸςτὴνΖΑ,οὕτωςἐδείχθηκαὶἡΒΕπρὸςτὴνΕΑ· GA [Prop. 6.2]. But, as CF (is) to FA, so it was also
καὶὡς ἄραἡΒΕπρὸςτὴνΕΑ, οὕτωςἡΔΗ πρὸςτὴν shown (is) BE to EA. And thus as BE (is) to EA, so
ΗΑ,καὶσυνθέντιἄραὡςἡΒΑπρὸςΑΕ,οὕτωςἡΔΑ DG (is) to GA. And, thus, compounding, as BA (is) to
πρὸςΑΗ,καὶἐναλλὰξὡςἡΒΑπρὸςτὴνΑΔ,οὕτωςἡ AE, so DA (is) to AG [Prop. 5.18]. And, alternately, as
ΕΑπρὸςτὴνΑΗ.τῶνἄραΑΒΓΔ,ΕΗπαραλληλογράμμων BA (is) to AD, so EA (is) to AG [Prop. 5.16]. Thus,
ἀνάλογόνεἰσιναἱπλευραὶαἱπερὶτὴνκοινὴνγωνίαντὴνὑπὸ in parallelograms ABCD and EG the sides about the
ΒΑΔ.καὶἐπεὶπαράλληλόςἐστινἡΗΖτῇΔΓ,ἴσηἐστὶν common angle BAD are proportional. And since GF is
ἡμὲνὑπὸΑΖΗγωνίατῇὑπὸΔΓΑ·καὶκοινὴτῶνδύο parallel to DC, angle AFG is equal to DCA [Prop. 1.29].
K
L
M
B
E
C
F
G
181

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
τριγώνωντῶνΑΔΓ,ΑΗΖἡὑπὸΔΑΓγωνία·ἰσογώνιον And angle DAC (is) common to the two triangles ADC
ἄραἐστὶτὸΑΔΓτρίγωνοντῷΑΗΖτριγώνῳ.διὰτὰαὐτὰ and AGF . Thus, triangle ADC is equiangular to triangle
δὴκαὶτὸΑΓΒτρίγωνονἰσογώνιόνἐστιτῷΑΖΕτριγώνῳ, AGF [Prop. 1.32]. So, for the same (reasons), triangle
καὶὅλοντὸΑΒΓΔπαραλληλόγραμμοντῷΕΗπαραλλη- ACB is equiangular to triangle AFE, and the whole parλογράμμῳἰσογώνιόνἐστιν.
ἀνάλογονἄραἐστὶνὡςἡΑΔ allelogram ABCD is equiangular to parallelogram EG.
πρὸςτὴνΔΓ,οὕτωςἡΑΗπρὸςτὴνΗΖ,ὡςδὲἡΔΓπρὸς Thus, proportionally, as AD (is) to DC, so AG (is) to
τὴνΓΑ,οὕτωςἡΗΖπρὸςτὴνΖΑ,ὡςδὲἡΑΓπρὸςτὴν GF , and as DC (is) to CA, so GF (is) to FA, and as AC
ΓΒ,οὕτωςἡΑΖπρὸςτὴνΖΕ,καὶἔτιὡςἡΓΒπρὸςτὴν (is) to CB, so AF (is) to FE, and, further, as CB (is)
ΒΑ,οὕτωςἡΖΕπρὸςτὴνΕΑ.καὶἐπεὶἐδείχθηὡςμὲν to BA, so FE (is) to EA [Prop. 6.4]. And since it was
ἡΔΓπρὸςτὴνΓΑ,οὕτωςἡΗΖπρὸςτὴνΖΑ,ὡςδὲἡ shown that as DC is to CA, so GF (is) to FA, and as
ΑΓπρὸςτὴνΓΒ,οὕτωςἡΑΖπρὸςτὴνΖΕ,δι᾿ἴσουἄρα AC (is) to CB, so AF (is) to FE, thus, via equality, as
ἐστὶνὡςἡΔΓπρὸςτὴνΓΒ,οὕτωςἡΗΖπρὸςτὴνΖΕ. DC is to CB, so GF (is) to FE [Prop. 5.22]. Thus, in
τῶν ἄρα ΑΒΓΔ, ΕΗ παραλληλογράμμωνἀνάλογόν εἰσιν parallelograms ABCD and EG the sides about the equal
αἱ πλευραὶαἱ περὶτὰς ἴσας γωνίας· ὅμοιον ἄρα ἐστὶ τὸ angles are proportional. Thus, parallelogram ABCD is
ΑΒΓΔπαραλληλογράμμοντῷΕΗπαραλληλογράμμῳ. διὰ similar to parallelogram EG [Def. 6.1]. So, for the same
τὰαὐτὰδὴτὸΑΒΓΔπαραλληλόγραμμονκαὶτῷΚΘπα- (reasons), parallelogram ABCD is also similar to parραλληλογράμμῳὅμοιόνἐστιν·ἑκάτερονἄρατῶνΕΗ,ΘΚ
allelogram KH. Thus, parallelograms EG and HK are
παραλληλογράμμωντῷΑΒΓΔ[παραλληλογράμμῳ]ὅμοιόν each similar to [parallelogram] ABCD. And (rectilin-
ἐστιν.τὰδὲτῷαὐτῷεὐθυγράμμῳὅμοιακαὶἀλλήλοιςἐστὶν ear figures) similar to the same rectilinear figure are also
ὅμοια·καὶτὸΕΗἄραπαραλληλόγραμμοντῷΘΚπαραλλη- similar to one another [Prop. 6.21]. Thus, parallelogram
λογράμμῳὅμοιόνἐστιν.
EG is also similar to parallelogram HK.
Α
Ε
Β
A
E
B
Η
Ζ
Θ
G
F
H
∆
Κ
Γ
D K C
Παντὸςἄραπαραλληλογράμμουτὰπερὶτὴνδιάμετρον Thus, in any parallelogram the parallelograms about
παραλληλόγραμμαὅμοιάἐστιτῷτεὅλῳκαὶἀλλήλοις·ὅπερ the diagonal are similar to the whole, and to one another.
ἔδειδεῖξαι.
(Which is) the very thing it was required to show.
. Proposition 25
Τῷδοθέντιεὐθυγράμμῳὅμοιονκαὶἄλλῳτῷδοθέντι To construct a single (rectilinear figure) similar to a
ἴσοντὸαὐτὸσυστήσασθαι.
given rectilinear figure, and equal to a different given rectilinear
figure.
182

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
Α
Κ
A
K
∆
D
Β
Γ
Ζ
Η
Θ
B
C
F
G
H
Λ Ε Μ
L E
M
῎Εστω τὸ μὲν δοθὲν εὐθύγραμμον, ᾧ δεῖ ὅμοιον Let ABC be the given rectilinear figure to which it is
συστήσασθαι, τὸΑΒΓ,ᾧδὲδεῖἴσον, τὸΔ·δεῖδὴτῷ required to construct a similar (rectilinear figure), and D
μὲνΑΒΓὅμοιον,τῷδὲΔἴσοντὸαὐτὸσυστήσασθαι. the (rectilinear figure) to which (the constructed figure)
ΠαραβεβλήσθωγὰρπαρὰμὲντὴνΒΓτῷΑΒΓτριγώνῳ is required (to be) equal. So it is required to construct
ἴσονπαραλληλόγραμμοντὸΒΕ,παρὰδὲτὴνΓΕτῷΔἴσον a single (rectilinear figure) similar to ABC, and equal to
παραλληλόγραμμοντὸΓΜἐνγωνίᾳτῇὑπὸΖΓΕ,ἥἐστιν D.
ἴσητῇὑπὸΓΒΛ.ἐπ᾿εὐθείαςἄραἐστὶνἡμὲνΒΓτῇΓΖ,ἡ For let the parallelogram BE, equal to triangle ABC,
δὲΛΕτῇΕΜ.καὶεἰλήφθωτῶνΒΓ,ΓΖμέσηἀνάλογονἡ have been applied to (the straight-line) BC [Prop. 1.44],
ΗΘ,καὶἀναγεγράφθωἀπὸτῆςΗΘτῷΑΒΓὅμοιόντεκαὶ and the parallelogram CM, equal to D, (have been ap-
ὁμοίωςκείμενοντὸΚΗΘ.
plied) to (the straight-line) CE, in the angle FCE, which
ΚαὶἐπείἐστινὡςἡΒΓπρὸςτὴνΗΘ,οὕτωςἡΗΘ is equal to CBL [Prop. 1.45]. Thus, BC is straight-on to
πρὸςτὴνΓΖ,ἐὰνδὲτρεῖςεὐθεῖαιἀνάλογονὦσιν, ἔστιν CF , and LE to EM [Prop. 1.14]. And let the mean pro-
ὡςἡπρώτηπρὸςτὴντρίτην, οὕτωςτὸἀπὸτῆςπρώτης portion GH have been taken of BC and CF [Prop. 6.13].
εἶδοςπρὸςτὸἀπὸτῆςδευτέραςτὸὅμοιονκαὶὁμοίωςἀνα- And let KGH, similar, and similarly laid out, to ABC
γραφόμενον,ἔστινἄραὡςἡΒΓπρὸςτὴνΓΖ,οὕτωςτὸ have been described on GH [Prop. 6.18].
ΑΒΓτρίγωνονπρὸςτὸΚΗΘτρίγωνον.ἀλλὰκαὶὡςἡΒΓ And since as BC is to GH, so GH (is) to CF , and if
πρὸςτὴνΓΖ,οὕτωςτὸΒΕπαραλληλόγραμμονπρὸςτὸΕΖ three straight-lines are proportional then as the first is to
παραλληλόγραμμον.καὶὡςἄρατὸΑΒΓτρίγωνονπρὸςτὸ the third, so the figure (described) on the first (is) to the
ΚΗΘτρίγωνον,οὕτωςτὸΒΕπαραλληλόγραμμονπρὸςτὸ similar, and similarly described, (figure) on the second
ΕΖπαραλληλόγραμμον·ἐναλλὰξἄραὡςτὸΑΒΓτρίγωνον [Prop. 6.19 corr.], thus as BC is to CF , so triangle ABC
πρὸςτὸΒΕπαραλληλόγραμμον,οὕτωςτὸΚΗΘτρίγωνον (is) to triangle KGH. But, also, as BC (is) to CF , so
πρὸςτὸΕΖπαραλληλόγραμμον.ἴσονδὲτὸΑΒΓτρίγωνον parallelogram BE (is) to parallelogram EF [Prop. 6.1].
τῷΒΕπαραλληλογράμμῳ·ἴσονἄρακαὶτὸΚΗΘτρίγωνον And, thus, as triangle ABC (is) to triangle KGH, so parτῷΕΖπαραλληλογράμμῳ.ἀλλὰτὸΕΖπαραλληλόγραμμον
allelogram BE (is) to parallelogram EF . Thus, alterτῷΔἐστινἴσον·καὶτὸΚΗΘἄρατῷΔἐστινἴσον.
ἔστι nately, as triangle ABC (is) to parallelogram BE, so triδὲτὸΚΗΘκαὶτῷΑΒΓὅμοιον.
angle KGH (is) to parallelogram EF [Prop. 5.16]. And
ΤῷἄραδοθέντιεὐθυγράμμῳτῷΑΒΓὅμοιονκαὶἄλλῳ triangle ABC (is) equal to parallelogram BE. Thus, triτῷδοθέντιτῷΔἴσοντὸαὐτὸσυνέσταταιτὸΚΗΘ·ὅπερ
angle KGH (is) also equal to parallelogram EF . But,
ἔδειποιῆσαι.
parallelogram EF is equal to D. Thus, KGH is also equal
to D. And KGH is also similar to ABC.
Thus, a single (rectilinear figure) KGH has been constructed
(which is) similar to the given rectilinear figure
ABC, and equal to a different given (rectilinear figure)
D. (Which is) the very thing it was required to do.
¦. Proposition 26
᾿Εὰνἀπὸπαραλληλογράμμουπαραλληλόγραμμονἀφαι- If from a parallelogram a(nother) parallelogram is
ρεθῇὅμοιόντετῷὅλῳκαὶὁμοίωςκείμενονκοινὴνγωνίαν subtracted (which is) similar, and similarly laid out, to
ἔχοναὐτῷ,περὶτὴναὐτὴνδιάμετρόνἐστιτῷὅλῳ. the whole, having a common angle with it, then (the sub-
ἈπὸγὰρπαραλληλογράμμουτοῦΑΒΓΔπαραλληλόγρα- tracted parallelogram) is about the same diagonal as the
μμον ἀφῃρήσθω τὸ ΑΖ ὅμοιον τῷ ΑΒΓΔ καὶ ὁμοίως whole.
κείμενονκοινὴνγωνίανἔχοναὐτῷτὴνὑπὸΔΑΒ·λέγω, For, from parallelogram ABCD, let (parallelogram)
183

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
ὅτιπερὶτὴναὐτὴνδιάμετρόνἐστιτὸΑΒΓΔτῷΑΖ.
Α
Η
∆
AF have been subtracted (which is) similar, and similarly
laid out, to ABCD, having the common angle DAB with
it. I say that ABCD is about the same diagonal as AF .
A G D
Ε
Ζ
E
F
Κ
Θ
K
H
Β
Γ
Μὴ γάρ, ἀλλ᾿ εἰ δυνατόν, ἔστω [αὐτῶν] διάμετροςἡ For (if) not, then, if possible, let AHC be [ABCD’s]
ΑΘΓ,καὶἐκβληθεῖσαἡΗΖδιήχθωἐπὶτὸΘ,καὶἤχθω diagonal. And producing GF , let it have been drawn
διὰτοῦΘὁπορέρᾳτῶνΑΔ,ΒΓπαράλληλοςἡΘΚ. through to (point) H. And let HK have been drawn
᾿ΕπεὶοὖνπερὶτὴναὐτὴνδιάμετρόνἐστιτὸΑΒΓΔτῷ through (point) H, parallel to either of AD or BC
ΚΗ,ἔστινἄραὡςἡΔΑπρὸςτὴνΑΒ,οὕτωςἡΗΑπρὸς [Prop. 1.31].
τὴνΑΚ.ἔστιδὲκαὶδιὰτὴνὁμοιότητατῶνΑΒΓΔ,ΕΗκαὶ Therefore, since ABCD is about the same diagonal as
ὡςἡΔΑπρὸςτὴνΑΒ,οὕτωςἡΗΑπρὸςτὴνΑΕ·καὶὡς KG, thus as DA is to AB, so GA (is) to AK [Prop. 6.24].
ἄραἡΗΑπρὸςτὴνΑΚ,οὕτωςἡΗΑπρὸςτὴνΑΕ.ἡΗΑ And, on account of the similarity of ABCD and EG, also,
ἄραπρὸςἑκατέραντῶνΑΚ,ΑΕτὸναὐτὸνἔχειλόγον.ἴση as DA (is) to AB, so GA (is) to AE. Thus, also, as GA
ἄραἐστὶνἡΑΕτῇΑΚἡἐλάττωντῇμείζονι·ὅπερἐστὶν (is) to AK, so GA (is) to AE. Thus, GA has the same
ἀδύνατον. οὐκἄραοὔκἐστιπερὶτὴναὐτὴνδιάμετροντὸ ratio to each of AK and AE. Thus, AE is equal to AK
ΑΒΓΔτῷΑΖ·περὶτὴναὐτὴνἄραἐστὶδιάμετροντὸΑΒΓΔ [Prop. 5.9], the lesser to the greater. The very thing is
παραλληλόγραμμοντῷΑΖπαραλληλογράμμῳ.
impossible. Thus, ABCD is not not about the same di-
᾿Εὰν ἄρα ἀπὸ παραλληλογράμμου παραλληλόγραμμον agonal as AF . Thus, parallelogram ABCD is about the
ἀφαιρεθῇὅμοιόντετῷ ὅλῳκαὶὁμοίωςκείμενονκοινὴν same diagonal as parallelogram AF .
γωνίανἔχοναὐτῷ,περὶτὴναὐτὴνδιάμετρόνἐστιτῷὅλῳ· Thus, if from a parallelogram a(nother) parallelogram
ὅπερἔδειδεῖξαι.
is subtracted (which is) similar, and similarly laid out,
to the whole, having a common angle with it, then (the
subtracted parallelogram) is about the same diagonal as
the whole. (Which is) the very thing it was required to
show.
Þ. Proposition 27
Πάντωντῶνπαρὰτὴναὐτὴνεὐθεῖανπαραβαλλομένων Of all the parallelograms applied to the same straightπαραλληλογράμμωνκαὶἐλλειπόντωνεἴδεσιπαραλληλογράμ-
line, and falling short by parallelogrammic figures similar,
μοιςὁμοίοιςτεκαὶὁμοίωςκειμένοιςτῷἀπὸτῆςἡμισείας and similarly laid out, to the (parallelogram) described
ἀναγραφομένῳμέγιστόνἐστιτὸἀπὸτῆςἡμισείαςπαρα- on half (the straight-line), the greatest is the [paralleloβαλλόμενον[παραλληλόγραμμον]ὅμοιονὂντῷἐλλείμμαντι.
gram] applied to half (the straight-line) which (is) similar
῎Εστω εὐθεῖα ἡ ΑΒ καὶ τετμήσθω δίχα κατὰ τὸ Γ, to (that parallelogram) by which it falls short.
καὶπαραβεβλήσθωπαρὰτὴνΑΒεὐθεῖαντὸΑΔπαραλ- Let AB be a straight-line, and let it have been cut in
ληλόγραμμονἐλλεῖπονεἴδειπαραλληλογράμμῳτῷΔΒἀνα- half at (point) C [Prop. 1.10]. And let the parallelogram
γραφέντιἀπὸτῆςἡμισείαςτῆςΑΒ,τουτέστιτῆςΓΒ·λέγω, AD have been applied to the straight-line AB, falling
ὅτιπάντωντῶνπαρὰτὴνΑΒπαραβαλλομένωνπαραλλη- short by the parallelogrammic figure DB (which is) apλογράμμων
καὶ ἐλλειπόντων εἴδεσι [παραλληλογράμμοις] plied to half of AB—that is to say, CB. I say that of all
ὁμοίοιςτεκαὶὁμοίωςκειμένοιςτῷΔΒμέγιστόνἐστιτὸ the parallelograms applied to AB, and falling short by
B
C
184

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
ΑΔ.παραβεβλήσθωγὰρπαρὰτὴνΑΒεὐθεῖαντὸΑΖπα- [parallelogrammic] figures similar, and similarly laid out,
ραλληλόγραμμονἐλλεῖπονεἴδειπαραλληλογράμμῳτῷΖΒ to DB, the greatest is AD. For let the parallelogram AF
ὁμοίῳτεκαὶὁμοίωςκειμένῳτῷΔΒ·λέγω,ὅτιμεῖζόνἐστι have been applied to the straight-line AB, falling short by
τὸΑΔτοῦΑΖ.
the parallelogrammic figure FB (which is) similar, and
similarly laid out, to DB. I say that AD is greater than
AF .
Η
∆
Ν
Ζ Μ
Λ
Ε
Θ
G
D
L
N
F M
E
H
Α
Γ Κ
Β
A C K
᾿ΕπεὶγὰρὅμοιόνἐστιτὸΔΒπαραλληλόγραμμοντῷΖΒ For since parallelogram DB is similar to paralleloπαραλληλογράμμῳ,περὶτὴναὐτήνεἰσιδιάμετρον.
ἤχθω gram FB, they are about the same diagonal [Prop. 6.26].
αὐτῶνδιάμετροςἡΔΒ,καὶκαταγεγράφθωτὸσχῆμα. Let their (common) diagonal DB have been drawn, and
᾿ΕπεὶοὖνἴσονἐστὶτὸΓΖτῷΖΕ,κοινὸνδὲτὸΖΒ, let the (rest of the) figure have been described.
ὅλονἄρατὸΓΘὅλῳτῷΚΕἐστινἴσον. ἀλλὰτὸΓΘτῷ Therefore, since (complement) CF is equal to (com-
ΓΗἐστινἴσον,ἐπεὶκαὶἡΑΓτῇΓΒ.καὶτὸΗΓἄρατῷΕΚ plement) FE [Prop. 1.43], and (parallelogram) FB is
ἐστινἴσον.κοινὸνπροσκείσθωτὸΓΖ·ὅλονἄρατὸΑΖτῷ common, the whole (parallelogram) CH is thus equal
ΛΜΝγνώμονίἐστινἴσον·ὥστετὸΔΒπαραλληλόγραμμον, to the whole (parallelogram) KE. But, (parallelogram)
τουτέστιτὸΑΔ,τοῦΑΖπαραλληλογράμμουμεῖζόνἐστιν. CH is equal to CG, since AC (is) also (equal) to CB
Πάντων ἄρα τῶν παρὰ τὴν αὐτὴν εὐθεῖαν παραβαλ- [Prop. 6.1]. Thus, (parallelogram) GC is also equal
λομένωνπαραλληλογράμμωνκαὶἐλλειπόντωνεἴδεσιπαραλ- to EK. Let (parallelogram) CF have been added to
ληλογράμμοιςὁμοίοιςτεκαὶὁμοίωςκειμένοιςτῷἀπὸτῆς both. Thus, the whole (parallelogram) AF is equal to
ἡμισείαςἀναγραφομένῳμέγιστόνἐστιτὸἀπὸτῆςἡμισείας the gnomon LMN. Hence, parallelogram DB—that is to
παραβληθέν·ὅπερἔδειδεῖξαι. say, AD—is greater than parallelogram AF .
Thus, for all parallelograms applied to the same
straight-line, and falling short by a parallelogrammic
figure similar, and similarly laid out, to the (parallelogram)
described on half (the straight-line), the greatest
is the [parallelogram] applied to half (the straight-line).
(Which is) the very thing it was required to show.
. Proposition 28 †
Παρὰ τὴν δοθεῖσαν εὐθεῖαν τῷ δοθέντι εὐθυγράμμῳ To apply a parallelogram, equal to a given rectilin-
ἴσον παραλληλόγραμμον παραβαλεῖν ἐλλεῖπον εἴδει πα- ear figure, to a given straight-line, (the applied parallelραλληλογράμμῳ
ὁμοίῳ τῷ δοθέντι· δεῖ δὲ τὸ διδόμενον ogram) falling short by a parallelogrammic figure similar
εὐθύγραμμον[ᾧδεῖἴσονπαραβαλεῖν]μὴμεῖζονεἶναιτοῦ to a given (parallelogram). It is necessary for the given
ἀπὸτῆςἡμισείαςἀναγραφομένουὁμοίουτῷἐλλείμματι[τοῦ rectilinear figure [to which it is required to apply an equal
τεἀπὸτῆςἡμισείαςκαὶᾧδεῖὅμοιονἐλλείπειν].
(parallelogram)] not to be greater than the (parallelo-
῎Εστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν gram) described on half (of the straight-line) and similar
εὐθύγραμμον,ᾧδεῖἴσονπαρὰτὴνΑΒπαραβαλεῖν,τὸΓμὴ to the deficit.
μεῖζον[ὂν]τοῦἀπὸτῆςἡμισείαςτῆςΑΒἀναγραφομένου Let AB be the given straight-line, and C the given
ὁμοίουτῷἐλλείμματι,ᾧδὲδεῖὅμοιονἐλλείπειν,τὸΔ·δεῖδὴ rectilinear figure to which the (parallelogram) applied to
B
185

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
παρὰτὴνδοθεῖσανεὐθεῖαντὴνΑΒτῷδοθέντιεὐθυγράμμῳ AB is required (to be) equal, [being] not greater than
τῷΓἴσονπαραλληλόγραμμονπαραβαλεῖνἐλλεῖπονεἴδειπα- the (parallelogram) described on half of AB and similar
ραλληλογράμμῳὁμοίῳὄντιτῷΔ.
to the deficit, and D the (parallelogram) to which the
deficit is required (to be) similar. So it is required to apply
a parallelogram, equal to the given rectilinear figure C, to
 À Ç
the straight-line AB, falling short by a parallelogrammic
figure which is similar to D.
H G P F
Ì È
Ë Ê Ä Å
C
L M
U
O V
T
Q
W R
à Æ
D
A E S B
K N
Τετμήσθω ἡ ΑΒ δίχα κατὰ τὸ Ε σημεῖον, καὶ ἀνα- Let AB have been cut in half at point E [Prop. 1.10],
γεγράφθωἀπὸτῆςΕΒτῷΔὅμοιονκαὶὁμοίωςκείμενον and let (parallelogram) EBFG, (which is) similar, and
τὸΕΒΖΗ,καὶσυμπεπληρώσθωτὸΑΗπαραλληλόγραμμον. similarly laid out, to (parallelogram) D, have been de-
ΕἰμὲνοὖνἴσονἐστὶτὸΑΗτῷΓ,γεγονὸςἂνεἴητὸἐπι- scribed on EB [Prop. 6.18]. And let parallelogram AG
ταχθέν·παραβέβληταιγὰρπαρὰτὴνδοθεῖσανεὐθεῖαντὴν have been completed.
ΑΒτῷδοθέντιεὐθυγράμμῳτῷΓἴσονπαραλληλόγραμμον Therefore, if AG is equal to C then the thing preτὸΑΗἐλλεῖπονεἴδειπαραλληλογράμμῳτῷΗΒὁμοίῳὄντι
scribed has happened. For a parallelogram AG, equal
τῷΔ.εἰδὲοὔ,μεῖζόνἔστωτὸΘΕτοῦΓ.ἴσονδὲτὸΘΕ to the given rectilinear figure C, has been applied to the
τῷΗΒ·μεῖζονἄρακαὶτὸΗΒτοῦΓ.ᾧδὴμεῖζόνἐστι given straight-line AB, falling short by a parallelogramτὸΗΒτοῦΓ,ταύτῃτῇὑπεροχῇἴσον,τῷδὲΔὅμοιονκαὶ
mic figure GB which is similar to D. And if not, let HE
ὁμοίωςκείμενοντὸαὐτὸσυνεστάτωτὸΚΛΜΝ.ἀλλὰτὸΔ be greater than C. And HE (is) equal to GB [Prop. 6.1].
τῷΗΒ[ἐστιν]ὅμοιον·καὶτὸΚΜἄρατῷΗΒἐστινὅμοιον. Thus, GB (is) also greater than C. So, let (parallelo-
ἔστωοὖνὁμόλογοςἡμὲνΚΛτῂΗΕ,ἡδὲΛΜτῇΗΖ. gram) KLMN have been constructed (so as to be) both
καὶἐπεὶἴσονἐστὶτὸΗΒτοῖςΓ,ΚΜ,μεῖζονἄραἐστὶτὸ similar, and similarly laid out, to D, and equal to the ex-
ΗΒτοῦΚΜ·μείζωνἄραἐστὶκαὶἡμὲνΗΕτῆςΚΛ,ἡδὲ cess by which GB is greater than C [Prop. 6.25]. But,
ΗΖτῆςΛΜ.κείσθωτῇμὲνΚΛἴσηἡΗΞ,τῇδὲΛΜἴση GB [is] similar to D. Thus, KM is also similar to GB
ἡΗΟ,καὶσυμπεπληρώσθωτὸΞΗΟΠπαραλληλόγραμμον· [Prop. 6.21]. Therefore, let KL correspond to GE, and
ἴσονἄρακαὶὅμοιονἐστι[τὸΗΠ]τῷΚΜ[ἀλλὰτὸΚΜτῷ LM to GF . And since (parallelogram) GB is equal to
ΗΒὅμοιόνἐστιν].καὶτὸΗΠἄρατῷΗΒὅμοιόνἐστιν·περὶ (figure) C and (parallelogram) KM, GB is thus greater
τὴναὐτὴνἄραδιάμετρόνἐστιτὸΗΠτῷΗΒ.ἔστωαὐτῶν than KM. Thus, GE is also greater than KL, and GF
διάμετροςἡΗΠΒ,καὶκαταγεγράφθωτὸσχῆμα. than LM. Let GO be made equal to KL, and GP to LM
᾿ΕπεὶοὖνἴσονἐστὶτὸΒΗτοῖςΓ,ΚΜ,ὧντὸΗΠτῷ [Prop. 1.3]. And let the parallelogram OGPQ have been
ΚΜἐστινἴσον,λοιπὸςἄραὁΥΧΦγνόμωνλοιπῷτῷΓἴσος completed. Thus, [GQ] is equal and similar to KM [but,
ἐστίν.καὶἐπεὶἴσονἐστὶτὸΟΡτῷΞΣ,κοινὸνπροσκείσθω KM is similar to GB]. Thus, GQ is also similar to GB
τὸΠΒ·ὅλονἄρατὸΟΒὅλῳτῷΞΒἴσονἐστίν.ἀλλὰτὸΞΒ [Prop. 6.21]. Thus, GQ and GB are about the same diagτῷΤΕἐστινἴσον,ἐπεὶκαὶπλευρὰἡΑΕπλευρᾷτῇΕΒἐστιν
onal [Prop. 6.26]. Let GQB be their (common) diagonal,
ἴση·καὶτὸΤΕἄρατῷΟΒἐστινἴσον.κοινὸνπροσκείσθω and let the (remainder of the) figure have been described.
τὸΞΣ·ὅλονἄρατὸΤΣὅλῳτῷΦΧΥγνώμονίἐστινἴσον. Therefore, since BG is equal to C and KM, of which
ἀλλ᾿ὁΦΧΥγνώμωντῷΓἐδείχθηἴσος·καὶτὸΤΣἄρατῷ GQ is equal to KM, the remaining gnomon UWV is thus
Γἐστινἴσον.
equal to the remainder C. And since (the complement)
Παρὰ τὴν δοθεῖσανἄραεὐθεῖαντὴν ΑΒ τῷ δοθέντι PR is equal to (the complement) OS [Prop. 1.43], let
εὐθυγράμμῳτῷΓἴσονπαραλληλόγραμμονπαραβέβληται (parallelogram) QB have been added to both. Thus, the
τὸΣΤἐλλεῖπονεἴδειπαραλληλογράμμῳτῷΠΒὁμοίῳὄντι whole (parallelogram) PB is equal to the whole (par-
Í É
186

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
τῷΔ[ἐπειδήπερτὸΠΒτῷΗΠὅμοιόνἐστιν]·ὅπερἔδει allelogram) OB. But, OB is equal to TE, since side
ποιῆσαι.
AE is equal to side EB [Prop. 6.1]. Thus, TE is also
equal to PB. Let (parallelogram) OS have been added
to both. Thus, the whole (parallelogram) TS is equal to
the gnomon V WU. But, gnomon V WU was shown (to
be) equal to C. Therefore, (parallelogram) TS is also
equal to (figure) C.
Thus, the parallelogram ST , equal to the given rectilinear
figure C, has been applied to the given straightline
AB, falling short by the parallelogrammic figure QB,
which is similar to D [inasmuch as QB is similar to GQ
[Prop. 6.24] ]. (Which is) the very thing it was required
to do.
† This proposition is a geometric solution of the quadratic equation x 2 −αx+β = 0. Here, x is the ratio of a side of the deficit to the corresponding
side of figure D, α is the ratio of the length of AB to the length of that side of figure D which corresponds to the side of the deficit running along
AB, and β is the ratio of the areas of figures C and D. The constraint corresponds to the condition β < α 2 /4 for the equation to have real roots.
Only the smaller root of the equation is found. The larger root can be found by a similar method.
. Proposition 29 †
Παρὰ τὴν δοθεῖσαν εὐθεῖαν τῷ δοθέντι εὐθυγράμμῳ To apply a parallelogram, equal to a given rectilin-
ἴσονπαραλληλόγραμμονπαραβαλεῖνὑπερβάλλονεἴδειπα- ear figure, to a given straight-line, (the applied paralleloραλληλογράμμῳὁμοίῳτῷδοθέντι.
gram) overshooting by a parallelogrammic figure similar
to a given (parallelogram).
Ä Å
Ã
Â
C
F
L
M
K
H
Æ
É
Ç
È
D
V
X B
P
W
A E
À
N Q O G
῎Εστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν Let AB be the given straight-line, and C the given
εὐθύγραμμον,ᾧδεῖἴσονπαρὰτὴνΑΒπαραβαλεῖν,τὸΓ, rectilinear figure to which the (parallelogram) applied to
ᾧδὲδεῖὅμοιονὑπερβάλλειν,τὸΔ·δεῖδὴπαρὰτὴνΑΒ AB is required (to be) equal, and D the (parallelogram)
εὐθεῖαντῷΓεὐθυγράμμῳἴσονπαραλληλόγραμμονπαρα- to which the excess is required (to be) similar. So it is
βαλεῖνὑπερβάλλονεἴδειπαραλληλογράμμῳὁμοίῳτῷΔ. required to apply a parallelogram, equal to the given rec-
Τετμήσθω ἡ ΑΒ δίχα κατὰ τὸ Ε, καὶ ἀναγεγράθω tilinear figure C, to the given straight-line AB, overshoot-
ἀπὸ τὴς ΕΒ τῷ Δ ὅμοιον καὶ ὁμοίως κείμενον παραλ- ing by a parallelogrammic figure similar to D.
ληλόγραμμοντὸΒΖ,καὶσυναμφοτέροιςμὲντοῖςΒΖ,Γ Let AB have been cut in half at (point) E [Prop. 1.10],
ἴσον, τῷδὲΔὅμοιονκαὶὁμοίωςκείμενοντὸαὐτὸσυ- and let the parallelogram BF , (which is) similar, and
νεστάτωτὸΗΘ.ὁμόλογοςδὲἔστωἡμὲνΚΘτῇΖΛ,ἡδὲ similarly laid out, to D, have been described on EB
ΚΗτῇΖΕ.καὶἐπεὶμεῖζόνἐστιτὸΗΘτοῦΖΒ,μείζωνἄρα [Prop. 6.18]. And let (parallelogram) GH have been con-
ἐστὶκαὶἡμὲνΚΘτῆςΖΛ,ἡδὲΚΗτῇΖΕ.ἐκβεβλήσθωσαν structed (so as to be) both similar, and similarly laid out,
αἱΖΛ,ΖΕ,καὶτῇμὲνΚΘἴσηἔστωἡΖΛΜ,τῇδὲΚΗἴση to D, and equal to the sum of BF and C [Prop. 6.25].
ἡΖΕΝ,καὶσυμπεπληρώσθωτὸΜΝ·τὸΜΝἄρατῷΗΘ And let KH correspond to FL, and KG to FE. And since
ἴσοντέἐστικαὶὅμοιον.ἀλλὰτὸΗΘτῷΕΛἐστινὅμοιον· (parallelogram) GH is greater than (parallelogram) FB,
187

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
καὶτὸΜΝἄρατῷΕΛὅμοιόνἐστιν·περὶτὴναὐτὴνἄρα KH is thus also greater than FL, and KG than FE.
διάμετρόνἐστιτὸΕΛτῷΜΝ.ἤχθωαὐτῶνδιάμετροςἡ Let FL and FE have been produced, and let FLM be
ΖΞ,καὶκαταγεγράφθωτὸσχῆμα.
(made) equal to KH, and FEN to KG [Prop. 1.3]. And
᾿ΕπεὶἴσονἐστὶτὸΗΘτοῖςΕΛ,Γ,ἀλλὰτὸΗΘτῷΜΝ let (parallelogram) MN have been completed. Thus,
ἴσονἐστίν,καὶτὸΜΝἄρατοῖςΕΛ,Γἴσονἐστίν. κοινὸν MN is equal and similar to GH. But, GH is similar to
ἀφῃρήσθωτὸΕΛ·λοιπὸςἄραὁΨΧΦγνώμωντῷΓἐστιν EL. Thus, MN is also similar to EL [Prop. 6.21]. EL is
ἴσος.καὶἐπεὶἴσηἐστὶνἡΑΕτῇΕΒ,ἴσονἐστὶκαὶτὸΑΝτῷ thus about the same diagonal as MN [Prop. 6.26]. Let
ΝΒ,τουτέστιτῷΛΟ.κοινὸνπροσκείσθωτὸΕΞ·ὅλονἄρα their (common) diagonal FO have been drawn, and let
τὸΑΞἴσονἐστὶτῷΦΧΨγνώμονι. ἀλλὰὁΦΧΨγνώμων the (remainder of the) figure have been described.
τῷΓἴσοςἐστίν·καὶτὸΑΞἄρατῷΓἴσονἐστίν.
And since (parallelogram) GH is equal to (parallel-
Παρὰ τὴν δοθεῖσανἄραεὐθεῖαντὴν ΑΒ τῷ δοθέντι ogram) EL and (figure) C, but GH is equal to (paralεὐθυγράμμῳτῷΓἴσονπαραλληλόγραμμονπαραβέβληται
lelogram) MN, MN is thus also equal to EL and C.
τὸΑΞὑπερβάλλονεἴδειπαραλληλογράμμῳτῷΠΟὁμοίῳ Let EL have been subtracted from both. Thus, the re-
ὄντιτῷΔ,ἐπεὶκαὶτῷΕΛἐστινὅμοιοντὸΟΠ·ὅπερἔδει maining gnomon XWV is equal to (figure) C. And since
ποιῆσαι.
AE is equal to EB, (parallelogram) AN is also equal to
(parallelogram) NB [Prop. 6.1], that is to say, (parallelogram)
LP [Prop. 1.43]. Let (parallelogram) EO have
been added to both. Thus, the whole (parallelogram) AO
is equal to the gnomon V WX. But, the gnomon V WX
is equal to (figure) C. Thus, (parallelogram) AO is also
equal to (figure) C.
Thus, the parallelogram AO, equal to the given rectilinear
figure C, has been applied to the given straightline
AB, overshooting by the parallelogrammic figure QP
which is similar to D, since PQ is also similar to EL
[Prop. 6.24]. (Which is) the very thing it was required
to do.
† This proposition is a geometric solution of the quadratic equation x 2 +α x−β = 0. Here, x is the ratio of a side of the excess to the corresponding
side of figure D, α is the ratio of the length of AB to the length of that side of figure D which corresponds to the side of the excess running along
AB, and β is the ratio of the areas of figures C and D. Only the positive root of the equation is found.
Ð. Proposition 30 †
Τὴνδοθεῖσανεὐθεῖανπεπερασμένηνἄκρονκαὶμέσον To cut a given finite straight-line in extreme and mean
λόγοντεμεῖν.
ratio.
Γ
Ζ
Θ
C
F
H
Α
Ε Β
A
E
B
∆
῎ΕστωἡδοθεῖσαεὐθεῖαπεπερασμένηἡΑΒ·δεῖδὴτὴν Let AB be the given finite straight-line. So it is re-
ΑΒεὐθεῖανἄκρονκαὶμέσονλόγοντεμεῖν.
quired to cut the straight-line AB in extreme and mean
D
188

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
ἈναγεγράφθωἀπὸτῆςΑΒτετράγωνοντὸΒΓ,καὶπα- ratio.
ραβεβλήσθωπαρὰτὴνΑΓτῷΒΓἴσονπαραλληλόγραμμον Let the square BC have been described on AB [Prop.
τὸΓΔὑπερβάλλονεἴδειτῷΑΔὁμοίῳτῷΒΓ.
1.46], and let the parallelogram CD, equal to BC, have
ΤετράγωνονδέἐστιτὸΒΓ·τετράγωνονἄραἐστικαὶ been applied to AC, overshooting by the figure AD
τὸΑΔ.καὶἐπεὶἴσονἐστὶτὸΒΓτῷΓΔ,κοινὸνἀφῃρήσθω (which is) similar to BC [Prop. 6.29].
τὸΓΕ·λοιπὸνἄρατὸΒΖλοιπῷτῷΑΔἐστινἴσον. ἔστι And BC is a square. Thus, AD is also a square.
δὲαὐτῷκαὶἰσογώνιον·τῶνΒΖ,ΑΔἄραἀντιπεπόνθασιναἱ And since BC is equal to CD, let (rectangle) CE have
πλευραὶαἱπερὶτὰςἴσαςγωνίας·ἔστινἄραὡςἡΖΕπρὸς been subtracted from both. Thus, the remaining (rectτὴνΕΔ,οὕτωςἡΑΕπρὸςτὴνΕΒ.ἴσηδὲἡμὲνΖΕτῇΑΒ,
angle) BF is equal to the remaining (square) AD. And
ἡδὲΕΔτῇΑΕ.ἔστινἄραὡςἡΒΑπρὸςτὴνΑΕ,οὕτωςἡ it is also equiangular to it. Thus, the sides of BF and
ΑΕπρὸςτὴνΕΒ.μείζωνδὲἡΑΒτῆςΑΕ·μείζωνἄρακαὶ AD about the equal angles are reciprocally proportional
ἡΑΕτῆςΕΒ.
[Prop. 6.14]. Thus, as FE is to ED, so AE (is) to EB.
῾ΗἄραΑΒεὐθεῖαἄκρονκαὶμέσονλόγοντέτμηταικατὰ And FE (is) equal to AB, and ED to AE. Thus, as BA is
τὸΕ,καὶτὸμεῖζοναὐτῆςτμῆμάἐστιτὸΑΕ·ὅπερἔδει to AE, so AE (is) to EB. And AB (is) greater than AE.
ποιῆσαι. Thus, AE (is) also greater than EB [Prop. 5.14].
Thus, the straight-line AB has been cut in extreme
and mean ratio at E, and AE is its greater piece. (Which
is) the very thing it was required to do.
† This method of cutting a straight-line is sometimes called the “Golden Section”—see Prop. 2.11.
Ð. Proposition 31
᾿Εντοῖςὀρθογωνίοιςτριγώνοιςτὸἀπὸτῆςτὴνὀρθὴν In right-angled triangles, the figure (drawn) on the
γωνίανὑποτεινούσηςπλευρᾶςεἶδοςἴσονἐστὶτοῖςἀπὸτῶν side subtending the right-angle is equal to the (sum of
τὴνὀρθὴνγωνίανπεριεχουσῶνπλευρῶνεἴδεσιτοῖςὁμοίοις the) similar, and similarly described, figures on the sides
τεκαὶὁμοίωςἀναγραφομένοις.
surrounding the right-angle.
Α
A
Β ∆ Γ B D
C
῎ΕστωτρίγωνονὀρθογώνιοντὸΑΒΓὀρθὴνἔχοντὴν Let ABC be a right-angled triangle having the angle
ὑπὸΒΑΓγωνίαν·λέγω,ὅτιτὸἀπὸτῆςΒΓεἶδοςἴσονἐστὶ BAC a right-angle. I say that the figure (drawn) on BC is
τοῖςἀπὸτῶνΒΑ,ΑΓεἴδεσιτοῖςὁμοίοιςτεκαὶὁμοίως equal to the (sum of the) similar, and similarly described,
ἀναγραφομένοις.
figures on BA and AC.
῎ΗχθωκάθετοςἡΑΔ.
Let the perpendicular AD have been drawn [Prop.
᾿ΕπεὶοὖνἐνὀρθογωνίῳτριγώνῳτῷΑΒΓἀπὸτῆςπρὸς 1.12].
τῷΑὀρθῆςγωνίαςἐπὶτὴνΒΓβάσινκάθετοςἦκταιἡΑΔ, Therefore, since, in the right-angled triangle ABC,
τὰΑΒΔ,ΑΔΓπρὸςτῇκαθέτῳτρίγωναὅμοιάἐστιτῷτε the (straight-line) AD has been drawn from the right-
ὅλῳτῷΑΒΓκαὶἀλλήλοις.καὶἐπεὶὅμοιόνἐστιτὸΑΒΓτῷ angle at A perpendicular to the base BC, the trian-
ΑΒΔ,ἔστινἄραὡςἡΓΒπρὸςτὴνΒΑ,οὕτωςἡΑΒπρὸς gles ABD and ADC about the perpendicular are simτὴνΒΔ.καὶἐπεὶτρεῖςεὐθεῖαιἀνάλογόνεἰσιν,ἔστινὡςἡ
ilar to the whole (triangle) ABC, and to one another
πρώτηπρὸςτὴντρίτην,οὕτωςτὸἀπὸτῆςπρώτηςεἶδοςπρὸς [Prop. 6.8]. And since ABC is similar to ABD, thus
189

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
τὸἀπὸτῆςδευτέραςτὸὅμοιονκαὶὁμοίωςἀναγραφόμενον. as CB is to BA, so AB (is) to BD [Def. 6.1]. And
ὡςἄραἡΓΒπρὸςτὴνΒΔ,οὕτωςτὸἀπὸτῆςΓΒεἶδος since three straight-lines are proportional, as the first is
πρὸςτὸἀπὸτῆςΒΑτὸὅμοιονκαὶὁμοίωςἀναγραφόμενον. to the third, so the figure (drawn) on the first is to the
διὰτὰαὐτὰδὴκαὶὡςἡΒΓπρὸςτὴνΓΔ,οὕτωςτὸἀπὸτῆς similar, and similarly described, (figure) on the second
ΒΓεἶδοςπρὸςτὸἀπὸτῆςΓΑ.ὥστεκαὶὡςἡΒΓπρὸςτὰς [Prop. 6.19 corr.]. Thus, as CB (is) to BD, so the fig-
ΒΔ,ΔΓ,οὕτωςτὸἀπὸτῆςΒΓεἶδοςπρὸςτὰἀπὸτῶνΒΑ, ure (drawn) on CB (is) to the similar, and similarly de-
ΑΓτὰὅμοιακαὶὁμοίωςἀναγραφόμενα. ἴσηδὲἡΒΓταῖς scribed, (figure) on BA. And so, for the same (reasons),
ΒΔ,ΔΓ·ἴσονἄρακαὶτὸἄπὸτῆςΒΓεἶδοςτοῖςἀπὸτῶν as BC (is) to CD, so the figure (drawn) on BC (is) to
ΒΑ,ΑΓεἴδεσιτοῖςὁμοίοιςτεκαὶὁμοίωςἀναγραφομένοις. the (figure) on CA. Hence, also, as BC (is) to BD and
᾿Εν ἄρα τοῖς ὀρθογωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν DC, so the figure (drawn) on BC (is) to the (sum of the)
ὀρθὴνγωνίανὑποτεινούσηςπλευρᾶςεἶδοςἴσονἐστὶτοῖς similar, and similarly described, (figures) on BA and AC
ἀπὸτῶντὴνὀρθὴνγωνίανπεριεχουσῶνπλευρῶνεἴδεσιτοῖς [Prop. 5.24]. And BC is equal to BD and DC. Thus, the
ὁμοίοιςτεκαὶὁμοίωςἀναγραφομένοις·ὅπερἔδειδεῖξαι. figure (drawn) on BC (is) also equal to the (sum of the)
similar, and similarly described, figures on BA and AC
[Prop. 5.9].
Thus, in right-angled triangles, the figure (drawn) on
the side subtending the right-angle is equal to the (sum of
the) similar, and similarly described, figures on the sides
surrounding the right-angle. (Which is) the very thing it
was required to show.
Ð. Proposition 32
᾿Εὰνδύοτρίγωνασυντεθῇκατὰμίανγωνίαντὰςδύο If two triangles, having two sides proportional to two
πλευρὰς ταῖς δυσὶ πλευραῖς ἀνάλογον ἔχοντα ὥστε τὰς sides, are placed together at a single angle such that the
ὁμολόγουςαὐτῶνπλευρὰςκαὶπαραλλήλουςεἶναι,αἱλοιπαὶ corresponding sides are also parallel, then the remaining
τῶντριγώνωνπλευραὶἐπ᾿εὐθείαςἔσονται.
sides of the triangles will be straight-on (with respect to
one another).
∆
D
Α
A
Β
Γ
Ε
῎ΕστωδύοτρίγωνατὰΑΒΓ,ΔΓΕτὰςδύοπλευρὰςτὰς Let ABC and DCE be two triangles having the two
ΒΑ,ΑΓταῖςδυσὶπλευραῖςταῖςΔΓ,ΔΕἀνάλογονἔχοντα, sides BA and AC proportional to the two sides DC and
ὡςμὲντὴνΑΒπρὸςτὴνΑΓ,οὕτωςτὴνΔΓπρὸςτὴνΔΕ, DE—so that as AB (is) to AC, so DC (is) to DE—and
παράλληλονδὲτὴνμὲνΑΒτῇΔΓ,τὴνδὲΑΓτῇΔΕ·λέγω, (having side) AB parallel to DC, and AC to DE. I say
ὅτιἐπ᾿εὐθείαςἐστὶνἡΒΓτῇΓΕ.
that (side) BC is straight-on to CE.
᾿ΕπεὶγὰρπαράλληλόςἐστινἡΑΒτῇΔΓ,καὶεἰςαὐτὰς For since AB is parallel to DC, and the straight-line
ἐμπέπτωκενεὐθεῖαἡΑΓ,αἱἐναλλὰξγωνίαιαἱὑπὸΒΑΓ, AC has fallen across them, the alternate angles BAC and
ΑΓΔἴσαιἀλλήλαιςεἰσίν.διὰτὰαὐτὰδὴκαὶἡὑπὸΓΔΕτῇ ACD are equal to one another [Prop. 1.29]. So, for the
ὑπὸΑΓΔἴσηἐστίν.ὥστεκαὶἡὑπὸΒΑΓτῇὑπὸΓΔΕἐστιν same (reasons), CDE is also equal to ACD. And, hence,
ἴση.καὶἐπεὶδύοτρίγωνάἐστιτὰΑΒΓ,ΔΓΕμίανγωνίαν BAC is equal to CDE. And since ABC and DCE are
τὴνπρὸςτῷΑμιᾷγωνίᾳτῇπρὸςτῷΔἴσηνἔχοντα,περὶ two triangles having the one angle at A equal to the one
B
C
E
190

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
δὲτὰςἴσαςγωνίαςτὰςπλευρὰςἀνάλογον,ὡςτὴνΒΑπρὸς angle at D, and the sides about the equal angles proτὴνΑΓ,οὕτωςτὴνΓΔπρὸςτὴνΔΕ,ἰσογώνιονἄραἐστὶ
portional, (so that) as BA (is) to AC, so CD (is) to
τὸΑΒΓτρίγωνοντῷΔΓΕτριγώνῳ·ἴσηἄραἡὑπὸΑΒΓ DE, triangle ABC is thus equiangular to triangle DCE
γωνίατῇὑπὸΔΓΕ.ἐδείχθηδὲκαὶἡὑπὸΑΓΔτῇὑπὸΒΑΓ [Prop. 6.6]. Thus, angle ABC is equal to DCE. And (an-
ἴση·ὅληἄραἡὑπὸΑΓΕδυσὶταῖςὑπὸΑΒΓ,ΒΑΓἴσηἐστίν. gle) ACD was also shown (to be) equal to BAC. Thus,
κοινὴπροσκείσθωἡὑπὸΑΓΒ·αἱἄραὑπὸΑΓΕ,ΑΓΒταῖς the whole (angle) ACE is equal to the two (angles) ABC
ὑπὸΒΑΓ,ΑΓΒ,ΓΒΑἴσαιεἰσίν. ἀλλ᾿αἱὑπὸΒΑΓ,ΑΒΓ, and BAC. Let ACB have been added to both. Thus,
ΑΓΒδυσὶνὀρθαῖςἴσαιεἰσίν· καὶαἱὑπὸΑΓΕ,ΑΓΒἄρα ACE and ACB are equal to BAC, ACB, and CBA.
δυσὶνὀρθαῖςἴσαιεἰσίν. πρὸςδήτινιεὐθείᾳτῇΑΓκαὶτῷ But, BAC, ABC, and ACB are equal to two right-angles
πρὸςαὐτῇσημείῳτῷΓδύοεὐθεῖαιαἱΒΓ,ΓΕμὴἐπὶτὰ [Prop. 1.32]. Thus, ACE and ACB are also equal to two
αὐτὰμέρηκείμεναιτὰςἐφεξῆςγωνάιςτὰςὑπὸΑΓΕ,ΑΓΒ right-angles. Thus, the two straight-lines BC and CE,
δυσὶνὀρθαῖςἴσαςποιοῦσιν·ἐπ᾿εὐθείαςἄραἐστὶνἡΒΓτῇ not lying on the same side, make adjacent angles ACE
ΓΕ.
and ACB (whose sum is) equal to two right-angles with
᾿Εὰνἄραδύοτρίγωνασυντεθῇκατὰμίανγωνίαντὰς some straight-line AC, at the point C on it. Thus, BC is
δύοπλευρὰςταῖςδυσὶπλευραῖςἀνάλογονἔχονταὥστετὰς straight-on to CE [Prop. 1.14].
ὁμολόγουςαὐτῶνπλευρὰςκαὶπαραλλήλουςεἶναι,αἱλοιπαὶ Thus, if two triangles, having two sides proportional
τῶντριγώνωνπλευραὶἐπ᾿εὐθείαςἔσονται·ὅπερἔδειδεῖξαι. to two sides, are placed together at a single angle such
that the corresponding sides are also parallel, then the
remaining sides of the triangles will be straight-on (with
respect to one another). (Which is) the very thing it was
required to show.
Ð. Proposition 33
᾿Εν τοῖς ἴσοις κύκλοις αἱ γωνίαι τὸν αὐτὸν ἔχουσι In equal circles, angles have the same ratio as the (raλόγον
ταῖς περιφερείαις, ἐφ᾿ ὧν βεβήκασιν, ἐάν τε πρὸς tio of the) circumferences on which they stand, whether
τοῖςκέντροιςἐάντεπρὸςταῖςπεριφερείαιςὦσιβεβηκυῖαι. they are standing at the centers (of the circles) or at the
circumferences.
À Â
D
A
Ä
G
B
H
à ŠÆ
L E
C
N
K
F
M
῎ΕστωσανἴσοικύκλοιοἱΑΒΓ,ΔΕΖ,καὶπρὸςμὲντοῖς Let ABC and DEF be equal circles, and let BGC and
κέντροιςαὐτῶντοῖςΗ,ΘγωνίαιἔστωσαναἱὑπὸΒΗΓ, EHF be angles at their centers, G and H (respectively),
ΕΘΖ,πρὸςδὲταῖςπεριφερείαιςαἱὑπὸΒΑΓ,ΕΔΖ·λέγω, and BAC and EDF (angles) at their circumferences. I
ὅτιἐστὶνὡςἡΒΓπεριφέρειαπρὸςτὴνΕΖπεριφέρειαν, say that as circumference BC is to circumference EF , so
οὕτωςἥτεὑπὸΒΗΓγωνίαπρὸςτὴνὑπὸΕΘΖκαὶἡὑπὸ angle BGC (is) to EHF , and (angle) BAC to EDF .
ΒΑΓπρὸςτὴνὑπὸΕΔΖ.
For let any number whatsoever of consecutive (cir-
ΚείσθωσανγὰρτῇμὲνΒΓπεριφερείᾳἴσαικατὰτὸἑξῆς cumferences), CK and KL, be made equal to circumfer-
ὁσαιδηποτοῦναἱΓΚ,ΚΛ,τῇδὲΕΖπεριφερείᾳἴσαιὁσαι- ence BC, and any number whatsoever, FM and MN, to
δηποτοῦναἱΖΜ,ΜΝ,καὶἐπεζεύχθωσαναἱΗΚ,ΗΛ,ΘΜ, circumference EF . And let GK, GL, HM, and HN have
ΘΝ.
been joined.
᾿ΕπεὶοὖνἴσαιεἰσὶναἱΒΓ,ΓΚ,ΚΛπεριφέρειαιἀλλήλαις, Therefore, since circumferences BC, CK, and KL are
ἴσαι εἰσὶκαὶαἱὑπὸΒΗΓ, ΓΗΚ, ΚΗΛ γωνίαι ἀλλήλαις· equal to one another, angles BGC, CGK, and KGL are
ὁσαπλασίωνἄραἐστὶνἡΒΛπεριφέρειατῆςΒΓ,τοσαυτα- also equal to one another [Prop. 3.27]. Thus, as many
πλασίωνἐστὶκαὶἡὑπὸΒΗΛγωνίατῆςὑπὸΒΗΓ.διὰτὰ times as circumference BL is (divisible) by BC, so many
191

ËÌÇÁÉÁÏƦº ELEMENTS
BOOK 6
αὐτὰδὴκαὶὁσαπλασίωνἐστὶνἡΝΕπεριφέρειατῆςΕΖ,το- times is angle BGL also (divisible) by BGC. And so, for
σαυταπλασίωνἐστὶκαὶἡὑπὸΝΘΕγωνίατῆςὑπὸΕΘΖ.εἰ the same (reasons), as many times as circumference NE
ἄραἴσηἐστὶνἡΒΛπεριφέρειατῇΕΝπεριφερείᾳ,ἴσηἐστὶ is (divisible) by EF , so many times is angle NHE also
καὶγωνίαἡὑπὸΒΗΛτῇὑπὸΕΘΝ,καὶεἰμείζωνἐστὶνἡΒΛ (divisible) by EHF . Thus, if circumference BL is equal
περιφέρειατῆςΕΝπεριφερείας,μείζωνἐστὶκαὶἡὑπὸΒΗΛ to circumference EN then angle BGL is also equal to
γωνίατῆςὑπὸΕΘΝ,καὶεἰἐλάσσων,ἐλάσσων.τεσσάρων EHN [Prop. 3.27], and if circumference BL is greater
δὴὄντωνμεγεθῶν,δύομὲνπεριφερειῶντῶνΒΓ,ΕΖ,δύο than circumference EN then angle BGL is also greater
δὲγωνιῶντῶνὑπὸΒΗΓ,ΕΘΖ,εἴληπταιτῆςμὲνΒΓπερι- than EHN, † and if (BL is) less (than EN then BGL is
φερείαςκαὶτῆςὑπὸΒΗΓγωνίαςἰσάκιςπολλαπλασίωνἥτε also) less (than EHN). So there are four magnitudes,
ΒΛπεριφέρειακαὶἡὑπὸΒΗΛγωνία,τῆςδὲΕΖπεριφερείας two circumferences BC and EF , and two angles BGC
καὶτῆςὑπὸΕΘΖγωνίαςἥτεΕΝπεριφέριακαὶἡὑπὸΕΘΝ and EHF . And equal multiples have been taken of cirγωνία.
καὶδέδεικται,ὅτιεἰὑπερέχειἡΒΛπεριφέρειατῆς cumference BC and angle BGC, (namely) circumference
ΕΝπεριφερείας,ὑπερέχεικαὶἡὑπὸΒΗΛγωνίατῆςὑπο BL and angle BGL, and of circumference EF and an-
ΕΘΝγωνίας, καὶεἰἴση, ἴση,καὶεἰἐλάσσων, ἐλάσσων. gle EHF , (namely) circumference EN and angle EHN.
ἔστινἄρα,ὡςἡΒΓπεριφέρειαπρὸςτὴνΕΖ,οὕτωςἡὑπὸ And it has been shown that if circumference BL exceeds
ΒΗΓγωνίαπρὸςτὴνὑπὸΕΘΖ.ἀλλ᾿ὡςἡὑπὸΒΗΓγωνία circumference EN then angle BGL also exceeds angle
πρὸςτὴνὑπὸΕΘΖ,οὕτωςἡὑπὸΒΑΓπρὸςτὴνὑπὸΕΔΖ. EHN, and if (BL is) equal (to EN then BGL is also)
διπλασίαγὰρἑκατέραἑκατέρας.καὶὡςἄραἡΒΓπεριφέρεια equal (to EHN), and if (BL is) less (than EN then BGL
πρὸςτὴνΕΖπεριφέρειαν,οὕτωςἥτεὑπὸΒΗΓγωνίαπρὸς is also) less (than EHN). Thus, as circumference BC
τὴνὑπὸΕΘΖκαὶἡὑπὸΒΑΓπρὸςτὴνὑπὸΕΔΖ. (is) to EF , so angle BGC (is) to EHF [Def. 5.5]. But as
᾿Ενἄρατοῖςἴσοιςκύκλοιςαἱγωνίαιτὸναὐτὸνἔχουσι angle BGC (is) to EHF , so (angle) BAC (is) to EDF
λόγονταῖςπεριφερείαις,ἐφ᾿ὧνβεβήκασιν,ἐάντεπρὸςτοῖς [Prop. 5.15]. For the former (are) double the latter (reκέντροιςἐάντεπρὸςταῖςπεριφερείαιςὦσιβεβηκυῖαι·ὅπερ
spectively) [Prop. 3.20]. Thus, also, as circumference BC
ἔδειδεῖξαι. (is) to circumference EF , so angle BGC (is) to EHF ,
and BAC to EDF .
Thus, in equal circles, angles have the same ratio as
the (ratio of the) circumferences on which they stand,
whether they are standing at the centers (of the circles)
or at the circumferences. (Which is) the very thing it was
required to show.
† This is a straight-forward generalization of Prop. 3.27
192

ELEMENTS BOOK 7
Elementary Number Theory †
† The propositions contained in Books 7–9 are generally attributed to the school of Pythagoras.
193

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
ÎÇÖÓ.
αʹ.Μονάςἐστιν,καθ᾿ἣνἕκαστοντῶνὄντωνἓνλέγεται. 1. A unit is (that) according to which each existing
Definitions
βʹ.Ἀριθμὸςδὲτὸἐκμονάδωνσυγκείμενονπλῆθος. (thing) is said (to be) one.
γʹ. Μέρος ἐστὶν ἀριθμὸς ἀριθμοῦ ὁ ἐλάσσων τοῦ 2. And a number (is) a multitude composed of units. †
μείζονος,ὅτανκαταμετρῇτὸνμείζονα.
3. A number is part of a(nother) number, the lesser of
δʹ.Μέρηδέ,ὅτανμὴκαταμετρῇ. the greater, when it measures the greater. ‡
εʹ.Πολλαπλάσιοςδὲὁμείζωντοῦἐλάσσονος,ὅτανκα- 4. But (the lesser is) parts (of the greater) when it
ταμετρῆταιὑπὸτοῦἐλάσσονος. does not measure it. §
ϛʹ.῎Αρτιοςἀριθμόςἐστινὁδίχαδιαιρούμενος.
5. And the greater (number is) a multiple of the lesser
ζʹ.Περισσὸς δὲ ὁμὴ διαιρούμενοςδίχαἢ[ὁ] μονάδι when it is measured by the lesser.
διαφέρωνἀρτίουἀριθμοῦ.
6. An even number is one (which can be) divided in
ηʹ.Ἀρτιάκιςἄρτιοςἀριθμόςἐστινὁὑπὸἀρτίουἀριθμοῦ half.
μετρούμενοςκατὰἄρτιονἀριθμόν.
7. And an odd number is one (which can)not (be)
θʹ.῎Αρτιάκιςδὲπερισσόςἐστινὁὑπὸἀρτίουἀριθμοῦ divided in half, or which differs from an even number by
μετρούμενοςκατὰπερισσὸνἀριθμόν.
a unit.
ιʹ.Περισσάκιςδὲπερισσὸςἀριθμόςἐστινὁὑπὸπερισσοῦ 8. An even-times-even number is one (which is) mea-
ἀριθμοῦμετρούμενοςκατὰπερισσὸνἀριθμόν. sured by an even number according to an even number.
ιαʹ.Πρῶτοςἀριθμόςἐστινὁμονάδιμόνῃμετρούμενος. 9. And an even-times-odd number is one (which
ιβʹ.Πρῶτοιπρὸςἀλλήλουςἀριθμοίεἰσινοἱμονάδιμόνῃ is) measured by an even number according to an odd
μετρούμενοικοινῷμέτρῳ.
number. ∗
ιγʹ.Σύνθετοςἀριθμόςἐστινὁἀριθμῷτινιμετρούμενος. 10. And an odd-times-odd number is one (which
ιδʹ.Σύνθετοιδὲπρὸςἀλλήλουςἀριθμοίεἰσινοἱἀριθμῷ is) measured by an odd number according to an odd
τινιμετρούμενοικοινῷμέτρῳ. number. $
ιεʹ. Ἀριθμὸς ἀριθμὸν πολλαπλασιάζειν λέγεται, ὅταν, 11. A prime ‖ number is one (which is) measured by a
ὅσαιεἰσὶνἐναὐτῷμονάδες, τοσαυτάκιςσυντεθῇὁπολ- unit alone.
λαπλασιαζόμενος,καὶγένηταίτις.
12. Numbers prime to one another are those (which
ιϛʹ.῞Οτανδὲδύοἀριθμοὶπολλαπλασιάσαντεςἀλλήλους are) measured by a unit alone as a common measure.
ποιῶσί τινα, ὁ γενόμενος ἐπίπεδος καλεῖται, πλευραὶ δὲ 13. A composite number is one (which is) measured
αὐτοῦοἱπολλαπλασιάσαντεςἀλλήλουςἀριθμοί.
by some number.
ιζʹ.῞Οτανδὲτρεῖςἀριθμοὶπολλαπλασιάσαντεςἀλλήλους 14. And numbers composite to one another are those
ποιῶσίτινα,ὁγενόμενοςστερεόςἐστιν,πλευραὶδὲαὐτοῦ (which are) measured by some number as a common
οἱπολλαπλασιάσαντεςἀλλήλουςἀριθμοί.
measure.
ιηʹ.Τετράγωνοςἀριθμόςἐστινὁἰσάκιςἴσοςἢ[ὁ]ὑπὸ 15. A number is said to multiply a(nother) number
δύοἴσωνἀριθμῶνπεριεχόμενος.
when the (number being) multiplied is added (to itself)
ιθʹ.Κύβοςδὲὁἰσάκιςἴσοςἰσάκιςἢ[ὁ]ὑπὸτριῶνἴσων as many times as there are units in the former (number),
ἀριθμῶνπεριεχόμενος.
and (thereby) some (other number) is produced.
κʹ.Ἀριθμοὶἀνάλογόνεἰσιν,ὅτανὁπρῶτοςτοῦδευτέρου 16. And when two numbers multiplying one another
καὶὁτρίτοςτοῦτετάρτουἰσάκιςᾖπολλαπλάσιοςἢτὸαὐτὸ make some (other number) then the (number so) creμέροςἢτὰαὐτὰμέρηὦσιν.
ated is called plane, and its sides (are) the numbers which
καʹ. ῞Ομοιοι ἐπίπεδοι καὶ στερεοὶ ἀριθμοί εἰσιν οἱ multiply one another.
ανάλογονἔχοντεςτὰςπλευράς.
17. And when three numbers multiplying one another
κβʹ.Τέλειοςἀριθμόςἐστινὁτοῖςἑαυτοῦμέρεσινἴσος make some (other number) then the (number so) created
ὤν.
is (called) solid, and its sides (are) the numbers which
multiply one another.
18. A square number is an equal times an equal, or (a
plane number) contained by two equal numbers.
19. And a cube (number) is an equal times an equal
times an equal, or (a solid number) contained by three
equal numbers.
194

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
20. Numbers are proportional when the first is the
same multiple, or the same part, or the same parts, of
the second that the third (is) of the fourth.
21. Similar plane and solid numbers are those having
proportional sides.
22. A perfect number is that which is equal to its own
parts. ††
† In other words, a “number” is a positive integer greater than unity.
‡ In other words, a number a is part of another number b if there exists some number n such that n a = b.
§ In other words, a number a is parts of another number b (where a < b) if there exist distinct numbers, m and n, such that n a = m b.
In other words. an even-times-even number is the product of two even numbers.
∗ In other words, an even-times-odd number is the product of an even and an odd number.
$ In other words, an odd-times-odd number is the product of two odd numbers.
‖ Literally, “first”.
†† In other words, a perfect number is equal to the sum of its own factors.
. Proposition 1
Δύοἀριθμῶνἀνίσωνἐκκειμένων,ἀνθυφαιρουμένουδὲ Two unequal numbers (being) laid down, and the
ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος, ἐὰν ὁ λειπόμενος lesser being continually subtracted, in turn, from the
μηδέποτεκαταμετρῇτὸνπρὸἑαυτοῦ,ἕωςοὗλειφθῇμονάς, greater, if the remainder never measures the (number)
οἱἐξἀρχῆςἀριθμοὶπρῶτοιπρὸςἀλλὴλουςἔσονται. preceding it, until a unit remains, then the original numbers
will be prime to one another.
Α
Ζ
Θ
Γ
Η
A H
F
C
G
Β
∆
Ε
Δύο γὰρ [ἀνίσων] ἀριθμῶν τῶν ΑΒ, ΓΔ ἀνθυφαι- For two [unequal] numbers, AB and CD, the lesser
ρουμένουἀεὶτοῦἐλάσσονοςἀπὸτοῦμείζονοςὁλειπόμενος being continually subtracted, in turn, from the greater,
μηδέποτε καταμετρείτω τὸν πρὸ ἑαυτοῦ, ἕως οὗ λειφθῇ let the remainder never measure the (number) preceding
μονάς·λέγω,ὅτιοἱΑΒ,ΓΔπρῶτοιπρὸςἀλλήλουςεἰσίν, it, until a unit remains. I say that AB and CD are prime
τουτέστινὅτιτοὺςΑΒ,ΓΔμονὰςμόνημετρεῖ.
to one another—that is to say, that a unit alone measures
Εἰ γὰρ μή εἰσιν οἱ ΑΒ, ΓΔ πρῶτοι πρὸς ἀλλήλους, (both) AB and CD.
μετρήσειτιςαὐτοὺςἀριθμός.μετρείτω,καὶἔστωὁΕ·καὶὁ For if AB and CD are not prime to one another then
μὲνΓΔτὸνΒΖμετρῶνλειπέτωἑαυτοῦἐλάσσονατὸνΖΑ, some number will measure them. Let (some number)
ὁδὲΑΖτὸνΔΗμετρῶνλειπέτωἑαυτοῦἐλάσσονατὸνΗΓ, measure them, and let it be E. And let CD measuring
ὁδὲΗΓτὸνΖΘμετρῶνλειπέτωμονάδατὴνΘΑ. BF leave FA less than itself, and let AF measuring DG
᾿ΕπεὶοὖνὁΕτὸνΓΔμετρεῖ,ὁδὲΓΔτὸνΒΖμετρεῖ, leave GC less than itself, and let GC measuring FH leave
καὶὁΕἄρατὸνΒΖμετρεῖ·μετρεῖδὲκαὶὅλοντὸνΒΑ· a unit, HA.
καὶλοιπὸνἄρατὸνΑΖμετρήσει.ὁδὲΑΖτὸνΔΗμετρεῖ· In fact, since E measures CD, and CD measures BF ,
καὶὁΕἄρατὸνΔΗμετρεῖ·μετρεῖδὲκαὶὅλοντὸνΔΓ· E thus also measures BF . † And (E) also measures the
καὶλοιπὸνἄρατὸνΓΗμετρήσει. ὁδὲΓΗτὸνΖΘμετρεῖ· whole of BA. Thus, (E) will also measure the remainder
B
D
E
195

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
καὶὁΕἄρατὸνΖΘμετρεῖ·μετρεῖδὲκαὶὅλοντὸνΖΑ· AF . ‡ And AF measures DG. Thus, E also measures DG.
καὶλοιπὴνἄρατὴνΑΘμονάδαμετρήσειἀριθμὸςὤν·ὅπερ And (E) also measures the whole of DC. Thus, (E) will
ἐστὶνἀδύνατον. οὐκἄρατοὺςΑΒ,ΓΔἀριθμοὺςμετρήσει also measure the remainder CG. And CG measures FH.
τιςἀριθμός·οἱΑΒ,ΓΔἄραπρῶτοιπρὸςἀλλήλουςεἰσίν· Thus, E also measures FH. And (E) also measures the
ὅπερἔδειδεῖξαι.
whole of FA. Thus, (E) will also measure the remaining
unit AH, (despite) being a number. The very thing is
impossible. Thus, some number does not measure (both)
the numbers AB and CD. Thus, AB and CD are prime
to one another. (Which is) the very thing it was required
to show.
† Here, use is made of the unstated common notion that if a measures b, and b measures c, then a also measures c, where all symbols denote
numbers.
‡ Here, use is made of the unstated common notion that if a measures b, and a measures part of b, then a also measures the remainder of b, where
all symbols denote numbers.
.
Proposition 2
Δύοἀριθμῶνδοθέντωνμὴπρώτωνπρὸςἀλλήλουςτὸ To find the greatest common measure of two given
μέγιστοναὐτῶνκοινὸνμέτρονεὑρεῖν.
numbers (which are) not prime to one another.
Α
Ε
Γ
Ζ
A
E
C
F
Β
∆
Η
῎Εστωσαν οἱ δοθέντες δύο ἀριθμοὶ μὴ πρῶτοι πρὸς Let AB and CD be the two given numbers (which
ἀλλήλουςοἱΑΒ,ΓΔ.δεῖδὴτῶνΑΒ,ΓΔτὸμέγιστονκοινὸν are) not prime to one another. So it is required to find
μέτρονεὑρεῖν.
the greatest common measure of AB and CD.
ΕἰμὲνοὖνὁΓΔτὸνΑΒμετρεῖ,μετρεῖδὲκαὶἑαυτόν,ὁ In fact, if CD measures AB, CD is thus a common
ΓΔἄρατῶνΓΔ,ΑΒκοινὸνμέτρονἐστίν.καὶφανερόν,ὅτι measure of CD and AB, (since CD) also measures itself.
καὶμέγιστον·οὐδεὶςγὰρμείζωντοῦΓΔτὸνΓΔμετρήσει. And (it is) manifest that (it is) also the greatest (com-
ΕἰδὲοὐμετρεῖὁΓΔτὸνΑΒ,τῶνΑΒ,ΓΔἀνθυφαι- mon measure). For nothing greater than CD can meaρουμένουἀεὶτοῦἐλάσσονοςἀπὸτοῦμείζονοςλειφθήσεταί
sure CD.
τις ἀριθμός, ὃς μετρήσει τὸν πρὸ ἑαυτοῦ. μονὰς μὲν But if CD does not measure AB then some number
γὰροὐλειφθήσεται·εἰδὲμή,ἔσονταιοἱΑΒ,ΓΔπρῶτοι will remain from AB and CD, the lesser being continπρὸςἀλλήλους·ὅπεροὐχὑπόκειται.
λειφθήσεταίτιςἄρα ually subtracted, in turn, from the greater, which will
ἀριθμὸς,ὃςμετρήσειτὸνπρὸἑαυτοῦ. καὶὁμὲνΓΔτὸν measure the (number) preceding it. For a unit will not be
ΒΕμετρῶνλειπέτωἑαυτοῦἐλάσσονατὸνΕΑ,ὁδὲΕΑτὸν left. But if not, AB and CD will be prime to one another
ΔΖμετρῶνλειπέτωἑαυτοῦἐλάσσονατὸνΖΓ,ὁδὲΓΖτὸν [Prop. 7.1]. The very opposite thing was assumed. Thus,
ΑΕμετρείτω. ἐπεὶοὖνὁΓΖτὸνΑΕμετρεῖ,ὁδὲΑΕτὸν some number will remain which will measure the (num-
ΔΖμετρεῖ,καὶὁΓΖἄρατὸνΔΖμετρήσει. μετρεῖδὲκαὶ ber) preceding it. And let CD measuring BE leave EA
ἑαυτόν·καὶὅλονἄρατὸνΓΔμετρήσει. ὁδὲΓΔτὸνΒΕ less than itself, and let EA measuring DF leave FC less
μετρεῖ·καὶὁΓΖἄρατὸνΒΕμετρεῖ·μετρεῖδὲκαὶτὸνΕΑ· than itself, and let CF measure AE. Therefore, since CF
καὶὅλονἄρατὸνΒΑμετρήσει·μετρεῖδὲκαὶτὸνΓΔ·ὁΓΖ measures AE, and AE measures DF , CF will thus also
ἄρατοὺςΑΒ,ΓΔμετρεῖ. ὁΓΖἄρατῶνΑΒ,ΓΔκοινὸν measure DF . And it also measures itself. Thus, it will
B
D
G
196

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
μέτρονἐστίν.λέγωδή,ὅτικαὶμέγιστον.εἰγὰρμήἐστινὁ also measure the whole of CD. And CD measures BE.
ΓΖτῶνΑΒ,ΓΔμέγιστονκοινὸνμέτρον,μετρήσειτιςτοὺς Thus, CF also measures BE. And it also measures EA.
ΑΒ,ΓΔἀριθμοὺςἀριθμὸςμείζωνὢντοῦΓΖ.μετρείτω, Thus, it will also measure the whole of BA. And it also
καὶἔστωὁΗ.καὶἐπεὶὁΗτὸνΓΔμετρεῖ,ὁδὲΓΔτὸν measures CD. Thus, CF measures (both) AB and CD.
ΒΕμετρεῖ,καὶὁΗἄρατὸνΒΕμετρεῖ·μετρεῖδὲκαὶὅλον Thus, CF is a common measure of AB and CD. So I say
τὸνΒΑ·καὶλοιπὸνἄρατὸνΑΕμετρήσει. ὁδὲΑΕτὸν that (it is) also the greatest (common measure). For if
ΔΖμετρεῖ·καὶὁΗἄρατὸνΔΖμετρήσει·μετρεῖδὲκαὶ CF is not the greatest common measure of AB and CD
ὅλοντὸνΔΓ·καὶλοιπὸνἄρατὸνΓΖμετρήσειὁμείζων then some number which is greater than CF will meaτὸνἐλάσσονα·ὅπερἐστὶνἀδύνατον·οὐκἄρατοὺςΑΒ,ΓΔ
sure the numbers AB and CD. Let it (so) measure (AB
ἀριθμοὺςἀριθμόςτιςμετρήσειμείζωνὢντοῦΓΖ·ὁΓΖἄρα and CD), and let it be G. And since G measures CD,
τῶνΑΒ,ΓΔμέγιστόνἐστικοινὸνμέτρον[ὅπερἔδειδεῖξαι]. and CD measures BE, G thus also measures BE. And it
also measures the whole of BA. Thus, it will also measure
the remainder AE. And AE measures DF . Thus, G
will also measure DF . And it also measures the whole
of DC. Thus, it will also measure the remainder CF ,
the greater (measuring) the lesser. The very thing is impossible.
Thus, some number which is greater than CF
cannot measure the numbers AB and CD. Thus, CF is
the greatest common measure of AB and CD. [(Which
is) the very thing it was required to show].
ÈÖ×Ñ.
Corollary
᾿Εκδὴτούτουφανερόν,ὅτιἐὰνἀριθμὸςδύοἀριθμοὺς So it is manifest, from this, that if a number measures
μετρῇ,καὶτὸμέγιστοναὐτῶνκοινὸνμέτρονμετρήσει·ὅπερ two numbers then it will also measure their greatest com-
ἔδειδεῖξαι.
mon measure. (Which is) the very thing it was required
to show.
. Proposition 3
Τριῶνἀριθμῶνδοθέντωνμὴπρώτωνπρὸςἀλλήλουςτὸ To find the greatest common measure of three given
μέγιστοναὐτῶνκοινὸνμέτρονεὑρεῖν.
numbers (which are) not prime to one another.
Α Β Γ ∆ Ε Ζ A B C D E F
῎Εστωσαν οἱ δοθέντες τρεῖς ἀριθμοὶ μὴ πρῶτοι πρὸς Let A, B, and C be the three given numbers (which
ἀλλήλουςοἱΑ,Β,Γ·δεῖδὴτῶνΑ,Β,Γτὸμέγιστονκοινὸν are) not prime to one another. So it is required to find
μέτρονεὑρεῖν. the greatest common measure of A, B, and C.
ΕἰλήφθωγὰρδύοτῶνΑ,Βτὸμέγιστονκοινὸνμέτρονὁ For let the greatest common measure, D, of the two
Δ·ὁδὴΔτὸνΓἤτοιμετρεῖἢοὐμετρεῖ.μετρείτωπρότερον· (numbers) A and B have been taken [Prop. 7.2]. So D
μετρεῖδέκαὶτοὺςΑ,Β·ὁΔἄρατοὺςΑ,Β,Γμετρεῖ·ὁ either measures, or does not measure, C. First of all, let
ΔἄρατῶνΑ,Β,Γκοινὸνμέτρονἐστίν. λέγωδή,ὅτικαὶ it measure (C). And it also measures A and B. Thus, D
197

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
μέγιστον.εἰγὰρμήἐστινὁΔτῶνΑ,Β,Γμέγιστονκοινὸν measures A, B, and C. Thus, D is a common measure
μέτρον,μετρήσειτιςτοὺςΑ,Β,Γἀριθμοὺςἀριθμὸςμείζων of A, B, and C. So I say that (it is) also the greatest
ὢντοῦΔ.μετρείτω,καὶἔστωὁΕ.ἐπεὶοὖνὁΕτοὺςΑ,Β, (common measure). For if D is not the greatest common
Γμετρεῖ,καὶτοὺςΑ,Βἄραμετρήσει·καὶτὸτῶνΑ,Βἄρα measure of A, B, and C then some number greater than
μέγιστονκοινὸνμέτρονμετρήσει.τὸδὲτῶνΑ,Βμέγιστον D will measure the numbers A, B, and C. Let it (so)
κοινὸνμέτρονἐστὶνὁΔ·ὁΕἄρατὸνΔμετρεῖὁμείζων measure (A, B, and C), and let it be E. Therefore, since
τὸνἐλάσσονα·ὅπερἐστὶνἀδύνατον.οὐκἄρατοὺςΑ,Β,Γ E measures A, B, and C, it will thus also measure A and
ἀριθμοὺςἀριθμόςτιςμετρήσειμείζωνὢντοῦΔ·ὁΔἄρα B. Thus, it will also measure the greatest common meaτῶνΑ,Β,Γμέγιστόνἐστικοινὸνμέτρον.
sure of A and B [Prop. 7.2 corr.]. And D is the greatest
ΜὴμετρείτωδὴὁΔτὸνΓ·λέγωπρῶτον,ὅτιοἱΓ,Δ common measure of A and B. Thus, E measures D, the
οὔκεἰσιπρῶτοιπρὸςἀλλήλους. ἐπεὶγὰροἱΑ,Β,Γοὔκ greater (measuring) the lesser. The very thing is impossiεἰσιπρῶτοιπρὸςἀλλήλους,μετρήσειτιςαὐτοὺςἀριθμός.ὁ
ble. Thus, some number which is greater than D cannot
δὴτοὺςΑ,Β,ΓμετρῶνκαὶτοὺςΑ,Βμετρήσει,καὶτὸ measure the numbers A, B, and C. Thus, D is the greatτῶνΑ,ΒμέγιστονκοινὸνμέτροντὸνΔμετρήσει·μετρεῖ
est common measure of A, B, and C.
δὲκαὶτὸνΓ·τοὺςΔ,Γἄραἀριθμοὺςἀριθμόςτιςμετρήσει· So let D not measure C. I say, first of all, that C
οἱΔ,Γἄραοὔκεἰσιπρῶτοιπρὸςἀλλήλους.εἰλήφθωοὖν and D are not prime to one another. For since A, B, C
αὐτῶντὸμέγιστονκοινὸνμέτρονὁΕ.καὶἐπεὶὁΕτὸνΔ are not prime to one another, some number will measure
μετρεῖ,ὁδὲΔτοὺςΑ,Βμετρεῖ,καὶὁΕἄρατοὺςΑ,Β them. So the (number) measuring A, B, and C will also
μετρεῖ·μετρεῖδὲκαὶτὸνΓ·ὁΕἄρατοὺςΑ,Β,Γμετρεῖ. measure A and B, and it will also measure the greatest
ὁΕἄρατῶνΑ,Β,Γκοινόνἐστιμέτρον.λέγωδή,ὅτικαὶ common measure, D, of A and B [Prop. 7.2 corr.]. And
μέγιστον. εἰγὰρμήἐστινὁΕτῶνΑ,Β,Γτὸμέγιστον it also measures C. Thus, some number will measure the
κοινὸνμέτρον,μετρήσειτιςτοὺςΑ,Β,Γἀριθμοὺςἀριθμὸς numbers D and C. Thus, D and C are not prime to one
μείζωνὢντοῦΕ.μετρείτω,καὶἔστωὁΖ.καὶἐπεὶὁΖτοὺς another. Therefore, let their greatest common measure,
Α,Β,Γμετρεῖ,καὶτοὺςΑ,Βμετρεῖ·καὶτὸτῶνΑ,Βἄρα E, have been taken [Prop. 7.2]. And since E measures
μέγιστονκοινὸνμέτρονμετρήσει.τὸδὲτῶνΑ,Βμέγιστον D, and D measures A and B, E thus also measures A
κοινὸνμέτρονἐστὶνὁΔ·ὁΖἄρατὸνΔμετρεῖ·μετρεῖδὲ and B. And it also measures C. Thus, E measures A, B,
καὶτὸνΓ·ὁΖἄρατοὺςΔ,Γμετρεῖ·καὶτὸτῶνΔ,Γἄρα and C. Thus, E is a common measure of A, B, and C. So
μέγιστονκοινὸνμέτρονμετρήσει.τὸδὲτῶνΔ,Γμέγιστον I say that (it is) also the greatest (common measure). For
κοινὸνμέτρονἐστὶνὁΕ·ὁΖἄρατὸνΕμετρεῖὁμείζων if E is not the greatest common measure of A, B, and C
τὸνἐλάσσονα·ὅπερἐστὶνἀδύνατον.οὐκἄρατοὺςΑ,Β,Γ then some number greater than E will measure the num-
ἀριθμοὺςἀριθμόςτιςμετρήσειμείζωνὢντοῦΕ·ὁΕἄρα bers A, B, and C. Let it (so) measure (A, B, and C), and
τῶνΑ,Β,Γμέγιστόνἐστικοινὸνμέτρον·ὅπερἔδειδεῖξαι. let it be F . And since F measures A, B, and C, it also
measures A and B. Thus, it will also measure the greatest
common measure of A and B [Prop. 7.2 corr.]. And
D is the greatest common measure of A and B. Thus, F
measures D. And it also measures C. Thus, F measures
D and C. Thus, it will also measure the greatest common
measure of D and C [Prop. 7.2 corr.]. And E is the
greatest common measure of D and C. Thus, F measures
E, the greater (measuring) the lesser. The very thing is
impossible. Thus, some number which is greater than E
does not measure the numbers A, B, and C. Thus, E is
the greatest common measure of A, B, and C. (Which
is) the very thing it was required to show.
. Proposition 4
῞Απαςἀριθμὸςπαντὸςἀριθμοῦὁἐλάσσωντοῦμείζονος Any number is either part or parts of any (other) num-
ἤτοιμέροςἐστὶνἢμέρη.
ber, the lesser of the greater.
῎ΕστωσανδύοἀριθμοὶοἱΑ,ΒΓ,καὶἔστωἐλάσσωνὁ Let A and BC be two numbers, and let BC be the
ΒΓ·λέγω,ὅτιὁΒΓτοῦΑἤτοιμέροςἐστὶνἢμέρη. lesser. I say that BC is either part or parts of A.
198

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
ΟἱΑ,ΒΓγὰρἤτοιπρῶτοιπρὸςἀλλήλουςεἰσὶνἢοὔ. For A and BC are either prime to one another, or not.
ἔστωσανπρότερονοἱΑ,ΒΓπρῶτοιπρὸςἀλλήλους. διαι- Let A and BC, first of all, be prime to one another. So
ρεθέντοςδὴτοῦΒΓεἰςτὰςἐναὐτῷμονάδαςἔσταιἑκάστη separating BC into its constituent units, each of the units
μονὰςτῶνἐντῷΒΓμέροςτιτοῦΑ·ὥστεμέρηἐστὶνὁΒΓ in BC will be some part of A. Hence, BC is parts of A.
τοῦΑ.
Β
Ε
Ζ
B
E
F
Γ
C
Α ∆ A D
ΜὴἔστωσανδὴοἱΑ,ΒΓπρῶτοιπρὸςἀλλήλους·ὁδὴ So let A and BC be not prime to one another. So BC
ΒΓτὸνΑἤτοιμετρεῖἢοὐμετρεῖ. εἰμὲνοὖνὁΒΓτὸν either measures, or does not measure, A. Therefore, if
Αμετρεῖ,μέροςἐστὶνὁΒΓτοῦΑ.εἰδὲοὔ,εἰλήφθωτῶν BC measures A then BC is part of A. And if not, let the
Α,ΒΓμέγιστονκοινὸνμέτρονὁΔ,καὶδιῃρήσθωὁΒΓεἰς greatest common measure, D, of A and BC have been
τοὺςτῷΔἴσουςτοὺςΒΕ,ΕΖ,ΖΓ.καὶἐπεὶὁΔτὸνΑ taken [Prop. 7.2], and let BC have been divided into BE,
μετρεῖ,μέροςἐστὶνὁΔτοῦΑ·ἴσοςδὲὁΔἑκάστῳτῶν EF , and FC, equal to D. And since D measures A, D is
ΒΕ,ΕΖ,ΖΓ·καὶἕκαστοςἄρατῶνΒΕ,ΕΖ,ΖΓτοῦΑμέρος a part of A. And D is equal to each of BE, EF , and FC.
ἐστίν·ὥστεμέρηἐστὶνὁΒΓτοῦΑ.
Thus, BE, EF , and FC are also each part of A. Hence,
῞Απας ἄρα ἀριθμὸς παντὸς ἀριθμοῦ ὁ ἐλάσσων τοῦ BC is parts of A.
μείζονοςἤτοιμέροςἐστὶνἢμέρη·ὅπερἔδειδεῖξαι.
Thus, any number is either part or parts of any (other)
number, the lesser of the greater. (Which is) the very
thing it was required to show.
. Proposition 5 †
᾿Εὰνἀριθμὸςἀριθμοῦμέροςᾖ, καὶἕτεροςἑτέρουτὸ If a number is part of a number, and another (numαὐτὸμέροςᾖ,
καὶσυναμφότεροςσυναμφοτέρουτὸαὐτὸ ber) is the same part of another, then the sum (of the
μέροςἔσται,ὅπερὁεἷςτοῦἑνός.
leading numbers) will also be the same part of the sum
(of the following numbers) that one (number) is of another.
Β
Ε
B
E
Η
Θ
G
H
Α
Γ
∆
Ζ
ἈριθμὸςγὰρὁΑ[ἀριθμοῦ]τοῦΒΓμέροςἔστω,καὶ For let a number A be part of a [number] BC, and
A
C
D
F
199

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
ἕτεροςὁΔἑτέρουτοῦΕΖτὸαὐτὸμέρος,ὅπερὁΑτοῦ another (number) D (be) the same part of another (num-
ΒΓ·λέγω,ὅτικαὶσυναμφότεροςὁΑ,Δσυναμφοτέρουτοῦ ber) EF that A (is) of BC. I say that the sum A, D is also
ΒΓ,ΕΖτὸαὐτὸμέροςἐστίν,ὅπερὁΑτοῦΒΓ. the same part of the sum BC, EF that A (is) of BC.
᾿Επεὶγάρ,ὃμέροςἐστὶνὁΑτοῦΒΓ,τὸαὐτὸμέροςἐστὶ For since which(ever) part A is of BC, D is the same
καὶὁΔτοῦΕΖ,ὅσοιἄραεἰσὶνἐντῷΒΓἀριθμοὶἴσοιτῷ part of EF , thus as many numbers as are in BC equal
Α,τοσοῦτοίεἰσικαὶἐντῷΕΖἀριθμοὶἴσοιτῷΔ.διῇρήσθω to A, so many numbers are also in EF equal to D. Let
ὁμὲνΒΓεἰςτοὺςτῷΑἴσουςτοὺςΒΗ,ΗΓ,ὁδὲΕΖεἰς BC have been divided into BG and GC, equal to A, and
τοὺςτῷΔἴσουςτοὺςΕΘ,ΘΖ·ἔσταιδὴἴσοντὸπλῆθος EF into EH and HF , equal to D. So the multitude of
τῶνΒΗ,ΗΓτῷπλήθειτῶνΕΘ,ΘΖ.καὶἐπεὶἴσοςἐστὶν (divisions) BG, GC will be equal to the multitude of (di-
ὁμὲνΒΗτῷΑ,ὁδὲΕΘτῷΔ,καὶοἱΒΗ,ΕΘἄρατοῖς visions) EH, HF . And since BG is equal to A, and EH
Α,Δἴσοι. διὰτὰαὐτὰδὴκαὶοἱΗΓ,ΘΖτοῖςΑ,Δ.ὅσοι to D, thus BG, EH (is) also equal to A, D. So, for the
ἄρα[εἰσὶν]ἐντῷΒΓἀριθμοὶἴσοιτῷΑ,τοσοῦτοίεἰσικαὶ same (reasons), GC, HF (is) also (equal) to A, D. Thus,
ἐντοῖςΒΓ,ΕΖἴσοιτοῖςΑ,Δ.ὁσαπλασίωνἄραἐστὶνὁΒΓ as many numbers as [are] in BC equal to A, so many are
τοῦΑ,τοσαυταπλασίωνἐστὶκαὶσυναμφότεροςὁΒΓ,ΕΖ also in BC, EF equal to A, D. Thus, as many times as
συναμφοτέρουτοῦΑ,Δ.ὃἄραμέροςἐστὶνὁΑτοῦΒΓ,τὸ BC is (divisible) by A, so many times is the sum BC, EF
αὐτὸμέροςἐστὶκαὶσυναμφότεροςὁΑ,Δσυναμφοτέρου also (divisible) by the sum A, D. Thus, which(ever) part
τοῦΒΓ,ΕΖ·ὅπερἔδειδεῖξαι.
A is of BC, the sum A, D is also the same part of the
sum BC, EF . (Which is) the very thing it was required
to show.
† In modern notation, this proposition states that if a = (1/n) b and c = (1/n) d then (a+c) = (1/n) (b+d), where all symbols denote numbers.
¦. Proposition 6 †
᾿Εὰν ἀριθμὸς ἀριθμοῦ μέρη ᾖ, καὶ ἕτερος ἑτέρου τὰ If a number is parts of a number, and another (numαὐτὰ
μέρη ᾖ, καὶ συναμφότερος συναμφοτέρου τὰ αὐτὰ ber) is the same parts of another, then the sum (of the
μέρηἔσται,ὅπερὁεἷςτοῦἑνός.
leading numbers) will also be the same parts of the sum
(of the following numbers) that one (number) is of another.
Α
Η
∆
Θ
A
G
D
H
Β Ε
B E
Γ Ζ C
ἈριθμὸςγὰρὁΑΒἀριθμοῦτοῦΓμέρηἔστω,καὶἕτερος For let a number AB be parts of a number C, and an-
ὁΔΕἑτέρουτοῦΖτὰαὐτὰμέρη,ἅπερὁΑΒτοῦΓ·λέγω, other (number) DE (be) the same parts of another (num-
ὅτικαὶσυναμφότεροςὁΑΒ,ΔΕσυναμφοτέρουτοῦΓ,Ζ ber) F that AB (is) of C. I say that the sum AB, DE is
τὰαὐτὰμέρηἐστίν,ἅπερὁΑΒτοῦΓ. also the same parts of the sum C, F that AB (is) of C.
᾿Επεὶγάρ,ἃμέρηἐστὶνὁΑΒτοῦΓ,τὰαὐτὰμέρηκαὶ For since which(ever) parts AB is of C, DE (is) also
ὁΔΕτοῦΖ,ὅσαἄραἐστὶνἐντῷΑΒμέρητοῦΓ,τοσαῦτά the same parts of F , thus as many parts of C as are in AB,
ἐστικαὶἐντῷΔΕμέρητοῦΖ.διῃρήσθωὁμὲνΑΒεἰςτὰ so many parts of F are also in DE. Let AB have been
τοῦΓμέρητὰΑΗ,ΗΒ,ὁδὲΔΕεἰςτὰτοῦΖμέρητὰ divided into the parts of C, AG and GB, and DE into the
ΔΘ,ΘΕ·ἔσταιδὴἴσοντὸπλῆθοςτῶνΑΗ,ΗΒτῷπλήθει parts of F , DH and HE. So the multitude of (divisions)
τῶν ΔΘ,ΘΕ.καὶἐπεί, ὃμέροςἐστὶν ὁΑΗ τοῦΓ,τὸ AG, GB will be equal to the multitude of (divisions) DH,
F
200

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
αὑτὸμέροςἐστὶκαὶὁΔΘτοῦΖ,ὃἄραμέροςἐστὶνὁΑΗ HE. And since which(ever) part AG is of C, DH is also
τοῦΓ,τὸαὐτὸμέροςἐστὶκαὶσυναμφότεροςὁΑΗ,ΔΘ the same part of F , thus which(ever) part AG is of C,
συναμφοτέρουτοῦΓ,Ζ.διὰτὰαὐτὰδὴκαὶὃμέροςἐστὶνὁ the sum AG, DH is also the same part of the sum C, F
ΗΒτοῦΓ,τὸαὐτὸμέροςἐστὶκαὶσυναμφότεροςὁΗΒ,ΘΕ [Prop. 7.5]. And so, for the same (reasons), which(ever)
συναμφοτέρουτοῦΓ,Ζ.ἃἄραμέρηἐστὶνὁΑΒτοῦΓ,τὰ part GB is of C, the sum GB, HE is also the same part
αὐτὰμέρηἐστὶκαὶσυναμφότεροςὁΑΒ,ΔΕσυναμφοτέρου of the sum C, F . Thus, which(ever) parts AB is of C,
τοῦΓ,Ζ·ὅπερἔδειδεῖξαι. the sum AB, DE is also the same parts of the sum C, F .
(Which is) the very thing it was required to show.
† In modern notation, this proposition states that if a = (m/n) b and c = (m/n) d then (a + c) = (m/n) (b + d), where all symbols denote
numbers.
Þ.
Proposition 7 †
᾿Εὰνἀριθμὸςἀριθμοῦμέροςᾖ,ὅπερἀφαιρεθεὶςἀφαι- If a number is that part of a number that a (part)
ρεθέντος,καὶὁλοιπὸςτοῦλοιποῦτὸαὐτὸμέροςἔσται, taken away (is) of a (part) taken away then the remain-
ὅπερὁὅλοςτοῦὅλου.
der will also be the same part of the remainder that the
whole (is) of the whole.
Α
Ε
Β
A
E
B
Η
Γ Ζ ∆
G C F D
ἈριθμὸςγὰρὁΑΒἀριθμοῦτοῦΓΔμέροςἔστω,ὅπερ For let a number AB be that part of a number CD
ἀφαιρεθεὶςὁΑΕἀφαιρεθέντοςτοῦΓΖ·λέγω,ὅτικαὶλοιπὸς that a (part) taken away AE (is) of a part taken away
ὁΕΒλοιποῦτοῦΖΔτὸαὐτὸμέροςἐστίν,ὅπερὅλοςὁΑΒ CF . I say that the remainder EB is also the same part of
ὅλουτοῦΓΔ.
the remainder FD that the whole AB (is) of the whole
῝ΟγὰρμέροςἐστὶνὁΑΕτοῦΓΖ,τὸαὐτὸμέροςἔστω CD.
καὶὁΕΒτοῦΓΗ.καὶἐπεί,ὃμέροςἐστὶνὁΑΕτοῦΓΖ,τὸ For which(ever) part AE is of CF , let EB also be the
αὐτὸμέροςἐστὶκαὶὁΕΒτοῦΓΗ,ὃἄραμέροςἐστὶνὁΑΕ same part of CG. And since which(ever) part AE is of
τοῦΓΖ,τὸαὐτὸμέροςἐστὶκαὶὁΑΒτοῦΗΖ.ὃδὲμέρος CF , EB is also the same part of CG, thus which(ever)
ἐστὶνὁΑΕτοῦΓΖ,τὸαὐτὸμέροςὑπόκειταικαὶὁΑΒτοῦ part AE is of CF , AB is also the same part of GF
ΓΔ·ὃἄραμέροςἐστὶκαὶὁΑΒτοῦΗΖ,τὸαὐτὸμέροςἐστὶ [Prop. 7.5]. And which(ever) part AE is of CF , AB is
καὶτοῦΓΔ·ἴσοςἄραἐστὶνὁΗΖτῷΓΔ.κοινὸςἀφῃρήσθω also assumed (to be) the same part of CD. Thus, also,
ὁΓΖ·λοιπὸςἄραὁΗΓλοιπῷτῷΖΔἐστινἴσος.καὶἐπεί, which(ever) part AB is of GF , (AB) is also the same
ὃμέροςἐστὶνὁΑΕτοῦΓΖ,τὸαὐτὸμέρος[ἐστὶ]καὶὁΕΒ part of CD. Thus, GF is equal to CD. Let CF have been
τοῦΗΓ,ἴσοςδὲὁΗΓτῷΖΔ,ὃἄραμέροςἐστὶνὁΑΕτοῦ subtracted from both. Thus, the remainder GC is equal
ΓΖ,τὸαὐτὸμέροςἐστὶκαὶὁΕΒτοῦΖΔ.ἀλλὰὃμέρος to the remainder FD. And since which(ever) part AE is
ἐστὶνὁΑΕτοῦΓΖ,τὸαὐτὸμέροςἐστὶκαὶὁΑΒτοῦΓΔ· of CF , EB [is] also the same part of GC, and GC (is)
καὶλοιπὸςἄραὁΕΒλοιποῦτοῦΖΔτὸαὐτὸμέροςἐστίν, equal to FD, thus which(ever) part AE is of CF , EB is
ὅπερὅλοςὁΑΒὅλουτοῦΓΔ·ὅπερἔδειδεῖξαι.
also the same part of FD. But, which(ever) part AE is of
CF , AB is also the same part of CD. Thus, the remainder
EB is also the same part of the remainder FD that
the whole AB (is) of the whole CD. (Which is) the very
thing it was required to show.
† In modern notation, this proposition states that if a = (1/n) b and c = (1/n) d then (a −c) = (1/n) (b −d), where all symbols denote numbers.
. Proposition 8 †
᾿Εὰν ἀριθμὸς ἀριθμοῦ μέρη ᾖ, ἅπερ ἀφαιρεθεὶςἀφαι- If a number is those parts of a number that a (part)
ρεθέντος, καὶὁλοιπὸςτοῦλοιποῦτὰ αὐτὰμέρη ἔσται, taken away (is) of a (part) taken away then the remain-
ἅπερὁὅλοςτοῦὅλου.
der will also be the same parts of the remainder that the
201

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
Γ Ζ ∆
whole (is) of the whole.
C
F
D
Η
Μ Κ
ΝΘ
G
M
K
N
H
Α Λ Ε
Β
A
L
E
B
ἈριθμὸςγὰρὁΑΒἀριθμοῦτοῦΓΔμέρηἔστω,ἅπερ For let a number AB be those parts of a number CD
ἀφαιρεθεὶςὁΑΕἀφαιρεθέντοςτοῦΓΖ·λέγω,ὅτικαὶλοιπὸς that a (part) taken away AE (is) of a (part) taken away
ὁΕΒλοιποῦτοῦΖΔτὰαὐτὰμέρηἐστίν,ἅπερὅλοςὁΑΒ CF . I say that the remainder EB is also the same parts
ὅλουτοῦΓΔ.
of the remainder FD that the whole AB (is) of the whole
ΚείσθωγὰρτῷΑΒἴσοςὁΗΘ,ἃἄραμέρηἐστὶνὁΗΘ CD.
τοῦΓΔ,τὰαὐτὰμέρηἐστὶκαὶὁΑΕτοῦΓΖ.διῃρήσθωὁ For let GH be laid down equal to AB. Thus,
μὲνΗΘεἰςτὰτοῦΓΔμέρητὰΗΚ,ΚΘ,ὁδὲΑΕεἰςτὰτοῦ which(ever) parts GH is of CD, AE is also the same
ΓΖμέρητὰΑΛ,ΛΕ·ἔσταιδὴἴσοντὸπλῆθοςτῶνΗΚ,ΚΘ parts of CF . Let GH have been divided into the parts
τῷπλήθειτῶνΑΛ,ΛΕ.καὶἐπεί,ὃμέροςἐστὶνὁΗΚτοῦ of CD, GK and KH, and AE into the part of CF , AL
ΓΔ,τὸαὐτὸμέροςἐστὶκαὶὁΑΛτοῦΓΖ,μείζωνδὲὁΓΔ and LE. So the multitude of (divisions) GK, KH will be
τοῦΓΖ,μείζωνἄρακαὶὁΗΚτοῦΑΛ.κείσθωτῷΑΛἴσος equal to the multitude of (divisions) AL, LE. And since
ὁΗΜ.ὃἄραμέροςἐστὶνὁΗΚτοῦΓΔ,τὸαὐτὸμέροςἐστὶ which(ever) part GK is of CD, AL is also the same part
καὶὁΗΜτοῦΓΖ·καὶλοιπὸςἄραὁΜΚλοιποῦτοῦΖΔ of CF , and CD (is) greater than CF , GK (is) thus also
τὸαὐτὸμέροςἐστίν,ὅπερὅλοςὁΗΚὅλουτοῦΓΔ.πάλιν greater than AL. Let GM be made equal to AL. Thus,
ἐπεί,ὃμέροςἐστὶνὁΚΘτοῦΓΔ,τὸαὐτὸμέροςἐστὶκαὶὁ which(ever) part GK is of CD, GM is also the same part
ΕΛτοῦΓΖ,μείζωνδὲὁΓΔτοῦΓΖ,μείζωνἄρακαὶὁΘΚ of CF . Thus, the remainder MK is also the same part of
τοῦΕΛ.κείσθωτῷΕΛἴσοςὁΚΝ.ὃἄραμέροςἐστὶνὁΚΘ the remainder FD that the whole GK (is) of the whole
τοῦΓΔ,τὸαὐτὸμέροςἐστὶκαὶὁΚΝτοῦΓΖ·καὶλοιπὸς CD [Prop. 7.5]. Again, since which(ever) part KH is of
ἄραὁΝΘλοιποῦτοῦΖΔτὸαὐτὸμέροςἐστίν,ὅπερὅλοςὁ CD, EL is also the same part of CF , and CD (is) greater
ΚΘὅλουτοῦΓΔ.ἐδείχθηδὲκαὶλοιπὸςὁΜΚλοιποῦτοῦ than CF , HK (is) thus also greater than EL. Let KN be
ΖΔτὸαὐτὸμέροςὤν,ὅπερὅλοςὁΗΚὅλουτοῦΓΔ·καὶ made equal to EL. Thus, which(ever) part KH (is) of
συναμφότεροςἄραὁΜΚ,ΝΘτοῦΔΖτὰαὐτὰμέρηἐστίν, CD, KN is also the same part of CF . Thus, the remain-
ἅπερὅλοςὁΘΗὅλουτοῦΓΔ.ἴσοςδὲσυναμφότεροςμὲν der NH is also the same part of the remainder FD that
ὁΜΚ,ΝΘτῷΕΒ,ὁδὲΘΗτῷΒΑ·καὶλοιπὸςἄραὁΕΒ the whole KH (is) of the whole CD [Prop. 7.5]. And the
λοιποῦτοῦΖΔτὰαὐτὰμέρηἐστίν,ἅπερὅλοςὁΑΒὅλου remainder MK was also shown to be the same part of
τοῦΓΔ·ὅπερἔδειδεῖξαι.
the remainder FD that the whole GK (is) of the whole
CD. Thus, the sum MK, NH is the same parts of DF
that the whole HG (is) of the whole CD. And the sum
MK, NH (is) equal to EB, and HG to BA. Thus, the
remainder EB is also the same parts of the remainder
FD that the whole AB (is) of the whole CD. (Which is)
the very thing it was required to show.
† In modern notation, this proposition states that if a = (m/n) b and c = (m/n) d then (a − c) = (m/n) (b − d), where all symbols denote
numbers.
. Proposition 9 †
᾿Εὰνἀριθμὸςἀριθμοῦμέροςᾖ, καὶἕτεροςἑτέρουτὸ If a number is part of a number, and another (numαὐτὸμέροςᾖ,καὶἐναλλάξ,ὃμέροςἐστὶνἢμέρηὁπρῶτος
ber) is the same part of another, also, alternately,
τοῦ τρίτου, τὸ αὐτὸ μέρος ἔσται ἢ τὰ αὐτὰμέρη καὶ ὁ which(ever) part, or parts, the first (number) is of the
δεύτεροςτοῦτετάρτου.
third, the second (number) will also be the same part, or
202

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
Β
Ε
the same parts, of the fourth.
B
E
Α Γ Η
∆
Θ
Ζ
ἈριθμὸςγὰρὁΑἀριθμοῦτοῦΒΓμέροςἔστω,καὶἕτε- For let a number A be part of a number BC, and anροςὁΔἑτέρουτοῦΕΖτὸαὐτὸμέρος,ὅπερὁΑτοῦΒΓ·
other (number) D (be) the same part of another EF that
λέγω,ὅτικαὶἐναλλάξ,ὃμέροςἐστὶνὁΑτοῦΔἢμέρη,τὸ A (is) of BC. I say that, also, alternately, which(ever)
αὐτὸμέροςἐστὶκαὶὁΒΓτοῦΕΖἢμέρη.
part, or parts, A is of D, BC is also the same part, or
᾿ΕπεὶγὰρὃμέροςἐστὶνὁΑτοῦΒΓ,τὸαὐτὸμέροςἐστὶ parts, of EF .
καὶὁΔτοῦΕΖ,ὅσοιἄραεἰσὶνἐντῷΒΓἀριθμοὶἴσοιτῷ For since which(ever) part A is of BC, D is also the
Α,τοσοῦτοίεἰσικαὶἐντῷΕΖἴσοιτῷΔ.διῃρήσθωὁμὲν same part of EF , thus as many numbers as are in BC
ΒΓεἰςτοὺςτῷΑἴσουςτοὺςΒΗ,ΗΓ,ὁδὲΕΖεἰςτοὺςτῷ equal to A, so many are also in EF equal to D. Let BC
ΔἴσουςτοὺςΕΘ,ΘΖ·ἔσταιδὴἴσοντὸπλῆθοςτῶνΒΗ, have been divided into BG and GC, equal to A, and EF
ΗΓτῷπλήθειτῶνΕΘ,ΘΖ.
into EH and HF , equal to D. So the multitude of (di-
ΚαὶἐπεὶἴσοιεἰσὶνοἱΒΗ,ΗΓἀριθμοὶἀλλήλοις,εἰσὶ visions) BG, GC will be equal to the multitude of (diviδὲκαὶοἱΕΘ,ΘΖἀριθμοὶἴσοιἀλλήλοις,καίἐστινἴσοντὸ
sions) EH, HF .
πλῆθοςτῶνΒΗ,ΗΓτῷπλήθειτῶνΕΘ,ΘΖ,ὃἄραμέρος And since the numbers BG and GC are equal to one
ἐστὶνὁΒΗτοῦΕΘἢμέρη,τὸαὐτὸμέροςἐστὶκαὶὁΗΓ another, and the numbers EH and HF are also equal to
τοῦΘΖἢτὰαὐτὰμέρη·ὥστεκαὶὃμέροςἐστὶνὁΒΗτοῦ one another, and the multitude of (divisions) BG, GC
ΕΘἢμέρη,τὸαὐτὸμέροςἐστὶκαὶσυναμφότεροςὁΒΓ is equal to the multitude of (divisions) EH, HC, thus
συναμφοτέρουτοῦΕΖἢτὰαὐτὰμέρη. ἴσοςδὲὁμὲνΒΗ which(ever) part, or parts, BG is of EH, GC is also
τῷΑ,ὁδὲΕΘτῷΔ·ὃἄραμέροςἐστὶνὁΑτοῦΔἢμέρη, the same part, or the same parts, of HF . And hence,
τὸαὐτὸμέροςἐστὶκαὶὁΒΓτοῦΕΖἢτὰαὐτὰμέρη·ὅπερ which(ever) part, or parts, BG is of EH, the sum BC
ἔδειδεῖξαι.
is also the same part, or the same parts, of the sum EF
[Props. 7.5, 7.6]. And BG (is) equal to A, and EH to D.
Thus, which(ever) part, or parts, A is of D, BC is also
the same part, or the same parts, of EF . (Which is) the
very thing it was required to show.
† In modern notation, this proposition states that if a = (1/n) b and c = (1/n) d then if a = (k/l) c then b = (k/l) d, where all symbols denote
numbers.
. Proposition 10 †
᾿Εὰνἀριθμὸςἀριθμοῦμέρηᾖ,καὶἕτεροςἑτέρουτὰαὐτὰ If a number is parts of a number, and another (numμέρηᾖ,καὶἐναλλάξ,ἃμέρηἐστὶνὁπρῶτοςτοῦτρίτουἢ
ber) is the same parts of another, also, alternately,
μέρος,τὰαὐτὰμέρηἔσταικαὶὁδεύτεροςτοῦτετάρτουἢ which(ever) parts, or part, the first (number) is of the
τὸαὐτὸμέρος.
third, the second will also be the same parts, or the same
ἈριθμὸςγὰρὁΑΒἀριθμοῦτοῦΓμέρηἔστω,καὶἕτερος part, of the fourth.
ὁΔΕἑτέρουτοῦΖτὰαὐτὰμέρη·λέγω,ὅτικαὶἐναλλάξ, For let a number AB be parts of a number C, and
ἃμέρηἐστὶνὁΑΒτοῦΔΕἢμέρος,τὰαὐτὰμέρηἐστὶκαὶ another (number) DE (be) the same parts of another F .
ὁΓτοῦΖἢτὸαὐτὸμέρος.
I say that, also, alternately, which(ever) parts, or part,
A
C G
D
H
F
203

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
AB is of DE, C is also the same parts, or the same part,
of F .
Α
Η
∆
Θ
A
G
D
H
Β Γ
Ε
Ζ
᾿Επεὶγάρ,ἃμέρηἐστὶνὁΑΒτοῦΓ,τὰαὐτὰμέρηἐστὶ For since which(ever) parts AB is of C, DE is also
καὶὁΔΕτοῦΖ,ὅσαἄραἐστὶνἐντῷΑΒμέρητοῦΓ, the same parts of F , thus as many parts of C as are in
τοσαῦτακαὶἐντῷΔΕμέρητοῦΖ.διῃρήσθωὁμὲνΑΒεἰς AB, so many parts of F (are) also in DE. Let AB have
τὰτοῦΓμέρητὰΑΗ,ΗΒ,ὁδὲΔΕεἰςτὰτοῦΖμέρητὰ been divided into the parts of C, AG and GB, and DE
ΔΘ,ΘΕ·ἔσταιδὴἴσοντὸπλῆθοςτῶνΑΗ,ΗΒτῷπλήθει into the parts of F , DH and HE. So the multitude of
τῶνΔΘ,ΘΕ.καὶἐπεί,ὃμέροςἐστὶνὁΑΗτοῦΓ,τὸαὐτὸ (divisions) AG, GB will be equal to the multitude of (diμέροςἐστὶκαὶὁΔΘτοῦΖ,καὶἐναλλάξ,ὃμέροςἐστὶνὁ
visions) DH, HE. And since which(ever) part AG is
ΑΗτοῦΔΘἢμέρη,τὸαὐτὸμέροςἐστὶκαὶὁΓτοῦΖἢ of C, DH is also the same part of F , also, alternately,
τὰαὐτὰμέρη.διὰτὰαὐτὰδὴκαί,ὃμέροςἐστὶνὁΗΒτοῦ which(ever) part, or parts, AG is of DH, C is also the
ΘΕἢμέρη,τὸαὐτὸμέροςἐστὶκαὶὁΓτοῦΖἢτὰαὐτὰ same part, or the same parts, of F [Prop. 7.9]. And so,
μέρη·ὥστεκαί[ὃμέροςἐστὶνὁΑΗτοῦΔΘἢμέρη,τὸ for the same (reasons), which(ever) part, or parts, GB is
αὐτὸμέροςἐστὶκαὶὁΗΒτοῦΘΕἢτὰαὐτὰμέρη·καὶὃ of HE, C is also the same part, or the same parts, of F
ἄραμέροςἐστὶνὁΑΗτοῦΔΘἢμέρη,τὸαὐτὸμέροςἐστὶ [Prop. 7.9]. And so [which(ever) part, or parts, AG is of
καὶὁΑΒτοῦΔΕἢτὰαὐτὰμέρη·ἀλλ᾿ὃμέροςἐστὶνὁΑΗ DH, GB is also the same part, or the same parts, of HE.
τοῦΔΘἢμέρη,τὸαὐτὸμέροςἐδείχθηκαὶὁΓτοῦΖἢτὰ And thus, which(ever) part, or parts, AG is of DH, AB is
αὐτὰμέρη,καὶ]ἃ[ἄρα]μέρηἐστὶνὁΑΒτοῦΔΕἢμέρος, also the same part, or the same parts, of DE [Props. 7.5,
τὰαὐτὰμέρηἐστὶκαὶὁΓτοῦΖἢτὸαὐτὸμέρος·ὅπερἔδει 7.6]. But, which(ever) part, or parts, AG is of DH, C
δεῖξαι.
was also shown (to be) the same part, or the same parts,
of F . And, thus] which(ever) parts, or part, AB is of DE,
C is also the same parts, or the same part, of F . (Which
is) the very thing it was required to show.
† In modern notation, this proposition states that if a = (m/n) b and c = (m/n) d then if a = (k/l) c then b = (k/l) d, where all symbols denote
numbers.
.
Proposition 11
᾿Εανᾖὡςὅλοςπρὸςὅλον,οὕτωςἀφαιρεθεὶςπρὸςἀφαι- If as the whole (of a number) is to the whole (of anρεθέντα,καὶὁλοιπὸςπρὸςτὸνλοιπὸνἔσται,ὡςὅλοςπρὸς
other), so a (part) taken away (is) to a (part) taken away,
ὅλον.
then the remainder will also be to the remainder as the
῎ΕστωὡςὅλοςὁΑΒπρὸςὅλοντὸνΓΔ,οὕτωςἀφαι- whole (is) to the whole.
ρεθεὶςὁΑΕπρὸςἀφαιρεθέντατὸνΓΖ·λέγω,ὅτικαὶλοιπὸς Let the whole AB be to the whole CD as the (part)
ὁΕΒπρὸςλοιπὸντὸνΖΔἐστιν,ὡςὅλοςὁΑΒπρὸςὅλον taken away AE (is) to the (part) taken away CF . I say
τὸνΓΔ.
that the remainder EB is to the remainder FD as the
whole AB (is) to the whole CD.
B
C
E
F
204

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
Γ
C
Α
Ζ
A
F
Ε
E
Β
∆
᾿ΕπείἐστινὡςὁΑΒπρὸςτὸνΓΔ,οὕτωςὁΑΕπρὸς (For) since as AB is to CD, so AE (is) to CF , thus
τὸνΓΖ,ὃἄραμέροςἐστὶνὁΑΒτοῦΓΔἢμέρη,τὸαὐτὸ which(ever) part, or parts, AB is of CD, AE is also the
μέροςἐστὶκαὶὁΑΕτοῦΓΖἢτὰαὐτὰμέρη. καὶλοιπὸς same part, or the same parts, of CF [Def. 7.20]. Thus,
ἄραὁΕΒλοιποῦτοῦΖΔτὸαὐτὸμέροςἐστὶνἢμέρη,ἅπερ the remainder EB is also the same part, or parts, of the
ὁΑΒτοῦΓΔ.ἔστινἄραὡςὁΕΒπρὸςτὸνΖΔ,οὕτωςὁ remainder FD that AB (is) of CD [Props. 7.7, 7.8].
ΑΒπρὸςτὸνΓΔ·ὅπερἔδειδεῖξαι. Thus, as EB is to FD, so AB (is) to CD [Def. 7.20].
(Which is) the very thing it was required to show.
† In modern notation, this proposition states that if a : b :: c : d then a : b :: a − c : b − d, where all symbols denote numbers.
. Proposition 12 †
᾿Εὰνὦσινὁποσοιοῦνἀριθμοὶἀνάλογον,ἔσταιὡςεἷς If any multitude whatsoever of numbers are proporτῶνἡγουμένωνπρὸςἕνατῶνἑπομένων,οὕτωςἅπαντεςοἱ
tional then as one of the leading (numbers is) to one of
ἡγούμενοιπρὸςἅπανταςτοὺςἑπομένους.
the following so (the sum of) all of the leading (numbers)
will be to (the sum of) all of the following.
B
D
Α Β Γ ∆
A B C D
῎ΕστωσανὁποσοιοῦνἀριθμοὶἀνάλογονοἱΑ,Β,Γ,Δ, Let any multitude whatsoever of numbers, A, B, C,
ὡςὁΑπρὸςτὸνΒ,οὕτωςὁΓπρὸςτὸνΔ·λέγω,ὅτιἐστὶν D, be proportional, (such that) as A (is) to B, so C (is)
ὡςὁΑπρὸςτὸνΒ,οὕτωςοἱΑ,ΓπρὸςτοὺςΒ,Δ. to D. I say that as A is to B, so A, C (is) to B, D.
᾿ΕπεὶγάρἐστινὡςὁΑπρὸςτὸνΒ,οὕτωςὁΓπρὸς For since as A is to B, so C (is) to D, thus which(ever)
τὸνΔ,ὃἄραμέροςἐστὶνὁΑτοῦΒἢμέρη,τὸαὐτὸμέρος part, or parts, A is of B, C is also the same part, or parts,
ἐστὶκαὶὁΓτοῦΔἢμέρη. καὶσυναμφότεροςἄραὁΑ, of D [Def. 7.20]. Thus, the sum A, C is also the same
ΓσυναμφοτέρουτοῦΒ,Δτὸαὐτὸμέροςἐστὶνἢτὰαὐτὰ part, or the same parts, of the sum B, D that A (is) of B
μέρη,ἅπερὁΑτοῦΒ.ἔστινἄραὡςὁΑπρὸςτὸνΒ,οὕτως [Props. 7.5, 7.6]. Thus, as A is to B, so A, C (is) to B, D
οἱΑ,ΓπρὸςτοὺςΒ,Δ·ὅπερἔδειδεῖξαι.
[Def. 7.20]. (Which is) the very thing it was required to
show.
205

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
† In modern notation, this proposition states that if a : b :: c : d then a : b :: a + c : b + d, where all symbols denote numbers.
. Proposition 13 †
᾿Εὰν τέσσαρες ἀριθμοὶ ἀνάλογον ὦσιν, καὶ ἐναλλὰξ If four numbers are proportional then they will also
ἀνάλογονἔσονται.
be proportional alternately.
Α Β Γ ∆
A B C D
῎ΕστωσαντέσσαρεςἀριθμοὶἀνάλογονοἱΑ,Β,Γ,Δ, Let the four numbers A, B, C, and D be proportional,
ὡςὁΑπρὸςτὸνΒ,οὕτωςὁΓπρὸςτὸνΔ·λέγω,ὅτικαὶ (such that) as A (is) to B, so C (is) to D. I say that they
ἐναλλὰξἀνάλογονἔσονται,ὡςὁΑπρὸςτὸνΓ,οὕτωςὁΒ will also be proportional alternately, (such that) as A (is)
πρὸςτὸνΔ. to C, so B (is) to D.
᾿ΕπεὶγάρἐστινὡςὁΑπρὸςτὸνΒ,οὕτωςὁΓπρὸςτὸν For since as A is to B, so C (is) to D, thus which(ever)
Δ,ὃἄραμέροςἐστὶνὁΑτοῦΒἢμέρη,τὸαὐτὸμέροςἐστὶ part, or parts, A is of B, C is also the same part,
καὶὁΓτοῦΔἢτὰαὐτὰμέρη.ἐναλλὰξἄρα,ὃμέροςἐστὶν or the same parts, of D [Def. 7.20]. Thus, alterately,
ὁΑτοῦΓἢμέρη,τὸαὐτὸμέροςἐστὶκαὶὁΒτοῦΔἢτὰ which(ever) part, or parts, A is of C, B is also the same
αὐτὰμέρη.ἔστινἄραὡςὁΑπρὸςτὸνΓ,οὕτωςὁΒπρὸς part, or the same parts, of D [Props. 7.9, 7.10]. Thus, as
τὸνΔ·ὅπερἔδειδεῖξαι.
A is to C, so B (is) to D [Def. 7.20]. (Which is) the very
thing it was required to show.
† In modern notation, this proposition states that if a : b :: c : d then a : c :: b : d, where all symbols denote numbers.
. Proposition 14 †
᾿Εὰνὦσινὁποσοιοῦνἀριθμοὶκαὶἄλλοιαὐτοῖςἴσοιτὸ If there are any multitude of numbers whatsoever,
πλῆθοςσύνδυολαμβανόμενοικαὶἐντῷαὐτῷλόγῳ,καὶδι᾿ and (some) other (numbers) of equal multitude to them,
ἴσουἐντῷαὐτῷλόγῷἔσονται.
(which are) also in the same ratio taken two by two, then
they will also be in the same ratio via equality.
Α ∆
A
D
Β Ε
B
E
Γ Ζ
C
F
῎ΕστωσανὁποσοιοῦνἀριθμοὶοἱΑ,Β,Γκαὶἄλλοιαὐτοῖς Let there be any multitude of numbers whatsoever, A,
ἴσοιτὸπλῆθοςσύνδυολαμβανόμενοιἐντῷαὐτῷλόγῳοἱ B, C, and (some) other (numbers), D, E, F , of equal
Δ,Ε,Ζ,ὡςμὲνὁΑπρὸςτὸνΒ,οὕτωςὁΔπρὸςτὸνΕ, multitude to them, (which are) in the same ratio taken
ὡςδὲὁΒπρὸςτὸνΓ,οὕτωςὁΕπρὸςτὸνΖ·λέγω,ὅτι two by two, (such that) as A (is) to B, so D (is) to E,
καὶδι᾿ἴσουἐστὶνὡςὁΑπρὸςτὸνΓ,οὕτωςὁΔπρὸςτὸν and as B (is) to C, so E (is) to F . I say that also, via
Ζ. equality, as A is to C, so D (is) to F .
᾿ΕπεὶγάρἐστινὡςὁΑπρὸςτὸνΒ,οὕτωςὁΔπρὸς For since as A is to B, so D (is) to E, thus, alternately,
τὸνΕ,ἐναλλὰξἄραἐστὶνὡςὁΑπρὸςτὸνΔ,οὕτωςὁΒ as A is to D, so B (is) to E [Prop. 7.13]. Again, since
πρὸςτὸνΕ.πάλιν,ἐπείἐστινὡςὁΒπρὸςτὸνΓ,οὕτωςὁ as B is to C, so E (is) to F , thus, alternately, as B is
206

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
ΕπρὸςτὸνΖ,ἐναλλὰξἄραἐστὶνὡςὁΒπρὸςτὸνΕ,οὕτως to E, so C (is) to F [Prop. 7.13]. And as B (is) to E,
ὁΓπρὸςτὸνΖ.ὡςδὲὁΒπρὸςτὸνΕ,οὕτωςὁΑπρὸς so A (is) to D. Thus, also, as A (is) to D, so C (is) to F .
τὸνΔ·καὶὡςἄραὁΑπρὸςτὸνΔ,οὕτωςὁΓπρὸςτὸν Thus, alternately, as A is to C, so D (is) to F [Prop. 7.13].
Ζ·ἐναλλὰξἄραἐστὶνὡςὁΑπρὸςτὸνΓ,οὕτωςὁΔπρὸς (Which is) the very thing it was required to show.
τὸνΖ·ὅπερἔδειδεῖξαι.
† In modern notation, this proposition states that if a : b :: d : e and b : c :: e : f then a : c :: d : f, where all symbols denote numbers.
. Proposition 15
᾿Εὰνμονὰςἀριθμόντιναμετρῇ,ἰσακιςδὲἕτεροςἀριθμὸς If a unit measures some number, and another num-
ἄλλοντινὰἀριθμὸνμετρῇ,καὶἐναλλὰξἰσάκιςἡμονὰςτὸν ber measures some other number as many times, then,
τρίτονἀριθμὸνμετρήσεικαὶὁδεύτεροςτὸντέταρτον. also, alternately, the unit will measure the third number
as many times as the second (number measures) the
fourth.
Β Η Θ Γ
B G H C
Α
A
Ε Κ Λ Ζ
E K L F
∆
D
ΜονὰςγὰρἡΑἀριθμόντινατὸνΒΓμετρείτω,ἰσάκιςδὲ For let a unit A measure some number BC, and let
ἕτεροςἀριθμὸςὁΔἄλλοντινὰἀριθμὸντὸνΕΖμετρείτω· another number D measure some other number EF as
λέγω, ὅτικαὶἐναλλὰξἰσάκιςἡΑμονὰςτὸνΔἀριθμὸν many times. I say that, also, alternately, the unit A also
μετρεῖκαὶὁΒΓτὸνΕΖ.
measures the number D as many times as BC (measures)
᾿ΕπεὶγὰρἰσάκιςἡΑμονὰςτὸνΒΓἀριθμὸνμετρεῖκαὶὁ EF .
ΔτὸνΕΖ,ὅσαιἄραεἰσὶνἐντῷΒΓμονάδες,τοσοῦτοίεἰσι For since the unit A measures the number BC as
καὶἐντῷΕΖἀριθμοὶἴσοιτῷΔ.διῃρήσθωὁμὲνΒΓεἰςτὰς many times as D (measures) EF , thus as many units as
ἐνἑαυτῷμονάδαςτὰςΒΗ,ΗΘ,ΘΓ,ὁδὲΕΖεἰςτοὺςτῷΔ are in BC, so many numbers are also in EF equal to
ἴσουςτοὺςΕΚ,ΚΛ,ΛΖ.ἔσταιδὴἴσοντὸπλῆθοςτῶνΒΗ, D. Let BC have been divided into its constituent units,
ΗΘ,ΘΓτῷπλήθειτῶνΕΚ,ΚΛ,ΛΖ.καὶἐπεὶἴσαιεἰσὶναἱ BG, GH, and HC, and EF into the (divisions) EK, KL,
ΒΗ,ΗΘ,ΘΓμονάδεςἀλλήλαις,εἰσὶδὲκαὶοἱΕΚ,ΚΛ,ΛΖ and LF , equal to D. So the multitude of (units) BG,
ἀριθμοὶἴσοιἀλλήλοις,καίἐστινἴσοντὸπλῆθοςτῶνΒΗ, GH, HC will be equal to the multitude of (divisions)
ΗΘ,ΘΓμονάδωντῷπλήθειτῶνΕΚ,ΚΛ,ΛΖἀριθμῶν, EK, KL, LF . And since the units BG, GH, and HC
ἔσταιἄραὡςἡΒΗμονὰςπρὸςτὸνΕΚἀριθμόν,οὕτωςἡ are equal to one another, and the numbers EK, KL, and
ΗΘμονὰςπρὸςτὸνΚΛἀριθμὸνκαὶἡΘΓμονὰςπρὸςτὸν LF are also equal to one another, and the multitude of
ΛΖἀριθμόν.ἔσταιἄρακαὶὡςεἷςτῶνἡγουμένωνπρὸςἕνα the (units) BG, GH, HC is equal to the multitude of the
τῶνἑπομένων,οὕτωςἅπαντεςοἱἡγούμενοιπρὸςἅπαντας numbers EK, KL, LF , thus as the unit BG (is) to the
τοὺςἑπομένους·ἔστινἄραὡςἡΒΗμονὰςπρὸςτὸνΕΚ number EK, so the unit GH will be to the number KL,
ἀριθμόν,οὕτωςὁΒΓπρὸςτὸνΕΖ.ἴσηδὲἡΒΗμονὰςτῇ and the unit HC to the number LF . And thus, as one of
Αμονάδι,ὁδὲΕΚἀριθμὸςτῷΔἀριθμῷ. ἔστινἄραὡςἡ the leading (numbers is) to one of the following, so (the
ΑμονὰςπρὸςτὸνΔἀριθμόν,οὕτωςὁΒΓπρὸςτὸνΕΖ. sum of) all of the leading will be to (the sum of) all of
ἰσάκιςἄραἡΑμονὰςτὸνΔἀριθμὸνμετρεῖκαὶὁΒΓτὸν the following [Prop. 7.12]. Thus, as the unit BG (is) to
ΕΖ·ὅπερἔδειδεῖξαι.
the number EK, so BC (is) to EF . And the unit BG (is)
equal to the unit A, and the number EK to the number
D. Thus, as the unit A is to the number D, so BC (is) to
EF . Thus, the unit A measures the number D as many
times as BC (measures) EF [Def. 7.20]. (Which is) the
very thing it was required to show.
† This proposition is a special case of Prop. 7.9.
207

ËÌÇÁÉÁÏÆÞº ELEMENTS BOOK 7
¦. Proposition 16 †
Εὰνδύοἀριθμοὶπολλαπλασιάσαντεςἀλλήλουςποιῶσί If two numbers multiplying one another make some
τινας,οἱγενόμενοιἐξαὐτῶνἴσοιἀλλήλοιςἔσονται. (numbers) then the (numbers) generated from them will
be equal to one another.
Α
Β
Γ
∆
Ε
῎ΕστωσανδύοἀριθμοὶοἱΑ,Β,καὶὁμὲνΑτὸνΒπολ- Let A and B be two numbers. And let A make C (by)
λαπλασιάσαςτὸνΓποιείτω,ὁδὲΒτὸνΑπολλαπλασιάσας multiplying B, and let B make D (by) multiplying A. I
τὸνΔποιείτω·λέγω,ὅτιἴσοςἐστὶνὁΓτῷΔ. say that C is equal to D.
᾿ΕπεὶγὰρὁΑτὸνΒπολλαπλασιάσαςτὸνΓπεποίηκεν, For since A has made C (by) multiplying B, B thus
ὁΒἄρατὸνΓμετρεῖκατὰτὰςἐντῷΑμονάδας.μετρεῖδὲ measures C according to the units in A [Def. 7.15]. And
καὶἡΕμονὰςτὸνΑἀριθμὸνκατὰτὰςἐναὐτῷμονάδας· the unit E also measures the number A according to the
ἰσάκιςἄραἡΕμονὰςτὸνΑἀριθμὸνμετρεῖκαὶὁΒτὸν units in it. Thus, the unit E measures the number A as
Γ.ἐναλλὰξἄραἰσάκιςἡΕμονὰςτὸνΒἀριθμὸνμετρεῖκαὶ many times as B (measures) C. Thus, alternately, the
ὁΑτὸνΓ.πάλιν,ἐπεὶὁΒτὸνΑπολλαπλασιάσαςτὸνΔ unit E measures the number B as many times as A (meaπεποίηκεν,ὁΑἄρατὸνΔμετρεῖκατὰτὰςἐντῷΒμονάδας.
sures) C [Prop. 7.15]. Again, since B has made D (by)
μετρεῖδὲκαὶἡΕμονὰςτὸνΒκατὰτὰςἐναὐτῷμονάδας· multiplying A, A thus measures D according to the units
ἰσάκιςἄραἡΕμονὰςτὸνΒἀριθμὸνμετρεῖκαὶὁΑτὸνΔ. in B [Def. 7.15]. And the unit E also measures B ac-
ἰσάκιςδὲἡΕμονὰςτὸνΒἀριθμὸνἐμέτρεικαὶὁΑτὸνΓ· cording to the units in it. Thus, the unit E measures the
ἰσάκιςἄραὁΑἑκάτεροντῶνΓ,Δμετρεῖ. ἴσοςἄραἐστὶν number B as many times as A (measures) D. And the
ὁΓτῷΔ·ὅπερἔδειδεῖξαι.
unit E was measuring the number B as many times as
A (measures) C. Thus, A measures each of C and D an
equal number of times. Thus, C is equal to D. (Which is)
the very thing it was required to show.
† In modern notation, this proposition states that a b = b a, where all symbols denote numbers.
Þ. Proposition 17 †
᾿Εὰνἀριθμὸςδύοἀριθμοὺςπολλαπλασιάσαςποιῇτινας, If a number multiplying two numbers makes some
οἱγενόμενοιἐξαὐτῶντὸναὐτὸνἕξουσιλόγοντοῖςπολλα- (numbers) then the (numbers) generated from them will
πλασιασθεῖσιν.
have the same ratio as the multiplied (numbers).
Α
A
Β Γ
B
C
∆
Ζ
Ε
D
F
E
Ἀριθμὸς γὰρ ὁ Α δύο ἀριθμοὺς τοὺς Β, Γ πολλα- For let the number A make (the numbers) D and
πλασιάσαςτοὺςΔ,Εποιείτω·λέγω,ὅτιἐστὶνὡςὁΒπρὸς E (by) multiplying the two numbers B and C (respecτὸνΓ,οὕτωςὁΔπρὸςτὸνΕ.
tively). I say that as B is to C, so D (is) to E.
᾿ΕπεὶγὰρὁΑτὸνΒπολλαπλασιάσαςτὸνΔπεποίηκεν, For since A has made D (by) multiplying B, B thus
ὁΒἄρατὸνΔμετρεῖκατὰτὰςἐντῷΑμονάδας. μετρεῖ measures D according to the units in A [Def. 7.15]. And
A
B
C
D
E
208

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
δὲκαὶἡΖμονὰςτὸνΑἀριθμὸνκατὰτὰςἐναὐτῷμονάδας· the unit F also measures the number A according to the
ἰσάκιςἄραἡΖμονὰςτὸνΑἀριθμὸνμετρεῖκαὶὁΒτὸνΔ. units in it. Thus, the unit F measures the number A as
ἔστινἄραὡςἡΖμονὰςπρὸςτὸνΑἀριθμόν,οὕτωςὁΒ many times as B (measures) D. Thus, as the unit F is
πρὸςτὸνΔ.διὰτὰαὐτὰδὴκαὶὡςἡΖμονὰςπρὸςτὸνΑ to the number A, so B (is) to D [Def. 7.20]. And so, for
ἀριθμόν,οὕτωςὁΓπρὸςτὸνΕ·καὶὡςἄραὁΒπρὸςτὸν the same (reasons), as the unit F (is) to the number A,
Δ,οὕτωςὁΓπρὸςτὸνΕ.ἐναλλὰξἄραἐστὶνὡςὁΒπρὸς so C (is) to E. And thus, as B (is) to D, so C (is) to E.
τὸνΓ,οὕτωςὁΔπρὸςτὸνΕ·ὅπερἔδειδεῖξαι. Thus, alternately, as B is to C, so D (is) to E [Prop. 7.13].
(Which is) the very thing it was required to show.
† In modern notation, this proposition states that if d = a b and e = a c then d : e :: b : c, where all symbols denote numbers.
. Proposition 18 †
᾿Εὰν δύο ἀριθμοὶ ἀριθμόν τινα πολλαπλασιάσαντες If two numbers multiplying some number make some
ποιῶσίτινας,οἱγενόμενοιἐξαὐτῶντὸναὐτὸνἕξουσιλόγον (other numbers) then the (numbers) generated from
τοῖςπολλαπλασιάσασιν.
them will have the same ratio as the multiplying (numbers).
Α
Β
Γ
∆
Ε
ΔύογὰρἀριθμοὶοἱΑ,ΒἀριθμόντινατὸνΓπολλα- For let the two numbers A and B make (the numbers)
πλασιάσαντεςτοὺςΔ,Εποιείτωσαν·λέγω,ὅτιἐστὶνὡςὁ D and E (respectively, by) multiplying some number C.
ΑπρὸςτὸνΒ,οὕτωςὁΔπρὸςτὸνΕ. I say that as A is to B, so D (is) to E.
᾿ΕπεὶγὰρὁΑτὸνΓπολλαπλασιάσαςτὸνΔπεποίηκεν, For since A has made D (by) multiplying C, C has
καὶὁΓἄρατὸνΑπολλαπλασιάσαςτὸνΔπεποίηκεν. διὰ thus also made D (by) multiplying A [Prop. 7.16]. So, for
τὰαὐτὰδὴκαὶὁΓτὸνΒπολλαπλασιάσαςτὸνΕπεποίηκεν. the same (reasons), C has also made E (by) multiplying
ἀριθμὸςδὴὁΓδύοἀριθμοὺςτοὺςΑ,Βπολλαπλασιάσας B. So the number C has made D and E (by) multiplying
τοὺςΔ,Επεποίηκεν.ἔστινἄραὡςὁΑπρὸςτὸνΒ,οὕτως the two numbers A and B (respectively). Thus, as A is to
ὁΔπρὸςτὸνΕ·ὅπερἔδειδεῖξαι.
B, so D (is) to E [Prop. 7.17]. (Which is) the very thing
it was required to show.
† In modern notation, this propositions states that if a c = d and b c = e then a : b :: d : e, where all symbols denote numbers.
. Proposition 19 †
᾿Εὰντέσσαρεςἀριθμοὶἀνάλογονὦσιν,ὁἐκπρώτουκαὶ If four number are proportional then the number creτετάρτουγενόμενοςἀριθμὸςἴσοςἔσταιτῷἐκδευτέρουκαὶ
ated from (multiplying) the first and fourth will be equal
τρίτουγενομένῳἀριθμῷ·καὶἐὰνὁἐκπρώτουκαὶτετάρτου to the number created from (multiplying) the second and
γενόμενοςἀριθμὸςἴσοςᾖτῷἐκδευτέρουκαὶτρίτου,οἱ third. And if the number created from (multiplying) the
τέσσασρεςἀριθμοὶἀνάλογονἔσονται.
first and fourth is equal to the (number created) from
῎ΕστωσαντέσσαρεςἀριθμοὶἀνάλογονοἱΑ,Β,Γ,Δ, (multiplying) the second and third then the four num-
ὡςὁΑπρὸςτὸνΒ,οὕτωςὁΓπρὸςτὸνΔ,καὶὁμὲνΑ bers will be proportional.
τὸνΔπολλαπλασιάσαςτὸνΕποιείτω,ὁδὲΒτὸνΓπολ- Let A, B, C, and D be four proportional numbers,
λαπλασιάσαςτὸνΖποιείτω·λέγω,ὅτιἴσοςἐστὶνὁΕτῷΖ. (such that) as A (is) to B, so C (is) to D. And let A make
E (by) multiplying D, and let B make F (by) multiplying
C. I say that E is equal to F .
A
B
C
D
E
209

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
Α Β Γ ∆ Ε Ζ Η
A B C D E F G
῾ΟγὰρΑτὸνΓπολλαπλασιάσαςτὸνΗποιείτω. ἐπεὶ For let A make G (by) multiplying C. Therefore, since
οὖνὁΑτὸνΓπολλαπλασιάσαςτὸνΗπεποίηκεν,τὸνδὲ A has made G (by) multiplying C, and has made E (by)
ΔπολλαπλασιάσαςτὸνΕπεποίηκεν,ἀριθμὸςδὴὁΑδύο multiplying D, the number A has made G and E by mul-
ἀριθμοὺςτοὺςΓ,ΔπολλαπλασιάσαςτούςΗ,Επεποίηκεν. tiplying the two numbers C and D (respectively). Thus,
ἔστινἄραὡςὁΓπρὸςτὸνΔ,οὕτωςὁΗπρὸςτὸνΕ.ἀλλ᾿ as C is to D, so G (is) to E [Prop. 7.17]. But, as C (is) to
ὡςὁΓπρὸςτὸνΔ,οὕτωςὁΑπρὸςτὸνΒ·καὶὡςἄρα D, so A (is) to B. Thus, also, as A (is) to B, so G (is) to
ὁΑπρὸςτὸνΒ,οὕτωςὁΗπρὸςτὸνΕ.πάλιν,ἐπεὶὁΑ E. Again, since A has made G (by) multiplying C, but,
τὸνΓπολλαπλασιάσαςτὸνΗπεποίηκεν,ἀλλὰμὴνκαὶὁ in fact, B has also made F (by) multiplying C, the two
ΒτὸνΓπολλαπλασιάσαςτὸνΖπεποίηκεν,δύοδὴἀριθμοὶ numbers A and B have made G and F (respectively, by)
οἱΑ,ΒἀριθμόντινατὸνΓπολλαπλασιάσαντεςτοὺςΗ,Ζ multiplying some number C. Thus, as A is to B, so G (is)
πεποιήκασιν.ἔστινἄραὡςὁΑπρὸςτὸνΒ,οὕτωςὁΗπρὸς to F [Prop. 7.18]. But, also, as A (is) to B, so G (is) to
τὸνΖ.ἀλλὰμὴνκαὶὡςὁΑπρὸςτὸνΒ,οὕτωςὁΗπρὸς E. And thus, as G (is) to E, so G (is) to F . Thus, G has
τὸνΕ·καὶὡςἄραὁΗπρὸςτὸνΕ,οὕτωςὁΗπρὸςτὸν the same ratio to each of E and F . Thus, E is equal to F
Ζ.ὁΗἄραπρὸςἑκάτεροντῶνΕ,Ζτὸναὐτὸνἔχειλόγον· [Prop. 5.9].
ἴσοςἄραἐστὶνὁΕτῷΖ. So, again, let E be equal to F . I say that as A is to B,
῎ΕστωδὴπάλινἴσοςὁΕτῷΖ·λέγω,ὅτιἐστὶνὡςὁΑ so C (is) to D.
πρὸςτὸνΒ,οὕτωςὁΓπρὸςτὸνΔ. For, with the same construction, since E is equal to F ,
Τῶνγὰραὐτῶνκατασκευασθέντων,ἐπεὶἴσοςἐστὶνὁ thus as G is to E, so G (is) to F [Prop. 5.7]. But, as G
ΕτῷΖ,ἔστινἄραὡςὁΗπρὸςτὸνΕ,οὕτωςὁΗπρὸςτὸν (is) to E, so C (is) to D [Prop. 7.17]. And as G (is) to F ,
Ζ.ἀλλ᾿ὡςμὲνὁΗπρὸςτὸνΕ,οὕτωςὁΓπρὸςτὸνΔ,ὡς so A (is) to B [Prop. 7.18]. And, thus, as A (is) to B, so
δὲὁΗπρὸςτὸνΖ,οὕτωςὁΑπρὸςτὸνΒ.καὶὡςἄραὁΑ C (is) to D. (Which is) the very thing it was required to
πρὸςτὸνΒ,οὕτωςὁΓπρὸςτὸνΔ·ὅπερἔδειδεῖξαι. show.
† In modern notation, this proposition reads that if a : b :: c : d then a d = bc, and vice versa, where all symbols denote numbers.
. Proposition 20
Οἱ ἐλάχιστοιἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων The least numbers of those (numbers) having the
αὐτοῖςμετροῦσιτοὺςτὸναὐτὸνλόγονἔχονταςἰσάκιςὅ same ratio measure those (numbers) having the same raτεμείζωντὸνμείζονακαὶὁἐλάσσωντὸνἐλάσσονα.
tio as them an equal number of times, the greater (mea-
῎Εστωσανγὰρἐλάχιστοιἀριθμοὶτῶντὸναὐτὸνλόγον suring) the greater, and the lesser the lesser.
ἐχόντωντοῖςΑ,ΒοἱΓΔ,ΕΖ·λέγω,ὅτιἰσάκιςὁΓΔτὸν For let CD and EF be the least numbers having the
ΑμετρεῖκαὶὁΕΖτὸνΒ.
same ratio as A and B (respectively). I say that CD measures
A the same number of times as EF (measures) B.
210

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
Α Β Γ
Η
Ε
E
Θ
A B C
G
H
∆
Ζ
D
F
῾ΟΓΔγὰρτοῦΑοὔκἐστιμέρη.εἰγὰρδυνατόν,ἔστω· For CD is not parts of A. For, if possible, let it be
καὶὁΕΖἄρατοῦΒτὰαὐτὰμέρηἐστίν,ἅπερὁΓΔτοῦ (parts of A). Thus, EF is also the same parts of B that
Α.ὅσαἄραἐστὶνἐντῷΓΔμέρητοῦΑ,τοσαῦτάἐστικαὶ CD (is) of A [Def. 7.20, Prop. 7.13]. Thus, as many parts
ἐντῷΕΖμέρητοῦΒ.διῃρήσθωὁμὲνΓΔεἰςτὰτοῦΑ of A as are in CD, so many parts of B are also in EF . Let
μέρητὰΓΗ,ΗΔ,ὁδὲΕΖεἰςτὰτοῦΒμέρητὰΕΘ,ΘΖ· CD have been divided into the parts of A, CG and GD,
ἔσταιδὴἴσοντὸπλῆθοςτῶνΓΗ,ΗΔτῷπλήθειτῶνΕΘ, and EF into the parts of B, EH and HF . So the multi-
ΘΖ.καὶἐπεὶἴσοιεἰσὶνοἱΓΗ,ΗΔἀριθμοὶἀλλήλοις,εἰσὶ tude of (divisions) CG, GD will be equal to the multitude
δὲκαὶοἱΕΘ,ΘΖἀριθμοὶἴσοιἀλλήλοις,καίἐστινἴσοντὸ of (divisions) EH, HF . And since the numbers CG and
πλῆθοςτῶνΓΗ,ΗΔτῷπλήθειτῶνΕΘ,ΘΖ,ἔστινἄραὡς GD are equal to one another, and the numbers EH and
ὁΓΗπρὸςτὸνΕΘ,οὕτωςὁΗΔπρὸςτὸνΘΖ.ἔσταιἄρα HF are also equal to one another, and the multitude of
καὶὡςεἷςτῶνἡγουμένωνπρὸςἕνατῶνἑπομένων,οὕτως (divisions) CG, GD is equal to the multitude of (divi-
ἅπαντεςοἱἡγούμενοιπρὸςἅπανταςτοὺςἑπομένους.ἔστιν sions) EH, HF , thus as CG is to EH, so GD (is) to HF .
ἄραὡςὁΓΗπρὸςτὸνΕΘ,οὕτωςὁΓΔπρὸςτὸνΕΖ·οἱ Thus, as one of the leading (numbers is) to one of the
ΓΗ,ΕΘἄρατοῖςΓΔ,ΕΖἐντῷαὐτῷλόγῳεἰσὶνἐλάσσονες following, so will (the sum of) all of the leading (num-
ὄντεςαὐτῶν·ὅπερἐστὶνἀδύνατον·ὑπόκεινταιγὰροἱΓΔ, bers) be to (the sum of) all of the following [Prop. 7.12].
ΕΖἐλάχιστοιτῶντὸναὐτὸνλόγονἐχόντωναὐτοῖς. οὐκ Thus, as CG is to EH, so CD (is) to EF . Thus, CG
ἄραμέρηἐστὶνὁΓΔτοῦΑ·μέροςἄρα.καὶὁΕΖτοῦΒτὸ and EH are in the same ratio as CD and EF , being less
αὐτὸμέροςἐστίν,ὅπερὁΓΔτοῦΑ·ἰσάκιςἄραὁΓΔτὸν than them. The very thing is impossible. For CD and
ΑμετρεῖκαὶὁΕΖτὸνΒ·ὅπερἔδειδεῖξαι.
EF were assumed (to be) the least of those (numbers)
having the same ratio as them. Thus, CD is not parts of
A. Thus, (it is) a part (of A) [Prop. 7.4]. And EF is the
same part of B that CD (is) of A [Def. 7.20, Prop 7.13].
Thus, CD measures A the same number of times that EF
(measures) B. (Which is) the very thing it was required
to show.
. Proposition 21
Οἱπρῶτοιπρὸςἀλλήλουςἀριθμοὶἐλάχιστοίεἰσιτῶντὸν Numbers prime to one another are the least of those
αὐτὸνλόγονἐχόντωναὐτοῖς.
(numbers) having the same ratio as them.
῎ΕστωσανπρῶτοιπρὸςἀλλήλουςἀριθμοὶοἱΑ,Β·λέγω, Let A and B be numbers prime to one another. I say
ὅτιοἱΑ,Βἐλάχιστοίεἰσιτῶντὸναὐτὸνλόγονἐχόντων that A and B are the least of those (numbers) having the
αὐτοῖς.
same ratio as them.
Εἰγὰρμή,ἔσονταίτινεςτῶνΑ,Βἐλάσσονεςἀριθμοὶ For if not then there will be some numbers less than A
ἐντῷαὐτῷλόγῳὄντεςτοῖςΑ,Β.ἔστωσανοἱΓ,Δ. and B which are in the same ratio as A and B. Let them
be C and D.
211

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
Α Β Γ ∆ Ε
A B C D E
᾿Επεὶοὖνοἱἐλάχιστοιἀριθμοὶτῶντὸναὐτὸνλόγον Therefore, since the least numbers of those (num-
ἐχόντωνμετροῦσιτοὺςτὸναὐτὸνλόγονἔχονταςἰσάκις bers) having the same ratio measure those (numbers)
ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάττων τὸν ἐλάττονα, having the same ratio (as them) an equal number of
τουτέστινὅτεἡγούμενοςτὸνἡγούμενονκαὶὁἑπόμενος times, the greater (measuring) the greater, and the lesser
τὸνἑπόμενον,ἰσάκιςἄραὁΓτὸνΑμετρεῖκαὶὁΔτὸνΒ. the lesser—that is to say, the leading (measuring) the
ὁσάκιςδὴὁΓτὸνΑμετρεῖ,τοσαῦταιμονάδεςἔστωσανἐν leading, and the following the following—C thus meaτῷΕ.καὶὁΔἄρατὸνΒμετρεῖκατὰτὰςἐντῷΕμονάδας.
sures A the same number of times that D (measures) B
καὶἐπεὶὁΓτὸνΑμετρεῖκατὰτὰςἐντῷΕμονάδας,καίὁ [Prop. 7.20]. So as many times as C measures A, so many
ΕἄρατὸνΑμετρεῖκατὰτὰςἐντῷΓμονάδας.διὰτὰαὐτὰ units let there be in E. Thus, D also measures B accordδὴὁΕκαὶτὸνΒμετρεῖκατὰτὰςἐντῷΔμονάδας.
ὁΕ ing to the units in E. And since C measures A according
ἄρατοὺςΑ,Βμετρεῖπρώτουςὄνταςπρὸςἀλλήλους·ὅπερ to the units in E, E thus also measures A according to
ἐστὶνἀδύνατον.οὐκἄραἔσονταίτινεςτῶνΑ,Βἐλάσσονες the units in C [Prop. 7.16]. So, for the same (reasons), E
ἀριθμοὶἐντῷαὐτῷλόγῳὄντεςτοῖςΑ,Β.οἱΑ,Βἄρα also measures B according to the units in D [Prop. 7.16].
ἐλάχιστοίεἰσιτῶντὸναὐτὸνλόγονἐχόντωναὐτοῖς·ὅπερ Thus, E measures A and B, which are prime to one an-
ἔδειδεῖξαι.
other. The very thing is impossible. Thus, there cannot
be any numbers less than A and B which are in the same
ratio as A and B. Thus, A and B are the least of those
(numbers) having the same ratio as them. (Which is) the
very thing it was required to show.
. Proposition 22
Οἱ ἐλάχιστοιἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων The least numbers of those (numbers) having the
αὐτοῖςπρῶτοιπρὸςἀλλήλουςεἰσίν.
same ratio as them are prime to one another.
Α
Β
Γ
∆
Ε
῎Εστωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον Let A and B be the least numbers of those (numbers)
ἐχόντωναὐτοῖςοἱΑ,Β·λέγω,ὅτιοἱΑ,Βπρῶτοιπρὸς having the same ratio as them. I say that A and B are
ἀλλήλουςεἰσίν.
prime to one another.
Εἰ γὰρ μή εἰσι πρῶτοι πρὸς ἀλλήλους, μετρήσει τις For if they are not prime to one another then some
αὐτοὺςἀριθμός. μετρείτω,καὶἔστωὁΓ.καὶὁσάκιςμὲν number will measure them. Let it (so measure them),
ὁΓτὸνΑμετρεῖ, τοσαῦταιμονάδεςἔστωσανἐντῷΔ, and let it be C. And as many times as C measures A, so
A
B
C
D
E
212

A
ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
ὁσάκιςδὲὁΓτὸνΒμετρεῖ,τοσαῦταιμονάδεςἔστωσανἐν many units let there be in D. And as many times as C
τῷΕ. measures B, so many units let there be in E.
᾿ΕπεὶὁΓτὸνΑμετρεῖκατὰτὰςἐντῷΔμονάδας,ὁ Since C measures A according to the units in D, C
ΓἄρατὸνΔπολλαπλασιάσαςτὸνΑπεποίηκεν. διὰτὰ has thus made A (by) multiplying D [Def. 7.15]. So, for
αὐτὰδὴκαὶὁΓτὸνΕπολλαπλασιάσαςτὸνΒπεποίηκεν. the same (reasons), C has also made B (by) multiplying
ἀριθμὸςδὴὁΓδύοἀριθμοὺςτοὺςΔ,Επολλαπλασιάσας E. So the number C has made A and B (by) multiplying
τοὺςΑ,Βπεποίηκεν·ἔστινἄραὡςὁΔπρὸςτὸνΕ,οὕτως the two numbers D and E (respectively). Thus, as D is
ὁΑπρὸςτὸνΒ·οἱΔ,ΕἄρατοῖςΑ,Βἐντῷαὐτῷλόγῳ to E, so A (is) to B [Prop. 7.17]. Thus, D and E are in
εἰσὶνἐλάσσονεςὄντεςαὐτῶν· ὅπερἐστὶνἀδύνατον. οὐκ the same ratio as A and B, being less than them. The
ἄρατοὺςΑ,Βἀριθμοὺςἀριθμόςτιςμετρήσει.οἱΑ,Βἄρα very thing is impossible. Thus, some number does not
πρῶτοιπρὸςἀλλήλουςεἰσίν·ὅπερἔδειδεῖξαι.
measure the numbers A and B. Thus, A and B are prime
to one another. (Which is) the very thing it was required
to show.
. Proposition 23
᾿Εὰνδύοἀριθμοὶπρῶτοιπρὸςἀλλήλουςὦσιν,ὁτὸνἕνα If two numbers are prime to one another then a numαὐτῶνμετρῶνἀριθμὸςπρὸςτὸνλοιπὸνπρῶτοςἔσται.
ber measuring one of them will be prime to the remaining
(one).
Α Β Γ ∆
.
B C D
῎ΕστωσανδύοἀριθμοὶπρῶτοιπρὸςἀλλήλουςοἱΑ,Β, Let A and B be two numbers (which are) prime to
τὸνδὲΑμετρείτωτιςἀριθμὸςὁΓ·λέγω,ὅτικαὶοἱΓ,Β one another, and let some number C measure A. I say
πρῶτοιπρὸςἀλλήλουςεἰσίν.
that C and B are also prime to one another.
ΕἰγὰρμήεἰσινοἱΓ,Βπρῶτοιπρὸςἀλλήλους,μετρήσει For if C and B are not prime to one another then
[τις]τοὺςΓ,Βἀριθμός. μετρείτω,καὶἔστωὁΔ.ἐπεὶὁ [some] number will measure C and B. Let it (so) mea-
ΔτὸνΓμετρεῖ,ὁδὲΓτὸνΑμετρεῖ,καὶὁΔἄρατὸνΑ sure (them), and let it be D. Since D measures C, and C
μετρεῖ. μετρεῖδὲκαὶτὸνΒ·ὁΔἄρατοὺςΑ,Βμετρεῖ measures A, D thus also measures A. And (D) also meaπρώτουςὄνταςπρὸςἀλλήλους·ὅπερἐστὶνἀδύνατον.
οὐκ sures B. Thus, D measures A and B, which are prime
ἄρατοὺςΓ,Βἀριθμοὺςἀριθμόςτιςμετρήσει.οἱΓ,Βἄρα to one another. The very thing is impossible. Thus, some
πρῶτοιπρὸςἀλλήλουςεἰσίν·ὅπερἔδειδεῖξαι.
number does not measure the numbers C and B. Thus,
C and B are prime to one another. (Which is) the very
thing it was required to show.
. Proposition 24
᾿Εὰνδύοἀριθμοὶπρόςτιναἀριθμὸνπρῶτοιὦσιν,καὶὁ If two numbers are prime to some number then the
ἐξαὐτῶνγενόμενοςπρὸςτὸναὐτὸνπρῶτοςἔσται. number created from (multiplying) the former (two numbers)
will also be prime to the latter (number).
213

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
Α Β Γ ∆ Ε Ζ
A B C D E F
ΔύογὰρἀριθμοὶοἱΑ,ΒπρόςτιναἀριθμὸντὸνΓπρῶτοι For let A and B be two numbers (which are both)
ἔστωσαν,καὶὁΑτὸνΒπολλαπλασιάσαςτὸνΔποιείτω· prime to some number C. And let A make D (by) multiλέγω,ὅτιοἱΓ,Δπρῶτοιπρὸςἀλλήλουςεἰσίν.
plying B. I say that C and D are prime to one another.
ΕἰγὰρμήεἰσινοἱΓ,Δπρῶτοιπρὸςἀλλήλους,μετρήσει For if C and D are not prime to one another then
[τις]τοὺςΓ,Δἀριθμός. μετρείτω,καὶἔστωὁΕ.καὶἐπεὶ [some] number will measure C and D. Let it (so) meaοἱΓ,Δπρῶτοιπρὸςἀλλήλουςεἰσίν,τὸνδὲΓμετρεῖτις
sure them, and let it be E. And since C and A are prime
ἀριθμὸς ὁ Ε, οἱ Α, Ε ἄρα πρῶτοι πρὸς ἀλλήλουςεἰσίν. to one another, and some number E measures C, A and
ὁσάκιςδὴὁΕτὸνΔμετρεῖ,τοσαῦταιμονάδεςἔστωσανἐν E are thus prime to one another [Prop. 7.23]. So as
τῷΖ·καὶὁΖἄρατὸνΔμετρεῖκατὰτὰςἐντῷΕμονάδας. many times as E measures D, so many units let there
ὁΕἄρατὸνΖπολλαπλασιάσαςτὸνΔπεποίηκεν. ἀλλὰ be in F . Thus, F also measures D according to the units
μὴνκαὶὁΑτὸνΒπολλαπλασιάσαςτὸνΔπεποίηκεν·ἴσος in E [Prop. 7.16]. Thus, E has made D (by) multiply-
ἄραἐστὶνὁἑκτῶνΕ,ΖτῷἐκτῶνΑ,Β.ἐὰνδὲὁὑπὸ ing F [Def. 7.15]. But, in fact, A has also made D (by)
τῶνἄκρωνἴσοςᾖτῷὑπὸτῶνμέσων,οἱτέσσαρεςἀριθμοὶ multiplying B. Thus, the (number created) from (multi-
ἀνάλογόνεἰσιν·ἔστινἄραὡςὁΕπρὸςτὸνΑ,οὕτωςὁΒ plying) E and F is equal to the (number created) from
πρὸςτὸνΖ.οἱδὲΑ,Επρῶτοι,οἱδὲπρῶτοικαὶἐλάχιστοι,οἱ (multiplying) A and B. And if the (rectangle contained)
δὲἐλάχιστοιἀριθμοὶτῶντὸναὐτὸνλόγονἐχόντωναὐτοῖς by the (two) outermost is equal to the (rectangle conμετροῦσιτοὺςτὸναὐτὸνλόγονἔχονταςἰσάκιςὅτεμείζων
tained) by the middle (two) then the four numbers are
τὸνμείζονακαὶὁἐλάσσωντὸνἐλάσσονα,τουτέστινὅτε proportional [Prop. 6.15]. Thus, as E is to A, so B (is)
ἡγούμενοςτὸνἡγούμενονκαὶὁἑπόμενοςτὸνἑπόμενον·ὁ to F . And A and E (are) prime (to one another). And
ΕἄρατὸνΒμετρεῖ.μετρεῖδὲκαὶτὸνΓ·ὁΕἄρατοὺςΒ,Γ (numbers) prime (to one another) are also the least (of
μετρεῖπρώτουςὄνταςπρὸςἀλλήλους·ὅπερἐστὶνἀδύνατον. those numbers having the same ratio) [Prop. 7.21]. And
οὐκἄρατοὺςΓ,Δἀριθμοὺςἀριθμόςτιςμετρήσει.οἱΓ,Δ the least numbers of those (numbers) having the same
ἄραπρῶτοιπρὸςἀλλήλουςεἰσίν·ὅπερἔδειδεῖξαι. ratio measure those (numbers) having the same ratio as
them an equal number of times, the greater (measuring)
the greater, and the lesser the lesser—that is to say, the
leading (measuring) the leading, and the following the
following [Prop. 7.20]. Thus, E measures B. And it also
measures C. Thus, E measures B and C, which are prime
to one another. The very thing is impossible. Thus, some
number cannot measure the numbers C and D. Thus,
C and D are prime to one another. (Which is) the very
thing it was required to show.
. Proposition 25
᾿Εὰνδύοἀριθμοὶπρῶτοιπρὸςἀλλήλουςὦσιν,ὁἐκτοῦ If two numbers are prime to one another then the
ἑνὸςαὐτῶνγενόμενοςπρὸςτὸνλοιπὸνπρῶτοςἔσται. number created from (squaring) one of them will be
῎ΕστωσανδύοἀριθμοὶπρῶτοιπρὸςἀλλήλουςοἱΑ,Β, prime to the remaining (number).
καὶὁΑἑαυτὸνπολλαπλασιάσαςτὸνΓποιείτω·λέγω,ὅτι Let A and B be two numbers (which are) prime to
214

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
οἱΒ,Γπρῶτοιπρὸςἀλλὴλουςεἰσίν.
Α Β Γ
∆
one another. And let A make C (by) multiplying itself. I
say that B and C are prime to one another.
A B C
D
ΚείσθωγὰρτῷΑἴσοςὁΔ.ἐπεὶοἱΑ,Βπρῶτοιπρὸς For let D be made equal to A. Since A and B are
ἀλλήλουςεἰσίν,ἴσοςδὲὁΑτῷΔ,καίοἱΔ,Βἄραπρῶτοι prime to one another, and A (is) equal to D, D and B are
πρὸςἀλλήλουςεἰσίν·ἑκάτεροςἄρατῶνΔ,ΑπρὸςτὸνΒ thus also prime to one another. Thus, D and A are each
πρῶτόςἐστιν·καὶὁἐκτῶνΔ,ΑἄραγενόμενοςπρὸςτὸνΒ prime to B. Thus, the (number) created from (multilyπρῶτοςἔσται.ὁδὲἐκτῶνΔ,Αγενόμενοςἀριθμόςἐστινὁ
ing) D and A will also be prime to B [Prop. 7.24]. And C
Γ.οἱΓ,Βἄραπρῶτοιπρὸςἀλλήλουςεἰσίν·ὅπερἔδειδεῖξαι. is the number created from (multiplying) D and A. Thus,
C and B are prime to one another. (Which is) the very
thing it was required to show.
¦. Proposition 26
᾿Εὰνδύοἀριθμοὶπρὸςδύοἀριθμοὺςἀμφότεροιπρὸς If two numbers are both prime to each of two numbers
ἑκάτερονπρῶτοιὦσιν,καὶοἱἐξαὐτῶνγενόμενοιπρῶτοι then the (numbers) created from (multiplying) them will
πρὸςἀλλήλουςἔσονται.
also be prime to one another.
Α
Β
Ε
Ζ
Γ
∆
ΔύογὰρἀριθμοὶοἱΑ,ΒπρὸςδύοἀριθμοὺςτοὺςΓ, For let two numbers, A and B, both be prime to each
Δἀμφότεροιπρὸςἑκάτερονπρῶτοιἔστωσαν, καὶὁμὲν of two numbers, C and D. And let A make E (by) mul-
ΑτὸνΒπολλαπλασιάσαςτὸνΕποιείτω, ὁδὲΓτὸνΔ tiplying B, and let C make F (by) multiplying D. I say
πολλαπλασιάσαςτὸνΖποιείτω·λέγω,ὅτιοἱΕ,Ζπρῶτοι that E and F are prime to one another.
πρὸςἀλλήλουςεἰσίν.
For since A and B are each prime to C, the (num-
᾿ΕπεὶγὰρἑκάτεροςτῶνΑ,ΒπρὸςτὸνΓπρῶτόςἐστιν, ber) created from (multiplying) A and B will thus also
καὶὁἐκτῶνΑ,ΒἄραγενόμενοςπρὸςτὸνΓπρῶτοςἔσται. be prime to C [Prop. 7.24]. And E is the (number) cre-
ὁδὲἐκτῶνΑ,ΒγενόμενόςἐστινὁΕ·οἱΕ,Γἄραπρῶτοι ated from (multiplying) A and B. Thus, E and C are
πρὸςἀλλήλουςεἰσίν. διὰτὰαὐτὰδὴκαὶοἱΕ,Δπρῶτοι prime to one another. So, for the same (reasons), E and
πρὸςἀλλήλουςεἰσίν. ἑκάτεροςἄρατῶνΓ,ΔπρὸςτὸνΕ D are also prime to one another. Thus, C and D are each
πρῶτόςἐστιν.καὶὁἐκτῶνΓ,Δἄραγενόμενοςπρὸςτὸν prime to E. Thus, the (number) created from (multiply-
Επρῶτοςἔσται.ὁδὲἐκτῶνΓ,ΔγενόμενόςἐστινὁΖ.οἱ ing) C and D will also be prime to E [Prop. 7.24]. And
Ε,Ζἄραπρῶτοιπρὸςἀλλήλουςεἰσίν·ὅπερἔδειδεῖξαι. F is the (number) created from (multiplying) C and D.
Thus, E and F are prime to one another. (Which is) the
very thing it was required to show.
A
B
E
F
C
D
215

ËÌÇÁÉÁÏÆÞº ELEMENTS BOOK 7
Þ. Proposition 27 †
᾿Εὰνδύοἀριθμοὶπρῶτοιπρὸςἀλλήλουςὦσιν,καὶπολ- If two numbers are prime to one another and each
λαπλασιάσαςἑκάτεροςἑαυτὸνποιῇτινα,οἱγενόμενοιἐξ makes some (number by) multiplying itself then the numαὐτῶν
πρῶτοι πρὸς ἀλλήλουςἔσονται, κἂν οἱ ἐξ ἀρχῆς bers created from them will be prime to one another, and
τοὺςγενομένουςπολλαπλασιάσαντεςποιῶσίτινας,κἀκεῖνοι if the original (numbers) make some (more numbers by)
πρῶτοιπρὸςἀλλήλουςἔσονται[καὶἀεὶπερὶτοὺςἄκρους multiplying the created (numbers) then these will also be
τοῦτοσυμβαίνει].
prime to one another [and this always happens with the
extremes].
Α Β Γ ∆ Ε Ζ A B C D E F
῎ΕστωσανδύοἀριθμοὶπρῶτοιπρὸςἀλλήλουςοἱΑ,Β, Let A and B be two numbers prime to one another,
καὶὁΑἑαυτὸνμὲνπολλαπλασιάσαςτὸνΓποιείτω,τὸν and let A make C (by) multiplying itself, and let it make
δὲΓπολλαπλασιάσαςτὸνΔποιείτω,ὁδὲΒἑαυτὸνμὲν D (by) multiplying C. And let B make E (by) multiplying
πολλαπλασιάσαςτὸνΕποιείτω,τὸνδὲΕπολλαπλασιάσας itself, and let it make F by multiplying E. I say that C
τὸνΖποιείτω·λέγω,ὅτιοἵτεΓ,ΕκαὶοἱΔ,Ζπρῶτοιπρὸς and E, and D and F , are prime to one another.
ἀλλήλουςεἰσίν.
For since A and B are prime to one another, and A has
᾿ΕπεὶγὰροἱΑ,Βπρῶτοιπρὸςἀλλήλουςεἰσίν,καὶὁ made C (by) multiplying itself, C and B are thus prime
ΑἑαυτὸνπολλαπλασιάσαςτὸνΓπεποίηκεν,οἱΓ,Βἄρα to one another [Prop. 7.25]. Therefore, since C and B
πρῶτοιπρὸςἀλλήλουςεἰσίν.ἐπεὶοὖνοἱΓ,Βπρῶτοιπρὸς are prime to one another, and B has made E (by) mul-
ἀλλήλουςεἰσίν, καὶ ὁ Β ἑαυτὸν πολλαπλασιάσαςτὸν Ε tiplying itself, C and E are thus prime to one another
πεποίηκεν,οἱΓ,Εἄραπρῶτοιπρὸςἀλλήλουςεἰσίν.πάλιν, [Prop. 7.25]. Again, since A and B are prime to one an-
ἐπεὶοἱΑ,Βπρῶτοιπρὸςἀλλήλουςεἰσίν,καὶὁΒἑαυτὸν other, and B has made E (by) multiplying itself, A and
πολλαπλασιάσαςτὸνΕπεποίηκεν,οἱΑ,Εἄραπρῶτοιπρὸς E are thus prime to one another [Prop. 7.25]. Therefore,
ἀλλήλουςεἰσίν. ἐπεὶοὖνδύοἀριθμοὶοἱΑ,Γπρὸςδύο since the two numbers A and C are both prime to each
ἀριθμοὺςτοὺςΒ,Εἀμφότεροιπρὸςἑκάτερονπρῶτοίεἰσιν, of the two numbers B and E, the (number) created from
καὶὁἐκτῶνΑ,ΓἄραγενόμενοςπρὸςτὸνἐκτῶνΒ,Ε (multiplying) A and C is thus prime to the (number creπρῶτόςἐστιν.
καίἐστινὁμὲνἐκτῶνΑ,ΓὁΔ,ὁδὲἐκ ated) from (multiplying) B and E [Prop. 7.26]. And D is
τῶνΒ,ΕὁΖ.οἱΔ,Ζἄραπρῶτοιπρὸςἀλλήλουςεἰσίν· the (number created) from (multiplying) A and C, and F
ὅπερἔδειδεῖξαι.
the (number created) from (multiplying) B and E. Thus,
D and F are prime to one another. (Which is) the very
thing it was required to show.
† In modern notation, this proposition states that if a is prime to b, then a 2 is also prime to b 2 , as well as a 3 to b 3 , etc., where all symbols denote
numbers.
.
Proposition 28
᾿Εὰνδύοἀριθμοὶπρῶτοιπρὸςἀλλήλουςὦσιν,καὶσυ- If two numbers are prime to one another then their
ναμφότεροςπρὸςἑκάτεροναὐτῶνπρῶτοςἔσται· καὶἐὰν sum will also be prime to each of them. And if the sum
συναμφότεροςπρὸςἕνατινὰαὐτῶνπρῶτοςᾖ, καὶοἱἐξ (of two numbers) is prime to any one of them then the
ἀρχῆςἀριθμοὶπρῶτοιπρὸςἀλλήλουςἔσονται.
original numbers will also be prime to one another.
216

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
∆
Α Β Γ A
B C
Συγκείσθωσαν γὰρ δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλ- For let the two numbers, AB and BC, (which are)
ουςοἱΑΒ,ΒΓ·λέγω,ὅτικαὶσυναμφότεροςὁΑΓπρὸς prime to one another, be laid down together. I say that
ἑκάτεροντῶνΑΒ,ΒΓπρῶτόςἐστιν.
their sum AC is also prime to each of AB and BC.
Εἰ γὰρ μή εἰσιν οἱ ΓΑ, ΑΒ πρῶτοι πρὸς ἀλλήλους, For if CA and AB are not prime to one another then
μετρήσειτιςτοὺςΓΑ,ΑΒἀριθμός.μετρείτω,καὶἔστωὁΔ. some number will measure CA and AB. Let it (so) mea-
ἐπεὶοὖνὁΔτοὺςΓΑ,ΑΒμετρεῖ,καὶλοιπὸνἄρατὸνΒΓ sure (them), and let it be D. Therefore, since D measures
μετρήσει.μετρεῖδὲκαὶτὸνΒΑ·ὁΔἄρατοὺςΑΒ,ΒΓμε- CA and AB, it will thus also measure the remainder BC.
τρεῖπρώτουςὄνταςπρὸςἀλλήλους·ὅπερἐστὶνἀδύνατον. And it also measures BA. Thus, D measures AB and
οὐκἄρατοὺςΓΑ,ΑΒἀριθμοὺςἀριθμόςτιςμετρήσει·οἱ BC, which are prime to one another. The very thing is
ΓΑ,ΑΒἄραπρῶτοιπρὸςἀλλήλουςεἰσίν. διὰτὰαὐτὰδὴ impossible. Thus, some number cannot measure (both)
καὶοἱΑΓ,ΓΒπρῶτοιπρὸςἀλλήλουςεἰσίν.ὁΓΑἄραπρὸς the numbers CA and AB. Thus, CA and AB are prime
ἑκάτεροντῶνΑΒ,ΒΓπρῶτόςἐστιν.
to one another. So, for the same (reasons), AC and CB
῎ΕστωσανδὴπάλινοἱΓΑ,ΑΒπρῶτοιπρὸςἀλλήλους· are also prime to one another. Thus, CA is prime to each
λέγω,ὅτικαὶοἱΑΒ,ΒΓπρῶτοιπρὸςἀλλήλουςεἰσίν. of AB and BC.
Εἰ γὰρ μή εἰσιν οἱ ΑΒ, ΒΓ πρῶτοι πρὸς ἀλλήλους, So, again, let CA and AB be prime to one another. I
μετρήσειτιςτοὺςΑΒ,ΒΓἀριθμός.μετρείτω,καὶἔστωὁΔ. say that AB and BC are also prime to one another.
καὶἐπεὶὁΔἑκάτεροντῶνΑΒ,ΒΓμετρεῖ,καὶὅλονἄρατὸν For if AB and BC are not prime to one another then
ΓΑμετρήσει.μετρεῖδὲκαὶτὸνΑΒ·ὁΔἄρατοὺςΓΑ,ΑΒ some number will measure AB and BC. Let it (so) meaμετρεῖπρώτουςὄνταςπρὸςἀλλήλους·ὅπερἐστὶνἀδύνατον.
sure (them), and let it be D. And since D measures each
οὐκἄρατοὺςΑΒ,ΒΓἀριθμοὺςἀριθμόςτιςμετρήσει. οἱ of AB and BC, it will thus also measure the whole of
ΑΒ,ΒΓἄραπρῶτοιπρὸςἀλλήλουςεἰσίν·ὅπερἔδειδεῖξαι. CA. And it also measures AB. Thus, D measures CA
and AB, which are prime to one another. The very thing
is impossible. Thus, some number cannot measure (both)
the numbers AB and BC. Thus, AB and BC are prime
to one another. (Which is) the very thing it was required
to show.
. Proposition 29
῞Απαςπρῶτοςἀριθμὸςπρὸςἅπανταἀριθμόν,ὃνμὴμε- Every prime number is prime to every number which
τρεῖ,πρῶτόςἐστιν.
it does not measure.
Α
Β
Γ
῎ΕστωπρῶτοςἀριθμὸςὁΑκαὶτὸνΒμὴμετρείτω·λέγω, Let A be a prime number, and let it not measure B. I
ὅτιοἱΒ,Απρῶτοιπρὸςἀλλήλουςεἰσίν.
say that B and A are prime to one another. For if B and
ΕἰγὰρμήεἰσινοἱΒ,Απρῶτοιπρὸςἀλλήλους,μετρήσει A are not prime to one another then some number will
τιςαὐτοὺςἀριθμός. μετρείτωὁΓ.ἐπεὶὁΓτὸνΒμετρεῖ, measure them. Let C measure (them). Since C measures
ὁδὲΑτὸνΒοὐμετρεῖ,ὁΓἄρατῷΑοὔκἐστινὁαὐτός. B, and A does not measure B, C is thus not the same as
καὶἐπεὶὁΓτοὺςΒ,Αμετρεῖ,καὶτὸνΑἄραμετρεῖπρῶτον A. And since C measures B and A, it thus also measures
ὄνταμὴὢναὐτῷὁαὐτός·ὅπερἐστὶνἀδύνατον. οὐκἄρα A, which is prime, (despite) not being the same as it.
τοὺςΒ,Αμετρήσειτιςἀριθμός. οἱΑ,Βἄραπρῶτοιπρὸς The very thing is impossible. Thus, some number cannot
ἀλλήλουςεἰσίν·ὅπερἔδειδεῖξαι.
measure (both) B and A. Thus, A and B are prime to
one another. (Which is) the very thing it was required to
D
A
B
C
217

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
show.
Ð. Proposition 30
᾿Εὰνδύοἀριθμοὶπολλαπλασιάσαντεςἀλλήλουςποιῶσί If two numbers make some (number by) multiplying
τινα,τὸνδὲγενόμενονἐξαὐτῶνμετρῇτιςπρῶτοςἀριθμός, one another, and some prime number measures the numκαὶἕνατῶνἐξἀρχῆςμετρήσει.
ber (so) created from them, then it will also measure one
of the original (numbers).
Α
Β
Γ
∆
Ε
ΔύογὰρἀριθμοὶοἱΑ,Βπολλαπλασιάσαντεςἀλλήλους For let two numbers A and B make C (by) multiplying
τὸνΓποιείτωσαν,τὸνδὲΓμετρείτωτιςπρῶτοςἀριθμὸςὁ one another, and let some prime number D measure C. I
Δ·λέγω,ὅτιὁΔἕνατῶνΑ,Βμετρεῖ. say that D measures one of A and B.
ΤὸνγὰρΑμὴμετρείτω·καίἐστιπρῶτοςὁΔ·οἱΑ, For let it not measure A. And since D is prime, A
Δἄραπρῶτοιπρὸςἀλλήλουςεἰσίν. καὶὁσάκιςὁΔτὸνΓ and D are thus prime to one another [Prop. 7.29]. And
μετρεῖ,τοσαῦταιμονάδεςἔστωσανἐντῷΕ.ἐπεὶοὖνὁΔ as many times as D measures C, so many units let there
τὸνΓμετρεῖκατὰτὰςἐντῷΕμονάδας,ὁΔἄρατὸνΕ be in E. Therefore, since D measures C according to
πολλαπλασιάσαςτὸνΓπεποίηκεν. ἀλλὰμὴνκαὶὁΑτὸν the units E, D has thus made C (by) multiplying E
ΒπολλαπλασιάσαςτὸνΓπεποίηκεν·ἴσοςἄραἐστὶνὁἐκ [Def. 7.15]. But, in fact, A has also made C (by) multiτῶνΔ,ΕτῷἐκτῶνΑ,Β.ἔστινἄραὡςὁΔπρὸςτὸνΑ,
plying B. Thus, the (number created) from (multiplying)
οὕτωςὁΒπρὸςτὸνΕ.οἱδὲΔ,Απρῶτοι,οἱδὲπρῶτοικαὶ D and E is equal to the (number created) from (mul-
ἐλάχιστοι,οἱδὲἐλάχιστοιμετροῦσιτοὺςτὸναὐτὸνλόγον tiplying) A and B. Thus, as D is to A, so B (is) to E
ἔχονταςἰσάκιςὅτεμείζωντὸνμείζονακαὶὁἐλάσσωντὸν [Prop. 7.19]. And D and A (are) prime (to one another),
ἐλάσσονα,τουτέστινὅτεἡγούμενοςτὸνἡγούμενονκαὶὁ and (numbers) prime (to one another are) also the least
ἐπόμενοςτὸνἑπόμενον·ὁΔἄρατὸνΒμετρεῖ.ὁμοίωςδὴ (of those numbers having the same ratio) [Prop. 7.21],
δείξομεν,ὅτικαὶἐὰντὸνΒμὴμετρῇ,τὸνΑμετρήσει.ὁΔ and the least (numbers) measure those (numbers) hav-
ἄραἕνατῶνΑ,Βμετρεῖ·ὅπερἔδειδεῖξαι.
ing the same ratio (as them) an equal number of times,
the greater (measuring) the greater, and the lesser the
lesser—that is to say, the leading (measuring) the leading,
and the following the following [Prop. 7.20]. Thus,
D measures B. So, similarly, we can also show that if
(D) does not measure B then it will measure A. Thus,
D measures one of A and B. (Which is) the very thing it
was required to show.
Ð. Proposition 31
῞Απαςσύνθεντοςἀριθμὸςὑπὸπρώτουτινὸςἀριθμοῦμε- Every composite number is measured by some prime
τρεῖται.
number.
Let A be a composite number. I say that A is measured
῎ΕστωσύνθεντοςἀριθμὸςὁΑ·λέγω,ὅτιὁΑὑπὸπρώτου by some prime number.
τινὸςἀριθμοῦμετρεῖται.
For since A is composite, some number will measure
᾿Επεὶ γὰρ σύνθετός ἐστιν ὁ Α, μετρήσει τις αὐτὸν it. Let it (so) measure (A), and let it be B. And if B
A
B
C
D
E
218

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
ἀριθμός.μετρείτω,καὶἔστωὁΒ.καὶεἰμὲνπρῶτόςἐστινὁ is prime then that which was prescribed has happened.
Β,γεγονὸςἂνεἴητὸἐπιταχθέν.εἰδὲσύνθετος,μετρήσει And if (B is) composite then some number will measure
τιςαὐτὸνἀριθμός. μετρείτω,καὶἔστωὁΓ.καὶἐπεὶὁΓ it. Let it (so) measure (B), and let it be C. And since
τὸνΒμετρεῖ,ὁδὲΒτὸνΑμετρεῖ,καὶὁΓἄρατὸνΑ C measures B, and B measures A, C thus also measures
μετρεῖ. καὶεἰμὲνπρῶτόςἐστινὁΓ,γεγονὸςἂνεἴητὸ A. And if C is prime then that which was prescribed has
ἐπιταχθέν. εἰδὲσύνθετος, μετρήσειτιςαὐτὸνἀριθμός. happened. And if (C is) composite then some number
τοιαύτηςδὴγινομένηςἐπισκέψεωςληφθήσεταίτιςπρῶτος will measure it. So, in this manner of continued inves-
ἀριθμός,ὃςμετρήσει. εἰγὰροὐληφθήσεται,μετρήσουσι tigation, some prime number will be found which will
τὸνΑἀριθμὸνἄπειροιἀριθμοί,ὧνἕτεροςἑτέρουἐλάσσων measure (the number preceding it, which will also mea-
ἐστίν·ὅπερἐστὶνἀδύνατονἐνἀριθμοῖς.ληφθήσεταίτιςἄρα sure A). And if (such a number) cannot be found then an
πρῶτοςἀριθμός,ὃςμετρήσειτὸνπρὸἑαυτοῦ,ὃςκαὶτὸνΑ infinite (series of) numbers, each of which is less than the
μετρήσει.
preceding, will measure the number A. The very thing is
impossible for numbers. Thus, some prime number will
(eventually) be found which will measure the (number)
preceding it, which will also measure A.
Α
Β
Γ
῞Απαςἄρασύνθεντοςἀριθμὸςὑπὸπρώτουτινὸςἀριθμοῦ Thus, every composite number is measured by some
μετρεῖται·ὅπερἔδειδεῖξαι.
prime number. (Which is) the very thing it was required
to show.
Ð. Proposition 32
῞Απαςἀριθμὸςἤτοιπρῶτόςἐστινἢὑπὸπρώτουτινὸς Every number is either prime or is measured by some
ἀριθμοῦμετρεῖται.
prime number.
Α
῎ΕστωἀριθμὸςὁΑ·λέγω,ὅτιὁΑἤτοιπρῶτόςἐστινἢ Let A be a number. I say that A is either prime or is
ὑπὸπρώτουτινὸςἀριθμοῦμετρεῖται.
measured by some prime number.
Εἰ μὲν οὖν πρῶτός ἐστιν ὁ Α, γεγονὸς ἂν εἴη τό In fact, if A is prime then that which was prescribed
ἐπιταχθέν. εἰ δὲ σύνθετος, μετρήσει τις αὐτὸν πρῶτος has happened. And if (it is) composite then some prime
ἀριθμός. number will measure it [Prop. 7.31].
῞Απας ἄραἀριθμὸς ἤτοιπρῶτός ἐστιν ἢ ὑπὸ πρώτου Thus, every number is either prime or is measured
τινὸςἀριθμοῦμετρεῖται·ὅπερἔδειδεῖξαι.
by some prime number. (Which is) the very thing it was
required to show.
Ð. Proposition 33
Ἀριθμῶνδοθέντωνὁποσωνοῦνεὑρεῖντοὺςἐλαχίστους To find the least of those (numbers) having the same
τῶντὸναὐτὸνλόγονἐχόντωναὐτοῖς.
ratio as any given multitude of numbers.
῎ΕστωσανοἱδοθέντεςὁποσοιοῦνἀριθμοὶοἱΑ,Β,Γ·δεῖ Let A, B, and C be any given multitude of numbers.
δὴεὑρεῖντοὺςἐλαχίστουςτῶντὸναὐτὸνλόγονἐχόντων So it is required to find the least of those (numbers) havτοῖςΑ,Β,Γ.
ing the same ratio as A, B, and C.
ΟἱΑ,Β,Γγὰρἤτοιπρῶτοιπρὸςἀλλήλουςεἰσὶνἢοὔ. For A, B, and C are either prime to one another, or
εἰμὲνοὖνοἱΑ,Β,Γπρῶτοιπρὸςἀλλήλουςεἰσίν,ἐλάχιστοί not. In fact, if A, B, and C are prime to one another then
εἰσιτῶντὸναὐτὸνλόγονἐχόντωναὐτοῖς.
they are the least of those (numbers) having the same
ratio as them [Prop. 7.22].
A
B
C
A
219

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
Α Β Γ ∆ Ε Ζ Η Θ Κ Λ Μ A B C D E F G H K L M
Εἰ δὲοὔ, εἰλήφθωτῶν Α, Β,Γτὸμέγιστονκοινὸν And if not, let the greatest common measure, D, of
μέτρονὁΔ,καὶὁσάκιςὁΔἕκαστοντῶνΑ,Β,Γμετρεῖ, A, B, and C have be taken [Prop. 7.3]. And as many
τοσαῦταιμονάδεςἔστωσανἐνἑκάστῳτῶνΕ,Ζ,Η.καὶ times as D measures A, B, C, so many units let there
ἕκαστοςἄρατῶνΕ,Ζ,ΗἕκαστοντῶνΑ,Β,Γμετρεῖκατὰ be in E, F , G, respectively. And thus E, F , G meaτὰςἐντῷΔμονάδας.οἱΕ,Ζ,ΗἄρατοὺςΑ,Β,Γἰσάκις
sure A, B, C, respectively, according to the units in D
μετροῦσιν·οἱΕ,Ζ,ΗἄρατοῖςΑ,Β,Γἐντῷαὐτῷλόγῳ [Prop. 7.15]. Thus, E, F , G measure A, B, C (respecεἰσίν.
λέγωδή,ὅτικαὶἐλάχιστοι. εἰγὰρμήεἰσινοἱΕ,Ζ, tively) an equal number of times. Thus, E, F , G are in
ΗἐλάχιστοιτῶντὸναὐτὸνλόγονἐχόντωντοῖςΑ,Β,Γ, the same ratio as A, B, C (respectively) [Def. 7.20]. So I
ἔσονται[τινες]τῶνΕ,Ζ,Ηἐλάσσονεςἀριθμοὶἐντῷαὐτῷ say that (they are) also the least (of those numbers havλόγῳὄντεςτοῖςΑ,Β,Γ.ἔστωσανοἱΘ,Κ,Λ·ἰσάκιςἄραὁ
ing the same ratio as A, B, C). For if E, F , G are not
ΘτὸνΑμετρεῖκαὶἑκάτεροςτῶνΚ,ΛἑκάτεροντῶνΒ,Γ. the least of those (numbers) having the same ratio as A,
ὁσάκιςδὲὁΘτὸνΑμετρεῖ,τοσαῦταιμονάδεςἔστωσανἐν B, C (respectively), then there will be [some] numbers
τῷΜ·καὶἑκάτεροςἄρατῶνΚ,ΛἑκάτεροντῶνΒ,Γμετρεῖ less than E, F , G which are in the same ratio as A, B, C
κατὰτὰςἐντῷΜμονάδας.καὶἐπεὶὁΘτὸνΑμετρεῖκατὰ (respectively). Let them be H, K, L. Thus, H measures
τὰςἐντῷΜμονάδας,καὶὁΜἄρατὸνΑμετρεῖκατὰτὰς A the same number of times that K, L also measure B,
ἐντῷΘμονάδας.διὰτὰαὐτὰδὴὁΜκαὶἑκάτεροντῶνΒ, C, respectively. And as many times as H measures A, so
ΓμετρεῖκατὰτὰςἐνἑκατέρῳτῶνΚ,Λμονάδας·ὁΜἄρα many units let there be in M. Thus, K, L measure B,
τοὺςΑ,Β,Γμετρεῖ. καὶἐπεὶὁΘτὸνΑμετρεῖκατὰτὰς C, respectively, according to the units in M. And since
ἐντῷΜμονάδας,ὁΘἄρατὸνΜπολλαπλασιάσαςτὸνΑ H measures A according to the units in M, M thus also
πεποίηκεν.διὰτὰαὐτὰδὴκαὶὁΕτὸνΔπολλαπλασιάσας measures A according to the units in H [Prop. 7.15]. So,
τὸνΑπεποίηκεν. ἴσοςἄραἐστὶνὁἐκτῶνΕ,Δτῷἐκ for the same (reasons), M also measures B, C accordτῶνΘ,Μ.ἔστινἄραὡςὁΕπρὸςτὸνΘ,οὕτωςὁΜπρὸς
ing to the units in K, L, respectively. Thus, M measures
τὸνΔ.μείζωνδὲὁΕτοῦΘ·μείζωνἄρακαὶὁΜτοῦΔ. A, B, and C. And since H measures A according to the
καὶμετρεῖτοὺςΑ,Β,Γ·ὅπερἐστὶνἀδύνατον·ὑπόκειται units in M, H has thus made A (by) multiplying M. So,
γὰρὁΔτῶνΑ,Β,Γτὸμέγιστονκοινὸνμέτρον.οὐκἄρα for the same (reasons), E has also made A (by) multiply-
ἔσονταίτινεςτῶνΕ,Ζ,Ηἐλάσσονεςἀριθμοὶἐντῷαὐτῷ ing D. Thus, the (number created) from (multiplying)
λόγῳὄντεςτοῖςΑ,Β,Γ.οἱΕ,Ζ,Ηἄραἐλάχιστοίεἰσιτῶν E and D is equal to the (number created) from (multiτὸναὐτὸνλόγονἐχόντωντοῖςΑ,Β,Γ·ὅπερἔδειδεῖξαι.
plying) H and M. Thus, as E (is) to H, so M (is) to
D [Prop. 7.19]. And E (is) greater than H. Thus, M
(is) also greater than D [Prop. 5.13]. And (M) measures
A, B, and C. The very thing is impossible. For D was
assumed (to be) the greatest common measure of A, B,
and C. Thus, there cannot be any numbers less than E,
F , G which are in the same ratio as A, B, C (respectively).
Thus, E, F , G are the least of (those numbers)
having the same ratio as A, B, C (respectively). (Which
is) the very thing it was required to show.
Ð. Proposition 34
Δύοἀριθμῶνδοθέντωνεὑρεῖν,ὃνἐλάχιστονμετροῦσιν To find the least number which two given numbers
ἀριθμόν.
(both) measure.
῎ΕστωσανοἱδοθέντεςδύοἀριθμοὶοἱΑ,Β·δεῖδὴεὑρεῖν, Let A and B be the two given numbers. So it is re-
220

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
ὃνἐλάχιστονμετροῦσινἀριθμόν.
Α
Γ
∆
Ε
Β
Ζ
quired to find the least number which they (both) measure.
ΟἱΑ,Βγὰρἤτοιπρῶτοιπρὸςἀλλήλουςεἰσὶνἢοὔ. For A and B are either prime to one another, or not.
ἔστωσανπρότερονοἱΑ,Βπρῶτοιπρὸςἀλλήλους,καὶὁΑ Let them, first of all, be prime to one another. And let A
τὸνΒπολλαπλασιάσαςτὸνΓποιείτω·καὶὁΒἄρατὸνΑ make C (by) multiplying B. Thus, B has also made C
πολλαπλασιάσαςτὸνΓπεποίηκεν. οἱΑ,ΒἄρατὸνΓμε- (by) multiplying A [Prop. 7.16]. Thus, A and B (both)
τροῦσιν.λέγωδή,ὅτικαὶἐλάχιστον.εἰγὰρμή,μετρήσουσί measure C. So I say that (C) is also the least (numτιναἀριθμὸνοἱΑ,ΒἐλάσσοναὄντατοῦΓ.μετρείτωσαν
ber which they both measure). For if not, A and B will
τὸνΔ.καὶὁσάκιςὁΑτὸνΔμετρεῖ,τοσαῦταιμονάδες (both) measure some (other) number which is less than
ἔστωσανἐντῷΕ,ὁσάκιςδὲὁΒτὸνΔμετρεῖ,τοσαῦται C. Let them (both) measure D (which is less than C).
μονάδες ἔστωσαν ἐν τῷ Ζ. ὁ μὲν Α ἄρα τὸν Ε πολλα- And as many times as A measures D, so many units let
πλασιάσαςτὸνΔπεποίηκεν,ὁδὲΒτὸνΖπολλαπλασιάσας there be in E. And as many times as B measures D,
τὸνΔπεποίηκεν·ἴσοςἄραἐστὶνὁἐκτῶνΑ,Ετῷἐκτῶν so many units let there be in F . Thus, A has made D
Β,Ζ.ἔστινἄραὡςὁΑπρὸςτὸνΒ,οὕτωςὁΖπρὸςτὸν (by) multiplying E, and B has made D (by) multiply-
Ε.οἱδὲΑ,Βπρῶτοι,οἱδὲπρῶτοικαὶἐλάχιστοι,οἱδὲ ing F . Thus, the (number created) from (multiplying)
ἐλάχιστοιμετροῦσιτοὺςτὸναὐτὸνλόγονἔχονταςἰσάκιςὅ A and E is equal to the (number created) from (multiτεμείζωντὸνμείζονακαὶὁἐλάσσωντὸνἐλάσσονα·ὁΒ
plying) B and F . Thus, as A (is) to B, so F (is) to E
ἄρατὸνΕμετρεῖ,ὡςἑπόμενοςἑπόμενον.καὶἐπεὶὁΑτοὺς [Prop. 7.19]. And A and B are prime (to one another),
Β,ΕπολλαπλασιάσαςτοὺςΓ,Δπεποίηκεν,ἔστινἄραὡς and prime (numbers) are the least (of those numbers
ὁΒπρὸςτὸνΕ,οὕτωςὁΓπρὸςτὸνΔ.μετρεῖδὲὁΒτὸν having the same ratio) [Prop. 7.21], and the least (num-
Ε·μετρεῖἄρακαὶὁΓτὸνΔὁμείζωντὸνἐλάσσονα·ὅπερ bers) measure those (numbers) having the same ratio (as
ἐστὶνἀδύνατον. οὐκἄραοἱΑ,Βμετροῦσίτιναἀριθμὸν them) an equal number of times, the greater (measuring)
ἐλάσσοναὄντατοῦΓ.ὁΓἄραἐλάχιστοςὢνὑπὸτῶνΑ,Β the greater, and the lesser the lesser [Prop. 7.20]. Thus,
μετρεῖται.
B measures E, as the following (number measuring) the
following. And since A has made C and D (by) multiplying
B and E (respectively), thus as B is to E, so C
(is) to D [Prop. 7.17]. And B measures E. Thus, C also
measures D, the greater (measuring) the lesser. The very
thing is impossible. Thus, A and B do not (both) measure
some number which is less than C. Thus, C is the
least (number) which is measured by (both) A and B.
Α
Ζ
Γ
∆
Η
Β
Ε
Θ
Μὴ ἔστωσαν δὴ οἱ Α, Β πρῶτοι πρὸς ἀλλήλους, So let A and B be not prime to one another. And
καὶεἰλήφθωσανἐλάχιστοιἀριθμοὶτῶντὸναὐτὸνλόγον let the least numbers, F and E, have been taken having
ἐχόντωντοῖςΑ,ΒοἱΖ,Ε·ἴσοςἄραἐστὶνὁἐκτῶνΑ,Ετῷ the same ratio as A and B (respectively) [Prop. 7.33].
A
C
D
E
A
F
C
D
G
B
F
B
E
H
221

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
ἐκτῶνΒ,Ζ.καὶὁΑτὸνΕπολλαπλασιάσαςτὸνΓποιείτω· Thus, the (number created) from (multiplying) A and E
καὶὁΒἄρατὸνΖπολλαπλασιάσαςτὸνΓπεποίηκεν·οἱΑ, is equal to the (number created) from (multiplying) B
ΒἄρατὸνΓμετροῦσιν.λέγωδή,ὅτικαὶἐλάχιστον.εἰγὰρ and F [Prop. 7.19]. And let A make C (by) multiplying
μή,μετρήσουσίτιναἀριθμὸνοἱΑ,Βἐλάσσοναὄντατοῦ E. Thus, B has also made C (by) multiplying F . Thus,
Γ.μετρείτωσαντὸνΔ.καὶὁσάκιςμὲνὁΑτὸνΔμετρεῖ, A and B (both) measure C. So I say that (C) is also the
τοσαῦταιμονάδεςἔστωσανἐντῷΗ,ὁσάκιςδὲὁΒτὸνΔ least (number which they both measure). For if not, A
μετρεῖ,τοσαῦταιμονάδεςἔστωσανἐντῷΘ.ὁμὲνΑἄρα and B will (both) measure some number which is less
τὸνΗπολλαπλασιάσαςτὸνΔπεποίηκεν,ὁδὲΒτὸνΘ than C. Let them (both) measure D (which is less than
πολλαπλασιάσαςτὸνΔπεποίηκεν.ἴσοςἄραἐστὶνὁἐκτῶν C). And as many times as A measures D, so many units
Α,ΗτῷἐκτῶνΒ,Θ·ἔστινἄραὡςὁΑπρὸςτὸνΒ,οὕτως let there be in G. And as many times as B measures D,
ὁΘπρὸςτὸνΗ.ὡςδὲὁΑπρὸςτὸνΒ,οὕτωςὁΖπρὸς so many units let there be in H. Thus, A has made D
τὸνΕ·καὶὡςἄραὁΖπρὸςτὸνΕ,οὕτωςὁΘπρὸςτὸν (by) multiplying G, and B has made D (by) multiplying
Η.οἱδὲΖ,Εἐλάχιστοι,οἱδὲἐλάχιστοιμετροῦσιτοὺςτὸν H. Thus, the (number created) from (multiplying) A and
αὐτὸνλόγονἔχονταςἰσάκιςὅτεμείζωντὸνμείζονακαὶὁ G is equal to the (number created) from (multiplying) B
ἐλάσσωντὸνἐλάσσονα·ὁΕἄρατὸνΗμετρεῖ. καὶἐπεὶὁ and H. Thus, as A is to B, so H (is) to G [Prop. 7.19].
ΑτοὺςΕ,ΗπολλαπλασιάσαςτοὺςΓ,Δπεποίηκεν,ἔστιν And as A (is) to B, so F (is) to E. Thus, also, as F (is)
ἄραὡςὁΕπρὸςτὸνΗ,οὕτωςὁΓπρὸςτὸνΔ.ὁδὲΕτὸν to E, so H (is) to G. And F and E are the least (num-
Ημετρεῖ·καὶὁΓἄρατὸνΔμετρεῖὁμείζωντὸνἐλάσσονα· bers having the same ratio as A and B), and the least
ὅπερἐστὶνἀδύνατον. οὐκἄραοἱΑ,Βμετρήσουσίτινα (numbers) measure those (numbers) having the same ra-
ἀριθμὸνἐλάσσοναὄντατοῦΓ.ὁΓἄραἐλάχιστοςὢνὑπὸ tio an equal number of times, the greater (measuring)
τῶνΑ,Βμετρεῖται·ὅπερἔπειδεῖξαι.
the greater, and the lesser the lesser [Prop. 7.20]. Thus,
E measures G. And since A has made C and D (by) multiplying
E and G (respectively), thus as E is to G, so C
(is) to D [Prop. 7.17]. And E measures G. Thus, C also
measures D, the greater (measuring) the lesser. The very
thing is impossible. Thus, A and B do not (both) measure
some (number) which is less than C. Thus, C (is)
the least (number) which is measured by (both) A and
B. (Which is) the very thing it was required to show.
Ð. Proposition 35
᾿Εὰνδύοἀριθμοὶἀριθμόντιναμετρῶσιν,καὶὁἐλάχιστος If two numbers (both) measure some number then the
ὑπ᾿αὐτῶνμετρούμενοςτὸναὐτὸνμετρήσει.
least (number) measured by them will also measure the
same (number).
Α Β
A
B
Γ Ζ ∆ C F
D
Ε
Δύο γὰρ ἀριθμοὶ οἱ Α, Β ἀριθμόν τινα τὸν ΓΔ με- For let two numbers, A and B, (both) measure some
τρείτωσαν,ἐλάχιστονδὲτὸνΕ·λέγω,ὅτικαὶὁΕτὸνΓΔ number CD, and (let) E (be the) least (number meaμετρεῖ.
sured by both A and B). I say that E also measures CD.
ΕἰγὰροὐμετρεῖὁΕτὸνΓΔ,ὁΕτὸνΔΖμετρῶν For if E does not measure CD then let E leave CF
λειπέτωἑαυτοῦἐλάσσονατὸνΓΖ.καὶἐπεὶοἱΑ,ΒτὸνΕ less than itself (in) measuring DF . And since A and B
μετροῦσιν, ὁδὲΕτὸνΔΖμετρεῖ, καὶοἱΑ,Βἄρατὸν (both) measure E, and E measures DF , A and B will
ΔΖμετρήσουσιν.μετροῦσιδὲκαὶὅλοντὸνΓΔ·καὶλοιπὸν thus also measure DF . And (A and B) also measure the
ἄρατὸνΓΖμετρήσουσινἐλάσσοναὄντατοῦΕ·ὅπερἐστὶν whole of CD. Thus, they will also measure the remainder
ἀδύνατον.οὐκἄραοὐμετρεῖὁΕτὸνΓΔ·μετρεῖἄρα·ὅπερ CF , which is less than E. The very thing is impossible.
ἔδειδεῖξαι.
Thus, E cannot not measure CD. Thus, (E) measures
E
222

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
(CD). (Which is) the very thing it was required to show.
Ц. Proposition 36
Τριῶν ἀριθμῶν δοθέντων εὑρεῖν, ὃν ἐλάχιστον με- To find the least number which three given numbers
τροῦσινἀριθμόν.
(all) measure.
῎ΕστωσανοἱδοθέντεςτρεῖςἀριθμοὶοἱΑ,Β,Γ·δεῖδὴ Let A, B, and C be the three given numbers. So it is
εὑρεῖν,ὃνἐλάχιστονμετροῦσινἀριθμόν.
required to find the least number which they (all) measure.
Α
Β
Γ
∆
Ε
Ζ
ΕἰλήφθωγὰρὑπὸδύοτῶνΑ,Βἐλάχιστοςμετρούμενος For let the least (number), D, measured by the two
ὁ Δ. ὁ δὴΓτὸν Δ ἤτοιμετρεῖἢοὐ μετρεῖ. μετρείτω (numbers) A and B have been taken [Prop. 7.34]. So C
πρότερον. μετροῦσιδὲκαὶοἱΑ,ΒτὸνΔ·οἱΑ,Β,Γ either measures, or does not measure, D. Let it, first of
ἄρατὸνΔμετροῦσιν. λέγωδή,ὅτικαὶἐλάχιστον. εἰγὰρ all, measure (D). And A and B also measure D. Thus,
μή,μετρήσουσιν[τινα]ἀριθμὸνοἱΑ,Β,Γἐλάσσοναὄντα A, B, and C (all) measure D. So I say that (D is) also
τοῦΔ.μετρείτωσαντὸνΕ.ἐπεὶοἱΑ,Β,ΓτὸνΕμετροῦσιν, the least (number measured by A, B, and C). For if not,
καὶοἱΑ,ΒἄρατὸνΕμετροῦσιν.καὶὁἐλάχιστοςἄραὑπὸ A, B, and C will (all) measure [some] number which
τῶνΑ,Βμετρούμενος[τὸνΕ]μετρήσει.ἐλάχιστοςδὲὑπὸ is less than D. Let them measure E (which is less than
τῶνΑ,ΒμετρούμενόςἐστινὁΔ·ὁΔἄρατὸνΕμετρήσει D). Since A, B, and C (all) measure E then A and B
ὁμείζωντὸνἐλάσσονα·ὅπερἐστὶνἀδύνατον. οὐκἄραοἱ thus also measure E. Thus, the least (number) measured
Α,Β,ΓμετρήσουσίτιναἀριθμὸνἐλάσσοναὄντατοῦΔ·οἱ by A and B will also measure [E] [Prop. 7.35]. And D
Α,Β,ΓἄραἐλάχιστοντὸνΔμετροῦσιν.
is the least (number) measured by A and B. Thus, D
ΜὴμετρείτωδὴπάλινὁΓτὸνΔ,καὶεἰλήφθωὑπὸτῶν will measure E, the greater (measuring) the lesser. The
Γ,ΔἐλάχιστοςμετρούμενοςἀριθμὸςὁΕ.ἐπεὶοἱΑ,Β very thing is impossible. Thus, A, B, and C cannot (all)
τὸνΔμετροῦσιν,ὁδὲΔτὸνΕμετρεῖ,καὶοἱΑ,Βἄρα measure some number which is less than D. Thus, A, B,
τὸνΕμετροῦσιν. μετρεῖδὲκαὶὁΓ[τὸνΕ·καὶ]οἱΑ,Β, and C (all) measure the least (number) D.
ΓἄρατὸνΕμετροῦσιν. λέγωδή,ὅτικαὶἐλάχιστον. εἰ So, again, let C not measure D. And let the least
γὰρμή,μετρήσουσίτιναοἱΑ,Β,ΓἐλάσσοναὄντατοῦΕ. number, E, measured by C and D have been taken
μετρείτωσαντὸνΖ.ἐπεὶοἱΑ,Β,ΓτὸνΖμετροῦσιν,καὶοἱ [Prop. 7.34]. Since A and B measure D, and D measures
Α,ΒἄρατὸνΖμετροῦσιν·καὶὁἐλάχιστοςἄραὑπὸτῶν E, A and B thus also measure E. And C also measures
Α,ΒμετρούμενοςτὸνΖμετρήσει. ἐλάχιστοςδὲὑπὸτῶν [E]. Thus, A, B, and C [also] measure E. So I say that
Α,ΒμετρούμενόςἐστινὁΔ·ὁΔἄρατὸνΖμετρεῖ.μετρεῖ (E is) also the least (number measured by A, B, and C).
δὲκαὶὁΓτὸνΖ·οἱΔ,ΓἄρατὸνΖμετροῦσιν·ὥστεκαὶὁ For if not, A, B, and C will (all) measure some (number)
ἐλάχιστοςὑπὸτῶνΔ,ΓμετρούμενοςτὸνΖμετρήσει.ὁδὲ which is less than E. Let them measure F (which is less
ἐλάχιστοςὑπὸτῶνΓ,ΔμετρούμενόςἐστινὁΕ·ὁΕἄρα than E). Since A, B, and C (all) measure F , A and B
τὸνΖμετρεῖὁμείζωντὸνἐλάσσονα·ὅπερἐστὶνἀδύνατον. thus also measure F . Thus, the least (number) measured
οὐκἄραοἱΑ,Β,Γμετρήσουσίτιναἀριθμὸνἐλάσσοναὄντα by A and B will also measure F [Prop. 7.35]. And D
τοῦΕ.ὁΕἄραἐλάχιστοςὢνὑπὸτῶνΑ,Β,Γμετρεῖται· is the least (number) measured by A and B. Thus, D
ὅπερἔδειδεῖξαι.
measures F . And C also measures F . Thus, D and C
(both) measure F . Hence, the least (number) measured
by D and C will also measure F [Prop. 7.35]. And E
A
B
C
D
E
F
223

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
is the least (number) measured by C and D. Thus, E
measures F , the greater (measuring) the lesser. The very
thing is impossible. Thus, A, B, and C cannot measure
some number which is less than E. Thus, E (is) the least
(number) which is measured by A, B, and C. (Which is)
the very thing it was required to show.
ÐÞ. Proposition 37
᾿Εὰνἀριθμὸςὑπότινοςἀριθμοῦμετρῆται,ὁμετρούμενος If a number is measured by some number then the
ὁμώνυμονμέροςἕξειτῷμετροῦντι.
(number) measured will have a part called the same as
the measuring (number).
Α
Β
Γ
∆
ἈριθμὸςγάρὁΑὑπότινοςἀριθμοῦτοῦΒμετρείσθω· For let the number A be measured by some number
λέγω,ὅτιὁΑὁμώνυμονμέροςἔχειτῷΒ. B. I say that A has a part called the same as B.
῾ΟσάκιςγὰρὁΒτὸνΑμετρεῖ,τοσαῦταιμονάδεςἔστω- For as many times as B measures A, so many units
σανἐντῷΓ.ἐπεὶὁΒτὸνΑμετρεῖκατὰτὰςἐντῷΓ let there be in C. Since B measures A according to the
μονάδας,μετρεῖδὲκαὶἡΔμονὰςτὸνΓἀριθμὸνκατὰτὰς units in C, and the unit D also measures C according
ἐναὐτῷμονάδας,ἰσάκιςἄραἡΔμονὰςτὸνΓἀριθμὸνμε- to the units in it, the unit D thus measures the number
τρεῖκαὶὁΒτὸνΑ.ἐναλλὰξἄραἰσάκιςἡΔμονὰςτὸνΒ C as many times as B (measures) A. Thus, alternately,
ἀριθμὸνμετρεῖκαὶὁΓτὸνΑ·ὃἄραμέροςἐστὶνἡΔμονὰς the unit D measures the number B as many times as C
τοῦΒἀριθμοῦ,τὸαὐτὸμέροςἐστὶκαὶὁΓτοῦΑ.ἡδὲΔ (measures) A [Prop. 7.15]. Thus, which(ever) part the
μονὰςτοῦΒἀριθμοῦμέροςἐστὶνὁμώνυμοναὐτῷ·καὶὁΓ unit D is of the number B, C is also the same part of A.
ἄρατοῦΑμέροςἐστὶνὁμώνυμοντῷΒ.ὥστεὁΑμέρος And the unit D is a part of the number B called the same
ἔχειτὸνΓὁμώνυμονὄντατῷΒ·ὅπερἔδειδεῖξαι. as it (i.e., a Bth part). Thus, C is also a part of A called
the same as B (i.e., C is the Bth part of A). Hence, A has
a part C which is called the same as B (i.e., A has a Bth
part). (Which is) the very thing it was required to show.
Ð. Proposition 38
᾿Εὰνἀριθμοςμέροςἔχῃὁτιοῦν,ὑπὸὁμωνύμουἀριθμοῦ If a number has any part whatever then it will be meaμετρηθήσεταιτῷμέρει.
sured by a number called the same as the part.
Α
Β
Γ
∆
ἈριθμὸςγὰρὁΑμέροςἐχέτωὁτιοῦντὸνΒ,καὶτῷΒ For let the number A have any part whatever, B. And
μέρειὁμώνυμοςἔστω[ἀριθμὸς]ὁΓ·λέγω,ὅτιὁΓτὸνΑ let the [number] C be called the same as the part B (i.e.,
μετρεῖ. B is the Cth part of A). I say that C measures A.
᾿ΕπεὶγὰρὁΒτοῦΑμέροςἐστὶνὁμώνυμοντῷΓ,ἔστι For since B is a part of A called the same as C, and
δὲκαὶἡΔμονὰςτοῦΓμέροςὁμώνυμοναὐτῷ,ὃἄραμέρος the unit D is also a part of C called the same as it (i.e.,
A
B
C
A
B
C
D
D
224

ËÌÇÁÉÁÏÆÞº ELEMENTS
BOOK 7
ἐστὶνἡΔμονὰςτοῦΓἀριθμοῦ,τὸαὐτὸμέροςἐστὶκαὶὁΒ D is the Cth part of C), thus which(ever) part the unit D
τοῦΑ·ἰσάκιςἄραἡΔμονὰςτὸνΓἀριθμὸνμετρεῖκαὶὁΒ is of the number C, B is also the same part of A. Thus,
τὸνΑ.ἐναλλὰξἄραἰσάκιςἡΔμονὰςτὸνΒἀριθμὸνμετρεῖ the unit D measures the number C as many times as B
καὶὁΓτὸνΑ.ὁΓἄρατὸνΑμετρεῖ·ὅπερἔδειδεὶξαι. (measures) A. Thus, alternately, the unit D measures the
number B as many times as C (measures) A [Prop. 7.15].
Thus, C measures A. (Which is) the very thing it was
required to show.
Ð. Proposition 39
Ἀριθμὸνεὐρεῖν,ὃςἐλάχιστοςὢνἕξειτὰδοθένταμέρη. To find the least number that will have given parts.
Α Β Γ
A B C
∆
Ε
D
E
Ζ
F
Η
Θ
G
H
῎Εστω τὰ δοθέντα μέρη τὰ Α, Β, Γ· δεῖ δὴ ἀριθμὸν Let A, B, and C be the given parts. So it is required
εὑρεῖν,ὃςἐλάχιστοςὢνἕξειτὰΑ,Β,Γμέρη. to find the least number which will have the parts A, B,
῎ΕστωσανγὰρτοῖςΑ,Β,Γμέρεσινὁμώνυμοιἀριθμοὶ and C (i.e., an Ath part, a Bth part, and a Cth part).
οἱΔ,Ε,Ζ,καὶεἰλήφθωὑπὸτῶνΔ,Ε,Ζἐλάχιστοςμε- For let D, E, and F be numbers having the same
τρούμενοςἀριθμὸςὁΗ.
names as the parts A, B, and C (respectively). And let
῾ΟΗἄραὁμώνυμαμέρηἔχειτοῖςΔ,Ε,Ζ.τοῖςδὲΔ, the least number, G, measured by D, E, and F , have
Ε,ΖὁμώνυμαμέρηἐστὶτὰΑ,Β,Γ·ὁΗἄραἔχειτὰΑ,Β, been taken [Prop. 7.36].
Γμέρη.λέγωδή,ὅτικαὶἐλάχιστοςὤν,εἰγὰρμή,ἔσταιτις Thus, G has parts called the same as D, E, and F
τοῦΗἐλάσσωνἀριθμός,ὃςἕξειτὰΑ,Β,Γμέρη.ἔστωὁ [Prop. 7.37]. And A, B, and C are parts called the same
Θ.ἐπεὶὁΘἔχειτὰΑ,Β,Γμέρη,ὁΘἄραὑπὸὁμωνύμων as D, E, and F (respectively). Thus, G has the parts A,
ἀριθμῶνμετρηθήσεταιτοῖςΑ,Β,Γμέρεσιν.τοῖςδὲΑ,Β, B, and C. So I say that (G) is also the least (number
ΓμέρεσινὁμώνυμοιἀριθμοίεἰσινοἱΔ,Ε,Ζ·ὁΘἄραὑπὸ having the parts A, B, and C). For if not, there will be
τῶνΔ,Ε,Ζμετρεῖται. καίἐστινἐλάσσωντοῦΗ·ὅπερ some number less than G which will have the parts A,
ἐστὶνἀδύνατον.οὐκἄραἔσταιτιςτοῦΗἐλάσσωνἀριθμός, B, and C. Let it be H. Since H has the parts A, B, and
ὃςἕξειτὰΑ,Β,Γμέρη·ὅπερἔδειδεῖξαι.
C, H will thus be measured by numbers called the same
as the parts A, B, and C [Prop. 7.38]. And D, E, and
F are numbers called the same as the parts A, B, and C
(respectively). Thus, H is measured by D, E, and F . And
(H) is less than G. The very thing is impossible. Thus,
there cannot be some number less than G which will have
the parts A, B, and C. (Which is) the very thing it was
required to show.
225

226

ELEMENTS BOOK 8
Continued Proportion †
† The propositions contained in Books 7–9 are generally attributed to the school of Pythagoras.
227

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
. Proposition 1
᾿Εὰνὦσινὁσοιδηποτοῦνἀριθμοὶἑξῆςἀνάλογον,οἱδὲ If there are any multitude whatsoever of continuously
ἄκροιαὐτῶνπρῶτοιπρὸςἀλλήλουςὦσιν, ἐλάχιστοίεἰσι proportional numbers, and the outermost of them are
τῶντὸναὐτὸνλόγονἐχόντωναὐτοῖς.
prime to one another, then the (numbers) are the least
of those (numbers) having the same ratio as them.
Α
Ε
A
E
Β
Ζ
B
F
Γ
Η
C
G
∆
Θ
D
H
῎ΕστωσανὁποσοιοῦνἀριθμοὶἑξῆςἀνάλογονοἱΑ,Β, Let A, B, C, D be any multitude whatsoever of con-
Γ,Δ,οἱδὲἄκροιαὐτῶνοἱΑ,Δ,πρῶτοιπρὸςἀλλήλους tinuously proportional numbers. And let the outermost
ἔστωσαν·λέγω,ὅτιοἱΑ,Β,Γ,Δἐλάχιστοίεἰσιτῶντὸν of them, A and D, be prime to one another. I say that
αὐτὸνλόγονἐχόντωναὐτοῖς.
A, B, C, D are the least of those (numbers) having the
Εἰγὰρμή,ἔστωσανἐλάττονεςτῶνΑ,Β,Γ,ΔοἱΕ, same ratio as them.
Ζ,Η,Θἐντῷαὐτῷλόγῳὄντεςαὐτοῖς. καὶἐπεὶοἱΑ, For if not, let E, F , G, H be less than A, B, C, D
Β,Γ,ΔἐντῷαὐτῷλόγῳεἰσὶτοῖςΕ,Ζ,Η,Θ,καίἐστιν (respectively), being in the same ratio as them. And since
ἴσοντὸπλῆθος[τῶνΑ,Β,Γ,Δ]τῷπλήθει[τῶνΕ,Ζ,Η, A, B, C, D are in the same ratio as E, F , G, H, and the
Θ],δι᾿ἴσουἄραἐστὶνὡςὁΑπρὸςτὸνΔ,ὁΕπρὸςτὸν multitude [of A, B, C, D] is equal to the multitude [of E,
Θ.οἱδὲΑ,Δπρῶτοι,οἱδὲπρῶτοικαὶἐλάχιστοι,οἱδὲ F , G, H], thus, via equality, as A is to D, (so) E (is) to H
ἐλάχιστοιἀριθμοὶμετροῦσιτοὺςτὸναὐτὸνλόγονἔχοντας [Prop. 7.14]. And A and D (are) prime (to one another).
ἰσάκιςὅτεμείζωντὸνμείζονακαὶὁἐλάσσωντὸνἐλάσσονα, And prime (numbers are) also the least of those (numbers
τουτέστινὅτεἡγούμενοςτὸνἡγούμενονκαὶὁἑπόμενος having the same ratio as them) [Prop. 7.21]. And the
τὸνἑπόμενον.μετρεῖἄραὁΑτὸνΕὁμείζωντὸνἐλάσσονα· least numbers measure those (numbers) having the same
ὅπερἐστὶνἀδύνατον. οὐκἄραοἱΕ,Ζ,Η,Θἐλάσσονες ratio (as them) an equal number of times, the greater
ὄντεςτῶνΑ,Β,Γ,Δἐντῷαὐτῷλόγῳεἰσὶναὐτοῖς.οἱΑ, (measuring) the greater, and the lesser the lesser—that
Β,Γ,Δἄραἐλάχιστοίεἰσιτῶντὸναὐτὸνλόγονἐχόντων is to say, the leading (measuring) the leading, and the
αὐτοῖς·ὅπερἔδειδεῖξαι.
following the following [Prop. 7.20]. Thus, A measures
E, the greater (measuring) the lesser. The very thing is
impossible. Thus, E, F , G, H, being less than A, B, C,
D, are not in the same ratio as them. Thus, A, B, C, D
are the least of those (numbers) having the same ratio as
them. (Which is) the very thing it was required to show.
. Proposition 2
Αριθμοὺςεὑρεῖνἑξῆςἀνάλογονἐλαχίστους,ὅσουςἂν To find the least numbers, as many as may be pre-
ἐπιτάξῃτις,ἐντῷδοθέντιλόγῳ.
scribed, (which are) continuously proportional in a given
῎Εστω ὁ δοθεὶς λόγος ἐν ἐλάχίστοιςἀριθμοῖς ὁ τοῦ ratio.
Α πρὸς τὸν Β· δεῖ δὴ ἀριθμοὺς εὑρεῖν ἑξῆς ἀνάλογον Let the given ratio, (expressed) in the least numbers,
ἐλαχίστους,ὅσουςἄντιςἐπιτάξῃ,ἐντῷτοῦΑπρὸςτὸνΒ be that of A to B. So it is required to find the least numλόγῳ.
bers, as many as may be prescribed, (which are) in the
᾿Επιτετάχθωσανδὴτέσσαρες,καὶὁΑἑαυτὸνπολλα- ratio of A to B.
πλασιάσαςτὸνΓποιείτω,τὸνδὲΒπολλαπλασιάσαςτὸνΔ Let four (numbers) have been prescribed. And let A
ποιείτω,καὶἔτιὁΒἑαυτὸνπολλαπλασιάσαςτὸνΕποιείτω, make C (by) multiplying itself, and let it make D (by)
καὶἔτιὁΑτοὺςΓ,Δ,ΕπολλαπλασιάσαςτοὺςΖ,Η,Θ multiplying B. And, further, let B make E (by) multiplyποιείτω,ὁδὲΒτὸνΕπολλαπλασιάσαςτὸνΚποιείτω.
ing itself. And, further, let A make F , G, H (by) multiplying
C, D, E. And let B make K (by) multiplying
E.
228

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
Α
Β
Ζ
Η
Θ
Κ
Γ
∆
Ε
Καὶ ἐπεὶ ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Γ And since A has made C (by) multiplying itself, and
πεποίηκεν, τὸν δὲ Β πολλαπλασιάσαςτὸν Δ πεποίηκεν, has made D (by) multiplying B, thus as A is to B, [so] C
ἔστινἄραὡςὁΑπρὸςτὸνΒ,[οὕτως]ὁΓπρὸςτὸνΔ. (is) to D [Prop. 7.17]. Again, since A has made D (by)
πάλιν,ἐπεὶὁμὲνΑτὸνΒπολλαπλασιάσαςτὸνΔπεποίηκεν, multiplying B, and B has made E (by) multiplying itself,
ὁδὲΒἑαυτὸνπολλαπλασιάσαςτὸνΕπεποίηκεν,ἑκάτερος A, B have thus made D, E, respectively, (by) multiplying
ἄρατῶνΑ,ΒτὸνΒπολλαπλασιάσαςἑκάτεροντῶνΔ,Ε B. Thus, as A is to B, so D (is) to E [Prop. 7.18]. But, as
πεποίηκεν.ἔστινἄραὡςὁΑπρὸςτὸνΒ,οὕτωςὁΔπρὸς A (is) to B, (so) C (is) to D. And thus as C (is) to D, (so)
τὸνΕ.ἀλλ᾿ὡςὁΑπρὸςτὸνΒ,ὁΓπρὸςτὸνΔ·καὶὡς D (is) to E. And since A has made F , G (by) multiplying
ἄραὁΓπρὸςτὸνΔ,ὁΔπρὸςτὸνΕ.καὶἐπεὶὁΑτοὺςΓ, C, D, thus as C is to D, [so] F (is) to G [Prop. 7.17].
ΔπολλαπλασιάσαςτοὺςΖ,Ηπεποίηκεν,ἔστινἄραὡςὁΓ And as C (is) to D, so A was to B. And thus as A (is)
πρὸςτὸνΔ,[οὕτως]ὁΖπρὸςτὸνΗ.ὡςδὲὁΓπρὸςτὸν to B, (so) F (is) to G. Again, since A has made G, H
Δ,οὕτωςἦνὁΑπρὸςτὸνΒ·καὶὡςἄραὁΑπρὸςτὸνΒ,ὁ (by) multiplying D, E, thus as D is to E, (so) G (is) to
ΖπρὸςτὸνΗ.πάλιν,ἐπεὶὁΑτοὺςΔ,Επολλαπλασιάσας H [Prop. 7.17]. But, as D (is) to E, (so) A (is) to B.
τοὺςΗ,Θπεποίηκεν,ἔστινἄραὡςὁΔπρὸςτὸνΕ,ὁΗ And thus as A (is) to B, so G (is) to H. And since A, B
πρὸςτὸνΘ.ἀλλ᾿ὡςὁΔπρὸςτὸνΕ,ὁΑπρὸςτὸνΒ.καὶ have made H, K (by) multiplying E, thus as A is to B,
ὡςἄραὁΑπρὸςτὸνΒ,οὕτωςὁΗπρὸςτὸνΘ.καὶἐπεὶ so H (is) to K. But, as A (is) to B, so F (is) to G, and
οἱΑ,ΒτὸνΕπολλαπλασιάσαντεςτοὺςΘ,Κπεποιήκασιν, G to H. And thus as F (is) to G, so G (is) to H, and H
ἔστινἄραὡςὁΑπρὸςτὸνΒ,οὕτωςὁΘπρὸςτὸνΚ.ἀλλ᾿ to K. Thus, C, D, E and F , G, H, K are (both continu-
ὡςὁΑπρὸςτὸνΒ,οὕτωςὅτεΖπρὸςτὸνΗκαὶὁΗπρὸς ously) proportional in the ratio of A to B. So I say that
τὸνΘ.καὶὡςἄραὁΖπρὸςτὸνΗ,οὕτωςὅτεΗπρὸς (they are) also the least (sets of numbers continuously
τὸνΘκαὶὁΘπρὸςτὸνΚ·οἱΓ,Δ,ΕἄρακαὶοἱΖ,Η, proportional in that ratio). For since A and B are the
Θ,ΚἀνάλογόνεἰσινἐντῷτοῦΑπρὸςτὸνΒλόγῳ.λέγω least of those (numbers) having the same ratio as them,
δή,ὅτικαὶἐλάχιστοι.ἐπεὶγὰροἱΑ,Βἐλάχιστοίεἰσιτῶν and the least of those (numbers) having the same ratio
τὸναὐτὸνλόγονἐχόντωναὐτοῖς,οἱδὲἐλάχιστοιτῶντὸν are prime to one another [Prop. 7.22], A and B are thus
αὐτὸνλόγονἐχόντωνπρῶτοιπρὸςἀλλήλουςεἰσίν,οἱΑ,Β prime to one another. And A, B have made C, E, respec-
ἄραπρῶτοιπρὸςἀλλήλουςεἰσίν.καὶἑκάτεροςμὲντῶνΑ, tively, (by) multiplying themselves, and have made F , K
ΒἑαυτὸνπολλαπλασιάσαςἑκάτεροντῶνΓ,Επεποίηκεν, by multiplying C, E, respectively. Thus, C, E and F , K
ἑκάτερονδὲτῶνΓ,ΕπολλαπλασιάσαςἑκάτεροντῶνΖ,Κ are prime to one another [Prop. 7.27]. And if there are
πεποίηκεν·οἱΓ,ΕἄρακαὶοἱΖ,Κπρῶτοιπρὸςἀλλήλους any multitude whatsoever of continuously proportional
εἰσίν. ἐὰνδὲὦσινὁποσοιοῦνἀριθμοὶἑξῆςἀνάλογον,οἱ numbers, and the outermost of them are prime to one
δὲἄκροιαὐτῶνπρῶτοιπρὸςἀλλήλουςὦσιν,ἐλάχιστοίεἰσι another, then the (numbers) are the least of those (numτῶντὸναὐτὸνλόγονἐχόντωναὐτοῖς.
οἱΓ,Δ,Εἄρακαὶ bers) having the same ratio as them [Prop. 8.1]. Thus, C,
οἱΖ,Η,Θ,Κἐλάχιστοίεἰσιτῶντὸναὐτὸνλόγονἐχόντων D, E and F , G, H, K are the least of those (continuously
τοῖςΑ,Β·ὅπερἔδειδεῖξαι.
proportional sets of numbers) having the same ratio as A
and B. (Which is) the very thing it was required to show.
A
B
F
G
H
K
C
D
E
229

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
ÈÖ×Ñ.
᾿Εκ δὴ τούτου φανερόν, ὅτι ἐὰν τρεῖς ἀριθμοὶ ἑξῆς So it is clear, from this, that if three continuously pro-
Corollary
ἀνάλογον ἐλάχιστοι ὦσι τῶν τὸν αὐτὸν λόγον ἐχόντων portional numbers are the least of those (numbers) havαὐτοῖς,οἱἄκροναὐτῶντετράγωνοίεἰσιν,ἐὰνδὲτέσσαρες,
ing the same ratio as them then the outermost of them
κύβοι.
are square, and, if four (numbers), cube.
. Proposition 3
᾿Εὰνὦσινὁποσοιοῦνἀριθμοὶἑξῆςἀνάλογονἐλάχιστ- If there are any multitude whatsoever of continuοιτῶντὸναὐτὸνλόγονἐχόντωναὐτοῖς,οἱἄκροιαὐτῶν
ously proportional numbers (which are) the least of those
πρῶτοιπρὸςἀλλήλουςεἰσίν.
(numbers) having the same ratio as them then the outermost
of them are prime to one another.
Α
Β
Γ
∆
Λ
Μ
Ν
Ξ
Ε
Ζ
Η
Θ
Κ
῎Εστωσανὁποσοιοῦνἀριθμοὶἑξῆςἀνάλογονἐλάχιστοι Let A, B, C, D be any multitude whatsoever of conτῶντὸναὐτὸνλόγονἐχόντωναὐτοῖςοἱΑ,Β,Γ,Δ·λέγω,
tinuously proportional numbers (which are) the least of
ὅτιοἱἄκροιαὐτῶνοἱΑ,Δπρῶτοιπρὸςἀλλήλουςεἰσίν. those (numbers) having the same ratio as them. I say
Εἰλήφθωσανγὰρδύομὲνἀριθμοὶἐλάχιστοιἐντῷτῶν that the outermost of them, A and D, are prime to one
Α,Β,Γ,ΔλόγῳοἱΕ,Ζ,τρεῖςδὲοἱΗ,Θ,Κ,καὶἑξῆς another.
ἑνὶπλείους,ἕωςτὸλαμβανόμενονπλῆθοςἴσονγένηταιτῷ For let the two least (numbers) E, F (which are)
πλήθειτῶνΑ,Β,Γ,Δ.εἰλήφθωσανκαὶἔστωσανοἱΛ,Μ, in the same ratio as A, B, C, D have been taken
Ν,Ξ.
[Prop. 7.33]. And the three (least numbers) G, H, K
ΚαὶἐπεὶοἱΕ,Ζἐλάχιστοίεἰσιτῶντὸναὐτὸνλόγον [Prop. 8.2]. And (so on), successively increasing by one,
ἐχόντων αὐτοῖς, πρῶτοι πρὸς ἀλλήλους εἰσίν. καὶ ἐπεὶ until the multitude of (numbers) taken is made equal to
ἑκάτεροςτῶνΕ,Ζἑαυτὸνμὲνπολλαπλασιάσαςἑκάτερον the multitude of A, B, C, D. Let them have been taken,
τῶν Η, Κ πεποίηκεν, ἑκάτερον δὲ τῶν Η, Κ πολλα- and let them be L, M, N, O.
πλασιάσαςἑκάτεροντῶνΛ,Ξπεποίηκεν,καὶοἱΗ,Κἄρα And since E and F are the least of those (numbers)
καὶοἱΛ,Ξπρῶτοιπρὸςἀλλήλουςεἰσίν. καὶἐπεὶοἱΑ,Β, having the same ratio as them they are prime to one an-
Γ,Δἐλάχιστοίεἰσιτῶντὸναὐτὸνλόγονἐχόντωναὐτοῖς, other [Prop. 7.22]. And since E, F have made G, K, reεἰσὶδὲκαὶοἱΛ,Μ,Ν,Ξἐλάχιστοιἐντῷαὐτῷλόγῳὄντες
spectively, (by) multiplying themselves [Prop. 8.2 corr.],
τοῖςΑ,Β,Γ,Δ,καίἐστινἴσοντὸπλῆθοςτῶνΑ,Β,Γ, and have made L, O (by) multiplying G, K, respec-
ΔτῷπλήθειτῶνΛ,Μ,Ν,Ξ,ἕκαστοςἄρατῶνΑ,Β,Γ, tively, G, K and L, O are thus also prime to one another
ΔἑκάστῳτῶνΛ,Μ,Ν,Ξἴσοςἐστίν· ἴσοςἄραἐστὶνὁ [Prop. 7.27]. And since A, B, C, D are the least of those
μὲνΑτῷΛ,ὁδὲΔτῷΞ.καίεἰσινοἱΛ,Ξπρῶτοιπρὸς (numbers) having the same ratio as them, and L, M, N,
ἀλλήλους. καὶοἱΑ,Δἄραπρῶτοιπρὸςἀλλήλουςεἰσίν· O are also the least (of those numbers having the same
ὅπερἔδειδεῖξαι.
ratio as them), being in the same ratio as A, B, C, D, and
the multitude of A, B, C, D is equal to the multitude of
L, M, N, O, thus A, B, C, D are equal to L, M, N, O,
respectively. Thus, A is equal to L, and D to O. And L
and O are prime to one another. Thus, A and D are also
prime to one another. (Which is) the very thing it was
A
B
C
D
L
M
N
O
E
F
G
H
K
230

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
required to show.
. Proposition 4
Λόγων δοθέντων ὁποσωνοῦν ἐν ἐλαχίστοις ἀριθμοῖς For any multitude whatsoever of given ratios, (ex-
ἀριθμοὺςεὑρεῖνἑξῆςἀνάλογονἐλαχίστουςἐντοῖςδοθεῖσι pressed) in the least numbers, to find the least numbers
λόγοις.
continuously proportional in these given ratios.
Α Β
A
B
Γ ∆
C
D
Ε Ζ
E
F
Ν
Ξ
Μ
Ο
Θ
Η
Κ
Λ
῎Εστωσανοἱδοθέντεςλόγοιἐνἐλαχίστοιςἀριθμοῖςὅ Let the given ratios, (expressed) in the least numbers,
τετοῦΑπρὸςτὸνΒκαὶὁτοῦΓπρὸςτὸνΔκαὶἔτιὁ be the (ratios) of A to B, and of C to D, and, further,
τοῦΕπρὸςτὸνΖ·δεῖδὴἀριθμοὺςεὑρεῖνἑξῆςἀνάλογον of E to F . So it is required to find the least numbers
ἐλαχίστουςἔντετῷτοῦΑπρὸςτὸνΒλόγῳκαὶἐντῷτοῦ continuously proportional in the ratio of A to B, and of
ΓπρὸςτὸνΔκαὶἔτιτῷτοῦΕπρὸςτὸνΖ. C to B, and, further, of E to F .
ΕἰλήφθωγὰρὁὑπὸτῶνΒ,Γἐλάχιστοςμετρούμενος For let the least number, G, measured by (both) B and
ἀριθμὸςὁΗ.καὶὁσάκιςμὲνὁΒτὸνΗμετρεῖ,τοσαυτάκις C have be taken [Prop. 7.34]. And as many times as B
καὶὁΑτὸνΘμετρείτω,ὁσάκιςδὲὁΓτὸνΗμετρεῖ,το- measures G, so many times let A also measure H. And as
σαυτάκιςκαὶὁΔτὸνΚμετρείτω.ὁδὲΕτὸνΚἤτοιμετρεῖ many times as C measures G, so many times let D also
ἢοὐμετρεῖ.μετρείτωπρότερον.καὶὁσάκιςὁΕτὸνΚμε- measure K. And E either measures, or does not measure,
τρεῖ,τοσαυτάκιςκαὶὁΖτὸνΛμετρείτω.καὶἐπεὶἰσάκιςὁ K. Let it, first of all, measure (K). And as many times as
ΑτὸνΘμετρεῖκαὶὁΒτὸνΗ,ἔστινἄραὡςὁΑπρὸςτὸν E measures K, so many times let F also measure L. And
Β,οὕτωςὁΘπρὸςτὸνΗ.διὰτὰαὐτὰδὴκαὶὡςὁΓπρὸς since A measures H the same number of times that B also
τὸνΔ,οὕτωςὁΗπρὸςτὸνΚ,καὶἔτιὡςὁΕπρὸςτὸνΖ, (measures) G, thus as A is to B, so H (is) to G [Def. 7.20,
οὕτωςὁΚπρὸςτὸνΛ·οἱΘ,Η,Κ,Λἄραἑξῆςἀνάλογόν Prop. 7.13]. And so, for the same (reasons), as C (is) to
εἰσινἔντετῷτοῦΑπρὸςτὸνΒκαὶἐντῷτοῦΓπρὸςτὸν D, so G (is) to K, and, further, as E (is) to F , so K (is)
ΔκαὶἔτιἐντῷτοῦΕπρὸςτὸνΖλόγῳ.λέγωδή,ὅτικαὶ to L. Thus, H, G, K, L are continuously proportional in
ἐλάχιστοι. εἰγὰρμήεἰσινοἱΘ,Η,Κ,Λἑξῆςἀνάλογον the ratio of A to B, and of C to D, and, further, of E to
ἐλάχιστοιἔντετοῖςτοῦΑπρὸςτὸνΒκαὶτοῦΓπρὸςτὸν F . So I say that (they are) also the least (numbers con-
ΔκαὶἐντῷτοῦΕπρὸςτὸνΖλόγοις,ἔστωσανοἱΝ,Ξ, tinuously proportional in these ratios). For if H, G, K,
Μ,Ο.καὶἐπείἐστινὡςὁΑπρὸςτὸνΒ,οὕτωςὁΝπρὸς L are not the least numbers continuously proportional in
τὸνΞ,οἱδὲΑ,Βἐλάχιστοι,οἱδὲἐλάχιστοιμετροῦσιτοὺς the ratios of A to B, and of C to D, and of E to F , let N,
τὸναὐτὸνλόγονἔχονταςἰσάκιςὅτεμείζωντὸνμείζονα O, M, P be (the least such numbers). And since as A is
καὶὁἐλάσσωντὸνἐλάσσονα,τουτέστινὅτεἡγούμενος to B, so N (is) to O, and A and B are the least (numbers
τὸνἡγούμενονκαὶὁἑπόμενοςτὸνἑπόμενον,ὁΒἄρατὸν which have the same ratio as them), and the least (num-
Ξμετρεῖ. διὰτὰαὐτὰδὴκαὶὁΓτὸνΞμετρεῖ·οἱΒ,Γ bers) measure those (numbers) having the same ratio (as
ἄρατὸνΞμετροῦσιν·καὶὁἐλάχιστοςἄραὑπὸτῶνΒ,Γ them) an equal number of times, the greater (measurμετρούμενοςτὸνΞμετρήσει.
ἐλάχιστοςδὲὑπὸτῶνΒ,Γ ing) the greater, and the lesser the lesser—that is to say,
μετρεῖταιὁΗ·ὁΗἄρατὸνΞμετρεῖὁμείζωντὸνἐλάσσονα· the leading (measuring) the leading, and the following
ὅπερἐστὶνἀδύντατον.οὐκἄραἔσονταίτινεςτῶνΘ,Η,Κ, the following [Prop. 7.20], B thus measures O. So, for
ΛἐλάσσονεςἀριθμοὶἑξῆςἔντετῷτοῦΑπρὸςτὸνΒκαὶ the same (reasons), C also measures O. Thus, B and C
τῷτοῦΓπρὸςτὸνΔκαὶἔτιτῷτοῦΕπρὸςτὸνΖλόγῷ. (both) measure O. Thus, the least number measured by
(both) B and C will also measure O [Prop. 7.35]. And
G (is) the least number measured by (both) B and C.
N
O
M
P
H
G
K
L
231

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
Α
Γ
Ε
Β
∆
Ζ
Thus, G measures O, the greater (measuring) the lesser.
The very thing is impossible. Thus, there cannot be any
numbers less than H, G, K, L (which are) continuously
(proportional) in the ratio of A to B, and of C to D, and,
further, of E to F .
A
B
C
E
D
F
Ν Θ
N
H
Ξ Η
O
G
Μ Κ
M
K
Ο
P
Π
Q
Ρ
R
Σ
S
Τ
T
ΜὴμετρείτωδὴὁΕτὸνΚ,καὶεἰλήφθωὑπὸτῶνΕ, So let E not measure K. And let the least num-
Κ ἐλάχιστοςμετρούμενοςἀριθμὸς ὁ Μ. καὶὁσάκιςμὲν ber, M, measured by (both) E and K have been taken
ὁ Κ τὸν Μ μετρεῖ, τοσαυτάκιςκαὶ ἑκάτεροςτῶν Θ, Η [Prop. 7.34]. And as many times as K measures M, so
ἑκάτεροντῶνΝ,Ξμετρείτω,ὁσάακιςδὲὁΕτὸνΜμε- many times let H, G also measure N, O, respectively.
τρεῖ,τοσαυτάκιςκαὶὁΖτὸνΟμετρείτω. ἐπεὶἰσάκιςὁΘ And as many times as E measures M, so many times let
τὸνΝμετρεῖκαὶὁΗτὸνΞ,ἔστινἄραὡςὁΘπρὸςτὸν F also measure P . Since H measures N the same num-
Η,οὕτωςὁΝπρὸςτὸνΞ.ὡςδὲὁΘπρὸςτὸνΗ,οὕτως ber of times as G (measures) O, thus as H is to G, so
ὁΑπρὸςτὸνΒ·καὶὡςἄραὁΑπρὸςτὸνΒ,οὕτωςὁΝ N (is) to O [Def. 7.20, Prop. 7.13]. And as H (is) to G,
πρὸςτὸνΞ.διὰτὰαὐτὰδὴκαὶὡςὁΓπρὸςτὸνΔ,οὕτως so A (is) to B. And thus as A (is) to B, so N (is) to
ὁΞπρὸςτὸνΜ.πάλιν,ἐπεὶἰσάκιςὁΕτὸνΜμετρεῖκαὶ O. And so, for the same (reasons), as C (is) to D, so O
ὁΖτὸνΟ,ἔστινἄραὡςὁΕπρὸςτὸνΖ,οὕτωςὁΜπρὸς (is) to M. Again, since E measures M the same numτὸνΟ·οἱΝ,Ξ,Μ,Οἄραἑξῆςἀνάλογόνεἰσινἐντοῖςτοῦ
ber of times as F (measures) P , thus as E is to F , so
τεΑπρὸςτὸνΒκαὶτοῦΓπρὸςτὸνΔκαὶἔτιτοῦΕπρὸς M (is) to P [Def. 7.20, Prop. 7.13]. Thus, N, O, M, P
τὸνΖλόγοις. λέγωδή,ὅτικαὶἐλάχιστοιἐντοῖςΑΒ,Γ are continuously proportional in the ratios of A to B, and
Δ,ΕΖλόγοις. εἰγὰρμή,ἔσονταίτινεςτῶνΝ,Ξ,Μ,Ο of C to D, and, further, of E to F . So I say that (they
ἐλάσσονεςἀριθμοὶἑξῆςἀνάλογονἐντοῖςΑΒ,ΓΔ,ΕΖ are) also the least (numbers) in the ratios of A B, C D,
λόγοις.ἔστωσανοἱΠ,Ρ,Σ,Τ.καὶἐπείἐστινὡςὁΠπρὸς E F . For if not, then there will be some numbers less
τὸνΡ,οὕτωςὁΑπρὸςτὸνΒ,οἱδὲΑ,Βἐλάχιστοι,οἱδὲ than N, O, M, P (which are) continuously proportional
ἐλάχιστοιμετροῦσιτοὺςτὸναὐτὸνλόγονἔχονταςαὐτοῖς in the ratios of A B, C D, E F . Let them be Q, R, S,
ἰσάκιςὅτεἡγούμενοςτὸνἡγούμενονκαὶὁἑπόμενοςτὸν T . And since as Q is to R, so A (is) to B, and A and B
ἑπόμενον,ὁΒἄρατὸνΡμετρεῖ. διὰτὰαὐτὰδὴκαὶὁΓ (are) the least (numbers having the same ratio as them),
τὸνΡμετρεῖ·οἱΒ,ΓἄρατὸνΡμετροῦσιν.καὶὁἐλάχιστος and the least (numbers) measure those (numbers) hav-
ἄραὑπὸτῶνΒ,ΓμετούμενοςτὸνΡμετρήσει. ἐλάχιστος ing the same ratio as them an equal number of times,
δὲὑπὸτῶνΒ,ΓμετρούμενοςἐστινὁΗ·ὁΗἄρατὸνΡ the leading (measuring) the leading, and the following
μετρεῖ.καίἐστινὡςὁΗπρὸςτὸνΡ,οὕτωςὁΚπρὸςτὸν the following [Prop. 7.20], B thus measures R. So, for
Σ·καὶὁΚἄρατὸνΣμετρεῖ.μετρεῖδὲκαὶὁΕτὸνΣ·οἱΕ, the same (reasons), C also measures R. Thus, B and C
ΚἄρατὸνΣμετροῦσιν.καὶὁἐλάχιστοςἄραὑπὸτῶνΕ,Κ (both) measure R. Thus, the least (number) measured by
μετρούμενοςτὸνΣμετρήσει. ἐλάχιστοςδὲὑπὸτῶνΕ,Κ (both) B and C will also measure R [Prop. 7.35]. And G
μετρούμενόςἐστινὁΜ·ὁΜἄρατὸνΣμετρεῖὁμείζωντὸν is the least number measured by (both) B and C. Thus,
ἐλάσσονα·ὅπερἐστὶνἀδύνατον.οὐκἄραἔσονταίτινεςτῶν G measures R. And as G is to R, so K (is) to S. Thus,
232

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
Ν,Ξ,Μ,Οἐλάσσονεςἀριθμοὶἑξῆςἀνάλογονἔντετοῖς K also measures S [Def. 7.20]. And E also measures
τοῦΑπρὸςτὸνΒκαὶτοῦΓπρὸςτὸνΔκαὶἔτιτοῦΕπρὸς S [Prop. 7.20]. Thus, E and K (both) measure S. Thus,
τὸνΖλόγοις·οἱΝ,Ξ,Μ,Οἄραἑξῆςἀνάλογονἐλάχιστοί the least (number) measured by (both) E and K will also
εἰσινἐντοῖςΑΒ,ΓΔ,ΕΖλόγοις·ὅπερἔδειδεῖξαι. measure S [Prop. 7.35]. And M is the least (number)
measured by (both) E and K. Thus, M measures S, the
greater (measuring) the lesser. The very thing is impossible.
Thus there cannot be any numbers less than N, O,
M, P (which are) continuously proportional in the ratios
of A to B, and of C to D, and, further, of E to F . Thus,
N, O, M, P are the least (numbers) continuously proportional
in the ratios of A B, C D, E F . (Which is) the very
thing it was required to show.
. Proposition 5
Οἱἐπίπεδοιἀριθμοὶπρὸςἀλλήλουςλόγονἔχουσιτὸν Plane numbers have to one another the ratio compounσυγκείμενονἐκτῶνπλευρῶν.
ded † out of (the ratios of) their sides.
Α
A
Β
B
Γ
Ε
∆
Ζ
C
E
D
F
Η
Θ
Κ
G
H
K
Λ
L
῎Εστωσαν ἐπίπεδοι ἀριθμοὶ οἱ Α, Β, καὶ τοῦ μὲν Α Let A and B be plane numbers, and let the numbers
πλευραὶἔστωσανοἱΓ,Δἀριθμοί,τοῦδὲΒοἱΕ,Ζ·λέγω, C, D be the sides of A, and (the numbers) E, F (the
ὅτιὁΑπρὸςτὸνΒλόγονἔχειτὸνσυγκείμενονἐκτῶν sides) of B. I say that A has to B the ratio compounded
πλευρῶν.
out of (the ratios of) their sides.
ΛόγωνγὰρδοθέντωντοῦτεὃνἔχειὁΓπρὸςτὸνΕκαὶ For given the ratios which C has to E, and D (has) to
ὁΔπρὸςτὸνΖεἰλήφθωσανἀριθμοὶἑξῆςἐλάχιστοιἐντοῖς F , let the least numbers, G, H, K, continuously propor-
ΓΕ,ΔΖλόγοις,οἱΗ,Θ,Κ,ὥστεεἶναιὡςμὲντὸνΓπρὸς tional in the ratios C E, D F have been taken [Prop. 8.4],
τὸνΕ,οὕτωςτὸνΗπρὸςτὸνΘ,ὡςδὲτὸνΔπρὸςτὸνΖ, so that as C is to E, so G (is) to H, and as D (is) to F , so
οὕτωςτὸνΘπρὸςτὸνΚ.καὶὁΔτὸνΕπολλαπλασιάσας H (is) to K. And let D make L (by) multiplying E.
τὸνΛποιείτω.
And since D has made A (by) multiplying C, and has
Καὶ ἐπεὶ ὁ Δ τὸν μὲν Γ πολλαπλασιάσας τὸν Α made L (by) multiplying E, thus as C is to E, so A (is) to
πεποίηκεν, τὸν δὲ Ε πολλαπλασιάσας τὸν Λ πεποίηκεν, L [Prop. 7.17]. And as C (is) to E, so G (is) to H. And
ἔστινἄραὡςὁΓπρὸςτὸνΕ,οὕτωςὁΑπρὸςτὸνΛ.ὡς thus as G (is) to H, so A (is) to L. Again, since E has
δὲὁΓπρὸςτὸνΕ,οὕτωςὁΗπρὸςτὸνΘ·καὶὡςἄραὁ made L (by) multiplying D [Prop. 7.16], but, in fact, has
ΗπρὸςτὸνΘ,οὕτωςὁΑπρὸςτὸνΛ.πάλιν,ἐπεὶὁΕτὸν also made B (by) multiplying F , thus as D is to F , so L
ΔπολλαπλασιάσαςτὸνΛπεποίηκεν,ἀλλὰμὴνκαὶτὸνΖ (is) to B [Prop. 7.17]. But, as D (is) to F , so H (is) to
πολλαπλασιάσαςτὸνΒπεποίηκεν,ἔστινἄραὡςὁΔπρὸς K. And thus as H (is) to K, so L (is) to B. And it was
τὸνΖ,οὕτωςὁΛπρὸςτὸνΒ.ἀλλ᾿ὡςὁΔπρὸςτὸνΖ, also shown that as G (is) to H, so A (is) to L. Thus, via
οὕτωςὁΘπρὸςτὸνΚ·καὶὡςἄραὁΘπρὸςτὸνΚ,οὕτως equality, as G is to K, [so] A (is) to B [Prop. 7.14]. And
ὁΛπρὸςτὸνΒ.ἐδείχθηδὲκαὶὡςὁΗπρὸςτὸνΘ,οὕτως G has to K the ratio compounded out of (the ratios of)
ὁΑπρὸςτὸνΛ·δι᾿ἴσουἄραἐστὶνὡςὁΗπρὸςτὸνΚ, the sides (of A and B). Thus, A also has to B the ratio
[οὕτως]ὁΑπρὸςτὸνΒ.ὁδὲΗπρὸςτὸνΚλόγονἔχει compounded out of (the ratios of) the sides (of A and B).
233

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
τὸνσυγκείμενονἐκτῶνπλευρῶν·καὶὁΑἄραπρὸςτὸν (Which is) the very thing it was required to show.
Βλόγονἔχειτὸνσυγκείμενονἐκτῶνπλευρῶν·ὅπερἔδει
δεῖξαι.
† i.e., multiplied.
¦.
Proposition 6
᾿Εὰν ὦσιν ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, ὁ δὲ If there are any multitude whatsoever of continuously
πρῶτοςτὸνδεύτερονμὴμετρῇ,οὐδὲἄλλοςοὐδεὶςοὐδένα proportional numbers, and the first does not measure the
μετρήσει.
second, then no other (number) will measure any other
(number) either.
Α
A
Β
B
Γ
C
∆
D
Ε
E
Ζ
Η
Θ
῎ΕστωσανὁποσοιοῦνἀριθμοὶἑξῆςἀνάλογονοἱΑ,Β, Let A, B, C, D, E be any multitude whatsoever of
Γ,Δ,Ε,ὁδὲΑτὸνΒμὴμετρείτω·λέγω,ὅτιοὐδὲἄλλος continuously proportional numbers, and let A not meaοὐδεὶςοὐδέναμετρήσει.
sure B. I say that no other (number) will measure any
῞ΟτιμὲνοὖνοἱΑ,Β,Γ,Δ,Εἑξῆςἀλλήλουςοὐμε- other (number) either.
τροῦσιν, φανερόν· οὐδὲ γὰρ ὁ Α τὸν Β μετρεῖ. λέγω Now, (it is) clear that A, B, C, D, E do not succesδή,
ὅτι οὐδὲἄλλοςοὐδεὶςοὐδέναμετρήσει. εἰγὰρ δυ- sively measure one another. For A does not even meaνατόν,
μετρείτω ὁ Α τὸν Γ. καὶὅσοι εἰσὶν οἱ Α, Β, Γ, sure B. So I say that no other (number) will measure
τοσοῦτοι εἰλήφθωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν any other (number) either. For, if possible, let A measure
λόγονἐχόντωντοῖςΑ,Β,ΓοἱΖ,Η,Θ.καὶἐπεὶοἱΖ, C. And as many (numbers) as are A, B, C, let so many
Η,ΘἐντῷαὐτῷλόγῳεἰσὶτοῖςΑ,Β,Γ,καίἐστινἴσοντὸ of the least numbers, F , G, H, have been taken of those
πλῆθοςτῶνΑ,Β,ΓτῷπλήθειτῶνΖ,Η,Θ,δι᾿ἴσουἄρα (numbers) having the same ratio as A, B, C [Prop. 7.33].
ἐστὶνὡςὁΑπρὸςτὸνΓ,οὕτωςὁΖπρὸςτὸνΘ.καὶἐπεί And since F , G, H are in the same ratio as A, B, C, and
ἐστινὡςὁΑπρὸςτὸνΒ,οὕτωςὁΖπρὸςτὸνΗ,οὐμετρεῖ the multitude of A, B, C is equal to the multitude of F ,
δὲὁΑτὸνΒ,οὐμετρεῖἄραοὐδὲὁΖτὸνΗ·οὐκἄραμονάς G, H, thus, via equality, as A is to C, so F (is) to H
ἐστινὁΖ·ἡγὰρμονὰςπάνταἀριθμὸνμετρεῖ. καίεἰσινοἱ [Prop. 7.14]. And since as A is to B, so F (is) to G,
Ζ,Θπρῶτοιπρὸςἀλλήλους[οὐδὲὁΖἄρατὸνΘμετρεῖ]. and A does not measure B, F does not measure G either
καίἐστινὡςὁΖπρὸςτὸνΘ,οὕτωςὁΑπρὸςτὸνΓ·οὐδὲ [Def. 7.20]. Thus, F is not a unit. For a unit measures
ὁΑἄρατὸνΓμετρεῖ.ὁμοίωςδὴδείξομεν,ὅτιοὐδὲἄλλος all numbers. And F and H are prime to one another
οὐδεὶςοὐδέναμετρήσει·ὅπερἔδειδεῖξαι.
[Prop. 8.3] [and thus F does not measure H]. And as
F is to H, so A (is) to C. And thus A does not measure
C either [Def. 7.20]. So, similarly, we can show that no
other (number) can measure any other (number) either.
(Which is) the very thing it was required to show.
Þ. Proposition 7
᾿Εὰν ὦσιν ὁποσοιοῦν ἀριθμοὶ [ἑξῆς] ἀνάλογον, ὁ δὲ If there are any multitude whatsoever of [continuπρῶτοςτὸνἔσχατονμετρῇ,καὶτὸνδεύτερονμετρήσει.
ously] proportional numbers, and the first measures the
F
G
H
234

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
Α
Β
Γ
∆
last, then (the first) will also measure the second.
῎ΕστωσανὁποσοιοῦνἀριθμοὶἑξῆςἀνάλογονοἱΑ,Β,Γ, Let A, B, C, D be any number whatsoever of continu-
Δ,ὁδὲΑτὸνΔμετρείτω·λέγω,ὅτικαὶὁΑτὸνΒμετρεῖ. ously proportional numbers. And let A measure D. I say
ΕἰγὰροὐμετρεῖὁΑτὸνΒ,οὐδὲἄλλοςοὐδεὶςοὐδένα that A also measures B.
μετρήσει·μετρεῖδὲὁΑτὸνΔ.μετρεῖἄρακαὶὁΑτὸνΒ· For if A does not measure B then no other (number)
ὅπερἔδειδεῖξαι.
will measure any other (number) either [Prop. 8.6]. But
A measures D. Thus, A also measures B. (Which is) the
very thing it was required to show.
. Proposition 8
᾿Εὰνδύοἀριθμῶνμεταξὺκατὰτὸσυνεχὲςἀνάλογον If between two numbers there fall (some) numbers in
ἐμπίπτωσινἀριθμοί,ὅσοιεἰςαὐτοὺςμεταξὺκατὰτὸσυ- continued proportion then, as many numbers as fall in
νεχὲςἀνόλογονἐμπίπτουσινἀριθμοί,τοσοῦτοικαὶεἰςτοὺς between them in continued proportion, so many (numτὸναὐτὸνλόγονἔχοντας[αὐτοῖς]μεταξὺκατὰτὸσυνὲχες
bers) will also fall in between (any two numbers) having
ἀνάλογονἐμπεσοῦνται
the same ratio [as them] in continued proportion.
Α
Γ
∆
Β
Η
Θ
Κ
Λ
Ε
Μ
Ν
Ζ
ΔύογὰρἀριθμῶντῶνΑ,Βμεταξὺκατὰτὸσυνεχὲς For let the numbers, C and D, fall between two num-
ἀνάλογονἐμπιπτέτωσανἀριθμοὶοἱΓ,Δ,καὶπεποιήσθωὡς bers, A and B, in continued proportion, and let it have
ὁΑπρὸςτὸνΒ,οὕτωςὁΕπρὸςτὸνΖ·λέγω,ὅτιὅσοιεἰς been contrived (that) as A (is) to B, so E (is) to F . I say
τοὺςΑ,Βμεταξὺκατὰτὸσυνεχὲςἀνάλογονἐμπεπτώκασιν that, as many numbers as have fallen in between A and
ἀριθμοί,τοσοῦτοικαὶεἰςτοὺςΕ,Ζμεταξὺκατὰτὸσυνεχὲς B in continued proportion, so many (numbers) will also
ἀνάλογονἐμπεσοῦνται.
fall in between E and F in continued proportion.
῞Οσοι γάρ εἰσι τῷ πλήθει οἱ Α, Β, Γ, Δ, τοσοῦτοι For as many as A, B, C, D are in multitude, let so
εἰλήφθωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον many of the least numbers, G, H, K, L, having the same
ἐχόντων τοῖςΑ, Γ,Δ, ΒοἱΗ,Θ,Κ,Λ·οἱἄραἄκροι ratio as A, B, C, D, have been taken [Prop. 7.33]. Thus,
αὐτῶνοἱΗ,Λπρῶτοιπρὸςἀλλήλουςεἰσίν.καὶἐπεὶοἱΑ, the outermost of them, G and L, are prime to one another
Γ,Δ,ΒτοῖςΗ,Θ,Κ,Λἐντῷαὐτῷλόγῳεἰσίν,καίἐστιν [Prop. 8.3]. And since A, B, C, D are in the same ratio
ἴσοντὸπλῆθοςτῶνΑ,Γ,Δ,ΒτῷπλήθειτῶνΗ,Θ,Κ, as G, H, K, L, and the multitude of A, B, C, D is equal
Λ,δι᾿ἴσουἄραἐστὶνὡςὁΑπρὸςτὸνΒ,οὕτωςὁΗπρὸς to the multitude of G, H, K, L, thus, via equality, as A is
τὸνΛ.ὡςδὲὁΑπρὸςτὸνΒ,οὕτωςὁΕπρὸςτὸνΖ·καὶ to B, so G (is) to L [Prop. 7.14]. And as A (is) to B, so
A
C
D
B
G
H
K
L
A
B
C
D
E
M
N
F
235

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
ὡςἄραὁΗπρὸςτὸνΛ,οὕτωςὁΕπρὸςτὸνΖ.οἱδὲΗ,Λ E (is) to F . And thus as G (is) to L, so E (is) to F . And
πρῶτοι,οἱδὲπρῶτοικαὶἐλάχιστοι,οἱδὲἐλάχιστοιἀριθμοὶ G and L (are) prime (to one another). And (numbers)
μετροῦσιτοὺςτὸναὐτὸνλόγονἔχονταςἰσάκιςὅτεμείζων prime (to one another are) also the least (numbers havτὸνμείζονακαὶὁἐλάσσωντὸνἐλάσσονα,τουτέστινὅτε
ing the same ratio as them) [Prop. 7.21]. And the least
ἡγούμενοςτὸνἡγούμενονκαὶὁἑπόμενοςτὸνἑπόμενον. numbers measure those (numbers) having the same ratio
ἰσάκιςἄραὁΗτὸνΕμετρεῖκαὶὁΛτὸνΖ.ὁσάκιςδὴὁΗ (as them) an equal number of times, the greater (measurτὸνΕμετρεῖ,τοσαυτάκιςκαὶἑκάτεροςτῶνΘ,Κἑκάτερον
ing) the greater, and the lesser the lesser—that is to say,
τῶνΜ,Νμετρείτω·οἱΗ,Θ,Κ,ΛἄρατοὺςΕ,Μ,Ν,Ζ the leading (measuring) the leading, and the following
ἰσάκιςμετροῦσιν.οἱΗ,Θ,Κ,ΛἄρατοῖςΕ,Μ,Ν,Ζἐντῷ the following [Prop. 7.20]. Thus, G measures E the same
αὐτῷλόγῳεἰσίν.ἀλλὰοἱΗ,Θ,Κ,ΛτοῖςΑ,Γ,Δ,Βἐντῷ number of times as L (measures) F . So as many times as
αὐτῷλόγῳεἰσίν·καὶοἱΑ,Γ,Δ,ΒἄρατοῖςΕ,Μ,Ν,Ζἐν G measures E, so many times let H, K also measure M,
τῷαὐτῷλόγῳεἰσίν.οἱδὲΑ,Γ,Δ,Βἑξῆςἀνάλογόνεἰσιν· N, respectively. Thus, G, H, K, L measure E, M, N,
καὶοἱΕ,Μ,Ν,Ζἄραἑξῆςἀνάλογόνεἰσιν. ὅσοιἄραεἰς F (respectively) an equal number of times. Thus, G, H,
τοὺςΑ,Βμεταξὺκατὰτὸσυνεχὲςἀνάλογονἐμπεπτώκασιν K, L are in the same ratio as E, M, N, F [Def. 7.20].
ἀριθμοί,τοσοῦτοικαὶεἰςτοὺςΕ,Ζμεταξὺκατὰτὸσυνεχὲς But, G, H, K, L are in the same ratio as A, C, D, B.
ἀνάλογονἐμπεπτώκασινἀριθμοί·ὅπερἔδειδεῖξαι. Thus, A, C, D, B are also in the same ratio as E, M, N,
F . And A, C, D, B are continuously proportional. Thus,
E, M, N, F are also continuously proportional. Thus,
as many numbers as have fallen in between A and B in
continued proportion, so many numbers have also fallen
in between E and F in continued proportion. (Which is)
the very thing it was required to show.
. Proposition 9
᾿Εὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, καὶ If two numbers are prime to one another and there
εἰςαὐτοὺςμεταξὺκατὰτὸσυνεχὲςἀνάλογονἐμπίπτωσιν fall in between them (some) numbers in continued pro-
ἀριθμοί,ὅσοιεἰςαὐτοὺςμεταξὺκατὰτὸσυνεχὲςἀνάλογον portion then, as many numbers as fall in between them
ἐμπίπτουσιν ἀριθμοί, τοσοῦτοι καὶ ἑκατέρου αὐτῶν καὶ in continued proportion, so many (numbers) will also fall
μονάδοςμεταξὺκατὰτὸσυνεχὲςἀνάλογονἐμπεσοῦνται. between each of them and a unit in continued proportion.
Α
Γ
∆
Β
Ε
Ζ
Η
Θ
Κ
Λ
Μ
Ν
Ξ
Ο
῎Εστωσαν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους οἱ Α, Let A and B be two numbers (which are) prime to
Β, καὶ εἰς αὐτοὺς μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον one another, and let the (numbers) C and D fall in be-
ἐμπιπτέτωσαν οἱ Γ, Δ, καὶ ἐκκείσθω ἡ Ε μονάς· λέγω, tween them in continued proportion. And let the unit E
ὅτιὅσοιεἰςτοὺςΑ,Βμεταξὺκατὰτὸσυνεχὲςἀνάλογον be set out. I say that, as many numbers as have fallen
ἐμπεπτώκασιν ἀριθμοί, τοσοῦτοι καὶ ἑκατέρου τῶν Α, in between A and B in continued proportion, so many
Β καὶ τῆς μονάδος μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον (numbers) will also fall between each of A and B and
ἐμπεσοῦνται.
the unit in continued proportion.
Εἰλήφθωσανγὰρδύομὲνἀριθμοὶἐλάχιστοιἐντῷτῶν For let the least two numbers, F and G, which are in
Α,Γ,Δ,ΒλόγῳὄντεςοἱΖ,Η,τρεῖςδὲοἱΘ,Κ,Λ,καὶἀεὶ the ratio of A, C, D, B, have been taken [Prop. 7.33].
A
C
D
B
E
F
G
H
K
L
M
N
O
P
236

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
ἑξῆςἑνὶπλείους,ἕωςἂνἴσονγένηταιτὸπλῆθοςαὐτῶντῷ And the (least) three (numbers), H, K, L. And so on,
πλήθειτῶνΑ,Γ,Δ,Β.εἰλήφθωσαν,καὶἔστωσανοἱΜ,Ν, successively increasing by one, until the multitude of the
Ξ,Ο.φανερὸνδή,ὅτιὁμὲνΖἑαυτὸνπολλαπλασιάσαςτὸν (least numbers taken) is made equal to the multitude of
Θπεποίηκεν,τὸνδὲΘπολλαπλασιάσαςτὸνΜπεποίηκεν, A, C, D, B [Prop. 8.2]. Let them have been taken, and
καὶὁΗἑαυτὸνμὲνπολλαπλασιάσαςτὸνΛπεποίηκεν,τὸν let them be M, N, O, P . So (it is) clear that F has made
δὲΛπολλαπλασιάσαςτὸνΟπεποίηκεν. καὶἐπεὶοἱΜ,Ν, H (by) multiplying itself, and has made M (by) multi-
Ξ,ΟἐλάχιστοίεἰσιτῶντὸναὐτὸνλόγονἐχόντωντοῖςΖ, plying H. And G has made L (by) multiplying itself, and
Η,εἰσὶδὲκαὶοἱΑ,Γ,Δ,Βἐλάχιστοιτῶντὸναὐτὸνλόγον has made P (by) multiplying L [Prop. 8.2 corr.]. And
ἐχόντωντοῖςΖ,Η,καίἐστινἴσοντὸπλῆθοςτῶνΜ,Ν,Ξ, since M, N, O, P are the least of those (numbers) hav-
ΟτῷπλήθειτῶνΑ,Γ,Δ,Β,ἕκαστοςἄρατῶνΜ,Ν,Ξ,Ο ing the same ratio as F , G, and A, C, D, B are also the
ἑκάστῳτῶνΑ,Γ,Δ,Βἴσοςἐστίν·ἴσοςἄραἐστὶνὁμὲνΜ least of those (numbers) having the same ratio as F , G
τῷΑ,ὁδὲΟτῷΒ.καὶἐπεὶὁΖἑαυτὸνπολλαπλασιάσας [Prop. 8.2], and the multitude of M, N, O, P is equal
τὸνΘπεποίηκεν,ὁΖἄρατὸνΘμετρεῖκατὰτὰςἐντῷΖ to the multitude of A, C, D, B, thus M, N, O, P are
μονάδας.μετρεῖδὲκαὶἡΕμονὰςτὸνΖκατὰτὰςἐναὐτῷ equal to A, C, D, B, respectively. Thus, M is equal to
μονάδας·ἰσάκιςἄραἡΕμονὰςτὸνΖἀριθμὸνμετρεῖκαὶὁΖ A, and P to B. And since F has made H (by) multiplyτὸνΘ.ἔστινἄραὡςἡΕμονὰςπρὸςτὸνΖἀριθμόν,οὕτως
ing itself, F thus measures H according to the units in F
ὁΖπρὸςτὸνΘ.πάλιν,ἐπεὶὁΖτὸνΘπολλαπλασιάσας [Def. 7.15]. And the unit E also measures F according to
τὸνΜπεποίηκεν,ὁΘἄρατὸνΜμετρεῖκατὰτὰςἐντῷΖ the units in it. Thus, the unit E measures the number F
μονάδας.μετρεῖδὲκαὶἡΕμονὰςτὸνΖἀριθμὸνκατὰτὰςἐν as many times as F (measures) H. Thus, as the unit E is
αὐτῷμονάδας·ἰσάκιςἄραἡΕμονὰςτὸνΖἀριθμὸνμετρεῖ to the number F , so F (is) to H [Def. 7.20]. Again, since
καὶὁΘτὸνΜ.ἔστινἄραὡςἡΕμονὰςπρὸςτὸνΖἀριθμόν, F has made M (by) multiplying H, H thus measures M
οὕτωςὁΘπρὸςτὸνΜ.ἐδείχθηδὲκαὶὡςἡΕμονὰςπρὸς according to the units in F [Def. 7.15]. And the unit E
τὸνΖἀριθμόν,οὕτωςὁΖπρὸςτὸνΘ·καὶὡςἄραἡΕμονὰς also measures the number F according to the units in it.
πρὸςτὸνΖἀριθμόν,οὕτωςὁΖπρὸςτὸνΘκαὶὁΘπρὸς Thus, the unit E measures the number F as many times
τὸνΜ.ἴσοςδὲὁΜτῷΑ·ἔστινἄραὡςἡΕμονὰςπρὸςτὸν as H (measures) M. Thus, as the unit E is to the number
Ζἀριθμόν,οὕτωςὁΖπρὸςτὸνΘκαὶὁΘπρὸςτὸνΑ.διὰ F , so H (is) to M [Prop. 7.20]. And it was shown that as
τὰαὐτὰδὴκαὶὡςἡΕμονὰςπρὸςτὸνΗἀριθμόν,οὕτωςὁ the unit E (is) to the number F , so F (is) to H. And thus
ΗπρὸςτὸνΛκαὶὁΛπρὸςτὸνΒ.ὅσοιἄραεἰςτοὺςΑ,Β as the unit E (is) to the number F , so F (is) to H, and H
μεταξὺκατὰτὸσυνεχὲςἀνάλογονἐμπεπτώκασινἀριθμοί, (is) to M. And M (is) equal to A. Thus, as the unit E is
τοσοῦτοικαὶἑκατέρουτῶνΑ,ΒκαὶμονάδοςτῆςΕμεταξὺ to the number F , so F (is) to H, and H to A. And so, for
κατὰτὸσυνεχὲςἀνάλογονἐμπεπτώκασινἀριθμοί·ὅπερἔδει the same (reasons), as the unit E (is) to the number G,
δεῖξαι.
so G (is) to L, and L to B. Thus, as many (numbers) as
have fallen in between A and B in continued proportion,
so many numbers have also fallen between each of A and
B and the unit E in continued proportion. (Which is) the
very thing it was required to show.
. Proposition 10
᾿Εάνδύοἀριθμῶνἑκατέρουκαὶμονάδοςμεταξὺκατὰ If (some) numbers fall between each of two numbers
τὸσυνεχὲςἀνάλογονἐμπίπτωσινἀριθμοί,ὅσοιἑκατέρου and a unit in continued proportion then, as many (numαὐτῶν
καὶ μονάδος μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον bers) as fall between each of the (two numbers) and the
ἐμπίπτουσινἀριθμοί,τοσοῦτοικαὶεἰςαὐτοὺςμεταξὺκατὰ unit in continued proportion, so many (numbers) will
τὸσυνεχὲςἀνάλογονἐμπεσοῦνται.
also fall in between the (two numbers) themselves in con-
Δύο γὰρ ἀριθμῶν τῶν Α, Β καὶμονάδος τῆς Γ με- tinued proportion.
ταξύκατὰτὸσυνεχὲςἀνάλογονἐμπιπτέτωσανἀριθμοὶοἵ For let the numbers D, E and F , G fall between the
τε Δ, Ε καὶοἱ Ζ, Η· λέγω, ὅτι ὅσοι ἑκατέρουτῶν Α, numbers A and B (respectively) and the unit C in con-
ΒκαὶμονάδοςτῆςΓμεταξὺκατὰτὸσυνεχὲςἀνάλογον tinued proportion. I say that, as many numbers as have
ἐμπεπτώκασινἀριθμοί,τοσοῦτοικαὶεἰςτοὺςΑ,Βμεταξὺ fallen between each of A and B and the unit C in continκατὰτὸσυνεχὲςἀνάλογονἐμπεσοῦνται.
ued proportion, so many will also fall in between A and
B in continued proportion.
237

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
Γ
∆
Ε
Α
Γ
Ζ
Η
Β
C
D
E
A
C
F
G
B
Θ
Κ
Λ
῾Ο Δ γὰρ τὸν Ζ πολλαπλασιάσας τὸν Θ ποιείτω, For let D make H (by) multiplying F . And let D, F
ἑκάτεροςδὲτῶν Δ,ΖτὸνΘπολλαπλασιάσαςἑκάτερον make K, L, respectively, by multiplying H.
τῶνΚ,Λποιείτω.
As since as the unit C is to the number D, so D (is) to
ΚαὶἐπείἐστινὡςἡΓμονὰςπρὸςτὸνΔἀριθμόν,οὕτως E, the unit C thus measures the number D as many times
ὁΔπρὸςτὸνΕ,ἰσάκιςἄραἡΓμονὰςτὸνΔἀριθμὸνμετρεῖ as D (measures) E [Def. 7.20]. And the unit C measures
καὶὁΔτὸνΕ.ἡδὲΓμονὰςτὸνΔἀριθμὸνμετρεῖκατὰτὰς the number D according to the units in D. Thus, the
ἐντῷΔμονάδας·καὶὁΔἄραἀριθμὸςτὸνΕμετρεῖκατὰ number D also measures E according to the units in D.
τὰςἐντῷΔμονάδας·ὁΔἄραἑαυτὸνπολλαπλασιάσαςτὸν Thus, D has made E (by) multiplying itself. Again, since
Επεποίηκεν. πάλιν,ἐπείἐστινὡςἡΓ[μονὰς]πρὸςτὸν as the [unit] C is to the number D, so E (is) to A, the
Δἀριθμὸν,οὕτωςὁΕπρὸςτὸνΑ,ἰσάκιςἄραἡΓμονὰς unit C thus measures the number D as many times as E
τὸνΔἀριθμὸνμετρεῖκαὶὁΕτὸνΑ.ἡδὲΓμονὰςτὸν (measures) A [Def. 7.20]. And the unit C measures the
ΔἀριθμὸνμετρεῖκατὰτὰςἐντῷΔμονάδας·καὶὁΕἄρα number D according to the units in D. Thus, E also meaτὸνΑμετρεῖκατὰτὰςἐντῷΔμονάδας·ὁΔἄρατὸνΕ
sures A according to the units in D. Thus, D has made
πολλαπλασιάσαςτὸνΑπεποίηκεν. διὰτὰαὐτὰδὴκαὶὁ A (by) multiplying E. And so, for the same (reasons), F
μὲνΖἑαυτὸνπολλαπλασιάσαςτὸνΗπεποίηκεν,τὸνδὲΗ has made G (by) multiplying itself, and has made B (by)
πολλαπλασιάσαςτὸνΒπεποίηκεν.καὶἐπεὶὁΔἑαυτὸνμὲν multiplying G. And since D has made E (by) multiplying
πολλαπλασιάσαςτὸνΕπεποίηκεν,τὸνδὲΖπολλαπλασιάσας itself, and has made H (by) multiplying F , thus as D is to
τὸνΘπεποίηκεν,ἔστινἄραὡςὁΔπρὸςτὸνΖ,οὕτωςὁΕ F , so E (is) to H [Prop 7.17]. And so, for the same reaπρὸςτὸνΘ.διὰτὰαὐτὰδὴκαὶὡςὁΔπρὸςτὸνΖ,οὕτωςὁ
sons, as D (is) to F , so H (is) to G [Prop. 7.18]. And thus
ΘπρὸςτὸνΗ.καὶὡςἄραὁΕπρὸςτὸνΘ,οὕτωςὁΘπρὸς as E (is) to H, so H (is) to G. Again, since D has made
τὸνΗ.πάλιν,ἐπεὶὁΔἑκάτεροντῶνΕ,Θπολλαπλασιάσας A, K (by) multiplying E, H, respectively, thus as E is to
ἑκάτεροντῶνΑ,Κπεποίηκεν,ἔστινἄραὡςὁΕπρὸςτὸνΘ, H, so A (is) to K [Prop 7.17]. But, as E (is) to H, so D
οὕτωςὁΑπρὸςτὸνΚ.ἀλλ᾿ὡςὁΕπρὸςτὸνΘ,οὕτωςὁΔ (is) to F . And thus as D (is) to F , so A (is) to K. Again,
πρὸςτὸνΖ·καὶὡςἄραὁΔπρὸςτὸνΖ,οὕτωςὁΑπρὸςτὸν since D, F have made K, L, respectively, (by) multiply-
Κ.πάλιν,ἐπεὶἑκάτεροςτῶνΔ,ΖτὸνΘπολλαπλασιάσας ing H, thus as D is to F , so K (is) to L [Prop. 7.18]. But,
ἑκάτεροντῶνΚ,Λπεποίηκεν,ἕστινἄραὡςὁΔπρὸςτὸν as D (is) to F , so A (is) to K. And thus as A (is) to K,
Ζ,οὕτωςὁΚπρὸςτὸνΛ.ἀλλ᾿ὡςὁΔπρὸςτὸνΖ,οὕτωςὁ so K (is) to L. Further, since F has made L, B (by) mul-
ΑπρὸςτὸνΚ·καὶὡςἄραὁΑπρὸςτὸνΚ,οὕτωςὁΚπρὸς tiplying H, G, respectively, thus as H is to G, so L (is) to
τὸνΛ.ἔτιἐπεὶὁΖἑκάτεροντῶνΘ,Ηπολλαπλασιάσας B [Prop 7.17]. And as H (is) to G, so D (is) to F . And
ἑκάτεροντῶνΛ,Βπεποίηκεν,ἔστινἄραὡςὁΘπρὸςτὸν thus as D (is) to F , so L (is) to B. And it was also shown
Η,οὕτωςὁΛπρὸςτὸνΒ.ὡςδὲὁΘπρὸςτὸνΗ,οὕτως that as D (is) to F , so A (is) to K, and K to L. And thus
ὁΔπρὸςτὸνΖ·καὶὡςἄραὁΔπρὸςτὸνΖ,οὕτωςὁΛ as A (is) to K, so K (is) to L, and L to B. Thus, A, K,
πρὸςτὸνΒ.ἐδείχθηδὲκαὶὡςὁΔπρὸςτὸνΖ,οὕτωςὅτε L, B are successively in continued proportion. Thus, as
ΑπρὸςτὸνΚκαὶὁΚπρὸςτὸνΛ·καὶὡςἄραὁΑπρὸς many numbers as fall between each of A and B and the
τὸνΚ,οὕτωςὁΚπρὸςτὸνΛκαὶὁΛπρὸςτὸνΒ.οἱΑ, unit C in continued proportion, so many will also fall in
Κ,Λ,Βἄρακατὰτὸσυνεχὲςἑξῆςεἰσινἀνάλογον. ὅσοι between A and B in continued proportion. (Which is)
ἄραἑκατέρουτῶνΑ,ΒκαὶτῆςΓμονάδοςμεταξὺκατὰ the very thing it was required to show.
τὸσυνεχὲςἀνάλογονἐμπίπτουσινἀριθμοί,τοσοῦτοικαὶεἰς
τοὺςΑ,Βμεταξὺκατὰτὸσυνεχὲςἐμπεσοῦνται·ὅπερἔδει
H
K
L
238

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
δεῖξαι.
. Proposition 11
Δύο τετραγώνων ἀριθμῶν εἷς μέσος ἀνάλογόν ἐστιν There exists one number in mean proportion to two
ἀριθμός, καὶ ὁ τετράγωνος πρὸς τὸν τετράγωνον δι- (given) square numbers. † And (one) square (number)
πλασίοναλόγονἔχειἤπερἡπλευρὰπρὸςτὴνπλευράν. has to the (other) square (number) a squared ‡ ratio with
respect to (that) the side (of the former has) to the side
(of the latter).
Α
Β
Γ
Ε
∆
῎ΕστωσαντετράγωνοιἀριθμοὶοἱΑ,Β,καὶτοῦμὲνΑ Let A and B be square numbers, and let C be the side
πλευρὰἔστωὁΓ,τοῦδὲΒὁΔ·λέγω,ὅτιτῶνΑ,Βεἷς of A, and D (the side) of B. I say that there exists one
μέσοςἀνάλογόνἐστινἀριθμός, καὶὁΑπρὸςτὸνΒδι- number in mean proportion to A and B, and that A has
πλασίοναλόγονἔχειἤπερὁΓπρὸςτὸνΔ. to B a squared ratio with respect to (that) C (has) to D.
῾ΟΓγὰρτὸνΔπολλαπλασιάσαςτὸνΕποιείτω. καὶ For let C make E (by) multiplying D. And since A is
ἐπεὶτετράγωνόςἐστινὁΑ,πλευρὰδὲαὐτοῦἐστινὁΓ,ὁΓ square, and C is its side, C has thus made A (by) multi-
ἄραἑαυτὸνπολλαπλασιάσαςτὸνΑπεποίηκεν.διὰτὰαὐτὰ plying itself. And so, for the same (reasons), D has made
δὴκαὶὁΔἑαυτὸνπολλαπλασιάσαςτὸνΒπεποίηκεν.ἐπεὶ B (by) multiplying itself. Therefore, since C has made A,
οὖνὁΓἑκάτεροντῶνΓ,Δπολλαπλασιάσαςἑκάτεροντῶν E (by) multiplying C, D, respectively, thus as C is to D,
Α,Επεποίηκεν,ἔστινἄραὡςὁΓπρὸςτὸνΔ,οὕτωςὁΑ so A (is) to E [Prop. 7.17]. And so, for the same (reaπρὸςτὸνΕ.διὰτὰαὐτὰδὴκαὶὡςὁΓπρὸςτὸνΔ,οὕτωςὁ
sons), as C (is) to D, so E (is) to B [Prop. 7.18]. And
ΕπρὸςτὸνΒ.καὶὡςἄραὁΑπρὸςτὸνΕ,οὕτωςὁΕπρὸς thus as A (is) to E, so E (is) to B. Thus, one number
τὸνΒ.τῶνΑ,Βἄραεἷςμέσοςἀνάλογόνἐστινἀριθμός. (namely, E) is in mean proportion to A and B.
Λέγωδή,ὅτικαὶὁΑπρὸςτὸνΒδιπλασίοναλόγονἔχει So I say that A also has to B a squared ratio with
ἤπερὁΓπρὸςτὸνΔ.ἐπεὶγὰρτρεῖςἀριθμοὶἀνάλογόνεἰσιν respect to (that) C (has) to D. For since A, E, B are
οἱΑ,Ε,Β,ὁΑἄραπρὸςτὸνΒδιπλασίοναλόγονἔχειἤπερ three (continuously) proportional numbers, A thus has
ὁΑπρὸςτὸνΕ.ὡςδὲὁΑπρὸςτὸνΕ,οὕτωςὁΓπρὸς to B a squared ratio with respect to (that) A (has) to E
τὸνΔ.ὁΑἄραπρὸςτὸνΒδιπλασίοναλόγονἔχειἤπερἡ [Def. 5.9]. And as A (is) to E, so C (is) to D. Thus, A has
ΓπλευρὰπρὸςτὴνΔ·ὅπερἔδειδεῖξαι.
to B a squared ratio with respect to (that) side C (has)
to (side) D. (Which is) the very thing it was required to
show.
† In other words, between two given square numbers there exists a number in continued proportion.
‡ Literally, “double”.
.
Proposition 12
Δύοκύβωνἀριθμῶνδύομέσοιἀνάλογόνεἰσινἀριθμοί, There exist two numbers in mean proportion to two
καὶὁκύβοςπρὸςτὸνκύβοντριπλασίοναλόγονἔχειἤπερἡ (given) cube numbers. † And (one) cube (number) has to
πλευρὰπρὸςτὴνπλευράν.
the (other) cube (number) a cubed ‡ ratio with respect
῎ΕστωσανκύβοιἀριθμοὶοἱΑ,ΒκαὶτοῦμὲνΑπλευρὰ to (that) the side (of the former has) to the side (of the
ἔστωὁΓ,τοῦδὲΒὁΔ·λέγω,ὅτιτῶνΑ,Βδύομέσοι latter).
ἀνάλογόνεἰσινἀριθμοί,καὶὁΑπρὸςτὸνΒτριπλασίονα Let A and B be cube numbers, and let C be the side
λόγονἔχειἤπερὁΓπρὸςτὸνΔ.
of A, and D (the side) of B. I say that there exist two
numbers in mean proportion to A and B, and that A has
A
B
C
E
D
239

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
to B a cubed ratio with respect to (that) C (has) to D.
Α
Ε
A
E
Β
Ζ
B
F
Γ
Η
C
G
∆
Θ
D
H
Κ
K
῾ΟγὰρΓἑαυτὸνμὲνπολλαπλασιάσαςτὸνΕποιείτω, For let C make E (by) multiplying itself, and let it
τὸνδὲΔπολλαπλασιάσαςτὸνΖποιείτω,ὁδὲΔἑαυτὸν make F (by) multiplying D. And let D make G (by) mulπολλαπλασιάσαςτὸνΗποιείτω,ἑκάτεροςδὲτῶνΓ,Δτὸν
tiplying itself, and let C, D make H, K, respectively, (by)
ΖπολλαπλασιάσαςἑκάτεροντῶνΘ,Κποιείτω. multiplying F .
ΚαὶἐπεὶκύβοςἐστὶνὁΑ,πλευρὰδὲαὐτοῦὁΓ,καὶὁ And since A is cube, and C (is) its side, and C has
ΓἑαυτὸνμὲνπολλαπλασιάσαςτὸνΕπεποίηκεν,ὁΓἄρα made E (by) multiplying itself, C has thus made E (by)
ἑαυτὸνμὲν πολλαπλασιάσαςτὸν Ε πεποίηκεν, τὸν δὲΕ multiplying itself, and has made A (by) multiplying E.
πολλαπλασιάσαςτὸνΑπεποίηκεν. διὰτὰαὐτὰδὴκαὶὁ And so, for the same (reasons), D has made G (by) mul-
ΔἑαυτὸνμὲνπολλαπλασιάσαςτὸνΗπεποίηκεν,τὸνδὲΗ tiplying itself, and has made B (by) multiplying G. And
πολλαπλασιάσαςτὸνΒπεποίηκεν. καὶἐπεὶὁΓἑκάτερον since C has made E, F (by) multiplying C, D, respecτῶνΓ,ΔπολλαπλασιάσαςἑκάτεροντῶνΕ,Ζπεποίηκεν,
tively, thus as C is to D, so E (is) to F [Prop. 7.17]. And
ἔστιν ἄραὡςὁΓπρὸςτὸνΔ,οὕτωςὁΕπρὸςτὸνΖ. so, for the same (reasons), as C (is) to D, so F (is) to G
διὰτὰαὐτὰδὴκαὶὡςὁΓπρὸςτὸνΔ,οὕτωςὁΖπρὸς [Prop. 7.18]. Again, since C has made A, H (by) multiτὸνΗ.πάλιν,ἐπεὶὁΓἑκάτεροντῶνΕ,Ζπολλαπλασιάσας
plying E, F , respectively, thus as E is to F , so A (is) to
ἑκάτεροντῶνΑ,Θπεποίηκεν,ἔστινἄραὡςὁΕπρὸςτὸν H [Prop. 7.17]. And as E (is) to F , so C (is) to D. And
Ζ,οὕτωςὁΑπρὸςτὸνΘ.ὡςδὲὁΕπρὸςτὸνΖ,οὕτωςὁΓ thus as C (is) to D, so A (is) to H. Again, since C, D
πρὸςτὸνΔ·καὶὡςἄραὁΓπρὸςτὸνΔ,οὕτωςὁΑπρὸςτὸν have made H, K, respectively, (by) multiplying F , thus
Θ.πάλιν,ἐπεὶἑκάτεροςτῶνΓ,ΔτὸνΖπολλαπλασιάσας as C is to D, so H (is) to K [Prop. 7.18]. Again, since D
ἑκάτεροντῶνΘ,Κπεποίηκεν,ἔστινἄραὡςὁΓπρὸςτὸν has made K, B (by) multiplying F , G, respectively, thus
Δ,οὕτωςὁΘπρὸςτὸνΚ.πάλιν,ἐπεὶὁΔἑκάτεροντῶν as F is to G, so K (is) to B [Prop. 7.17]. And as F (is)
Ζ,ΗπολλαπλασιάσαςἑκάτεροντῶνΚ,Βπεποίηκεν,ἔστιν to G, so C (is) to D. And thus as C (is) to D, so A (is)
ἄραὡςὁΖπρὸςτὸνΗ,οὕτωςὁΚπρὸςτὸνΒ.ὡςδὲὁΖ to H, and H to K, and K to B. Thus, H and K are two
πρὸςτὸνΗ,οὕτωςὁΓπρὸςτὸνΔ·καὶὡςἄραὁΓπρὸς (numbers) in mean proportion to A and B.
τὸνΔ,οὕτωςὅτεΑπρὸςτὸνΘκαὶὁΘπρὸςτὸνΚκαὶ So I say that A also has to B a cubed ratio with re-
ὁΚπρὸςτὸνΒ.τῶνΑ,Βἄραδύομέσοιἀνάλογόνεἰσιν spect to (that) C (has) to D. For since A, H, K, B are
οἱΘ,Κ.
four (continuously) proportional numbers, A thus has
Λέγωδή,ὅτικαὶὁΑπρὸςτὸνΒτριπλασίοναλόγονἔχει to B a cubed ratio with respect to (that) A (has) to H
ἤπερὁΓπρὸςτὸνΔ.ἐπεὶγὰρτέσσαρεςἀριθμοὶἀνάλογόν [Def. 5.10]. And as A (is) to H, so C (is) to D. And
εἰσινοἱΑ,Θ,Κ,Β,ὁΑἄραπρὸςτὸνΒτριπλασίοναλόγον [thus] A has to B a cubed ratio with respect to (that) C
ἔχειἤπερὁΑπρὸςτὸνΘ.ὡςδὲὁΑπρὸςτὸνΘ,οὕτωςὁ (has) to D. (Which is) the very thing it was required to
ΓπρὸςτὸνΔ·καὶὁΑ[ἄρα]πρὸςτὸνΒτριπλασίοναλόγον show.
ἔχειἤπερὁΓπρὸςτὸνΔ·ὅπερἔδειδεῖξαι.
† In other words, between two given cube numbers there exist two numbers in continued proportion.
‡ Literally, “triple”.
.
Proposition 13
᾿Εὰν ὦσιν ὁσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, καὶ If there are any multitude whatsoever of continuously
πολλαπλασιάσαςἕκαστοςἑαυτὸνποιῇτινα, οἱγενόμενοι proportional numbers, and each makes some (number
ἐξ αὐτῶν ἀνάλογον ἔσονται· καὶ ἐὰν οἱ ἐξ ἀρχῆς τοὺς by) multiplying itself, then the (numbers) created from
γενομένους πολλαπλασιάσαντες ποιῶσί τινας, καὶ αὐτοὶ them will (also) be (continuously) proportional. And if
ἀνάλογονἔσονται[καὶἀεὶπερὶτοὺςἄκρουςτοῦτοσυμβαίνει]. the original (numbers) make some (more numbers by)
῎Εστωσανὁποσοιοῦνἀριθμοὶἑξῆςἀνάλογον,οἱΑ,Β, multiplying the created (numbers) then these will also
240

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
Γ, ὡςὁΑπρὸςτὸνΒ, οὕτωςὁΒπρὸςτὸν Γ,καὶοἱ be (continuously) proportional [and this always happens
Α, Β, Γ ἑαυτοὺςμὲν πολλαπλασιάσαντεςτοὺς Δ, Ε, Ζ with the extremes].
ποιείτωσαν,τοὺςδὲΔ,Ε,ΖπολλαπλασιάσαντεςτοὺςΗ, Let A, B, C be any multitude whatsoever of contin-
Θ,Κποιείτωσαν·λέγω,ὅτιοἵτεΔ,Ε,ΖκαὶοἱΗ,Θ,Κ uously proportional numbers, (such that) as A (is) to B,
ἑξῆςἀνάλογονεἰσιν.
so B (is) to C. And let A, B, C make D, E, F (by)
multiplying themselves, and let them make G, H, K (by)
multiplying D, E, F . I say that D, E, F and G, H, K are
continuously proportional.
Α
Λ
A
L
Β
Γ
∆
Ε
Ζ
Η
Ξ
Μ
Ν
Ο
Π
Θ
H
Κ
K
῾ΟμὲνγὰρΑτὸνΒπολλαπλασιάσαςτὸνΛποιείτω, For let A make L (by) multiplying B. And let A, B
ἑκάτεροςδὲτῶνΑ,ΒτὸνΛπολλαπλασιάσαςἑκάτεροντῶν make M, N, respectively, (by) multiplying L. And, again,
Μ,Νποιείτω.καὶπάλινὁμὲνΒτὸνΓπολλαπλασιάσαςτὸν let B make O (by) multiplying C. And let B, C make P ,
Ξποιείτω,ἑκάτεροςδὲτῶνΒ,ΓτὸνΞπολλαπλασιάσας Q, respectively, (by) multplying O.
ἑκάτεροντῶνΟ,Πποιείτω. So, similarly to the above, we can show that D, L,
῾Ομοίωςδὴτοῖςἐπάνωδεῖξομεν,ὅτιοἱΔ,Λ,Εκαὶοἱ E and G, M, N, H are continuously proportional in the
Η,Μ,Ν,ΘἑξῆςεἰσινἀνάλογονἐντῷτοῦΑπρὸςτὸν ratio of A to B, and, further, (that) E, O, F and H, P , Q,
Βλόγῳ,καὶἔτιοἱΕ,Ξ,ΖκαὶοἱΘ,Ο,Π,Κἑξῆςεἰσιν K are continuously proportional in the ratio of B to C.
ἀνάλογονἐντῷτοῦΒπρὸςτὸνΓλόγῳ.καίἐστινὡςὁΑ And as A is to B, so B (is) to C. And thus D, L, E are in
πρὸςτὸνΒ,οὕτωςὁΒπρὸςτὸνΓ·καὶοἱΔ,Λ,Εἄρατοῖς the same ratio as E, O, F , and, further, G, M, N, H (are
Ε,Ξ,ΖἐντῷαὐτῷλόγῳεἰσὶκαὶἔτιοἱΗ,Μ,Ν,Θτοῖς in the same ratio) as H, P , Q, K. And the multitude of
Θ,Ο,Π,Κ.καίἐστινἴσοντὸμὲντῶνΔ,Λ,Επλῆθοςτῷ D, L, E is equal to the multitude of E, O, F , and that of
τῶνΕ,Ξ,Ζπλήθει,τὸδὲτῶνΗ,Μ,Ν,ΘτῷτῶνΘ,Ο, G, M, N, H to that of H, P , Q, K. Thus, via equality, as
Π,Κ·δι᾿ἴσουἄραἐστὶνὡςμὲνὁΔπρὸςτὸνΕ,οὕτωςὁ D is to E, so E (is) to F , and as G (is) to H, so H (is) to
ΕπρὸςτὸνΖ,ὡςδὲὁΗπρὸςτὸνΘ,οὕτωςὁΘπρὸςτὸν K [Prop. 7.14]. (Which is) the very thing it was required
Κ·ὅπερἔδειδεῖξαι.
to show.
. Proposition 14
᾿Εὰντετράγωνοςτετράγωνονμετρῇ,καὶἡπλευρὰτὴν If a square (number) measures a(nother) square
πλευρὰνμετρήσει·καὶἐὰνἡπλευρὰτὴνπλευρὰνμετρῇ,καὶ (number) then the side (of the former) will also mea-
ὁτετράγωνοςτὸντετράγωνονμετρήσει.
sure the side (of the latter). And if the side (of a square
῎ΕστωσαντετράγωνοιἀριθμοὶοἱΑ,Β,πλευραὶδὲαὐτῶν number) measures the side (of another square number)
ἔστωσανοἱΓ,Δ,ὁδὲΑτὸνΒμετρείτω·λέγω,ὅτικαὶὁ then the (former) square (number) will also measure the
ΓτὸνΔμετρεῖ.
(latter) square (number).
Let A and B be square numbers, and let C and D be
their sides (respectively). And let A measure B. I say that
C also measures D.
B
C
D
E
F
G
O
M
N
P
Q
241

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
Α
Β
Γ
∆
A
B
C
D
Ε
῾ΟΓγὰρτὸνΔπολλαπλασιάσαςτὸνΕποιείτω·οἱΑ,Ε, For let C make E (by) multiplying D. Thus, A, E,
ΒἄραἑξῆςἀνάλογόνεἰσινἐντῷτοῦΓπρὸςτὸνΔλόγῳ. B are continuously proportional in the ratio of C to D
καὶἐπεὶοἱΑ,Ε,Βἐξῆςἀνάλογόνεἰσιν,καὶμετρεῖὁΑτὸν [Prop. 8.11]. And since A, E, B are continuously pro-
Β,μετρεῖἄρακαὶὁΑτὸνΕ.καίἐστινὡςὁΑπρὸςτὸνΕ, portional, and A measures B, A thus also measures E
οὕτωςὁΓπρὸςτὸνΔ·μετρεῖἄρακαὶὁΓτὸνΔ. [Prop. 8.7]. And as A is to E, so C (is) to D. Thus, C
ΠάλινδὴὁΓτὸνΔμετρείτω·λέγω,ὅτικαὶὁΑτὸνΒ also measures D [Def. 7.20].
μετρεῖ.
So, again, let C measure D. I say that A also measures
Τῶνγὰραὐτῶνκατασκευασθέντωνὁμοίωςδείξομεν, B.
ὅτιοἱΑ,Ε,ΒἑξῆςἀνάλογόνεἰσινἐντῷτοῦΓπρὸςτὸνΔ For similarly, with the same construction, we can
λόγῳ.καὶἐπείἐστινὡςὁΓπρὸςτὸνΔ,οὕτωςὁΑπρὸς show that A, E, B are continuously proportional in the
τὸνΕ,μετρεῖδὲὁΓτὸνΔ,μετρεῖἄρακαὶὁΑτὸνΕ.καί ratio of C to D. And since as C is to D, so A (is) to E,
εἰσινοἱΑ,Ε,Βἑξῆςἀνάλογον·μετρεῖἄρακαὶὁΑτὸνΒ. and C measures D, A thus also measures E [Def. 7.20].
᾿Εὰνἄρατετράγωνοςτετράγωνονμετρῇ,καὶἡπλευρὰ And A, E, B are continuously proportional. Thus, A also
τὴνπλευρὰνμετρήσει·καὶἐὰνἡπλευρὰτὴνπλευρὰνμετρῇ, measures B.
καὶ ὁ τετράγωνος τὸν τετράγωνον μετρήσει· ὅπερ ἔδει Thus, if a square (number) measures a(nother) square
δεῖξαι.
(number) then the side (of the former) will also measure
the side (of the latter). And if the side (of a square number)
measures the side (of another square number) then
the (former) square (number) will also measure the (latter)
square (number). (Which is) the very thing it was
required to show.
. Proposition 15
᾿Εὰνκύβοςἀριθμὸςκύβονἀριθμὸνμετρῇ,καὶἡπλευρὰ If a cube number measures a(nother) cube number
τὴνπλευρὰνμετρήσει·καὶἐὰνἡπλευρὰτὴνπλευρὰνμετρῇ, then the side (of the former) will also measure the side
καὶὁκύβοςτὸνκύβονμετρήσει.
(of the latter). And if the side (of a cube number) mea-
ΚύβοςγὰρἀριθμὸςὁΑκύβοντὸνΒμετρείτω,καὶτοῦ sures the side (of another cube number) then the (forμὲνΑπλευρὰἔστωὁΓ,τοῦδὲΒὁΔ·λέγω,ὅτιὁΓτὸν
mer) cube (number) will also measure the (latter) cube
Δμετρεῖ.
(number).
For let the cube number A measure the cube (number)
B, and let C be the side of A, and D (the side) of B.
I say that C measures D.
Α
Β
Γ
∆
A
B
E
C
D
Ε
Η
Θ
Κ
E
G
H
K
Ζ
῾ΟΓγὰρἑαυτὸνπολλαπλασιάσαςτὸνΕποιείτω,ὁδὲΔ For let C make E (by) multiplying itself.
F
And let
242

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
ἑαυτὸνπολλαπλασιάσαςτὸνΗποιείτω,καὶἔτιὁΓτὸνΔ D make G (by) multiplying itself. And, further, [let] C
πολλαπλασιάσαςτὸνΖ[ποιείτω],ἑκάτεροςδὲτῶνΓ,Δτὸν [make] F (by) multiplying D, and let C, D make H, K,
ΖπολλαπλασιάσαςἑκάτεροντῶνΘ,Κποιείτω. φανερὸν respectively, (by) multiplying F . So it is clear that E, F ,
δή,ὅτιοἱΕ,Ζ,ΗκαὶοἱΑ,Θ,Κ,Βἑξῆςἀνάλογόνεἰσιν G and A, H, K, B are continuously proportional in the
ἐντῷτοῦΓπρὸςτὸνΔλόγῳ.καὶἐπεὶοἱΑ,Θ,Κ,Βἑξῆς ratio of C to D [Prop. 8.12]. And since A, H, K, B are
ἀνάλογόνεἰσιν,καὶμετρεῖὁΑτὸνΒ,μετρεῖἄρακαὶτὸν continuously proportional, and A measures B, (A) thus
Θ.καίἐστινὡςὁΑπρὸςτὸνΘ,οὕτωςὁΓπρὸςτὸνΔ· also measures H [Prop. 8.7]. And as A is to H, so C (is)
μετρεῖἄρακαὶὁΓτὸνΔ. to D. Thus, C also measures D [Def. 7.20].
ἈλλὰδὴμετρείτωὁΓτὸνΔ·λέγω,ὅτικαὶὁΑτὸνΒ And so let C measure D. I say that A will also meaμετρήσει.
sure B.
Τῶνγὰραὐτῶνκατασκευασθέντωνὁμοίωςδὴδείξομεν, For similarly, with the same construction, we can
ὅτιοἱΑ,Θ,Κ,ΒἑξῆςἀνάλογόνεἰσινἐντῷτοῦΓπρὸς show that A, H, K, B are continuously proportional in
τὸνΔλόγῳ. καὶἐπεὶὁΓτὸνΔμετρεῖ,καίἐστινὡςὁΓ the ratio of C to D. And since C measures D, and as C is
πρὸςτὸνΔ,οὕτωςὁΑπρὸςτὸνΘ,καὶὁΑἄρατὸνΘ to D, so A (is) to H, A thus also measures H [Def. 7.20].
μετρεῖ·ὥστεκαὶτὸνΒμετρεῖὁΑ·ὅπερἔδειδεῖξαι. Hence, A also measures B. (Which is) the very thing it
was required to show.
¦. Proposition 16
᾿Εὰν τετράγωνος ἀριθμὸς τετράγωνον ἀριθμὸν μὴ If a square number does not measure a(nother)
μετρῇ,οὐδὲἡπλευρὰτὴνπλευρὰνμετρήσει·κἂνἡπλευρὰ square number then the side (of the former) will not
τὴνπλευρὰνμὴμετρῇ,οὐδὲὁτετράγωνοςτὸντετράγωνον measure the side (of the latter) either. And if the side (of
μετρήσει.
a square number) does not measure the side (of another
square number) then the (former) square (number) will
not measure the (latter) square (number) either.
Α
Β
Γ
.
∆
῎ΕστωσαντετρὰγωνοιἀριθμοὶοἱΑ,Β,πλευραὶδὲαὐτῶν Let A and B be square numbers, and let C and D be
ἔστωσανοἱΓ,Δ,καὶμὴμετρείτωὁΑτὸνΒ·λὲγω,ὅτιοὐδὲ their sides (respectively). And let A not measure B. I say
ὁΓτὸνΔμετρεῖ.
that C does not measure D either.
ΕἰγὰρμετρεῖὁΓτὸνΔ,μετρήσεικαὶὁΑτὸνΒ.οὐ For if C measures D then A will also measure B
μετρεῖδὲὁΑτὸνΒ·οὐδὲἄραὁΓτὸνΔμετρήσει. [Prop. 8.14]. And A does not measure B. Thus, C will
Μὴμετρείτω[δὴ]πάλινὁΓτὸνΔ·λέγω,ὅτιοὐδὲὁΑ not measure D either.
τὸνΒμετρήσει.
[So], again, let C not measure D. I say that A will not
ΕἰγὰρμετρεῖὁΑτὸνΒ,μετρήσεικαὶὁΓτὸνΔ.οὐ measure B either.
μετρεῖδὲὁΓτὸνΔ·οὐδ᾿ἄραὁΑτὸνΒμετρήσει·ὅπερ For if A measures B then C will also measure D
ἔδειδεῖξαι.
[Prop. 8.14]. And C does not measure D. Thus, A will
not measure B either. (Which is) the very thing it was
required to show.
Þ. Proposition 17
᾿Εὰνκύβοςἀριθμὸςκύβονἀριθμὸνμὴμετρῇ,οὐδὲἡ If a cube number does not measure a(nother) cube
πλευρὰτὴνπλευρὰνμετρήσει·κἂνἡπλευρὰτὴνπλευρὰν number then the side (of the former) will not measure the
μὴμετρῇ,οὐδὲὁκύβοςτὸνκύβονμετρήσει.
side (of the latter) either. And if the side (of a cube number)
does not measure the side (of another cube number)
then the (former) cube (number) will not measure the
(latter) cube (number) either.
A
B
C
D
243

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
Α
Β
Γ
.
∆
ΚύβοςγὰρἀριθμὸςὁΑκύβονἀριθμὸντὸνΒμὴμε- For let the cube number A not measure the cube numτρείτω,καὶτοῦμὲνΑπλευρὰἔστωὁΓ,τοῦδὲΒὁΔ·
ber B. And let C be the side of A, and D (the side) of B.
λέγω,ὅτιὁΓτὸνΔοὐμετρήσει. I say that C will not measure D.
ΕἰγὰρμετρεῖὁΓτὸνΔ,καὶὁΑτὸνΒμετρήσει. οὐ For if C measures D then A will also measure B
μετρεῖδὲὁΑτὸνΒ·οὐδ᾿ἄραὁΓτὸνΔμετρεῖ. [Prop. 8.15]. And A does not measure B. Thus, C does
ἈλλὰδὴμὴμετρείτωὁΓτὸνΔ·λέγω,ὅτιοὐδὲὁΑ not measure D either.
τὸνΒμετρήσει.
And so let C not measure D. I say that A will not
ΕἰγὰρὁΑτὸνΒμετρεῖ,καὶὁΓτὸνΔμετρήσει. οὐ measure B either.
μετρεῖδὲὁΓτὸνΔ·οὐδ᾿ἄραὁΑτὸνΒμετρήσει·ὅπερ For if A measures B then C will also measure D
ἔδειδεῖξαι.
[Prop. 8.15]. And C does not measure D. Thus, A will
not measure B either. (Which is) the very thing it was
required to show.
. Proposition 18
Δύοὁμοίωνἐπιπέδωνἀριθμῶνεἷςμέσοςἀνάλογόνἐστιν There exists one number in mean proportion to two
ἀριθμός·καὶὁἐπίπεδοςπρὸςτὸνἐπίπεδονδιπλασίοναλόγον similar plane numbers. And (one) plane (number) has to
ἔχειἤπερἡὁμόλογοςπλευρὰπρὸςτὴνὁμόλογονπλευράν. the (other) plane (number) a squared † ratio with respect
to (that) a corresponding side (of the former has) to a
corresponding side (of the latter).
Α
Γ
∆
Η
Β
Ε
Ζ
῎ΕστωσανδύοὅμοιοιἐπίπεδοιἀριθμοὶοἱΑ,Β,καὶτοῦ Let A and B be two similar plane numbers. And let
μὲνΑπλευραὶἔστωσανοἱΓ,Δἀριθμοί,τοῦδὲΒοἱΕ, the numbers C, D be the sides of A, and E, F (the sides)
Ζ.καὶἐπεὶὅμοιοιἐπίπεδοίεἰσινοἱἀνάλογονἔχοντεςτὰς of B. And since similar numbers are those having proπλευράς,ἔστινἄραὡςὁΓπρὸςτὸνΔ,οὕτωςὁΕπρὸς
portional sides [Def. 7.21], thus as C is to D, so E (is) to
τὸνΖ.λέγωοὖν,ὅτιτῶνΑ,Βεἷςμέσοςἀνάλογόνἐστιν F . Therefore, I say that there exists one number in mean
ἀριθμός,καὶὁΑπρὸςτὸνΒδιπλασίοναλόγονἔχειἤπερὁ proportion to A and B, and that A has to B a squared
ΓπρὸςτὸνΕἢὁΔπρὸςτὸνΖ,τουτέστινἤπερἡὁμόλογος ratio with respect to that C (has) to E, or D to F —that is
πλευρὰπρὸςτὴνὁμόλογον[πλευράν].
to say, with respect to (that) a corresponding side (has)
ΚαὶἐπείἐστινὡςὁΓπρὸςτὸνΔ,οὕτωςὁΕπρὸςτὸν to a corresponding [side].
Ζ,ἐναλλὰξἄραἐστὶνὡςὁΓπρὸςτὸνΕ,ὁΔπρὸςτὸνΖ. For since as C is to D, so E (is) to F , thus, alternately,
καὶἐπεὶἐπίπεδόςἐστινὁΑ,πλευραὶδὲαὐτοῦοἱΓ,Δ,ὁΔ as C is to E, so D (is) to F [Prop. 7.13]. And since A is
ἄρατὸνΓπολλαπλασιάσαςτὸνΑπεποίηκεν. διὰτὰαὐτὰ plane, and C, D its sides, D has thus made A (by) mulδὴκαὶὁΕτὸνΖπολλαπλασιάσαςτὸνΒπεποίηκεν.
ὁΔ tiplying C. And so, for the same (reasons), E has made
δὴτὸνΕπολλαπλασιάσαςτὸνΗποιείτω.καὶἐπεὶὁΔτὸν B (by) multiplying F . So let D make G (by) multiplying
μὲνΓπολλαπλασιάσαςτὸνΑπεποίηκεν,τὸνδὲΕπολλα- E. And since D has made A (by) multiplying C, and has
πλασιάσαςτὸνΗπεποίηκεν,ἔστινἄραὡςὁΓπρὸςτὸνΕ, made G (by) multiplying E, thus as C is to E, so A (is) to
οὕτωςὁΑπρὸςτὸνΗ.ἀλλ᾿ὡςὁΓπρὸςτὸνΕ,[οὕτως] G [Prop. 7.17]. But as C (is) to E, [so] D (is) to F . And
ὁΔπρὸςτὸνΖ·καὶὡςἄραὁΔπρὸςτὸνΖ,οὕτωςὁΑ thus as D (is) to F , so A (is) to G. Again, since E has
πρὸςτὸνΗ.πάλιν,ἐπεὶὁΕτὸνμὲνΔπολλαπλασιάσαςτὸν made G (by) multiplying D, and has made B (by) multi-
Ηπεποίηκεν,τὸνδὲΖπολλαπλασιάσαςτὸνΒπεποίηκεν, plying F , thus as D is to F , so G (is) to B [Prop. 7.17].
ἔστινἄραὡςὁΔπρὸςτὸνΖ,οὕτωςὁΗπρὸςτὸνΒ. And it was also shown that as D (is) to F , so A (is) to G.
ἐδείχθηδὲκαὶὡςὁΔπρὸςτὸνΖ,οὕτωςὁΑπρὸςτὸν And thus as A (is) to G, so G (is) to B. Thus, A, G, B are
A
A
C
D
G
B
B
E
F
C
D
244

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
Η·καὶὡςἄραὁΑπρὸςτὸνΗ,οὕτωςὁΗπρὸςτὸνΒ.οἱ continously proportional. Thus, there exists one number
Α,Η,Βἄραἑξῆςἀνάλογόνεἰσιν.τῶνΑ,Βἄραεἷςμέσος (namely, G) in mean proportion to A and B.
ἀνάλογόνἐστινἀριθμός.
So I say that A also has to B a squared ratio with
Λέγωδή, ὅτικαὶὁΑπρὸςτὸνΒδιπλασίοναλόγον respect to (that) a corresponding side (has) to a corre-
ἔχειἤπερἡὁμόλογοςπλευρὰπρὸςτὴνὁμόλογονπλευράν, sponding side—that is to say, with respect to (that) C
τουτέστινἤπερὁΓπρὸςτὸνΕἢὁΔπρὸςτὸνΖ.ἐπεὶγὰρ (has) to E, or D to F . For since A, G, B are continuously
οἱΑ,Η,Βἑξῆςἀνάλογόνεἰσιν,ὁΑπρὸςτὸνΒδιπλασίονα proportional, A has to B a squared ratio with respect to
λόγονἔχειἤπερπρὸςτὸνΗ.καίἐστινὡςὁΑπρὸςτὸνΗ, (that A has) to G [Prop. 5.9]. And as A is to G, so C (is)
οὕτωςὅτεΓπρὸςτὸνΕκαὶὁΔπρὸςτὸνΖ.καὶὁΑἄρα to E, and D to F . And thus A has to B a squared ratio
πρὸςτὸνΒδιπλασίοναλόγονἔχειἤπερὁΓπρὸςτὸνΕἢ with respect to (that) C (has) to E, or D to F . (Which
ὁΔπρὸςτὸνΖ·ὅπερἔδειδεῖξαι.
is) the very thing it was required to show.
† Literally, “double”.
.
Proposition 19
Δύο ὁμοίων στερεῶν ἀριθμῶν δύο μέσοι ἀνάλογον Two numbers fall (between) two similar solid num-
ἐμπίπτουσινἀριθμοί·καὶὁστερεὸςπρὸςτὸνὅμοιονστερεὸν bers in mean proportion. And a solid (number) has to
τριπλασίοναλόγονἔχειἤπερἡὁμόλογοςπλευρὰπρὸςτὴν a similar solid (number) a cubed † ratio with respect to
ὁμόλογονπλευράν.
(that) a corresponding side (has) to a corresponding side.
Α
Γ
A
C
∆
D
Ε
E
Β
Κ
Μ
Ζ
Η
Θ
Ν
N
Λ
Ξ
O
῎ΕστωσανδύοὅμοιοιστερεοὶοἱΑ,Β,καὶτοῦμὲνΑ Let A and B be two similar solid numbers, and let
πλευραὶἔστωσανοἱΓ,Δ,Ε,τοῦδὲΒοἱΖ,Η,Θ.καὶἐπεὶ C, D, E be the sides of A, and F , G, H (the sides) of
ὅμοιοιστερεοίεἰσινοἱἀνάλογονἔχοντεςτὰςπλευράς,ἔστιν B. And since similar solid (numbers) are those having
ἄραὡςμὲνὁΓπρὸςτὸνΔ,οὕτωςὁΖπρὸςτὸνΗ,ὡςδὲ proportional sides [Def. 7.21], thus as C is to D, so F
ὁΔπρὸςτὸνΕ,οὕτωςὁΗπρὸςτὸνΘ.λέγω,ὅτιτῶνΑ, (is) to G, and as D (is) to E, so G (is) to H. I say that
Βδύομέσοιἀνάλογόνἐμπίπτουσινἀριθμοί,καὶὁΑπρὸς two numbers fall (between) A and B in mean proportion,
τὸνΒτριπλασίοναλόγονἔχειἤπερὁΓπρὸςτὸνΖκαὶὁ and (that) A has to B a cubed ratio with respect to (that)
ΔπρὸςτὸνΗκαὶἔτιὁΕπρὸςτὸνΘ. C (has) to F , and D to G, and, further, E to H.
῾ΟΓγὰρτὸνΔπολλαπλασιάσαςτὸνΚποιείτω,ὁδὲ For let C make K (by) multiplying D, and let F make
ΖτὸνΗπολλαπλασιάσαςτὸνΛποιείτω. καὶἐπεὶοἱΓ, L (by) multiplying G. And since C, D are in the same
ΔτοὶςΖ,Ηἐντῷαὐτῷλόγῳεἰσίν,καὶἐκμὲντῶνΓ, ratio as F , G, and K is the (number created) from (mul-
ΔἐστινὁΚ,ἐκδὲτῶνΖ,ΗὁΛ,οἱΚ,Λ[ἄρα]ὅμοιοι tiplying) C, D, and L the (number created) from (multi-
ἐπίπεδοίεἰσινἀριθμοί·τῶνΚ,Λἄραεἷςμέσοςἀνάλογόν plying) F , G, [thus] K and L are similar plane numbers
ἐστινἀριθμός. ἔστωὁΜ.ὁΜἄραἐστὶνὁἐκτῶνΔ, [Def. 7.21]. Thus, there exits one number in mean pro-
Ζ,ὡςἐντῷπρὸτούτουθεωρήματιἐδείχθη. καὶἐπεὶὁ portion to K and L [Prop. 8.18]. Let it be M. Thus, M is
ΔτὸνμὲνΓπολλαπλασιάσαςτὸνΚπεποίηκεν,τὸνδὲΖ the (number created) from (multiplying) D, F , as shown
πολλαπλασιάσαςτὸνΜπεποίηκεν,ἔστινἄραὡςὁΓπρὸς in the theorem before this (one). And since D has made
τὸνΖ,οὕτωςὁΚπρὸςτὸνΜ.ἀλλ᾿ὡςὁΚπρὸςτὸνΜ, K (by) multiplying C, and has made M (by) multiplying
ὁΜπρὸςτὸνΛ.οἱΚ,Μ,Λἄραἑξῆςεἰσινἀνάλογονἐν F , thus as C is to F , so K (is) to M [Prop. 7.17]. But, as
B
K
M
L
F
G
H
245

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
τῷτοῦΓπρὸςτὸνΖλόγῷ.καὶἐπείἐστινὡςὁΓπρὸςτὸν K (is) to M, (so) M (is) to L. Thus, K, M, L are contin-
Δ,οὕτωςὁΖπρὸςτὸνΗ,ἐναλλὰξἄραἐστὶνὡςὁΓπρὸς uously proportional in the ratio of C to F . And since as
τὸνΖ,οὕτωςὁΔπρὸςτὸνΗ.διὰτὰαὐτὰδὴκαὶὡςὁΔ C is to D, so F (is) to G, thus, alternately, as C is to F , so
πρὸςτὸνΗ,οὕτωςὁΕπρὸςτὸνΘ.οἱΚ,Μ,Λἄραἑξῆς D (is) to G [Prop. 7.13]. And so, for the same (reasons),
εἰσινἀνάλογονἔντετῷτοῦΓπρὸςτὸνΖλόγῳκαὶτῷ as D (is) to G, so E (is) to H. Thus, K, M, L are continτοῦΔπρὸςτὸνΗκαὶἔτιτῷτοῦΕπρὸςτὸνΘ.ἑκατερος
uously proportional in the ratio of C to F , and of D to G,
δὴτῶνΕ,ΘτὸνΜπολλαπλασιάσαςἑκάτεροντῶνΝ,Ξ and, further, of E to H. So let E, H make N, O, respecποιείτω.καὶἐπεὶστερεόςἐστινὁΑ,πλευραὶδὲαὐτοῦεἰσιν
tively, (by) multiplying M. And since A is solid, and C,
οἱΓ,Δ,Ε,ὁΕἄρατὸνἐκτῶνΓ,Δπολλαπλασιάσαςτὸν D, E are its sides, E has thus made A (by) multiplying
Απεποίηκεν.ὁδὲἐκτῶνΓ,ΔἐστινὁΚ·ὁΕἄρατὸνΚ the (number created) from (multiplying) C, D. And K
πολλαπλασιάσαςτὸνΑπεποίηκεν.διὰτὰαὐτὰδὴκαὶὁΘ is the (number created) from (multiplying) C, D. Thus,
τὸνΛπολλαπλασιάσαςτὸνΒπεποίηκεν.καὶἐπεὶὁΕτὸν E has made A (by) multiplying K. And so, for the same
ΚπολλαπλασιάσαςτὸνΑπεποίηκεν,ἀλλὰμὴνκαὶτὸνΜ (reasons), H has made B (by) multiplying L. And since
πολλαπλασιάσαςτὸνΝπεποίηκεν,ἔστινἄραὡςὁΚπρὸς E has made A (by) multiplying K, but has, in fact, also
τὸνΜ,οὕτωςὁΑπρὸςτὸνΝ.ὡςδὲὁΚπρὸςτὸνΜ, made N (by) multiplying M, thus as K is to M, so A (is)
οὕτωςὅτεΓπρὸςτὸνΖκαὶὁΔπρὸςτὸνΗκαὶἔτιὁΕ to N [Prop. 7.17]. And as K (is) to M, so C (is) to F ,
πρὸςτὸνΘ·καὶὡςἄραὁΓπρὸςτὸνΖκαὶὁΔπρὸςτὸν and D to G, and, further, E to H. And thus as C (is) to
ΗκαὶὁΕπρὸςτὸνΘ,οὕτωςὁΑπρὸςτὸνΝ.πάλιν,ἐπεὶ F , and D to G, and E to H, so A (is) to N. Again, since
ἑκάτεροςτῶνΕ,ΘτὸνΜπολλαπλασιάσαςἑκάτεροντῶν E, H have made N, O, respectively, (by) multiplying M,
Ν,Ξπεποίηκεν,ἔστινἄραὡςὁΕπρὸςτὸνΘ,οὕτωςὁΝ thus as E is to H, so N (is) to O [Prop. 7.18]. But, as
πρὸςτὸνΞ.ἀλλ᾿ὡςὁΕπρὸςτὸνΘ,οὕτωςὅτεΓπρὸςτὸν E (is) to H, so C (is) to F , and D to G. And thus as C
ΖκαὶὁΔπρὸςτὸνΗ·καὶὡςἄραὁΓπρὸςτὸνΖκαὶὁΔ (is) to F , and D to G, and E to H, so (is) A to N, and
πρὸςτὸνΗκαὶὁΕπρὸςτὸνΘ,οὕτωςὅτεΑπρὸςτὸνΝ N to O. Again, since H has made O (by) multiplying M,
καὶὁΝπρὸςτὸνΞ.πάλιν,ἐπεὶὁΘτὸνΜπολλαπλασιάσας but has, in fact, also made B (by) multiplying L, thus as
τὸνΞπεποίηκεν,ἀλλὰμὴνκαὶτὸνΛπολλαπλασιάσαςτὸν M (is) to L, so O (is) to B [Prop. 7.17]. But, as M (is)
Βπεποίηκεν,ἔστινἄραὡςὁΜπρὸςτὸνΛ,οὕτωςὁΞπρὸς to L, so C (is) to F , and D to G, and E to H. And thus
τὸνΒ.ἀλλ᾿ὡςὁΜπρὸςτὸνΛ,οὕτωςὅτεΓπρὸςτὸνΖ as C (is) to F , and D to G, and E to H, so not only (is)
καὶὁΔπρὸςτὸνΗκαὶὁΕπρὸςτὸνΘ.καὶὡςἄραὁΓ O to B, but also A to N, and N to O. Thus, A, N, O,
πρὸςτὸνΖκαὶὁΔπρὸςτὸνΗκαὶὁΕπρὸςτὸνΘ,οὕτως B are continuously proportional in the aforementioned
οὐμόνονὁΞπρὸςτὸνΒ,ἀλλὰκαὶὁΑπρὸςτὸνΝκαὶὁ ratios of the sides.
ΝπρὸςτὸνΞ.οἱΑ,Ν,Ξ,Βἄραἑξῆςεἰσινἀνάλογονἐν So I say that A also has to B a cubed ratio with respect
τοῖςεἰρημένοιςτῶνπλευρῶνλόγοις.
to (that) a corresponding side (has) to a corresponding
Λέγω, ὅτι καὶ ὁ Α πρὸς τὸν Β τριπλασίονα λόγον side—that is to say, with respect to (that) the number C
ἔχειἤπερἡὁμόλογοςπλευρὰπρὸςτὴνὁμόλογονπλευράν, (has) to F , or D to G, and, further, E to H. For since A,
τουτέστινἤπερὁΓἀριθμὸςπρὸςτὸνΖἢὁΔπρὸςτὸν N, O, B are four continuously proportional numbers, A
ΗκαὶἔτιὁΕπρὸςτὸνΘ.ἐπεὶγὰρτέσσαρεςἀριθμοὶἑξῆς thus has to B a cubed ratio with respect to (that) A (has)
ἀνάλογόνεἰσινοἱΑ,Ν,Ξ,Β,ὁΑἄραπρὸςτὸνΒτρι- to N [Def. 5.10]. But, as A (is) to N, so it was shown (is)
πλασίοναλόγονἔχειἤπερὁΑπρὸςτὸνΝ.ἀλλ᾿ὡςὁΑ C to F , and D to G, and, further, E to H. And thus A has
πρὸςτὸνΝ,οὕτωςἐδείχθηὅτεΓπρὸςτὸνΖκαὶὁΔπρὸς to B a cubed ratio with respect to (that) a corresponding
τὸνΗκαὶἔτιὁΕπρὸςτὸνΘ.καὶὁΑἄραπρὸςτὸνΒ side (has) to a corresponding side—that is to say, with
τριπλασίοναλόγονἔχειἤπερἡομόλογοςπλευρὰπρὸςτὴν respect to (that) the number C (has) to F , and D to G,
ὁμόλογονπλευράν,τουτέστινἤπερὁΓἀριθμὸςπρὸςτὸν and, further, E to H. (Which is) the very thing it was
ΖκαὶὁΔπρὸςτὸνΗκαὶἔτιὁΕπρὸςτὸνΘ·ὅπερἔδει required to show.
δεῖξαι.
† Literally, “triple”.
.
Proposition 20
᾿Εὰνδύοἀριθμῶνεἷςμέσοςἀνάλογονἐμπίπτῇἀριθμός, If one number falls between two numbers in mean
ὅμοιοιἐπίπεδοιἔσονταιοἱἀριθμοί.
proportion then the numbers will be similar plane (num-
246

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
Δύο γὰρ ἀριθμῶν τῶν Α, Β εἷς μέσος ἀνάλογον bers).
ἐμπιπτέτωἀριθμὸςὁΓ·λέγω,ὅτιοἱΑ,Βὅμοιοιἐπίπεδοί For let one number C fall between the two numbers A
εἰσινἀριθμοί.
and B in mean proportion. I say that A and B are similar
plane numbers.
Α
Γ
Β
∆
Ζ
Ε
Η
Εἰλήφθωσαν [γὰρ] ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν [For] let the least numbers, D and E, having the
λόγονἐχόντωντοῖςΑ,ΓοἱΔ,Ε·ἰσάκιςἄραὁΔτὸνΑ same ratio as A and C have been taken [Prop. 7.33].
μετρεῖκαὶὁΕτὸνΓ.ὁσάκιςδὴὁΔτὸνΑμετρεῖ,τοσαῦται Thus, D measures A as many times as E (measures) C
μονάδεςἔστωσανἐντῷΖ·ὁΖἄρατὸνΔπολλαπλασιάσας [Prop. 7.20]. So as many times as D measures A, so
τὸνΑπεποίηκεν. ὥστεὁΑἐπίπεδόςἐστιν, πλευραὶδὲ many units let there be in F . Thus, F has made A (by)
αὐτοῦοἱΔ,Ζ.πάλιν,ἐπεὶοἱΔ,Εἐλάχιστοίεἰσιτῶντὸν multiplying D [Def. 7.15]. Hence, A is plane, and D,
αὐτὸνλόγονἐχόντωντοῖςΓ,Β,ἰσάκιςἄραὁΔτὸνΓμε- F (are) its sides. Again, since D and E are the least
τρεῖκαὶὁΕτὸνΒ.ὁσάκιςδὴὁΕτὸνΒμετρεῖ,τοσαῦται of those (numbers) having the same ratio as C and B,
μονάδεςἔστωσανἐντῷΗ.ὁΕἄρατὸνΒμετρεῖκατὰτὰς D thus measures C as many times as E (measures) B
ἐντῷΗμονάδας·ὁΗἄρατὸνΕπολλαπλασιάσαςτὸνΒ [Prop. 7.20]. So as many times as E measures B, so
πεποίηκεν. ὁΒἄραἐπίπεδοςἐστι,πλευραὶδὲαὐτοῦεἰσιν many units let there be in G. Thus, E measures B acοἱΕ,Η.οἱΑ,Βἄραἐπίπεδοίεἰσινἀριθμοί.
λέγωδή,ὅτι cording to the units in G. Thus, G has made B (by) mulκαὶὅμοιοι.
ἐπεὶγὰρὁΖτὸνμὲνΔπολλαπλασιάσαςτὸν tiplying E [Def. 7.15]. Thus, B is plane, and E, G are
Απεποίηκεν,τὸνδὲΕπολλαπλασιάσαςτὸνΓπεποίηκεν, its sides. Thus, A and B are (both) plane numbers. So I
ἔστινἄραὡςὁΔπρὸςτὸνΕ,οὕτωςὁΑπρὸςτὸνΓ, say that (they are) also similar. For since F has made A
τουτέστινὁΓπρὸςτὸνΒ.πάλιν,ἐπεὶὁΕἑκάτεροντῶνΖ, (by) multiplying D, and has made C (by) multiplying E,
ΗπολλαπλασιάσαςτοὺςΓ,Βπεποίηκεν,ἔστινἄραὡςὁΖ thus as D is to E, so A (is) to C—that is to say, C to B
πρὸςτὸνΗ,οὕτωςὁΓπρὸςτὸνΒ.ὡςδὲὁΓπρὸςτὸνΒ, [Prop. 7.17]. † Again, since E has made C, B (by) multiοὕτωςὁΔπρὸςτὸνΕ·καὶὡςἄραὁΔπρὸςτὸνΕ,οὕτως
plying F , G, respectively, thus as F is to G, so C (is) to
ὁΖπρὸςτὸνΗ·καὶἐναλλὰξὡςὁΔπρὸςτὸνΖ,οὕτωςὁ B [Prop. 7.17]. And as C (is) to B, so D (is) to E. And
ΕπρὸςτὸνΗ.οἱΑ,Βἄραὅμοιοιἐπίπεδοιἀριθμοίεἰσιν·αἱ thus as D (is) to E, so F (is) to G. And, alternately, as D
γὰρπλευραὶαὐτῶνἀνάλογόνεἰσιν·ὅπερἔδειδεῖξαι. (is) to F , so E (is) to G [Prop. 7.13]. Thus, A and B are
similar plane numbers. For their sides are proportional
[Def. 7.21]. (Which is) the very thing it was required to
show.
† This part of the proof is defective, since it is not demonstrated that F × E = C. Furthermore, it is not necessary to show that D : E :: A : C,
A
C
B
D
F
E
G
because this is true by hypothesis.. Proposition 21
᾿Εὰν δύο ἀριθμῶν δύο μέσοι ἀνάλογον ἐμπίπτωσιν If two numbers fall between two numbers in mean
ἀριθμοί,ὅμοιοιστερεοίεἰσινοἱἀριθμοί.
proportion then the (latter) are similar solid (numbers).
Δύο γὰρ ἀριθμῶν τῶν Α, Β δύο μέσοι ἀνάλογον For let the two numbers C and D fall between the two
ἐμπιπτέτωσανἀριθμοὶοἱΓ,Δ·λέγω,ὅτιοἱΑ,Βὅμοιοι numbers A and B in mean proportion. I say that A and
στερεοίεἰσιν.
B are similar solid (numbers).
Εἰλήφθωσαν γὰρ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν For let the three least numbers E, F , G having the
λόγονἐχόντωντοῖςΑ,Γ,ΔτρεῖςοἱΕ,Ζ,Η·οἱἄραἄκροι same ratio as A, C, D have been taken [Prop. 8.2]. Thus,
αὐτῶνοἱΕ,Ηπρῶτοιπρὸςἀλλήλουςεἰσίν. καὶἐπεὶτῶν the outermost of them, E and G, are prime to one an-
Ε,ΗεἷςμέσοςἀνάλογονἐμπέπτωκενἀριθμὸςὁΖ,οἱΕ, other [Prop. 8.3]. And since one number, F , has fallen
Ηἄραἀριθμοὶὅμοιοιἐπίπεδοίεἰσιν. ἔστωσανοὖντοῦμὲν (between) E and G in mean proportion, E and G are
247

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
ΕπλευραὶοἱΘ,Κ,τοῦδὲΗοἱΛ,Μ.φανερὸνἄραἐστὶν thus similar plane numbers [Prop. 8.20]. Therefore, let
ἐκτοῦπρὸτούτου,ὅτιοἱΕ,Ζ,Ηἑξῆςεἰσινἀνάλογονἔν H, K be the sides of E, and L, M (the sides) of G. Thus,
τετῷτοῦΘπρὸςτὸνΛλόγῳκαὶτῷτοῦΚπρὸςτὸνΜ. it is clear from the (proposition) before this (one) that E,
καὶἐπεὶοἱΕ,Ζ,Ηἐλάχιστοίεἰσιτῶντὸναὐτὸνλόγον F , G are continuously proportional in the ratio of H to
ἐχόντωντοῖςΑ,Γ,Δ,καίἐστινἴσοντὸπλῆθοςτῶνΕ, L, and of K to M. And since E, F , G are the least (num-
Ζ,ΗτῷπλήθειτῶνΑ,Γ,Δ,δι᾿ἴσουἄραἐστὶνὡςὁΕ bers) having the same ratio as A, C, D, and the multitude
πρὸςτὸνΗ,οὕτωςὁΑπρὸςτὸνΔ.οἱδὲΕ,Ηπρῶτοι,οἱ of E, F , G is equal to the multitude of A, C, D, thus,
δὲπρῶτοικαὶἐλάχιστοι,οἱδὲἐλάχιστοιμετροῦσιτοὺςτὸν via equality, as E is to G, so A (is) to D [Prop. 7.14].
αὐτὸνλόγονἔχονταςαὐτοῖςἰσάκιςὅτεμείζωντὸνμείζονα And E and G (are) prime (to one another), and prime
καὶὁἐλάσσωντὸνἐλάσσονα,τουτέστινὅτεἡγούμενος (numbers) are also the least (of those numbers having
τὸνἡγούμενονκαὶὁἑπόμενοςτὸνἑπόμενον· ἰσάκιςἄρα the same ratio as them) [Prop. 7.21], and the least (num-
ὁΕτὸνΑμετρεῖκαὶὁΗτὸνΔ.ὁσάκιςδὴὁΕτὸνΑ bers) measure those (numbers) having the same ratio as
μετρεῖ,τοσαῦταιμονάδεςἔστωσανἐντῷΝ.ὁΝἄρατὸνΕ them an equal number of times, the greater (measuring)
πολλαπλασιάσαςτὸνΑπεποίηκεν. ὁδὲΕἐστινὁἐκτῶν the greater, and the lesser the lesser—that is to say, the
Θ,Κ·ὁΝἄρατὸνἐκτῶνΘ,ΚπολλαπλασιάσαςτὸνΑ leading (measuring) the leading, and the following the
πεποίηκεν. στερεὸςἄραἐστὶνὁΑ,πλευραὶδὲαὐτοῦεἰσιν following [Prop. 7.20]. Thus, E measures A the same
οἱΘ,Κ,Ν.πάλιν,ἐπεὶοἱΕ,Ζ,Ηἐλάχιστοίεἰσιτῶντὸν number of times as G (measures) D. So as many times as
αὐτὸνλόγονἐχόντωντοῖςΓ,Δ,Β,ἰσάκιςἄραὁΕτὸνΓ E measures A, so many units let there be in N. Thus, N
μετρεῖκαὶὁΗτὸνΒ.ὁσάκιςδὴὁΕτὸνΓμετρεῖ,τοσαῦται has made A (by) multiplying E [Def. 7.15]. And E is the
μονάδεςἔστωσανἐντῷΞ.ὁΗἄρατὸνΒμετρεῖκατὰτὰς (number created) from (multiplying) H and K. Thus, N
ἐντῷΞμονάδας·ὁΞἄρατὸνΗπολλαπλασιάσαςτὸνΒ has made A (by) multiplying the (number created) from
πεποίηκεν. ὁδὲΗἐστινὁἐκτῶνΛ,Μ·ὁΞἄρατὸνἐκ (multiplying) H and K. Thus, A is solid, and its sides are
τῶνΛ,ΜπολλαπλασιάσαςτὸνΒπεποίηκεν. στερεὸςἄρα H, K, N. Again, since E, F , G are the least (numbers)
ἐστὶνὁΒ,πλευραὶδὲαὐτοῦεἰσινοἱΛ,Μ,Ξ·οἱΑ,Βἄρα having the same ratio as C, D, B, thus E measures C the
στερεοίεἰσιν. same number of times as G (measures) B [Prop. 7.20].
So as many times as E measures C, so many units let
there be in O. Thus, G measures B according to the units
in O. Thus, O has made B (by) multiplying G. And
G is the (number created) from (multiplying) L and M.
Thus, O has made B (by) multiplying the (number created)
from (multiplying) L and M. Thus, B is solid, and
its sides are L, M, O. Thus, A and B are (both) solid.
Α
Γ
∆
Β
Ε
Ζ
Η
Θ
Κ
Ν
Λ
Μ
Ξ
Λέγω[δή],ὅτικαὶὅμοιοι.ἐπεὶγὰροἱΝ,ΞτὸνΕπολ- [So] I say that (they are) also similar. For since N, O
λαπλασιάσαντεςτοὺςΑ,Γπεποιήκασιν,ἔστινἄραὡςὁΝ have made A, C (by) multiplying E, thus as N is to O, so
πρὸςτὸνΞ,ὁΑπρὸςτὸνΓ,τουτέστινὁΕπρὸςτὸνΖ. A (is) to C—that is to say, E to F [Prop. 7.18]. But, as
ἀλλ᾿ὡςὁΕπρὸςτὸνΖ,ὁΘπρὸςτὸνΛκαὶὁΚπρὸςτὸν E (is) to F , so H (is) to L, and K to M. And thus as H
Μ·καὶὡςἄραὁΘπρὸςτὸνΛ,οὕτωςὁΚπρὸςτὸνΜκαὶ (is) to L, so K (is) to M, and N to O. And H, K, N are
ὁΝπρὸςτὸνΞ.καίεἰσινοἱμὲνΘ,Κ,ΝπλευραὶτοῦΑ, the sides of A, and L, M, O the sides of B. Thus, A and
A
C
D
B
E
F
G
H
K
N
L
M
O
248

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
οἱδὲΞ,Λ,ΜπλευραὶτοῦΒ.οἱΑ,Βἄραἀριθμοὶὅμοιοι B are similar solid numbers [Def. 7.21]. (Which is) the
στερεοίεἰσιν·ὅπερἔδειδεῖξαι.
very thing it was required to show.
† The Greek text has “O, L, M”, which is obviously a mistake.
. Proposition 22
᾿Εὰντρεῖςἀριθμοὶἑξῆςἀνάλογονὦσιν, ὁδὲπρῶτος If three numbers are continuously proportional, and
τετράγωνοςᾖ,καὶὁτρίτοςτετράγωνοςἔσται.
the first is square, then the third will also be square.
Α
Β
Γ
῎ΕστωσαντρεῖςἀριθμοὶἑξῆςἀνάλογονοἱΑ,Β,Γ,ὁδὲ Let A, B, C be three continuously proportional numπρῶτοςὁΑτετράγωνοςἔστω·λέγω,ὅτικαὶὁτρίτοςὁΓ
bers, and let the first A be square. I say that the third C
τετράγωνόςἐστιν.
is also square.
᾿ΕπεὶγὰρτῶνΑ,Γεἷςμέσοςἀνάλογόνἐστινἀριθμὸς For since one number, B, is in mean proportion to
ὁΒ,οἱΑ,Γἄραὅμοιοιἐπίπεδοίεἰσιν.τετράγωνοςδὲὁΑ· A and C, A and C are thus similar plane (numbers)
τετράγωνοςἄρακαὶὁΓ·ὅπερἔδειδεῖξαι.
[Prop. 8.20]. And A is square. Thus, C is also square
[Def. 7.21]. (Which is) the very thing it was required to
show.
. Proposition 23
᾿Εὰντέσσαρεςἀριθμοὶἑξῆςἀνάλογονὦσιν,ὁδὲπρῶτος If four numbers are continuously proportional, and
κύβοςᾖ,καὶὁτέταρτοςκύβοςἔσται.
the first is cube, then the fourth will also be cube.
Α
Β
Γ
∆
῎ΕστωσαντέσσαρεςἀριθμοὶἑξῆςἀνάλογονοἱΑ,Β,Γ, Let A, B, C, D be four continuously proportional
Δ,ὁδὲΑκύβοςἔστω·λέγω,ὅτικαὶὁΔκύβοςἐστίν. numbers, and let A be cube. I say that D is also cube.
᾿ΕπεὶγὰρτῶνΑ,Δδύομέσοιἀνάλογόνεἰσινἀριθμοὶ For since two numbers, B and C, are in mean proporοἱΒ,Γ,οἱΑ,Δἄραὅμοιοίεἰσιστερεοὶἀριθμοί.κύβοςδὲ
tion to A and D, A and D are thus similar solid numbers
ὁΑ·κύβοςἄρακαὶὁΔ·ὅπερἔδειδεῖξαι.
[Prop. 8.21]. And A (is) cube. Thus, D (is) also cube
[Def. 7.21]. (Which is) the very thing it was required to
show.
. Proposition 24
᾿Εὰν δύο ἀριθμοὶ πρὸς ἀλλήλους λόγον ἔχωσιν, ὃν If two numbers have to one another the ratio which
τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, ὁ δὲ a square number (has) to a(nother) square number, and
πρῶτοςτετράγωνοςᾖ,καὶὁδεύτεροςτετράγωνοςἔσται. the first is square, then the second will also be square.
Α
Β
Γ
∆
Δύο γὰρ ἀριθμοὶ οἱ Α, Β πρὸς ἀλλήλους λόγον For let two numbers, A and B, have to one another
A
B
A
B
C
. A
B
C
D
C
D
249

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
ἐχέτωσαν,ὃντετράγωνοςἀριθμὸςὁΓπρὸςτετράγωνον the ratio which the square number C (has) to the square
ἀριθμὸντὸνΔ,ὁδὲΑτετράγωνοςἔστω·λέγω,ὅτικαὶὁ number D. And let A be square. I say that B is also
Βτετράγωνόςἐστιν.
square.
᾿ΕπεὶγὰροἱΓ,Δτετράγωνοίεἰσιν,οἱΓ,Δἄραὅμοιοι For since C and D are square, C and D are thus sim-
ἐπίπεδοίεἰσιν. τῶνΓ,Δἄραεἷςμέσοςἀνάλογονἐμπίπτει ilar plane (numbers). Thus, one number falls (between)
ἀριθμός. καίἐστινὡςὁΓπρὸςτὸνΔ,ὁΑπρὸςτὸνΒ· C and D in mean proportion [Prop. 8.18]. And as C is
καὶτῶνΑ,Βἄραεἷςμέσοςἀνάλογονἐμπίπτειἀριθμός.καί to D, (so) A (is) to B. Thus, one number also falls (be-
ἐστινὁΑτετράγωνος·καὶὁΒἄρατετράγωνόςἐστιν·ὅπερ tween) A and B in mean proportion [Prop. 8.8]. And A
ἔδειδεῖξαι.
is square. Thus, B is also square [Prop. 8.22]. (Which is)
the very thing it was required to show.
. Proposition 25
᾿Εὰν δύο ἀριθμοὶ πρὸς ἀλλήλους λόγον ἔχωσιν, ὃν If two numbers have to one another the ratio which
κύβοςἀριθμὸςπρὸςκύβονἀριθμόν,ὁδὲπρῶτοςκύβοςᾖ, a cube number (has) to a(nother) cube number, and the
καὶὁδεύτεροςκύβοςἔσται.
first is cube, then the second will also be cube.
Α
Ε
Ζ
Β
Γ
∆
Δύο γὰρ ἀριθμοὶ οἱ Α, Β πρὸς ἀλλήλους λόγον For let two numbers, A and B, have to one another
ἐχέτωσαν,ὃνκύβοςἀριθμὸςὁΓπρὸςκύβονἀριθμὸντὸν the ratio which the cube number C (has) to the cube
Δ,κύβοςδὲἔστωὁΑ·λέγω[δή],ὅτικαὶὁΒκύβοςἐστίν. number D. And let A be cube. [So] I say that B is also
᾿Επεὶ γὰρ οἱ Γ, Δ κύβοι εἰσίν, οἱ Γ, Δ ὅμοιοι στε- cube.
ρεοίεἰσιν·τῶνΓ,Δἄραδύομέσοιἀνάλογονἐμπίπτουσιν For since C and D are cube (numbers), C and D are
ἀριθμοί. ὅσοιδὲεἰςτοὺςΓ,Δμεταξὺκατὰτὸσυνεχὲς (thus) similar solid (numbers). Thus, two numbers fall
ἀνάλογον ἐμπίπτουσιν, τοσοῦτοικαὶ εἰς τοὺς τὸν αὐτὸν (between) C and D in mean proportion [Prop. 8.19].
λόγον ἔχοντας αὐτοῖς· ὥστε καὶ τῶν Α, Β δύο μέσοι And as many (numbers) as fall in between C and D in
ἀνάλογονἐμπίπτουσινἀριθμοί.ἐμπιπτέτωσανοἱΕ,Ζ.ἐπεὶ continued proportion, so many also (fall) in (between)
οὖντέσσαρεςἀριθμοὶοἱΑ,Ε,Ζ,Βἑξῆςἀνάλογόνεἰσιν, those (numbers) having the same ratio as them (in conκαίἐστικύβοςὁΑ,κύβοςἄρακαὶὁΒ·ὅπερἔδειδεῖξαι.
tinued proportion) [Prop. 8.8]. And hence two numbers
fall (between) A and B in mean proportion. Let E and F
(so) fall. Therefore, since the four numbers A, E, F , B
are continuously proportional, and A is cube, B (is) thus
also cube [Prop. 8.23]. (Which is) the very thing it was
required to show.
¦. Proposition 26
Οἱὅμοιοιἐπίπεδοιἀριθμοὶπρὸςἀλλήλουςλόγονἔχου- Similar plane numbers have to one another the ratio
σιν,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν. which (some) square number (has) to a(nother) square
number.
Α
Γ
Β
∆
Ε
Ζ
῎ΕστωσανὅμοιοιἐπίπεδοιἀριθμοὶοἱΑ,Β·λέγω,ὅτι Let A and B be similar plane numbers. I say that A
ὁΑπρὸςτὸνΒλόγονἔχει,ὃντετράγωνοςἀριθμὸςπρὸς has to B the ratio which (some) square number (has) to
A
E
F
B
A
C
B
C
D
D
E
F
250

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 8
τετράγωνονἀριθμόν.
a(nother) square number.
᾿ΕπεὶγὰροἱΑ,Βὅμοιοιἐπίπεδοίεἰσιν,τῶνΑ,Βἄρα For since A and B are similar plane numbers, one
εἷςμέσοςἀνάλογονἐμπίπτειἀριθμός.ἐμπιπτέτωκαὶἔστωὁ number thus falls (between) A and B in mean propor-
Γ,καὶεἰλήφθωσανἐλάχιστοιἀριθμοὶτῶντὸναὐτὸνλόγον tion [Prop. 8.18]. Let it (so) fall, and let it be C. And
ἐχόντωντοῖςΑ,Γ,ΒοἱΔ,Ε,Ζ·οἱἄραἄκροιαὐτῶνοἱ let the least numbers, D, E, F , having the same ratio
Δ,Ζτετράγωνοίεἰσιν.καὶἐπείἐστινὡςὁΔπρὸςτὸνΖ, as A, C, B have been taken [Prop. 8.2]. The outermost
οὕτωςὁΑπρὸςτὸνΒ,καίεἰσινοἱΔ,Ζτετράγωνοι,ὁΑ of them, D and F , are thus square [Prop. 8.2 corr.]. And
ἄραπρὸςτὸνΒλόγονἔχει,ὃντετράγωνοςἀριθμὸςπρὸς since as D is to F , so A (is) to B, and D and F are square,
τετράγωνονἀριθμόν·ὅπερἔδειδεῖξαι.
A thus has to B the ratio which (some) square number
(has) to a(nother) square number. (Which is) the very
thing it was required to show.
Þ. Proposition 27
Οἱὅμοιοιστερεοὶἀριθμοὶπρὸςἀλλήλουςλόγονἔχου- Similar solid numbers have to one another the ratio
σιν,ὃνκύβοςἀριθμὸςπρὸςκύβονἀριθμόν.
which (some) cube number (has) to a(nother) cube number.
Α
Γ
∆
Β
Ε
Ζ
Η
Θ
῎ΕστωσανὅμοιοιστερεοὶἀριθμοὶοἱΑ,Β·λέγω,ὅτιὁ Let A and B be similar solid numbers. I say that A
ΑπρὸςτὸνΒλόγονἔχει,ὃνκύβοςἀριθμὸςπρὸςκύβον has to B the ratio which (some) cube number (has) to
ἀριθμόν.
a(nother) cube number.
᾿ΕπεὶγὰροἱΑ,Βὅμοιοιστερεοίεἰσιν,τῶνΑ,Βἄραδύο For since A and B are similar solid (numbers), two
μέσοιἀνάλογονἐμπίπτουσινἀριθμοί. ἐμπιπτέτωσανοἱΓ, numbers thus fall (between) A and B in mean proportion
Δ,καὶεἰλήφθωσανἐλάχιστοιἀριθμοὶτῶντὸναὐτὸνλόγον [Prop. 8.19]. Let C and D have (so) fallen. And let the
ἐχόντωντοῖςΑ,Γ,Δ,ΒἴσοιαὐτοῖςτὸπλῆθοςοἱΕ,Ζ,Η, least numbers, E, F , G, H, having the same ratio as A,
Θ·οἱἄραἄκροιαὐτῶνοἱΕ,Θκύβοιεἰσίν.καίἐστινὡςὁ C, D, B, (and) equal in multitude to them, have been
ΕπρὸςτὸνΘ,οὕτωςὁΑπρὸςτὸνΒ·καὶὁΑἄραπρὸς taken [Prop. 8.2]. Thus, the outermost of them, E and
τὸνΒλόγονἔχει,ὃνκύβοςἀριθμὸςπρὸςκύβονἀριθμόν· H, are cube [Prop. 8.2 corr.]. And as E is to H, so A (is)
ὅπερἔδειδεῖξαι.
to B. And thus A has to B the ratio which (some) cube
number (has) to a(nother) cube number. (Which is) the
very thing it was required to show.
A
C
D
B
E
F
G
H
251

252

ELEMENTS BOOK 9
Applications of Number Theory †
† The propositions contained in Books 7–9 are generally attributed to the school of Pythagoras.
253

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
. Proposition 1
᾿Εὰν δύο ὅμοιοι ἐπίπεδοι ἀριθμοὶ πολλαπλασιάσαντες If two similar plane numbers make some (number by)
ἀλλήλουςποιῶσίτινα,ὁγενόμενοςτετράγωνοςἔσται. multiplying one another then the created (number) will
be square.
Α
Β
Γ
∆
῎ΕστωσανδύοὅμοιοιἐπίπεδοιἀριθμοὶοἱΑ,Β,καὶὁ Let A and B be two similar plane numbers, and let A
ΑτὸνΒπολλαπλασιάσαςτὸνΓποιείτω· λέγω, ὅτιὁΓ make C (by) multiplying B. I say that C is square.
τετράγωνόςἐστιν.
For let A make D (by) multiplying itself. D is thus
῾ΟγὰρΑἑαυτὸνπολλαπλασιάσαςτὸνΔποιείτω. ὁΔ square. Therefore, since A has made D (by) multiply-
ἄρατετράγωνόςἐστιν. ἐπεὶοὖνὁΑἑαυτὸνμὲνπολλα- ing itself, and has made C (by) multiplying B, thus as A
πλασιάσαςτὸνΔπεποίηκεν,τὸνδὲΒπολλαπλασιάσαςτὸν is to B, so D (is) to C [Prop. 7.17]. And since A and
Γπεποίηκεν,ἔστινἄραὡςὁΑπρὸςτὸνΒ,οὕτωςὁΔ B are similar plane numbers, one number thus falls (beπρὸςτὸνΓ.καὶἐπεὶοἱΑ,Βὅμοιοιἐπίπεδοίεἰσινἀριθμοί,
tween) A and B in mean proportion [Prop. 8.18]. And if
τῶνΑ,Βἄραεἷςμέσοςἀνάλογονἐμπίπτειἀριθμός.ἐὰνδὲ (some) numbers fall between two numbers in continued
δύοἀριθμῶνμεταξὺκατὰτὸσυνεχὲςἀνάλογονἐμπίπτωσιν proportion then, as many (numbers) as fall in (between)
ἀριθμοί,ὅσοιεἰςαὐτοὺςἐμπίπτουσι,τοσοῦτοικαὶεἰςτοὺς them (in continued proportion), so many also (fall) in
τὸναὐτὸνλόγονἔχοντας·ὥστεκαὶτῶνΔ,Γεἷςμέσος (between numbers) having the same ratio (as them in
ἀνάλογον ἐμπίπτει ἀριθμός. καί ἐστι τετράγωνος ὁ Δ· continued proportion) [Prop. 8.8]. And hence one numτετράγωνοςἄρακαὶὁΓ·ὅπερἔδειδεῖξαι.
ber falls (between) D and C in mean proportion. And D
is square. Thus, C (is) also square [Prop. 8.22]. (Which
is) the very thing it was required to show.
. Proposition 2
᾿Εὰνδύοἀριθμοὶπολλαπλασιάσαντεςἀλλήλουςποιῶσι If two numbers make a square (number by) multiplyτετράγωνον,ὅμοιοιἐπίπεδοίεἰσινἀριθμοί.
ing one another then they are similar plane numbers.
Α
Β
Γ
∆
῎ΕστωσανδύοἀριθμοὶοἱΑ,Β,καὶὁΑτὸνΒπολλα- Let A and B be two numbers, and let A make the
πλασιάσαςτετράγωνοντὸνΓποιείτω·λέγω,ὅτιοἱΑ,Β square (number) C (by) multiplying B. I say that A and
ὅμοιοιἐπίπεδοίεἰσινἀριθμοί.
B are similar plane numbers.
῾ΟγὰρΑἑαυτὸνπολλαπλασιάσαςτὸνΔποιείτω·ὁΔ For let A make D (by) multiplying itself. Thus, D is
ἄρατετράγωνόςἐστιν. καὶἐπεὶὁΑἑαυτὸνμὲνπολλα- square. And since A has made D (by) multiplying itself,
πλασιάσαςτὸνΔπεποίηκεν,τὸνδὲΒπολλαπλασιάσαςτὸν and has made C (by) multiplying B, thus as A is to B, so
Γπεποίηκεν,ἔστινἄραὡςὁΑπρὸςτὸνΒ,ὁΔπρὸςτὸν D (is) to C [Prop. 7.17]. And since D is square, and C
Γ.καὶἐπεὶὁΔτετράγωνόςἐστιν,ἀλλὰκαὶὁΓ,οἱΔ,Γ (is) also, D and C are thus similar plane numbers. Thus,
ἄραὅμοιοιἐπίπεδοίεἰσιν.τῶνΔ,Γἄραεἷςμέσοςἀνάλογον one (number) falls (between) D and C in mean propor-
A
B
C
D
A
B
C
D
254

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
ἐμπίπτει.καίἐστινὡςὁΔπρὸςτὸνΓ,οὕτωςὁΑπρὸςτὸν tion [Prop. 8.18]. And as D is to C, so A (is) to B. Thus,
Β·καὶτῶνΑ,Βἄραεἷςμέσοςἀνάλογονἐμπίπτει. ἐὰνδὲ one (number) also falls (between) A and B in mean proδύοἀριθμῶνεἷςμέσοςἀνάλογονἐμπίπτῃ,ὅμοιοιἐπίπεδοί
portion [Prop. 8.8]. And if one (number) falls (between)
εἰσιν[οἱ]ἀριθμοί·οἱἄραΑ,Βὅμοιοίεἰσινἐπίπεδοι·ὅπερ two numbers in mean proportion then [the] numbers are
ἔδειδεῖξαι.
similar plane (numbers) [Prop. 8.20]. Thus, A and B are
similar plane (numbers). (Which is) the very thing it was
required to show.
. Proposition 3
᾿Εὰνκύβοςἀριθμὸςἑαυτὸνπολλαπλασιάσαςποιῇτινα, If a cube number makes some (number by) multiply-
ὁγενόμενοςκύβοςἔσται.
ing itself then the created (number) will be cube.
Α
Β
Γ
∆
ΚύβοςγὰρἀριθμὸςὁΑἑαυτὸνπολλαπλασιάσαςτὸνΒ For let the cube number A make B (by) multiplying
ποιείτω·λέγω,ὅτιὁΒκύβοςἐστίν.
itself. I say that B is cube.
ΕἰλήφθωγὰρτοῦΑπλευρὰὁΓ,καὶὁΓἑαυτὸνπολλα- For let the side C of A have been taken. And let C
πλασιάσαςτὸνΔποιείτω.φανερὸνδήἐστιν,ὅτιὁΓτὸνΔ make D by multiplying itself. So it is clear that C has
πολλαπλασιάσαςτὸνΑπεποίηκεν.καὶἐπεὶὁΓἑαυτὸνπολ- made A (by) multiplying D. And since C has made D
λαπλασιάσαςτὸνΔπεποίηκεν,ὁΓἄρατὸνΔμετρεῖκατὰ (by) multiplying itself, C thus measures D according to
τὰςἐναὑτῷμονάδας. ἀλλὰμὴνκαὶἡμονὰςτὸνΓμετρεῖ the units in it [Def. 7.15]. But, in fact, a unit also meaκατὰτὰςἐναὐτῷμονάδας·ἔστινἄραὡςἡμονὰςπρὸςτὸν
sures C according to the units in it [Def. 7.20]. Thus, as
Γ,ὁΓπρὸςτὸνΔ.πάλιν,ἐπεὶὁΓτὸνΔπολλαπλασιάσας a unit is to C, so C (is) to D. Again, since C has made A
τὸνΑπεποίηκεν,ὁΔἄρατὸνΑμετρεῖκατὰτὰςἐντῷΓ (by) multiplying D, D thus measures A according to the
μονάδας. μετρεῖδὲκαὶἡμονὰςτὸνΓκατὰτὰςἐναὐτῷ units in C. And a unit also measures C according to the
μονάδας·ἔστινἄραὡςἡμονὰςπρὸςτὸνΓ,ὁΔπρὸςτὸν units in it. Thus, as a unit is to C, so D (is) to A. But,
Α.ἀλλ᾿ὡςἡμονὰςπρὸςτὸνΓ,ὁΓπρὸςτὸνΔ·καὶὡς as a unit (is) to C, so C (is) to D. And thus as a unit (is)
ἄραἡμονὰςπρὸςτὸνΓ,οὕτωςὁΓπρὸςτὸνΔκαὶὁΔ to C, so C (is) to D, and D to A. Thus, two numbers, C
πρὸςτὸνΑ.τῆςἄραμονάδοςκαὶτοῦΑἀριθμοῦδύομέσοι and D, have fallen (between) a unit and the number A
ἀνάλογονκατὰτὸσυνεχὲςἐμπεπτώκασινἀριθμοὶοἱΓ,Δ. in continued mean proportion. Again, since A has made
πάλιν,ἐπεὶὁΑἑαυτὸνπολλαπλασιάσαςτὸνΒπεποίηκεν, B (by) multiplying itself, A thus measures B according
ὁΑἄρατὸνΒμετρεῖκατὰτὰςἐναὐτῷμονάδας·μετρεῖδὲ to the units in it. And a unit also measures A according
καὶἡμονὰςτὸνΑκατὰτὰςἐναὐτῷμονάδας·ἔστινἄραὡς to the units in it. Thus, as a unit is to A, so A (is) to B.
ἡμονὰςπρὸςτὸνΑ,ὁΑπρὸςτὸνΒ.τῆςδὲμονάδοςκαὶ And two numbers have fallen (between) a unit and A in
τοῦΑδύομέσοιἀνάλογονἐμπεπτώκασινἀριθμοί·καὶτῶν mean proportion. Thus two numbers will also fall (be-
Α,Βἄραδύομέσοιἀνάλογονἐμπεσοῦνταιἀριθμοί.ἐὰνδὲ tween) A and B in mean proportion [Prop. 8.8]. And if
δύοἀριθμῶνδύομέσοιἀνάλογονἐμπίπτωσιν,ὁδὲπρῶτος two (numbers) fall (between) two numbers in mean proκύβοςᾖ,καὶὁδεύτεροςκύβοςἔσται.καίἐστινὁΑκύβος·
portion, and the first (number) is cube, then the second
καὶὁΒἄρακύβοςἐστίν·ὅπερἔδειδεῖξαι.
will also be cube [Prop. 8.23]. And A is cube. Thus, B
is also cube. (Which is) the very thing it was required to
show.
. Proposition 4
᾿Εὰν κύβος ἀριθμὸς κύβον ἀριθμὸν πολλαπλασιάσας If a cube number makes some (number by) multiplyποιῇτινα,ὁγενόμενοςκύβοςἔσται.
ing a(nother) cube number then the created (number)
A
B
C
D
255

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
Α
Β
Γ
∆
will be cube.
ΚύβοςγὰρἀριθμὸςὁΑκύβονἀριθμὸντὸνΒπολλα- For let the cube number A make C (by) multiplying
πλασιάσαςτὸνΓποιείτω·λέγω,ὅτιὁΓκύβοςἐστίν. the cube number B. I say that C is cube.
῾Ο γὰρ Α ἑαυτὸν πολλαπλασιάσαςτὸν Δ ποιείτω· ὁ For let A make D (by) multiplying itself. Thus, D is
Δ ἄρα κύβος ἐστίν. καὶ ἐπεὶ ὁ Α ἑαυτὸν μὲν πολλα- cube [Prop. 9.3]. And since A has made D (by) multiπλασιάσαςτὸνΔπεποίηκεν,τὸνδὲΒπολλαπλασιάσαςτὸν
plying itself, and has made C (by) multiplying B, thus
Γπεποίηκεν,ἔστινἄραὡςὁΑπρὸςτὸνΒ,οὕτωςὁΔπρὸς as A is to B, so D (is) to C [Prop. 7.17]. And since A
τὸνΓ.καὶἐπεὶοἱΑ,Βκύβοιεἰσίν,ὅμοιοιστερεοίεἰσινοἱΑ, and B are cube, A and B are similar solid (numbers).
Β.τῶνΑ,Βἄραδύομέσοιἀνάλογονἐμπίπτουσινἀριθμοί· Thus, two numbers fall (between) A and B in mean pro-
ὥστε καὶ τῶν Δ, Γ δύο μέσοι ἀνάλογον ἐμπεσοῦνται portion [Prop. 8.19]. Hence, two numbers will also fall
ἀριθμοί. καίἐστικύβοςὁΔ·κύβοςἄρακαὶὁΓ·ὅπερ (between) D and C in mean proportion [Prop. 8.8]. And
ἔδειδεῖξαι.
D is cube. Thus, C (is) also cube [Prop. 8.23]. (Which
is) the very thing it was required to show.
. Proposition 5
᾿Εὰνκύβοςἀριθμὸςἀριθμόντιναπολλαπλασιάσαςκύβον If a cube number makes a(nother) cube number (by)
ποιῇ,καὶὁπολλαπλασιασθεὶςκύβοςἔσται.
multiplying some (number) then the (number) multiplied
will also be cube.
Α
Β
Γ
∆
Κύβοςγὰρ ἀριθμὸςὁΑἀριθμόν τινα τὸν Β πολλα- For let the cube number A make the cube (number)
πλασιάσαςκύβοντὸνΓποιείτω·λέγω,ὅτιὁΒκύβοςἐστίν. C (by) multiplying some number B. I say that B is cube.
῾ΟγὰρΑἑαυτὸνπολλαπλασιάσαςτὸνΔποιείτω·κύβος For let A make D (by) multiplying itself. D is thus
ἄραἐστίνὁΔ.καὶἐπεὶὁΑἑαυτὸνμὲνπολλαπλασιάσαςτὸν cube [Prop. 9.3]. And since A has made D (by) multiply-
Δπεποίηκεν,τὸνδὲΒπολλαπλασιάσαςτὸνΓπεποίηκεν, ing itself, and has made C (by) multiplying B, thus as A
ἔστινἀραὡςὁΑπρὸςτὸνΒ,ὁΔπρὸςτὸνΓ.καὶἐπεὶ is to B, so D (is) to C [Prop. 7.17]. And since D and C
οἱΔ,Γκύβοιεἰσίν,ὅμοιοιστερεοίεἰσιν. τῶνΔ,Γἄρα are (both) cube, they are similar solid (numbers). Thus,
δύομέσοιἀνάλογονἐμπίπτουσινἀριθμοί.καίἐστινὡςὁΔ two numbers fall (between) D and C in mean proportion
πρὸςτὸνΓ,οὕτωςὁΑπρὸςτὸνΒ·καὶτῶνΑ,Βἄραδύο [Prop. 8.19]. And as D is to C, so A (is) to B. Thus,
μέσοιἀνάλογονἐμπίπτουσινἀριθμοί. καίἐστικύβοςὁΑ· two numbers also fall (between) A and B in mean proκύβοςἄραἐστὶκαὶὁΒ·ὅπερἔδειδεῖξαι.
portion [Prop. 8.8]. And A is cube. Thus, B is also cube
[Prop. 8.23]. (Which is) the very thing it was required to
show.
¦. Proposition 6
᾿Εὰνἀριθμὸςἑαυτὸνπολλαπλασιάσαςκύβονποιῇ,καὶ If a number makes a cube (number by) multiplying
A
B
C
D
A
B
C
D
256

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
αὐτὸςκύβοςἔσται.
Α
Β
Γ
itself then it itself will also be cube.
ἈριθμὸςγὰρὁΑἑαυτὸνπολλαπλασιάσαςκύβοντὸνΒ For let the number A make the cube (number) B (by)
ποιείτω·λέγω,ὅτικαὶὁΑκύβοςἐστίν.
multiplying itself. I say that A is also cube.
῾ΟγὰρΑτὸνΒπολλαπλασιάσαςτὸνΓποιείτω.ἐπεὶοὖν For let A make C (by) multiplying B. Therefore, since
ὁΑἑαυτὸνμὲνπολλαπλασιάσαςτὸνΒπεποίηκεν,τὸνδὲ A has made B (by) multiplying itself, and has made C
ΒπολλαπλασιάσαςτὸνΓπεποίηκεν,ὁΓἄρακύβοςἐστίν. (by) multiplying B, C is thus cube. And since A has made
καὶἐπεὶὁΑἑαυτὸνπολλαπλασιάσαςτὸνΒπεποίηκεν,ὁ B (by) multiplying itself, A thus measures B according to
ΑἄρατὸνΒμετρεῖκατὰτὰςἐναὑτῷμονάδας. μετρεῖδὲ the units in (A). And a unit also measures A according
καὶἡμονὰςτὸνΑκατὰτὰςἐναὐτῷμονάδας. ἔστινἄρα to the units in it. Thus, as a unit is to A, so A (is) to
ὡςἡμονὰςπρὸςτὸνΑ,οὕτωςὁΑπρὸςτὸνΒ.καὶἐπεὶ B. And since A has made C (by) multiplying B, B thus
ὁΑτὸνΒπολλαπλασιάσαςτὸνΓπεποίηκεν,ὁΒἄρατὸν measures C according to the units in A. And a unit also
ΓμετρεῖκατὰτὰςἐντῷΑμονάδας.μετρεὶδὲκαὶἡμονὰς measures A according to the units in it. Thus, as a unit is
τὸνΑκατὰτὰςἐναὐτῷμονάδας. ἔστινἄραὡςἡμονὰς to A, so B (is) to C. But, as a unit (is) to A, so A (is) to
πρὸςτὸνΑ,οὕτωςὁΒπρὸςτὸνΓ.ἀλλ᾿ὡςἡμονὰςπρὸς B. And thus as A (is) to B, (so) B (is) to C. And since B
τὸνΑ,οὕτωςὁΑπρὸςτὸνΒ·καὶὡςἄραὁΑπρὸςτὸνΒ, and C are cube, they are similar solid (numbers). Thus,
ὁΒπρὸςτὸνΓ.καὶἐπεὶοἱΒ,Γκύβοιεἰσίν,ὅμοιοιστερεοί there exist two numbers in mean proportion (between)
εἰσιν.τῶνΒ,Γἄραδύομέσοιἀνάλογόνεἰσινἀριθμοί.καί B and C [Prop. 8.19]. And as B is to C, (so) A (is) to B.
ἐστινὡςὁΒπρὸςτὸνΓ,ὁΑπρὸςτὸνΒ.καὶτῶνΑ,Β Thus, there also exist two numbers in mean proportion
ἄραδύομέσοιἀνάλογόνεἰσινἀριθμοί. καίἐστινκύβοςὁ (between) A and B [Prop. 8.8]. And B is cube. Thus, A
Β·κύβοςἄραἐστὶκαὶὁΑ·ὅπερἔδειδεὶξαι.
is also cube [Prop. 8.23]. (Which is) the very thing it was
required to show.
Þ. Proposition 7
᾿Εὰνσύνθετοςἀριθμὸςἀριθμόντιναπολλαπλασιάσας If a composite number makes some (number by) mulποιῇτινα,ὁγενόμενοςστερεὸςἔσται.
tiplying some (other) number then the created (number)
will be solid.
Α
Β
Γ
∆
Ε
ΣύνθετοςγὰρἀριθμὸςὁΑἀριθμόντινατὸνΒπολλα- For let the composite number A make C (by) multiπλασιάσαςτὸνΓποιείτω·λέγω,ὅτιὁΓστερεόςἐστιν.
plying some number B. I say that C is solid.
᾿ΕπεὶγὰρὁΑσύνθετόςἐστιν,ὑπὸἀριθμοῦτινοςμε- For since A is a composite (number), it will be meaτρηθήσεται.
μετρείσθωὑπὸτοῦΔ,καὶὁσάκιςὁΔτὸνΑ sured by some number. Let it be measured by D. And,
μετρεῖ,τοσαῦταιμονάδεςἔστωσανἐντῷΕ.ἐπεὶοὖνὁΔ as many times as D measures A, so many units let there
τὸνΑμετρεῖκατὰτὰςἐντῷΕμονάδας,ὁΕἄρατὸνΔ be in E. Therefore, since D measures A according to
πολλαπλασιάσαςτὸνΑπεποίηκεν.καὶἐπεὶὁΑτὸνΒπολ- the units in E, E has thus made A (by) multiplying D
λαπλασιάσαςτὸνΓπεποίηκεν,ὁδὲΑἐστινὁἐκτῶνΔ,Ε, [Def. 7.15]. And since A has made C (by) multiplying B,
ὁἄραἐκτῶνΔ,ΕτὸνΒπολλαπλασιάσαςτὸνΓπεποίηκεν. and A is the (number created) from (multiplying) D, E,
ὁΓἄραστερεόςἐστιν,πλευραὶδὲαὐτοῦεἰσινοἱΔ,Ε,Β· the (number created) from (multiplying) D, E has thus
A
B
C
A
B
C
D
E
257

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
ὅπερἔδειδεῖξαι.
made C (by) multiplying B. Thus, C is solid, and its sides
are D, E, B. (Which is) the very thing it was required to
show.
. Proposition 8
᾿Εὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον If any multitude whatsoever of numbers is continu-
ὦσιν, ὁ μὲν τρίτος ἀπὸ τῆς μονάδος τετράγωνος ἔσται ously proportional, (starting) from a unit, then the third
καὶ οἱ ἕνα διαλείποντες, ὁ δὲ τέταρτος κύβος καὶ οἱ from the unit will be square, and (all) those (numbers
δύο διαλείποντες πάντες, ὁ δὲ ἕβδομος κύβος ἅμα καὶ after that) which leave an interval of one (number), and
τετράγωνοςκαὶοἱπέντεδιαλείποντες.
the fourth (will be) cube, and all those (numbers after
that) which leave an interval of two (numbers), and the
seventh (will be) both cube and square, and (all) those
(numbers after that) which leave an interval of five (numbers).
Α
Β
Γ
∆
Ε
Ζ
῎Εστωσανἀπὸμονάδοςὁποσοιοῦνἀριθμοὶἑξῆςἀνάλογ- Let any multitude whatsoever of numbers, A, B, C,
ον οἱ Α, Β, Γ, Δ, Ε, Ζ· λέγω, ὅτι ὁ μὲν τρίτος ἀπὸ D, E, F , be continuously proportional, (starting) from
τῆςμονάδοςὁΒτετράγωνόςἐστικαὶοἱἕναδιαλείποντες a unit. I say that the third from the unit, B, is square,
πάντες,ὁδὲτέταρτοςὁΓκύβοςκαὶοἱδύοδιαλείποντες and all those (numbers after that) which leave an interπάντες,ὁδὲἕβδομοςὁΖκύβοςἅμακαὶτετράγωνοςκαὶοἱ
val of one (number). And the fourth (from the unit), C,
πέντεδιαλείποντεςπάντες.
(is) cube, and all those (numbers after that) which leave
᾿ΕπεὶγάρἐστινὡςἡμονὰςπρὸςτὸνΑ,οὕτωςὁΑ an interval of two (numbers). And the seventh (from the
πρὸςτὸνΒ,ἰσάκιςἄραἡμονὰςτὸνΑἀριθμὸνμετρεῖκαὶ unit), F , (is) both cube and square, and all those (num-
ὁΑτὸνΒ.ἡδὲμονὰςτὸνΑἀριθμὸνμετρεῖκατὰτὰς bers after that) which leave an interval of five (numbers).
ἐναὐτῷμονάδας·καὶὁΑἄρατὸνΒμετρεῖκατὰτὰςἐν For since as the unit is to A, so A (is) to B, the unit
τῷΑμονάδας. ὁΑἄραἑαυτὸνπολλαπλασιάσαςτὸνΒ thus measures the number A the same number of times
πεποίηκεν·τετράγωνοςἄραἐστὶνὁΒ.καὶἐπεὶοἱΒ,Γ,Δ as A (measures) B [Def. 7.20]. And the unit measures
ἑξῆςἀνάλογόνεἰσιν,ὁδὲΒτετράγωνόςἐστιν,καὶὁΔἄρα the number A according to the units in it. Thus, A also
τετράγωνόςἐστιν. διὰτὰαὐτὰδὴκαὶὁΖτετράγωνός measures B according to the units in A. A has thus made
ἐστιν. ὁμοίως δὴ δείξομεν, ὅτι καὶ οἱ ἕνα διαλείποντες B (by) multiplying itself [Def. 7.15]. Thus, B is square.
πάντεςτετράγωνοίεἰσιν. λέγωδή,ὅτικαὶὁτέταρτοςἀπὸ And since B, C, D are continuously proportional, and B
τῆςμονάδοςὁΓκύβοςἐστὶκαὶοἱδύοδιαλείποντεςπάντες. is square, D is thus also square [Prop. 8.22]. So, for the
ἐπεὶγάρἐστινὡςἡμονὰςπρὸςτὸνΑ,οὕτωςὁΒπρὸςτὸν same (reasons), F is also square. So, similarly, we can
Γ,ἰσάκιςἄραἡμονὰςτὸνΑἀριθμὸνμετρεῖκαὶὁΒτὸνΓ.ἡ also show that all those (numbers after that) which leave
δὲμονὰςτὸνΑἀριθμὸνμετρεῖκατὰτὰςἐντῷΑμονάδας· an interval of one (number) are square. So I also say that
καὶὁΒἄρατὸνΓμετρεῖκατὰτὰςἐντῷΑμονάδας·ὁΑ the fourth (number) from the unit, C, is cube, and all
ἄρατὸνΒπολλαπλασιάσαςτὸνΓπεποίηκεν. ἐπεὶοὖνὁ those (numbers after that) which leave an interval of two
ΑἑαυτὸνμὲνπολλαπλασιάσαςτὸνΒπεποίηκεν,τὸνδὲΒ (numbers). For since as the unit is to A, so B (is) to C,
πολλαπλασιάσαςτὸνΓπεποίηκεν,κύβοςἄραἐστὶνὁΓ.καὶ the unit thus measures the number A the same number
ἐπεὶοἱΓ,Δ,Ε,Ζἑξῆςἀνάλογόνεἰσιν,ὁδὲΓκύβοςἐστίν, of times that B (measures) C. And the unit measures the
A
B
C
D
E
F
258

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
καὶὁΖἄρακύβοςἐστίν.ἐδείχθηδὲκαὶτετράγωνος·ὁἄρα number A according to the units in A. And thus B mea-
ἕβδομοςἀπὸτῆςμονάδοςκύβοςτέἐστικαὶτετράγωνος. sures C according to the units in A. A has thus made C
ὁμοίωςδὴδείξομεν,ὅτικαὶοἱπέντεδιαλείποντεςπάντες (by) multiplying B. Therefore, since A has made B (by)
κύβοιτέεἰσικαὶτετράγωνοι·ὅπερἔδειδεῖξαι.
multiplying itself, and has made C (by) multiplying B, C
is thus cube. And since C, D, E, F are continuously proportional,
and C is cube, F is thus also cube [Prop. 8.23].
And it was also shown (to be) square. Thus, the seventh
(number) from the unit is (both) cube and square. So,
similarly, we can show that all those (numbers after that)
which leave an interval of five (numbers) are (both) cube
and square. (Which is) the very thing it was required to
show.
. Proposition 9
᾿Εὰν ἀπὸ μονάδος ὁποσοιοῦν ἑξῆς κατὰ τὸ συνεχὲς If any multitude whatsoever of numbers is continu-
ἀριθμοὶἀνάλογονὦσιν,ὁδὲμετὰτὴνμονάδατετράγωνος ously proportional, (starting) from a unit, and the (num-
ᾖ,καὶοἱλοιποὶπάντεςτετράγωνοιἔσονται.καὶἐὰνὁμετὰ ber) after the unit is square, then all the remaining (numτὴνμονάδακύβοςᾖ,καὶοἱλοιποὶπάντεςκύβοιἔσονται.
bers) will also be square. And if the (number) after the
unit is cube, then all the remaining (numbers) will also
be cube.
Α
Β
Γ
∆
Ε
Ζ
῎Εστωσαν ἀπὸμονάδοςἑξῆςἀνάλογονὁσοιδηποτοῦν Let any multitude whatsoever of numbers, A, B, C,
ἀριθμοὶ οἱ Α, Β, Γ, Δ, Ε, Ζ, ὁ δὲ μετὰ τὴν μονάδα D, E, F , be continuously proportional, (starting) from a
ὁ Α τετράγωνος ἔστω· λέγω, ὅτι καὶ οἱ λοιποὶ πάντες unit. And let the (number) after the unit, A, be square. I
τετράγωνοιἔσονται.
say that all the remaining (numbers) will also be square.
῞ΟτιμὲνοὖνὁτρίτοςἀπὸτῆςμονάδοςὁΒτετράγωνός In fact, it has (already) been shown that the third
ἐστικαὶοἱἕναδιαπλείποντεςπάντες,δέδεικται·λέγω[δή], (number) from the unit, B, is square, and all those (num-
ὅτικαὶοἱλοιποὶπάντεςτετράγωνοίεἰσιν. ἐπεὶγὰροἱΑ, bers after that) which leave an interval of one (number)
Β,Γἑξῆςἀνάλογόνεἰσιν,καίἐστινὁΑτετράγωνος,καὶὁ [Prop. 9.8]. [So] I say that all the remaining (num-
Γ[ἄρα]τετράγωνοςἐστιν.πάλιν,ἐπεὶ[καὶ]οἱΒ,Γ,Δἑξῆς bers) are also square. For since A, B, C are continu-
ἀνάλογόνεἰσιν,καίἐστινὁΒτετράγωνος,καὶὁΔ[ἄρα] ously proportional, and A (is) square, C is [thus] also
τετράγωνόςἐστιν. ὁμοίωςδὴδείξομεν,ὅτικαὶοἱλοιποὶ square [Prop. 8.22]. Again, since B, C, D are [also] conπάντεςτετράγωνοίεἰσιν.
tinuously proportional, and B is square, D is [thus] also
ἈλλὰδὴἔστωὁΑκύβος·λέγω,ὅτικαὶοἱλοιποὶπάντες square [Prop. 8.22]. So, similarly, we can show that all
κύβοιεἰσίν.
the remaining (numbers) are also square.
῞ΟτιμὲνοὖνὁτέταρτοςἀπὸτῆςμονάδοςὁΓκύβοςἐστὶ And so let A be cube. I say that all the remaining
καὶοἱδύοδιαλείποντεςπάντες,δέδεικται·λέγω[δή],ὅτικαὶ (numbers) are also cube.
οἱλοιποὶπάντεςκύβοιεἰσίν.ἐπεὶγάρἐστινὡςἡμονὰςπρὸς In fact, it has (already) been shown that the fourth
τὸνΑ,οὕτωςὁΑπρὸςτὸνΒ,ἰσάκιςἀραἡμονὰςτὸνΑ (number) from the unit, C, is cube, and all those (numμετρεῖκαὶὁΑτὸνΒ.ἡδὲμονὰςτὸνΑμετρεῖκατὰτὰςἐν
bers after that) which leave an interval of two (numbers)
A
B
C
D
E
F
259

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
αὐτῷμονάδας·καὶὁΑἄρατὸνΒμετρεῖκατὰτὰςἐναὑτῷ [Prop. 9.8]. [So] I say that all the remaining (numbers)
μονάδας·ὁΑἄραἑαυτὸνπολλαπλασιάσαςτὸνΒπεποίηκεν. are also cube. For since as the unit is to A, so A (is) to
καίἐστινὁΑκύβος.ἐὰνδὲκύβοςἀριθμὸςἑαυτὸνπολλα- B, the unit thus measures A the same number of times
πλασιάσαςποιῇτινα,ὁγενόμενοςκύβοςἐστίν·καὶὁΒἄρα as A (measures) B. And the unit measures A according
κύβοςἐστίν.καὶἐπεὶτέσσαρεςἀριθμοὶοἱΑ,Β,Γ,Δἑξῆς to the units in it. Thus, A also measures B according to
ἀνάλογόνεἰσιν,καίἐστινὁΑκύβος,καὶὁΔἄρακύβος the units in (A). A has thus made B (by) multiplying it-
ἐστίν. διὰτὰαὐτὰδὴκαὶὁΕκύβοςἐστίν,καὶὁμοίωςοἱ self. And A is cube. And if a cube number makes some
λοιποὶπάντεςκύβοιεἰσίν·ὅπερἔδειδεῖξαι.
(number by) multiplying itself then the created (number)
is cube [Prop. 9.3]. Thus, B is also cube. And since the
four numbers A, B, C, D are continuously proportional,
and A is cube, D is thus also cube [Prop. 8.23]. So, for
the same (reasons), E is also cube, and, similarly, all the
remaining (numbers) are cube. (Which is) the very thing
it was required to show.
. Proposition 10
᾿Εὰνἀπὸμονάδοςὁποσοιοῦνἀριθμοὶ[ἑξῆς]ἀνάλογον If any multitude whatsoever of numbers is [continu-
ὦσιν,ὁδὲμετὰτὴνμονάδαμὴᾖτετράγωνος,οὐδ᾿ἄλλος ously] proportional, (starting) from a unit, and the (numοὐδεὶςτετράγωνοςἔσταιχωρὶςτοῦτρίτουἀπὸτῆςμονάδος
ber) after the unit is not square, then no other (number)
καὶ τῶν ἕνα διαλειπόντων πάντων. καὶ ἐὰν ὁ μετὰ τὴν will be square either, apart from the third from the unit,
μονάδακύβοςμὴᾖ,οὐδὲἄλλοςοὐδεὶςκύβοςἔσταιχωρὶς and all those (numbers after that) which leave an interτοῦτετάρτουἀπὸτῆςμονάδοςκαὶτῶνδύοδιαλειπόντων
val of one (number). And if the (number) after the unit
πάντων.
is not cube, then no other (number) will be cube either,
apart from the fourth from the unit, and all those (numbers
after that) which leave an interval of two (numbers).
Α
Β
Γ
∆
Ε
Ζ
῎Εστωσαν ἀπὸμονάδοςἑξῆςἀνάλογονὁσοιδηποτοῦν Let any multitude whatsoever of numbers, A, B, C,
ἀριθμοὶοἱΑ,Β,Γ,Δ,Ε,Ζ,ὁμετὰτὴνμονάδαὁΑμὴ D, E, F , be continuously proportional, (starting) from
ἔστωτετράγωνος·λέγω,ὅτιοὐδὲἄλλοςοὐδεὶςτετράγωνος a unit. And let the (number) after the unit, A, not be
ἔσταιχωρὶςτοῦτρίτουἀπὸτὴςμονάδος[καὶτῶνἕναδια- square. I say that no other (number) will be square eiλειπόντων].
ther, apart from the third from the unit [and (all) those
Εἰγὰρδυνατόν,ἔστωὁΓτετράγωνος. ἔστιδὲκαὶὁ (numbers after that) which leave an interval of one (num-
Βτετράγωνος· οἱΒ,Γἄραπρὸςἀλλήλουςλόγονἔχου- ber)].
σιν,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν.καί For, if possible, let C be square. And B is also
ἐστινὡςὁΒπρὸςτὸνΓ,ὁΑπρὸςτὸνΒ·οἱΑ,Βἄρα square [Prop. 9.8]. Thus, B and C have to one another
πρὸςἀλλήλουςλόγονἔχουσιν,ὃντετράγωνοςἀριθμὸςπρὸς (the) ratio which (some) square number (has) to (some
τετράγωνονἀριθμόν·ὥστεοἱΑ,Βὅμοιοιἐπίπεδοίεἰσιν. other) square number. And as B is to C, (so) A (is)
καίἐστιτετράγωνοςὁΒ·τετράγωνοςἄραἐστὶκαὶὁΑ· to B. Thus, A and B have to one another (the) ratio
ὅπεροὐχὑπέκειτο.οὐκἄραὁΓτετράγωνόςἐστιν.ὁμοίως which (some) square number has to (some other) square
δὴδείξομεν,ὅτιοὐδ᾿ἄλλοςοὐδεὶςτετράγωνόςἐστιχωρὶς number. Hence, A and B are similar plane (numbers)
A
B
C
D
E
F
260

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
τοῦτρίτουἀπὸτῆςμονάδοςκαὶτῶνἕναδιαλειπόντων. [Prop. 8.26]. And B is square. Thus, A is also square.
ἈλλὰδὴμὴἔστωὁΑκύβος. λέγω,ὅτιοὐδ᾿ἄλλος The very opposite thing was assumed. C is thus not
οὐδεὶςκύβοςἔσταιχωρὶςτοῦτετάρτουἀπὸτῆςμονάδος square. So, similarly, we can show that no other (number
καὶτῶνδύοδιαλειπόντων.
is) square either, apart from the third from the unit, and
Εἰγὰρδυνατόν,ἔστωὁΔκύβος.ἔστιδὲκαὶὁΓκύβος· (all) those (numbers after that) which leave an interval
τέταρτοςγάρἐστινἀπὸτῆςμονάδος.καίἐστινὡςὁΓπρὸς of one (number).
τὸνΔ,ὁΒπρὸςτὸνΓ·καὶὁΒἄραπρὸςτὸνΓλόγονἔχει, And so let A not be cube. I say that no other (num-
ὃνκύβοςπρὸςκύβον. καίἐστινὁΓκύβος·καὶὁΒἄρα ber) will be cube either, apart from the fourth from the
κύβοςἐστίν.καὶἐπείἐστινὡςἡμονὰςπρὸςτὸνΑ,ὁΑπρὸς unit, and (all) those (numbers after that) which leave an
τὸνΒ,ἡδὲμονὰςτὸνΑμετρεῖκατὰτὰςἐναὐτῷμονάδας, interval of two (numbers).
καὶὁΑἄρατὸνΒμετρεῖκατὰτὰςἐναὑτῷμονάδας·ὁΑ For, if possible, let D be cube. And C is also cube
ἄραἑαυτὸνπολλαπλασιάσαςκύβοντὸνΒπεποίηκεν. ἐὰν [Prop. 9.8]. For it is the fourth (number) from the unit.
δὲἀριθμὸςἑαυτὸνπολλαπλασιάσαςκύβονποιῇ,καὶαὐτὸς And as C is to D, (so) B (is) to C. And B thus has to
κύβοςἔσται.κύβοςἄρακαὶὁΑ·ὅπεροὐχὑπόκειται.οὐκ C the ratio which (some) cube (number has) to (some
ἄραὁΔκύβοςἐστίν. ὁμοίωςδὴδείξομεν,ὅτιοὐδ᾿ἄλλος other) cube (number). And C is cube. Thus, B is also
οὐδεὶςκύβοςἐστὶχωρὶςτοῦτετάρτουἀπὸτῆςμονάδοςκαὶ cube [Props. 7.13, 8.25]. And since as the unit is to
τῶνδύοδιαλειπόντων·ὅπερἔδειδεῖξαι.
A, (so) A (is) to B, and the unit measures A according
to the units in it, A thus also measures B according
to the units in (A). Thus, A has made the cube (number)
B (by) multiplying itself. And if a number makes a
cube (number by) multiplying itself then it itself will be
cube [Prop. 9.6]. Thus, A (is) also cube. The very opposite
thing was assumed. Thus, D is not cube. So, similarly,
we can show that no other (number) is cube either,
apart from the fourth from the unit, and (all) those (numbers
after that) which leave an interval of two (numbers).
(Which is) the very thing it was required to show.
. Proposition 11
᾿Εὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον If any multitude whatsoever of numbers is continu-
ὦσιν,ὁἐλάττωντὸνμείζοναμετρεῖκατάτινατῶνὑπαρχόντ- ously proportional, (starting) from a unit, then a lesser
ωνἐντοῖςἀνάλογονἀριθμοῖς.
(number) measures a greater according to some existing
(number) among the proportional numbers.
Α
Β
Γ
∆
Ε
῎ΕστωσανἀπὸμονάδοςτῆςΑὁποσοιοῦνἀριθμοὶἑξῆς Let any multitude whatsoever of numbers, B, C, D,
ἀνάλογονοἱΒ, Γ, Δ, Ε· λέγω, ὅτιτῶν Β, Γ, Δ, Ε ὁ E, be continuously proportional, (starting) from the unit
ἐλάχιστοςὁΒτὸνΕμετρεῖκατάτινατῶνΓ,Δ. A. I say that, for B, C, D, E, the least (number), B,
᾿ΕπεὶγάρἐστινὡςἡΑμονὰςπρὸςτὸνΒ,οὕτωςὁΔ measures E according to some (one) of C, D.
πρὸςτὸνΕ,ἰσάκιςἄραἡΑμονὰςτὸνΒἀριθμὸνμετρεῖ For since as the unit A is to B, so D (is) to E, the
καὶὁΔτὸνΕ·ἐναλλὰξἄραἰσάκιςἡΑμονὰςτὸνΔμετρεῖ unit A thus measures the number B the same number of
καὶὁΒτὸνΕ.ἡδὲΑμονὰςτὸνΔμετρεῖκατὰτὰςἐν times as D (measures) E. Thus, alternately, the unit A
A
B
C
D
E
261

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
αὐτῷμονάδας·καὶὁΒἄρατὸνΕμετρεῖκατὰτὰςἐντῷ measures D the same number of times as B (measures)
Δμονάδας·ὥστεὁἐλάσσωνὁΒτὸνμείζονατὸνΕμε- E [Prop. 7.15]. And the unit A measures D according to
τρεῖκατάτιναἀριθμὸντῶνὑπαρχόντωνἐντοῖςἀνάλογον the units in it. Thus, B also measures E according to the
ἀριθμοῖς.
units in D. Hence, the lesser (number) B measures the
greater E according to some existing number among the
proportional numbers (namely, D).
ÈÖ×Ñ.
Corollary
Καὶφανερόν,ὅτιἣνἔχειτάξινὁμετρῶνἀπὸμονάδος, And (it is) clear that what(ever relative) place the
τὴναὐτὴνἔχεικαὶὁκαθ᾿ὃνμετρεῖἀπὸτοῦμετρουμένου measuring (number) has from the unit, the (number)
ἐπὶτὸπρὸαὐτοῦ.ὅπερἔδειδεῖξαι.
according to which it measures has the same (relative)
place from the measured (number), in (the direction of
the number) before it. (Which is) the very thing it was
required to show.
. Proposition 12
᾿Εὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον If any multitude whatsoever of numbers is continu-
ὦσιν,ὑφ᾿ὅσωνἂνὁἔσχατοςπρώτωνἀριθμῶνμετρῆται, ously proportional, (starting) from a unit, then however
ὑπὸτῶναὐτῶνκαὶὁπαρὰτὴνμονάδαμετρηθήσεται. many prime numbers the last (number) is measured by,
the (number) next to the unit will also be measured by
the same (prime numbers).
Α
Β
Γ
∆
Ε
Ζ
Η
Θ
῎Εστωσανἀπὸμονάδοςὁποσοιδηποτοῦνἀριθμοὶἀνάλογ- Let any multitude whatsoever of numbers, A, B, C,
ονοἱΑ,Β,Γ,Δ·λέγω, ὅτιὑφ᾿ὅσωνἂνὁΔπρώτων D, be (continuously) proportional, (starting) from a unit.
ἀριθμῶνμετρῆται,ὑπὸτῶναὐτῶνκαὶὁΑμετρηθήσεται. I say that however many prime numbers D is measured
ΜετρείσθωγὰρὁΔὑπότινοςπρώτουἀριθμοῦτοῦΕ· by, A will also be measured by the same (prime numbers).
λέγω,ὅτιὁΕτὸνΑμετρεῖ.μὴγάρ·καίἐστινὁΕπρῶτος, For let D be measured by some prime number E. I
ἅπαςδὲπρῶτοςἀριθμὸςπρὸςἅπαντα,ὃνμὴμετρεῖ,πρῶτός say that E measures A. For (suppose it does) not. E is
ἐστιν·οἱΕ,Αἄραπρῶτοιπρὸςἀλλήλουςεἰσίν. καὶἐπεὶ prime, and every prime number is prime to every num-
ὁΕτὸνΔμετρεῖ,μετρείτωαὐτὸνκατὰτὸνΖ·ὁΕἄρα ber which it does not measure [Prop. 7.29]. Thus, E
τὸνΖπολλαπλασιάσαςτὸνΔπεποίηκεν. πάλιν,ἐπεὶὁΑ and A are prime to one another. And since E measures
τὸνΔμετρεῖκατὰτὰςἐντῷΓμονάδας,ὁΑἄρατὸνΓ D, let it measure it according to F . Thus, E has made
πολλαπλασιάσαςτὸνΔπεποίηκεν.ἀλλὰμὴνκαὶὁΕτὸνΖ D (by) multiplying F . Again, since A measures D acπολλαπλασιάσαςτὸνΔπεποίηκεν·ὁἄραἐκτῶνΑ,Γἴσος
cording to the units in C [Prop. 9.11 corr.], A has thus
ἐστὶτῷἐκτῶνΕ,Ζ.ἔστινἄραὡςὁΑπρὸςτὸνΕ,ὁΖ made D (by) multiplying C. But, in fact, E has also
πρὸςτὸνΓ.οἱδὲΑ,Επρῶτοι,οἱδὲπρῶτοικαὶἐλάχιστοι, made D (by) multiplying F . Thus, the (number creοἱδὲἐλάχιστοιμετροῦσιτοὺςτὸναὐτὸνλόγονἔχοντας
ated) from (multiplying) A, C is equal to the (number
ἰσάκιςὅτεἡγούμενοςτὸνἡγούμενονκαὶὁἑπόμενοςτὸν created) from (multiplying) E, F . Thus, as A is to E,
ἑπόμενον·μετρεῖἄραὁΕτὸνΓ.μετρείτωαὐτὸνκατὰτὸν (so) F (is) to C [Prop. 7.19]. And A and E (are) prime
Η·ὁΕἄρατὸνΗπολλαπλασιάσαςτὸνΓπεποίηκεν.ἀλλὰ (to one another), and (numbers) prime (to one another
μὴνδιὰτὸπρὸτούτουκαὶὁΑτὸνΒπολλαπλασιάσαςτὸν are) also the least (of those numbers having the same ra-
Γπεποίηκεν.ὁἄραἐκτῶνΑ,ΒἴσοςἐστὶτῷἐκτῶνΕ,Η. tio as them) [Prop. 7.21], and the least (numbers) mea-
ἔστινἄραὡςὁΑπρὸςτὸνΕ,ὁΗπρὸςτὸνΒ.οἱδὲΑ,Ε sure those (numbers) having the same ratio as them an
πρῶτοι,οἱδὲπρῶτοικαὶἐλάχιστοι,οἱδὲἐλάχιστοιἀριθμοὶ equal number of times, the leading (measuring) the lead-
A
B
C
D
E
F
G
H
262

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
μετροῦσιτοὺςτὸναὐτὸνλόγονἔχονταςαὐτοῖςἰσάκιςὅτε ing, and the following the following [Prop. 7.20]. Thus,
ἡγούμενοςτὸνἡγούμενονκαὶὁἑπόμενοςτὸνἑπόμενον· E measures C. Let it measure it according to G. Thus,
μετρεῖἄραὁΕτὸνΒ.μετρείτωαὐτὸνκατὰτὸνΘ·ὁΕἄρα E has made C (by) multiplying G. But, in fact, via the
τὸνΘπολλαπλασιάσαςτὸνΒπεποίηκεν.ἀλλὰμὴνκαὶὁΑ (proposition) before this, A has also made C (by) multi-
ἑαυτὸνπολλαπλασιάσαςτὸνΒπεποίηκεν·ὁἄραἐκτῶνΕ, plying B [Prop. 9.11 corr.]. Thus, the (number created)
ΘἴσοςἐστὶτῷἀπὸτοῦΑ.ἔστινἄραὡςὁΕπρὸςτὸνΑ,ὁΑ from (multiplying) A, B is equal to the (number created)
πρὸςτὸνΘ.οἱδὲΑ,Επρῶτοι,οἱδὲπρῶτοικαὶἐλάχιστοι,οἱ from (multiplying) E, G. Thus, as A is to E, (so) G
δὲἐλάχιστοιμετροῦσιτοὺςτὸναὐτὸνλόγονἔχονταςἰσάκις (is) to B [Prop. 7.19]. And A and E (are) prime (to
ὅἡγούμενοςτὸνἡγούμενονκαὶὁἑπόμενοςτὸνἑπόμενον· one another), and (numbers) prime (to one another are)
μετρεῖἄραὁΕτὸνΑὡςἡγούμενοςἡγούμενον.ἀλλὰμὴν also the least (of those numbers having the same ratio
καὶοὐμετρεῖ·ὅπερἀδύνατον.οὐκἄραοἱΕ,Απρῶτοιπρὸς as them) [Prop. 7.21], and the least (numbers) measure
ἀλλήλουςεἰσίν.σύνθετοιἄρα.οἱδὲσύνθετοιὑπὸ[πρώτου] those (numbers) having the same ratio as them an equal
ἀριθμοῦτινοςμετροῦνται.καὶἐπεὶὁΕπρῶτοςὑπόκειται,ὁ number of times, the leading (measuring) the leading,
δὲπρῶτοςὑπὸἑτέρουἀριθμοῦοὐμετρεῖταιἢὑφ᾿ἑαυτοῦ,ὁ and the following the following [Prop. 7.20]. Thus, E
ΕἄρατοὺςΑ,Εμετρεῖ·ὥστεὁΕτὸνΑμετρεῖ.μετρεῖδὲ measures B. Let it measure it according to H. Thus,
καὶτὸνΔ·ὁΕἄρατοὺςΑ,Δμετρεῖ.ὁμοίωςδὴδείξομεν, E has made B (by) multiplying H. But, in fact, A has
ὅτιὑφ᾿ὅσωνἂνὁΔπρώτωνἀριθμῶνμετρῆται,ὑπὸτῶν also made B (by) multiplying itself [Prop. 9.8]. Thus,
αὐτῶνκαὶὁΑμετρηθήσεται·ὅπερἔδειδεῖξαι.
the (number created) from (multiplying) E, H is equal
to the (square) on A. Thus, as E is to A, (so) A (is)
to H [Prop. 7.19]. And A and E are prime (to one another),
and (numbers) prime (to one another are) also
the least (of those numbers having the same ratio as
them) [Prop. 7.21], and the least (numbers) measure
those (numbers) having the same ratio as them an equal
number of times, the leading (measuring) the leading,
and the following the following [Prop. 7.20]. Thus, E
measures A, as the leading (measuring the) leading. But,
in fact, (E) also does not measure (A). The very thing
(is) impossible. Thus, E and A are not prime to one
another. Thus, (they are) composite (to one another).
And (numbers) composite (to one another) are (both)
measured by some [prime] number [Def. 7.14]. And
since E is assumed (to be) prime, and a prime (number)
is not measured by another number (other) than itself
[Def. 7.11], E thus measures (both) A and E. Hence, E
measures A. And it also measures D. Thus, E measures
(both) A and D. So, similarly, we can show that however
many prime numbers D is measured by, A will also be
measured by the same (prime numbers). (Which is) the
very thing it was required to show.
. Proposition 13
᾿Εὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον If any multitude whatsoever of numbers is continu-
ὦσιν, ὁ δὲ μετὰ τὴν μονάδα πρῶτος ᾖ, ὁ μέγιστος ὑπ᾿ ously proportional, (starting) from a unit, and the (numοὐδενὸς[ἄλλου]μετρηθήσεταιπαρὲξτῶνὑπαρχόντωνἐν
ber) after the unit is prime, then the greatest (number)
τοῖςἀνάλογονἀριθμοῖς.
will be measured by no [other] (numbers) except (num-
῎Εστωσανἀπὸμονάδοςὁποσοιοῦνἀριθμοὶἑξῆςἀνάλογον bers) existing among the proportional numbers.
οἱΑ,Β,Γ,Δ,ὁδὲμετὰτὴνμονάδαὁΑπρῶτοςἔστω· Let any multitude whatsoever of numbers, A, B, C,
λέγω,ὅτιὁμέγιστοςαὐτῶνὁΔὑπ᾿οὐδενὸςἄλλουμε- D, be continuously proportional, (starting) from a unit.
τρηθήσεταιπαρὲξτῶνΑ,Β,Γ.
And let the (number) after the unit, A, be prime. I say
263

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
Α
Β
Γ
∆
Ε
Ζ
Η
Θ
that the greatest of them, D, will be measured by no other
(numbers) except A, B, C.
Εἰγὰρδυνατόν,μετρείσθωὑπὸτοῦΕ,καὶὁΕμηδενὶ For, if possible, let it be measured by E, and let E not
τῶνΑ,Β,Γἔστωὁαὐτός. φανερὸνδή,ὅτιὁΕπρῶτος be the same as one of A, B, C. So it is clear that E is
οὔκἐστιν.εἰγὰρὁΕπρῶτόςἐστικαὶμετρεῖτὸνΔ,καὶτὸν not prime. For if E is prime, and measures D, then it will
Αμετρήσειπρῶτονὄνταμὴὢναὐτῷὁαὐτός·ὅπερἐστὶν also measure A, (despite A) being prime (and) not being
ἀδύνατον. οὐκἄραὁΕπρῶτόςἐστιν. σύνθετοςἄρα. πᾶς the same as it [Prop. 9.12]. The very thing is impossible.
δὲσύνθετοςἀριθμὸςὑπὸπρώτουτινὸςἀριθμοῦμετρεῖται· Thus, E is not prime. Thus, (it is) composite. And every
ὁΕἄραὑπὸπρώτουτινὸςἀριθμοῦμετρεῖται.λέγωδή,ὅτι composite number is measured by some prime number
ὑπ᾿οὐδενὸςἄλλουπρώτουμετρηθήσεταιπλὴντοῦΑ.εἰγὰρ [Prop. 7.31]. Thus, E is measured by some prime num-
ὑφ᾿ἑτέρουμετρεῖταιὁΕ,ὁδὲΕτὸνΔμετρεῖ,κἀκεῖνοςἄρα ber. So I say that it will be measured by no other prime
τὸνΔμετρήσει·ὥστεκαὶτὸνΑμετρήσειπρῶτονὄνταμὴ number than A. For if E is measured by another (prime
ὢναὐτῷὁαὐτός·ὅπερἐστὶνἀδύνατον. ὁΑἄρατὸνΕ number), and E measures D, then this (prime number)
μετρεῖ.καὶἐπεὶὁΕτὸνΔμετρεῖ,μετρείτωαὐτὸνκατὰτὸν will thus also measure D. Hence, it will also measure A,
Ζ.λέγω,ὅτιὁΖοὐδενὶτῶνΑ,Β,Γἐστινὁαὐτός.εἰγὰρὁ (despite A) being prime (and) not being the same as it
ΖἑνὶτῶνΑ,Β,ΓἐστινὁαὐτὸςκαὶμετρεῖτὸνΔκατὰτὸν [Prop. 9.12]. The very thing is impossible. Thus, A mea-
Ε,καὶεἷςἄρατῶνΑ,Β,ΓτὸνΔμετρεῖκατάτὸνΕ.ἀλλὰ sures E. And since E measures D, let it measure it acεἷςτῶνΑ,Β,ΓτὸνΔμετρεῖκατάτινατῶνΑ,Β,Γ·καὶὁ
cording to F . I say that F is not the same as one of A, B,
ΕἄραἑνὶτῶνΑ,Β,Γἐστινὁαὐτός·ὅπεροὐχὑπόκειται. C. For if F is the same as one of A, B, C, and measures
οὐκἄραὁΖἑνὶτῶνΑ,Β,Γἐστινὁαὐτός. ὁμοίωςδὴ D according to E, then one of A, B, C thus also measures
δείξομεν,ὅτιμετρεῖταιὁΖὑπὸτοῦΑ,δεικνύντεςπάλιν, D according to E. But one of A, B, C (only) measures
ὅτιὁΖοὔκἐστιπρῶτος.εἰγὰρ,καὶμετρεῖτὸνΔ,καὶτὸν D according to some (one) of A, B, C [Prop. 9.11]. And
Αμετρήσειπρῶτονὄνταμὴὢναὐτῷὁαὐτός·ὅπερἐστὶν thus E is the same as one of A, B, C. The very oppo-
ἀδύνατον·οὐκἄραπρῶτόςἐστινὁΖ·σύνθετοςἄρα.ἅπας site thing was assumed. Thus, F is not the same as one
δὲσύνθετοςἀριθμὸςὑπὸπρώτουτινὸςἀριθμοῦμετρεῖται·ὁ of A, B, C. Similarly, we can show that F is measured
Ζἄραὑπὸπρώτουτινὸςἀριθμοῦμετρεῖται.λέγωδή,ὅτιὑφ᾿ by A, (by) again showing that F is not prime. For if (F
ἑτέρουπρώτουοὐμετρηθήσεταιπλὴντοῦΑ.εἰγὰρἕτερός is prime), and measures D, then it will also measure A,
τιςπρῶτοςτὸνΖμετρεῖ,ὁδὲΖτὸνΔμετρεῖ,κἀκεῖνος (despite A) being prime (and) not being the same as it
ἄρατὸνΔμετρήσει·ὥστεκαὶτὸνΑμετρήσειπρῶτονὄντα [Prop. 9.12]. The very thing is impossible. Thus, F is
μὴὢναὐτῷὁαὐτός·ὅπερἐστὶνἀδύνατον.ὁΑἄρατὸνΖ not prime. Thus, (it is) composite. And every composite
μετρεῖ.καὶἐπεὶὁΕτὸνΔμετρεῖκατὰτὸνΖ,ὁΕἄρατὸν number is measured by some prime number [Prop. 7.31].
ΖπολλαπλασιάσαςτὸνΔπεποίηκεν.ἀλλὰμὴνκαὶὁΑτὸν Thus, F is measured by some prime number. So I say
ΓπολλαπλασιάσαςτὸνΔπεποίηκεν·ὁἄραἐκτῶνΑ,Γ that it will be measured by no other prime number than
ἴσοςἐστὶτῷἐκτῶνΕ,Ζ.ἀνάλογονἄραἐστὶνὡςὁΑπρὸς A. For if some other prime (number) measures F , and
τὸνΕ,οὕτωςὁΖπρὸςτὸνΓ.ὁδὲΑτὸνΕμετρεῖ·καὶὁΖ F measures D, then this (prime number) will thus also
ἄρατὸνΓμετρεῖ. μετρείτωαὐτὸνκατὰτὸνΗ.ὁμοίωςδὴ measure D. Hence, it will also measure A, (despite A)
δείξομεν,ὅτιὁΗοὐδενὶτῶνΑ,Βἐστινὁαὐτός,καὶὅτι being prime (and) not being the same as it [Prop. 9.12].
μετρεῖταιὑπὸτοῦΑ.καὶἐπεὶὁΖτὸνΓμετρεῖκατὰτὸνΗ, The very thing is impossible. Thus, A measures F . And
ὁΖἄρατὸνΗπολλαπλασιάσαςτὸνΓπεποίηκεν.ἀλλὰμὴν since E measures D according to F , E has thus made
καὶὁΑτὸνΒπολλαπλασιάσαςτὸνΓπεποίηκεν·ὁἄραἐκ D (by) multiplying F . But, in fact, A has also made D
τῶνΑ,ΒἴσοςἐστὶτῷἐκτῶνΖ,Η.ἀνάλογονἄραὡςὁΑ (by) multiplying C [Prop. 9.11 corr.]. Thus, the (number
πρὸςτὸνΖ,ὁΗπρὸςτὸνΒ.μετρεῖδὲὁΑτὸνΖ·μετρεῖ created) from (multiplying) A, C is equal to the (number
ἄρακαὶὁΗτὸνΒ.μετρείτωαὐτὸνκατὰτὸνΘ.ὁμοίωςδὴ created) from (multiplying) E, F . Thus, proportionally,
δείξομεν,ὅτιὁΘτῷΑοὐκἔστινὁαὐτός.καὶἐπεὶὁΗτὸν as A is to E, so F (is) to C [Prop. 7.19]. And A measures
A
B
C
D
E
F
G
H
264

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
ΒμετρεῖκατὰτὸνΘ,ὁΗἄρατὸνΘπολλαπλασιάσαςτὸν E. Thus, F also measures C. Let it measure it according
Βπεποίηκεν. ἀλλὰμὴνκαὶὁΑἑαυτὸνπολλαπλασιάσας to G. So, similarly, we can show that G is not the same
τὸνΒπεποίηκεν·ὁἄραὑπὸΘ,ΗἴσοςἐστὶτῷἀπὸτοῦΑ as one of A, B, and that it is measured by A. And since
τετραγώνῳ·ἔστινἄραὡςὁΘπρὸςτὸνΑ,ὁΑπρὸςτὸνΗ. F measures C according to G, F has thus made C (by)
μετρεῖδὲὁΑτὸνΗ·μετρεῖἄρακαὶὁΘτὸνΑπρῶτονὄντα multiplying G. But, in fact, A has also made C (by) mulμὴὢναὐτῷὁαὐτός·ὅπερἄτοπον.
οὐκἄραὁμέγιστοςὁ tiplying B [Prop. 9.11 corr.]. Thus, the (number created)
ΔὑπὸἑτέρουἀριθμοῦμετρηθήσεταιπαρὲξτῶνΑ,Β,Γ· from (multiplying) A, B is equal to the (number created)
ὅπερἔδειδεῖξαι.
from (multiplying) F , G. Thus, proportionally, as A (is)
to F , so G (is) to B [Prop. 7.19]. And A measures F .
Thus, G also measures B. Let it measure it according to
H. So, similarly, we can show that H is not the same as
A. And since G measures B according to H, G has thus
made B (by) multiplying H. But, in fact, A has also made
B (by) multiplying itself [Prop. 9.8]. Thus, the (number
created) from (multiplying) H, G is equal to the square
on A. Thus, as H is to A, (so) A (is) to G [Prop. 7.19].
And A measures G. Thus, H also measures A, (despite
A) being prime (and) not being the same as it. The very
thing (is) absurd. Thus, the greatest (number) D cannot
be measured by another (number) except (one of) A, B,
C. (Which is) the very thing it was required to show.
. Proposition 14
᾿Εὰνἐλάχιστοςἀριθμὸςὑπὸπρώτωνἀριθμῶνμετρῆται, If a least number is measured by (some) prime num-
ὑπ᾿ οὑδενὸς ἄλλου πρώτου ἀριθμοῦ μετρηθήσεται παρὲξ bers then it will not be measured by any other prime
τῶνἐξἀρχῆςμετρούντων.
number except (one of) the original measuring (numbers).
Α
Ε
Ζ
Β
Γ
∆
᾿ΕλάχιστοςγὰρἀριθμὸςὁΑὑπὸπρώτωνἀριθμῶντῶν For let A be the least number measured by the prime
Β, Γ, Δ μετρείσθω· λέγω, ὅτι ὁ Α ὑπ᾿ οὐδενὸς ἄλλου numbers B, C, D. I say that A will not be measured by
πρώτουἀριθμοῦμετρηθήσεταιπαρὲξτῶνΒ,Γ,Δ. any other prime number except (one of) B, C, D.
Εἰγὰρδυνατόν,μετρείσθωὑπὸπρώτουτοῦΕ,καὶὁ For, if possible, let it be measured by the prime (num-
ΕμηδενὶτῶνΒ,Γ,Δἔστωὁαὐτός. καὶἐπεὶὁΕτὸν ber) E. And let E not be the same as one of B, C, D.
Α μετρεῖ, μετρείτω αὐτὸν κατὰ τὸν Ζ· ὁ Ε ἄρα τὸν Ζ And since E measures A, let it measure it according to F .
πολλαπλασιάσαςτὸνΑπεποίηκεν. καὶμετρεῖταιὁΑὑπὸ Thus, E has made A (by) multiplying F . And A is meaπρώτωνἀριθμῶντῶνΒ,Γ,Δ.ἐὰνδὲδύοἀριθμοὶπολ-
sured by the prime numbers B, C, D. And if two numλαπλασιάσαντεςἀλλήλουςποιῶσίτινα,τὸνδὲγενόμενον
bers make some (number by) multiplying one another,
ἐξαὐτῶνμετρῇτιςπρῶτοςἀριθμός,καὶἕνατῶνἐξἀρχῆς and some prime number measures the number created
μετρήσει·οἱΒ,Γ,ΔἄραἕνατῶνΕ,Ζμετρήσουσιν. τὸν from them, then (the prime number) will also measure
μὲνοὖνΕοὐμετρήσουσιν·ὁγὰρΕπρῶτόςἐστικαὶοὐδενὶ one of the original (numbers) [Prop. 7.30]. Thus, B, C,
τῶνΒ,Γ,Δὁαὐτός.τὸνΖἄραμετροῦσινἐλάσσοναὄντα D will measure one of E, F . In fact, they do not measure
τοῦΑ·ὅπερἀδύνατον. ὁγὰρΑὑπόκειταιἐλάχιστοςὑπὸ E. For E is prime, and not the same as one of B, C, D.
τῶνΒ,Γ,Δμετρούμενος.οὐκἄρατὸνΑμετρήσειπρῶτος Thus, they (all) measure F , which is less than A. The
ἀριθμὸςπαρὲξτῶνΒ,Γ,Δ·ὅπερἔδειδεῖξαι.
very thing (is) impossible. For A was assumed (to be) the
least (number) measured by B, C, D. Thus, no prime
A
E
F
B
C
D
265

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
number can measure A except (one of) B, C, D. (Which
is) the very thing it was required to show.
. Proposition 15
᾿Εὰντρεῖςἀριθμοὶἑξῆςἀνάλογονὦσινἐλάχιστοιτῶν If three continuously proportional numbers are the
τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς, δύο ὁποιοιοῦν συν- least of those (numbers) having the same ratio as them
τεθέντεςπρὸςτὸνλοιπὸνπρῶτοίεἰσιν.
then two (of them) added together in any way are prime
to the remaining (one).
Α
Β
Γ
∆
Ε Ζ D E
A
῎Εστωσαντρεῖςἀριθμοὶἑξῆςἀνάλογονἐλάχιστοιτῶν Let A, B, C be three continuously proportional numτὸναὐτὸνλόγονἐχόντωναὐτοῖςοἱΑ,Β,Γ·λέγω,ὅτιτῶν
bers (which are) the least of those (numbers) having the
Α,Β,Γδύοὁποιοιοῦνσυντεθέντεςπρὸςτὸνλοιπὸνπρῶτοι same ratio as them. I say that two of A, B, C added toεἰσιν,οἱμὲνΑ,ΒπρὸςτὸνΓ,οἱδὲΒ,ΓπρὸςτὸνΑκαὶ
gether in any way are prime to the remaining (one), (that
ἔτιοἱΑ,ΓπρὸςτὸνΒ.
is) A and B (prime) to C, B and C to A, and, further, A
Εἰλήφθωσαν γὰρ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν and C to B.
λόγον ἐχόντων τοῖς Α, Β, Γ δύο οἱ ΔΕ, ΕΖ. φανερὸν Let the two least numbers, DE and EF , having the
δή,ὅτιὁμὲνΔΕἑαυτὸνπολλαπλασιάσαςτὸνΑπεποίηκεν, same ratio as A, B, C, have been taken [Prop. 8.2].
τὸνδὲΕΖπολλαπλασιάσαςτὸνΒπεποίηκεν,καὶἔτιὁΕΖ So it is clear that DE has made A (by) multiplying it-
ἑαυτὸνπολλαπλασιάσαςτὸνΓπεποίηκεν. καὶἐπεὶοἱΔΕ, self, and has made B (by) multiplying EF , and, fur-
ΕΖἐλάχιστοίεἰσιν,πρῶτοιπρὸςἀλλήλουςεἰσίν. ἐὰνδὲ ther, EF has made C (by) multiplying itself [Prop. 8.2].
δύοἀριθμοὶπρῶτοιπρὸςἀλλήλουςὦσιν,καὶσυναμφότερος And since DE, EF are the least (of those numbers havπρὸςἑκάτερονπρῶτόςἐστιν·καὶὁΔΖἄραπρὸςἑκάτερον
ing the same ratio as them), they are prime to one anτῶνΔΕ,ΕΖπρῶτόςἐστιν.
ἀλλὰμὴνκαὶὁΔΕπρὸςτὸν other [Prop. 7.22]. And if two numbers are prime to
ΕΖπρῶτόςἐστιν·οἱΔΖ,ΔΕἄραπρὸςτὸνΕΖπρῶτοίεἰσιν. one another then the sum (of them) is also prime to each
ἐὰνδὲδύοἀριθμοὶπρόςτιναἀριθμὸνπρῶτοιὦσιν,καὶὁ [Prop. 7.28]. Thus, DF is also prime to each of DE, EF .
ἐξαὐτῶνγενόμενοςπρὸςτὸνλοιπὸνπρῶτόςἐστιν·ὥστε But, in fact, DE is also prime to EF . Thus, DF , DE
ὁἐκτῶνΖΔ,ΔΕπρὸςτὸνΕΖπρῶτόςἐστιν·ὥστεκαὶὁ are (both) prime to EF . And if two numbers are (both)
ἐκτῶνΖΔ,ΔΕπρὸςτὸνἀπὸτοῦΕΖπρῶτόςἐστιν. [ἐὰν prime to some number then the (number) created from
γὰρδύοἀριθμοὶπρῶτοιπρὸςἀλλήλουςὦσιν,ὁἐκτοῦἑνὸς (multiplying) them is also prime to the remaining (numαὐτῶνγενόμενοςπρὸςτὸνλοιπὸνπρῶτόςἐστιν].
ἀλλ᾿ὁ ber) [Prop. 7.24]. Hence, the (number created) from
ἐκτῶνΖΔ,ΔΕὁἀπὸτοῦΔΕἐστιμετὰτοῦἐκτῶνΔΕ, (multiplying) FD, DE is prime to EF . Hence, the (num-
ΕΖ·ὁἄραἀπὸτοῦΔΕμετὰτοῦἐκτῶνΔΕ,ΕΖπρὸςτὸν ber created) from (multiplying) FD, DE is also prime
ἀπὸτοῦΕΖπρῶτόςἐστιν. καίἐστινὁμὲνἀπὸτοῦΔΕ to the (square) on EF [Prop. 7.25]. [For if two num-
ὁΑ,ὁδὲἐκτῶνΔΕ,ΕΖὁΒ,ὁδὲἀπὸτοῦΕΖὁΓ·οἱ bers are prime to one another then the (number) created
Α,ΒἄρασυντεθέντεςπρὸςτὸνΓπρῶτοίεἰσιν.ὁμοίωςδὴ from (squaring) one of them is prime to the remaining
δείξομεν,ὅτικαὶοἱΒ,ΓπρὸςτὸνΑπρῶτοίεἰσιν. λέγω (number).] But the (number created) from (multiplying)
δή,ὅτικαὶοἱΑ,ΓπρὸςτὸνΒπρῶτοίεἰσιν.ἐπεὶγὰρὁΔΖ FD, DE is the (square) on DE plus the (number creπρὸςἑκάτεροντῶνΔΕ,ΕΖπρῶτόςἐστιν,καὶὁἀπὸτοῦ
ated) from (multiplying) DE, EF [Prop. 2.3]. Thus, the
ΔΖπρὸςτὸνἐκτῶνΔΕ,ΕΖπρῶτόςἐστιν. ἀλλὰτῷἀπὸ (square) on DE plus the (number created) from (multiτοῦΔΖἴσοιεἰσὶνοἱἀπὸτῶνΔΕ,ΕΖμετὰτοῦδὶςἐκτῶν
plying) DE, EF is prime to the (square) on EF . And
ΔΕ,ΕΖ·καὶοἱἀπὸτῶνΔΕ,ΕΖἄραμετὰτοῦδὶςὑπὸτῶν the (square) on DE is A, and the (number created) from
ΔΕ,ΕΖπρὸςτὸνὑπὸτῶνΔΕ,ΕΖπρῶτοί[εἰσι].διελόντι (multiplying) DE, EF (is) B, and the (square) on EF
οἱἀπὸτῶνΔΕ,ΕΖμετὰτοῦἅπαξὑπὸΔΕ,ΕΖπρὸςτὸν (is) C. Thus, A, B summed is prime to C. So, similarly,
ὑπὸΔΕ,ΕΖπρῶτοίεἰσιν.ἔτιδιελόντιοἱἀπὸτῶνΔΕ,ΕΖ we can show that B, C (summed) is also prime to A. So
ἄραπρὸςτὸνὑπὸΔΕ,ΕΖπρῶτοίεἰσιν. καίἐστινὁμὲν I say that A, C (summed) is also prime to B. For since
B
C
F
266

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
ἀπὸτοῦΔΕὁΑ,ὁδὲὑπὸτῶνΔΕ,ΕΖὁΒ,ὁδὲἀπὸτοῦ DF is prime to each of DE, EF then the (square) on DF
ΕΖὁΓ.οἱΑ,ΓἄρασυντεθέντεςπρὸςτὸνΒπρῶτοίεἰσιν· is also prime to the (number created) from (multiplying)
ὅπερἔδειδεῖξαι.
DE, EF [Prop. 7.25]. But, the (sum of the squares) on
DE, EF plus twice the (number created) from (multiplying)
DE, EF is equal to the (square) on DF [Prop. 2.4].
And thus the (sum of the squares) on DE, EF plus twice
the (rectangle contained) by DE, EF [is] prime to the
(rectangle contained) by DE, EF . By separation, the
(sum of the squares) on DE, EF plus once the (rectangle
contained) by DE, EF is prime to the (rectangle
contained) by DE, EF . † Again, by separation, the (sum
of the squares) on DE, EF is prime to the (rectangle
contained) by DE, EF . And the (square) on DE is A,
and the (rectangle contained) by DE, EF (is) B, and
the (square) on EF (is) C. Thus, A, C summed is prime
to B. (Which is) the very thing it was required to show.
† Since if α β measures α 2 + β 2 + 2 α β then it also measures α 2 + β 2 + α β, and vice versa.
¦. Proposition 16
᾿Εὰνδύοἀριθμοὶπρῶτοιπρὸςἀλλήλουςὦσιν,οὐκἔσται If two numbers are prime to one another then as the
ὡςὁπρῶτοςπρὸςτὸνδεύτερον,οὕτωςὁδεύτεροςπρὸς first is to the second, so the second (will) not (be) to some
ἄλλοντινά.
other (number).
Α
Β
Γ
ΔύογὰρἀριθμοὶοἱΑ,Βπρῶτοιπρὸςἀλλήλουςἔστω- For let the two numbers A and B be prime to one
σαν·λέγω,ὅτιοὐκἔστινὡςὁΑπρὸςτὸνΒ,οὕτωςὁΒ another. I say that as A is to B, so B is not to some other
πρὸςἄλλοντινά.
(number).
Εἰγὰρδυνατόν,ἔστωὡςὁΑπρὸςτὸνΒ,ὁΒπρὸς For, if possible, let it be that as A (is) to B, (so)
τὸνΓ.οἱδὲΑ,Βπρῶτοι,οἱδὲπρῶτοικαὶἐλάχιστοι,οἱδὲ B (is) to C. And A and B (are) prime (to one an-
ἐλάχιστοιἀριθμοὶμετροῦσιτοὺςτὸναὐτὸνλόγονἔχοντας other). And (numbers) prime (to one another are) also
ἰσάκιςὅτεἡγούμενοςτὸνἡγούμενονκαὶὁἑπόμενοςτὸν the least (of those numbers having the same ratio as
ἑπόμενον·μετρεῖἄραὁΑτὸνΒὡςἡγούμενοςἡγούμενον. them) [Prop. 7.21]. And the least numbers measure
μετρεῖδὲκαὶἑαυτόν·ὁΑἄρατοὺςΑ,Βμετρεῖπρώτους those (numbers) having the same ratio (as them) an
ὄνταςπρὸςἀλλήλους·ὅπερἄτοπον.οὐκἄραἔσταιὡςὁΑ equal number of times, the leading (measuring) the leadπρὸςτὸνΒ,οὕτωςὁΒπρὸςτὸνΓ·ὅπερἔδειδεῖξαι.
ing, and the following the following [Prop. 7.20]. Thus,
A measures B, as the leading (measuring) the leading.
And (A) also measures itself. Thus, A measures A and B,
which are prime to one another. The very thing (is) absurd.
Thus, as A (is) to B, so B cannot be to C. (Which
is) the very thing it was required to show.
Þ. Proposition 17
᾿Εὰνὦσινὁσοιδηποτοῦνἀριθμοὶἑξῆςἀνάλογον,οἱδὲ If any multitude whatsoever of numbers is continu-
ἄκροιαὐτῶνπρῶτοιπρὸςἀλλήλουςὦσιν,οὐκἔσταιὡςὁ ously proportional, and the outermost of them are prime
πρῶτοςπρὸςτὸνδεύτερον,οὕτωςὁἔσχατοςπρὸςἄλλον to one another, then as the first (is) to the second, so the
A
B
C
267

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
τινά.
last will not be to some other (number).
῎ΕστωσανὁσοιδηποτοῦνἀριθμοὶἑξῆςἀνάλογονοἱΑ, Let A, B, C, D be any multitude whatsoever of con-
Β,Γ,Δ,οἱδὲἄκροιαὐτῶνοἱΑ,Δπρῶτοιπρὸςἀλλήλους tinuously proportional numbers. And let the outermost
ἔστωσαν·λέγω,ὅτιοὐκἔστινὡςὁΑπρὸςτὸνΒ,οὕτωςὁ of them, A and D, be prime to one another. I say that as
Δπρὸςἄλλοντινά.
A is to B, so D (is) not to some other (number).
Α
Β
Γ
∆
Ε
Εἰγὰρδυνατόν,ἔστωὡςὁΑπρὸςτὸνΒ,οὕτωςὁΔ For, if possible, let it be that as A (is) to B, so D
πρὸςτὸνΕ·ἐναλλὰξἄραἐστὶνὡςὁΑπρὸςτὸνΔ,ὁΒπρὸς (is) to E. Thus, alternately, as A is to D, (so) B (is)
τὸνΕ.οἱδὲΑ,Δπρῶτοι,οἱδὲπρῶτοικαὶἐλάχιστοι,οἱδὲ to E [Prop. 7.13]. And A and D are prime (to one
ἐλάχιστοιἀριθμοὶμετροῦσιτοὺςτὸναὐτὸνλόγονἔχοντας another). And (numbers) prime (to one another are)
ἰσάκιςὅτεἡγούμενοςτὸνἡγούμενονκαὶὁἑπόμενοςτὸν also the least (of those numbers having the same ra-
ἑπόμενον. μετρεῖἄραὁΑτὸνΒ.καίἐστινὡςὁΑπρὸς tio as them) [Prop. 7.21]. And the least numbers meaτὸνΒ,ὁΒπρὸςτὸνΓ.καὶὁΒἄρατὸνΓμετρεῖ΄·ὥστε
sure those (numbers) having the same ratio (as them) an
καὶὁΑτὸνΓμετρεῖ. καὶἐπείἐστινὡςὁΒπρὸςτὸνΓ, equal number of times, the leading (measuring) the lead-
ὁΓπρὸςτὸνΔ,μετρεῖδὲὁΒτὸνΓ,μετρεῖἄρακαὶὁΓ ing, and the following the following [Prop. 7.20]. Thus,
τὸνΔ.ἀλλ᾿ὁΑτὸνΓἐμέτρει·ὥστεὁΑκαὶτὸνΔμετρεῖ. A measures B. And as A is to B, (so) B (is) to C. Thus, B
μετρεῖδὲκαὶἑαυτόν. ὁΑἄρατοὺςΑ,Δμετρεῖπρώτους also measures C. And hence A measures C [Def. 7.20].
ὄνταςπρὸςἀλλήλους·ὅπερἐστὶνἀδύνατον.οὐκἄραἔσται And since as B is to C, (so) C (is) to D, and B mea-
ὡςὁΑπρὸςτὸνΒ,οὕτωςὁΔπρὸςἄλλοντινά·ὅπερἔδει sures C, C thus also measures D [Def. 7.20]. But, A was
δεῖξαι.
(found to be) measuring C. And hence A also measures
D. And (A) also measures itself. Thus, A measures A
and D, which are prime to one another. The very thing is
impossible. Thus, as A (is) to B, so D cannot be to some
other (number). (Which is) the very thing it was required
to show.
. Proposition 18
Δύοἀριθμῶνδοθέντωνἐπισκέψασθαι,εἰδυνατόνἐστιν For two given numbers, to investigate whether it is
αὐτοῖςτρίτονἀνάλογονπροσευρεῖν.
possible to find a third (number) proportional to them.
Α
Β
Γ
∆
῎Εστωσαν οἱ δοθέντες δύο ἀριθμοὶ οἱ Α, Β, καὶ Let A and B be the two given numbers. And let it
δέον ἔστωἐπισκέψασθαι, εἰδυνατόνἐστιναὐτοῖςτρίτον be required to investigate whether it is possible to find a
ἀνάλογονπροσευρεῖν.
third (number) proportional to them.
ΟἱδὴΑ,Βἤτοιπρῶτοιπρὸςἀλλήλουςεἰσὶνἢοὔ.καὶεἰ So A and B are either prime to one another, or not.
πρῶτοιπρὸςἀλλήλουςεἰσίν,δέδεικται,ὅτιἀδύνατόνἐστιν And if they are prime to one another then it has (already)
αὐτοῖςτρίτονἀνάλογονπροσευρεῖν.
been show that it is impossible to find a third (number)
ἈλλὰδὴμὴἔστωσανοἱΑ,Βπρῶτοιπρὸςἀλλήλους, proportional to them [Prop. 9.16].
καὶὁΒἑαυτονπολλαπλασιάσαςτὸνΓποιείτω.ὁΑδὴτὸν And so let A and B not be prime to one another. And
Γἤτοιμετρεῖἢοὐμετρεῖ.μετρείτωπρότερονκατὰτὸνΔ· let B make C (by) multiplying itself. So A either mea-
ὁΑἄρατὸνΔπολλαπλασιάσαςτὸνΓπεποίηκεν.ἀλλαμὴν sures, or does not measure, C. Let it first of all measure
καὶὁΒἑαυτὸνπολλαπλασιάσαςτὸνΓπεποίηκεν·ὁἄρα (C) according to D. Thus, A has made C (by) multiply-
A
B
A
B
C
D
E
C
D
268

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
ἐκτῶνΑ,ΔἴσοςἐστὶτῷἀπὸτοῦΒ.ἔστινἄραὡςὁΑ ing D. But, in fact, B has also made C (by) multiplying
πρὸςτὸνΒ,ὁΒπρὸςτὸνΔ·τοῖςΑ,Βἄρατρίτοςἀριθμὸς itself. Thus, the (number created) from (multiplying) A,
ἀνάλογονπροσηύρηταιὁΔ.
D is equal to the (square) on B. Thus, as A is to B, (so)
ἈλλὰδὴμὴμετρείτωὁΑτὸνΓ·λέγω,ὅτιτοῖςΑ,Β B (is) to D [Prop. 7.19]. Thus, a third number has been
ἀδύνατόνἐστιτρίτονἀνάλογονπροσευρεῖνἀριθμόν.εἰγὰρ found proportional to A, B, (namely) D.
δυνατόν,προσηυρήσθωὁΔ.ὁἄραἐκτῶνΑ,Δἴσοςἐστὶ And so let A not measure C. I say that it is impossiτῷἀπὸτοῦΒ.ὁδὲἀπὸτοῦΒἐστινὁΓ·ὁἄραἐκτῶνΑ,
ble to find a third number proportional to A, B. For, if
ΔἴσοςἐστὶτῷΓ.ὥστεὁΑτὸνΔπολλαπλασιάσαςτὸν possible, let it have been found, (and let it be) D. Thus,
Γπεποίηκεν·ὁΑἄρατὸνΓμετρεῖκατὰτὸνΔ.ἀλλαμὴν the (number created) from (multiplying) A, D is equal to
ὑπόκειταικαὶμὴμετρῶν·ὅπερἄτοπον. οὐκἄραδυνατόν the (square) on B [Prop. 7.19]. And the (square) on B
ἐστιτοῖςΑ,Βτρίτονἀνάλογονπροσευρεῖνἀριθμὸν,ὅταν is C. Thus, the (number created) from (multiplying) A,
ὁΑτὸνΓμὴμετρῇ·ὅπερἔδειδεῖξαι.
D is equal to C. Hence, A has made C (by) multiplying
D. Thus, A measures C according to D. But (A) was, in
fact, also assumed (to be) not measuring (C). The very
thing (is) absurd. Thus, it is not possible to find a third
number proportional to A, B when A does not measure
C. (Which is) the very thing it was required to show.
. Proposition 19 †
Τριῶνἀριθμῶνδοθέντωνἐπισκέψασθαι,πότεδυνατόν For three given numbers, to investigate when it is pos-
ἐστιναὐτοῖςτέταρτονἀνάλογονπροσευρεῖν.
sible to find a fourth (number) proportional to them.
Α
Β
Γ
∆
Ε
῎ΕστωσανοἱδοθέντεςτρεῖςἀριθμοὶοἱΑ,Β,Γ,καὶδέον Let A, B, C be the three given numbers. And let it be
ἔστω επισκέψασθαι, πότεδυνατόνἐστιν αὐτοῖςτέταρτον required to investigate when it is possible to find a fourth
ἀνάλογονπροσευρεῖν.
(number) proportional to them.
῎Ητοιοὖνοὔκεἰσινἑξῆςἀνάλογον,καὶοἱἄκροιαὐτῶν In fact, (A, B, C) are either not continuously proπρῶτοιπρὸςἀλλήλουςεἰσίν,ἢἑξῆςεἰσινἀνάλογον,καὶοἱ
portional and the outermost of them are prime to one
ἄκροιαὐτῶνοὔκεἰσιπρῶτοιπρὸςἀλλήλους,ἢοὕτεἑξῆς another, or are continuously proportional and the outerεἰσινἀνάλογον,οὔτεοἱἄκροιαὐτῶνπρῶτοιπρὸςἀλλήλους
most of them are not prime to one another, or are neither
εἰσίν,ἢκαὶἑξῆςεἰσινἀνάλογον,καὶοἱἄκροιαὐτῶνπρῶτοι continuously proportional nor are the outermost of them
πρὸςἀλλήλουςεἰσίν.
prime to one another, or are continuously proportional
Εἰ μὲν οὖν οἱ Α, Β, Γ ἑξῆς εἰσιν ἀνάλογον, καὶ οἱ and the outermost of them are prime to one another.
ἄκροιαὐτῶνοἱΑ,Γπρῶτοιπρὸςἀλλήλουςεἰσίν,δέδεικται, In fact, if A, B, C are continuously proportional, and
ὅτιἀδύνατόνἐστιναὐτοῖςτέταρτονἀνάλογονπροσευρεῖν the outermost of them, A and C, are prime to one an-
ἀριθμόν. μὴἔστωσανδὴοἱΑ,Β,Γἑξῆςἀνάλογοντῶν other, (then) it has (already) been shown that it is im-
ἀκρῶν πάλιν ὄντων πρώτων πρὸς ἀλλήλους. λέγω, ὅτι possible to find a fourth number proportional to them
καὶοὕτωςἀδύνατόνἐστιναὐτοῖςτέταρτονἀνάλογονπρο- [Prop. 9.17]. So let A, B, C not be continuously proporσευρεῖν.εἰγὰρδυνατόν,προσευρήσθωὁΔ,ὥστεεἶναιὡς
tional, (with) the outermost of them again being prime to
τὸνΑπρὸςτὸνΒ,τὸνΓπρὸςτὸνΔ,καὶγεγονέτωὡςὁΒ one another. I say that, in this case, it is also impossible
πρὸςτὸνΓ,ὁΔπρὸςτὸνΕ.καὶἐπείἐστινὡςμὲνὁΑπρὸς to find a fourth (number) proportional to them. For, if
τὸνΒ,ὁΓπρὸςτὸνΔ,ὡςδὲὁΒπρὸςτὸνΓ,ὁΔπρὸς possible, let it have been found, (and let it be) D. Hence,
τὸνΕ,δι᾿ἴσουἄραὡςὁΑπρὸςτὸνΓ,ὁΓπρὸςτὸνΕ.οἱ it will be that as A (is) to B, (so) C (is) to D. And let it be
δὲΑ,Γπρῶτοι,οἱδὲπρῶτοικαὶἐλάχιστοι,οἱδὲἐλάχιστοι contrived that as B (is) to C, (so) D (is) to E. And since
A
B
C
D
E
269

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
μετροῦσιτοὺςτὸναὐτὸνλόγονἔχονταςὅτεἡγούμενος as A is to B, (so) C (is) to D, and as B (is) to C, (so) D
τὸνἡγούμενονκαὶὁἑπόμενοςτὸνἑπόμενον.μετρεῖἄραὁ (is) to E, thus, via equality, as A (is) to C, (so) C (is) to E
ΑτὸνΓὡςἡγούμενοςἡγούμενον. μετρεῖδὲκαὶἑαυτόν· [Prop. 7.14]. And A and C (are) prime (to one another).
ὁΑἄρατοὺςΑ,Γμετρεῖπρώτουςὄνταςπρὸςἀλλήλους· And (numbers) prime (to one another are) also the least
ὅπερἐστὶνἀδύνατον. οὐκἄρατοῖςΑ,Β,Γδυνατόνἐστι (numbers having the same ratio as them) [Prop. 7.21].
τέταρτονἀνάλογονπροσευρεῖν.
And the least (numbers) measure those numbers having
ἈλλάδὴπάλινἔστωσανοἱΑ,Β,Γἑξῆςἀνάλογον,οἱδὲ the same ratio as them (the same number of times), the
Α,Γμὴἔστωσανπρῶτοιπρὸςἀλλήλους.λέγω,ὅτιδυνατόν leading (measuring) the leading, and the following the
ἐστιναὐτοῖςτέταρτονἀνάλογονπροσευρεῖν.ὁγὰρΒτὸνΓ following [Prop. 7.20]. Thus, A measures C, (as) the
πολλαπλασιάσαςτὸνΔποιείτω·ὁΑἄρατὸνΔἤτοιμετρεῖ leading (measuring) the leading. And it also measures
ἢοὐμετρεῖ.μετρείτωαὐτὸνπρότερονκατὰτὸνΕ·ὁΑἄρα itself. Thus, A measures A and C, which are prime to
τὸνΕπολλαπλασιάσαςτὸνΔπεποίηκεν. ἀλλὰμὴνκαὶὁ one another. The very thing is impossible. Thus, it is not
ΒτὸνΓπολλαπλασιάσαςτὸνΔπεποίηκεν·ὁἄραἐκτῶν possible to find a fourth (number) proportional to A, B,
Α,ΕἴσοςἐστὶτῷἐκτῶνΒ,Γ.ἀνάλογονἄρα[ἐστὶν]ὡςὁ C.
ΑπρὸςτὸνΒ,ὁΓπρὸςτὸνΕ·τοὶςΑ,Β,Γἄρατέταρτος And so let A, B, C again be continuously propor-
ἀνάλογονπροσηύρηταιὁΕ.
tional, and let A and C not be prime to one another. I
ἈλλὰδὴμὴμετρείτωὁΑτὸνΔ·λέγω,ὅτιἀδύνατόν say that it is possible to find a fourth (number) propor-
ἐστιτοῖςΑ,Β,Γτέταρτονἀνάλογονπροσευρεῖνἀριθμόν. tional to them. For let B make D (by) multiplying C.
εἰγὰρδυνατόν,προσευρήσθωὁΕ·ὁἄραἐκτῶνΑ,Εἴσος Thus, A either measures or does not measure D. Let it,
ἐστὶτῷἐκτῶνΒ,Γ.ἀλλὰὁἑκτῶνΒ,ΓἐστινὁΔ·καὶ first of all, measure (D) according to E. Thus, A has
ὁἐκτῶνΑ,ΕἄραἴσοςἐστὶτῷΔ.ὁΑἄρατὸνΕπολλα- made D (by) multiplying E. But, in fact, B has also made
πλασιάσαςτὸνΔπεποίηκεν·ὁΑἄρατὸνΔμετρεῖκατὰτὸν D (by) multiplying C. Thus, the (number created) from
Ε·ὥστεμετρεῖὁΑτὸνΔ.ἀλλὰκαὶοὐμετρεῖ·ὅπερἄτοπον. (multiplying) A, E is equal to the (number created) from
οὐκἄραδυνάτονἐστιτοῖςΑ,Β,Γτέταρτονἀνάλογονπρο- (multiplying) B, C. Thus, proportionally, as A [is] to B,
σευρεῖνἀριθμόν,ὅτανὁΑτὸνΔμὴμετρῇ.ἀλλὰδὴοἱΑ,Β, (so) C (is) to E [Prop. 7.19]. Thus, a fourth (number)
Γμήτεἑξῆςἔστωσανἀνάλογονμήτεοἱἄκροιπρῶτοιπρὸς proportional to A, B, C has been found, (namely) E.
ἀλλήλους. καὶὁΒτὸνΓπολλαπλασιάσαςτὸνΔποιείτω. And so let A not measure D. I say that it is impossible
ὁμοίωςδὴδειχθήσεται,ὅτιεἰμὲνμετρεῖὁΑτὸνΔ,δυ- to find a fourth number proportional to A, B, C. For, if
νατόνἐστιναὐτοῖςἀνάλογονπροσευρεῖν,εἰδὲοὐμετρεῖ, possible, let it have been found, (and let it be) E. Thus,
ἀδύνατον·ὅπερἔδειδεῖξαι.
the (number created) from (multiplying) A, E is equal to
the (number created) from (multiplying) B, C. But, the
(number created) from (multiplying) B, C is D. And thus
the (number created) from (multiplying) A, E is equal to
D. Thus, A has made D (by) multiplying E. Thus, A
measures D according to E. Hence, A measures D. But,
it also does not measure (D). The very thing (is) absurd.
Thus, it is not possible to find a fourth number proportional
to A, B, C when A does not measure D. And so
(let) A, B, C (be) neither continuously proportional, nor
(let) the outermost of them (be) prime to one another.
And let B make D (by) multiplying C. So, similarly, it
can be show that if A measures D then it is possible to
find a fourth (number) proportional to (A, B, C), and
impossible if (A) does not measure (D). (Which is) the
very thing it was required to show.
† The proof of this proposition is incorrect. There are, in fact, only two cases. Either A, B, C are continuously proportional, with A and C prime
to one another, or not. In the first case, it is impossible to find a fourth proportional number. In the second case, it is possible to find a fourth
proportional number provided that A measures B times C. Of the four cases considered by Euclid, the proof given in the second case is incorrect,
since it only demonstrates that if A : B :: C : D then a number E cannot be found such that B : C :: D : E. The proofs given in the other three
270

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
cases are correct.
.
Proposition 20
Οἱπρῶτοιἀριθμοὶπλείουςεἰσὶπαντὸςτοῦπροτεθέντος The (set of all) prime numbers is more numerous than
πλήθουςπρώτωνἀριθμῶν.
any assigned multitude of prime numbers.
Α
Β
Γ
Ε
Η
∆ Ζ
῎ΕστωσανοἱπροτεθέντεςπρῶτοιἀριθμοὶοἱΑ,Β,Γ· Let A, B, C be the assigned prime numbers. I say that
λέγω,ὅτιτῶνΑ,Β,Γπλείουςεἰσὶπρῶτοιἀριθμοί. the (set of all) primes numbers is more numerous than A,
ΕἰλήφθωγὰρὁὑπὸτῶνΑ,Β,Γἐλάχιστοςμετρούμενος B, C.
καὶἔστωΔΕ,καὶπροσκείσθωτῷΔΕμονὰςἡΔΖ.ὁδὴΕΖ For let the least number measured by A, B, C have
ἤτοιπρῶτόςἐστινἢοὔ.ἔστωπρότερονπρῶτος·εὐρημένοι been taken, and let it be DE [Prop. 7.36]. And let the
ἄραεἰσὶπρῶτοιἀριθμοὶοἱΑ,Β,Γ,ΕΖπλείουςτῶνΑ,Β, unit DF have been added to DE. So EF is either prime,
Γ. or not. Let it, first of all, be prime. Thus, the (set of)
ἈλλὰδὴμὴἔστωὁΕΖπρῶτος·ὑπὸπρώτουἄρατινὸς prime numbers A, B, C, EF , (which is) more numerous
ἀριθμοῦμετρεῖται. μετρείσθωὑπὸπρώτουτοῦΗ·λέγω, than A, B, C, has been found.
ὅτιὁΗοὐδενὶτῶνΑ,Β,Γἐστινὁαὐτός.εἰγὰρδυνατόν, And so let EF not be prime. Thus, it is measured by
ἔστω. οἱδὲΑ,Β,ΓτὸνΔΕμετροῦσιν·καὶὁΗἄρατὸν some prime number [Prop. 7.31]. Let it be measured by
ΔΕμετρήσει. μετρεῖδὲκαὶτὸνΕΖ·καὶλοιπὴντὴνΔΖ the prime (number) G. I say that G is not the same as
μονάδαμετρήσειὁΗἀριθμὸςὤν·ὄπερἄτοπον.οὐκἄραὁ any of A, B, C. For, if possible, let it be (the same). And
ΗἑνὶτῶνΑ,Β,Γἐστινὁαὐτός. καὶὑπόκειταιπρῶτος. A, B, C (all) measure DE. Thus, G will also measure
εὑρημένοιἄραεἰσὶπρῶτοιἀριθμοὶπλείουςτοῦπροτεθέντος DE. And it also measures EF . (So) G will also meaπλήθουςτῶνΑ,Β,ΓοἱΑ,Β,Γ,Η·ὅπερἔδειδεῖξαι.
sure the remainder, unit DF , (despite) being a number
[Prop. 7.28]. The very thing (is) absurd. Thus, G is not
the same as one of A, B, C. And it was assumed (to be)
prime. Thus, the (set of) prime numbers A, B, C, G,
(which is) more numerous than the assigned multitude
(of prime numbers), A, B, C, has been found. (Which is)
the very thing it was required to show.
. Proposition 21
᾿Εὰν ἄρτιοι ἀριθμοὶ ὁποσοιοῦν συντεθῶσιν, ὁ ὅλος If any multitude whatsoever of even numbers is added
ἄρτιόςἐστιν.
together then the whole is even.
Α Β Γ ∆ Ε A B C D E
Συγκείσθωσαν γὰρ ἄρτιοιἀριθμοὶὁποσοιοῦνοἱ ΑΒ, For let any multitude whatsoever of even numbers,
ΒΓ,ΓΔ,ΔΕ·λέγω,ὅτιὅλοςὁΑΕἄρτιόςἐστιν. AB, BC, CD, DE, lie together. I say that the whole,
᾿ΕπεὶγὰρἕκαστοςτῶνΑΒ,ΒΓ,ΓΔ,ΔΕἄρτιόςἐστιν, AE, is even.
ἔχειμέροςἥμισυ·ὥστεκαὶὅλοςὁΑΕἔχειμέροςἥμισυ. For since everyone of AB, BC, CD, DE is even, it
ἄρτιοςδὲἀριθμόςἐστινὁδίχαδιαιρούμενος· ἄρτιοςἄρα has a half part [Def. 7.6]. And hence the whole AE has
ἐστὶνὁΑΕ·ὅπερἔδειδεῖξαι.
a half part. And an even number is one (which can be)
divided in half [Def. 7.6]. Thus, AE is even. (Which is)
A
B
C
E
G
D
F
271

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
the very thing it was required to show.
. Proposition 22
᾿Εὰν περισσοὶ ἀριθμοὶ ὁποσοιοῦν συντεθῶσιν, τὸ δὲ If any multitude whatsoever of odd numbers is added
πλῆθοςαὐτῶνἄρτιονᾖ,ὁὅλοςἄρτιοςἔσται.
together, and the multitude of them is even, then the
whole will be even.
Α Β Γ ∆ Ε A B C D E
Συγκείσθωσαν γὰρ περισσοὶ ἀριθμοὶ ὁσοιδηποτοῦν For let any even multitude whatsoever of odd num-
ἄρτιοι τὸ πλῆθοςοἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ· λέγω, ὅτι ὅλος bers, AB, BC, CD, DE, lie together. I say that the
ὁΑΕἄρτιόςἐστιν.
whole, AE, is even.
᾿ΕπεὶγὰρἕκαστοςτῶνΑΒ,ΒΓ,ΓΔ,ΔΕπεριττόςἐστιν, For since everyone of AB, BC, CD, DE is odd then, a
ἀφαιρεθείσης μονάδος ἀφ᾿ ἑκάστου ἕκαστος τῶν λοιπῶν unit being subtracted from each, everyone of the remain-
ἄρτιος ἔσται· ὥστε καὶ ὁ συγκείμενος ἐξ αὐτῶν ἄρτιος ders will be (made) even [Def. 7.7]. And hence the sum
ἔσται. ἔστιδὲκαὶτὸπλῆθοςτῶνμονάδωνἄρτιον. καὶ of them will be even [Prop. 9.21]. And the multitude
ὅλοςἄραὁΑΕἄρτιόςἐστιν·ὅπερἔδειδεῖξαι.
of the units is even. Thus, the whole AE is also even
[Prop. 9.21]. (Which is) the very thing it was required to
show.
. Proposition 23
᾿Εὰν περισσοὶ ἀριθμοὶ ὁποσοιοῦν συντεθῶσιν, τὸ δὲ If any multitude whatsoever of odd numbers is added
πλῆθοςαὐτῶνπερισσὸνᾖ,καὶὁὅλοςπερισσὸςἔσται. together, and the multitude of them is odd, then the
whole will also be odd.
Α Β Γ Ε ∆ A B C E D
Συγκείσθωσανγὰρὁποσοιοῦνπερισσοὶἀριθμοί,ὧντὸ For let any multitude whatsoever of odd numbers,
πλῆθοςπερισσὸνἔστω,οἱΑΒ,ΒΓ,ΓΔ·λέγω,ὅτικαὶὅλος AB, BC, CD, lie together, and let the multitude of them
ὁΑΔπερισσόςἐστιν.
be odd. I say that the whole, AD, is also odd.
ἈφῃρήσθωἀπὸτοῦΓΔμονὰςἡΔΕ·λοιπὸςἄραὁΓΕ For let the unit DE have been subtracted from CD.
ἄρτιόςἐστιν. ἔστιδὲκαὶὁΓΑἄρτιος·καὶὅλοςἄραὁΑΕ The remainder CE is thus even [Def. 7.7]. And CA
ἄρτιόςἐστιν. καίἐστιμονὰςἡΔΕ.περισσὸςἄραἐστὶνὁ is also even [Prop. 9.22]. Thus, the whole AE is also
ΑΔ·ὅπερἔδειδεῖξαι.
even [Prop. 9.21]. And DE is a unit. Thus, AD is odd
[Def. 7.7]. (Which is) the very thing it was required to
show.
. Proposition 24
᾿Εὰν ἀπὸ ἀρτίου ἀριθμοῦ ἄρτιος ἀφαιρεθῇ, ὁ λοιπὸς If an even (number) is subtracted from an(other) even
ἄρτιοςἔσται.
number then the remainder will be even.
Α Γ Β A C B
ἈπὸγὰρἀρτίουτοῦΑΒἄρτιοςἀφῃρήσθωὁΒΓ·λέγω, For let the even (number) BC have been subtracted
ὅτιὁλοιπὸςὁΓΑἄρτιόςἐστιν.
from the even number AB. I say that the remainder CA
᾿ΕπεὶγὰρὁΑΒἄρτιόςἐστιν,ἔχειμέροςἥμισυ. διὰτὰ is even.
αὐτὰδὴκαὶὁΒΓἔχειμέροςἥμισυ·ὥστεκαὶλοιπὸς[ὁΓΑ For since AB is even, it has a half part [Def. 7.6]. So,
ἔχειμέροςἥμισυ]ἄρτιος[ἄρα]ἐστὶνὁΑΓ·ὅπερἔδειδεῖξαι. for the same (reasons), BC also has a half part. And
hence the remainder [CA has a half part]. [Thus,] AC is
even. (Which is) the very thing it was required to show.
272

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
. Proposition 25
᾿Εὰνἀπὸἀρτίουἀριθμοῦπερισσὸςἀφαιρεθῇ,ὁλοιπὸς If an odd (number) is subtracted from an even numπερισσὸςἔσται.
ber then the remainder will be odd.
Α
Γ
∆
Β
A C D
Ἀπὸ γὰρἀρτίουτοῦΑΒ περισσὸςἀφῃρήσθω ὁ ΒΓ· For let the odd (number) BC have been subtracted
λέγω,ὅτιὁλοιπὸςὁΓΑπερισσόςἐστιν.
from the even number AB. I say that the remainder CA
ἈφῃρήσθωγὰρἀπὸτοῦΒΓμονὰςἡΓΔ·ὁΔΒἄρα is odd.
ἄρτιόςἐστιν. ἔστιδὲκαὶὁΑΒἄρτιος·καὶλοιπὸςἄραὁ For let the unit CD have been subtracted from BC.
ΑΔἄρτιόςἐστιν.καίἐστιμονὰςἡΓΔ·ὁΓΑἄραπερισσός DB is thus even [Def. 7.7]. And AB is also even. And
ἐστιν·ὅπερἔδειδεῖξαι.
thus the remainder AD is even [Prop. 9.24]. And CD is
a unit. Thus, CA is odd [Def. 7.7]. (Which is) the very
thing it was required to show.
¦. Proposition 26
᾿Εὰνἀπὸπερισσοῦἀριθμοῦπερισσὸςἀφαιρεθῇ,ὁλοιπὸς If an odd (number) is subtracted from an odd number
ἄρτιοςἔσται.
then the remainder will be even.
Α Γ ∆ Β A C
D B
ἈπὸγὰρπερισσοῦτοῦΑΒπερισσὸςἀφῃρήσθωὁΒΓ· For let the odd (number) BC have been subtracted
λέγω,ὅτιὁλοιπὸςὁΓΑἄρτιόςἐστιν.
from the odd (number) AB. I say that the remainder CA
᾿ΕπεὶγὰρὁΑΒπερισσόςἐστιν,ἀφῃρήσθωμονὰςἡΒΔ· is even.
λοιπὸςἄραὁΑΔἄρτιόςἐστιν. διὰτὰαὐτὰδὴκαὶὁΓΔ For since AB is odd, let the unit BD have been
ἄρτιόςἐστιν·ὥστεκαὶλοιπὸςὁΓΑἄρτιόςἐστιν·ὅπερἔδει subtracted (from it). Thus, the remainder AD is even
δεῖξαι. [Def. 7.7]. So, for the same (reasons), CD is also
even. And hence the remainder CA is even [Prop. 9.24].
(Which is) the very thing it was required to show.
Þ. Proposition 27
᾿Εὰνἀπὸπερισσοῦἀριθμοῦἄρτιοςἀφαιρεθῇ,ὁλοιπὸς If an even (number) is subtracted from an odd numπερισσὸςἔσται.
ber then the remainder will be odd.
Α ∆ Γ Β A D
C
Ἀπὸ γὰρπερισσοῦτοῦΑΒ ἄρτιοςἀφῃρήσθω ὁ ΒΓ· For let the even (number) BC have been subtracted
λέγω,ὅτιὁλοιπὸςὁΓΑπερισσόςἐστιν.
from the odd (number) AB. I say that the remainder CA
Ἀφῃρήσθω[γὰρ]μονὰςἡΑΔ·ὁΔΒἄραἄρτιόςἐστιν. is odd.
ἔστιδὲκαὶὁΒΓἄρτιος·καὶλοιπὸςἄραὁΓΔἄρτιόςἐστιν. [For] let the unit AD have been subtracted (from AB).
περισσὸςἄραὁΓΑ·ὅπερἔδειδεῖξαι.
DB is thus even [Def. 7.7]. And BC is also even. Thus,
the remainder CD is also even [Prop. 9.24]. CA (is) thus
odd [Def. 7.7]. (Which is) the very thing it was required
to show.
. Proposition 28
᾿Εὰν περισσὸς ἀριθμὸς ἄρτιον πολλαπλασιάσας ποιῇ If an odd number makes some (number by) multiplyτινα,ὁγενόμενοςἄρτιοςἔσται.
ing an even (number) then the created (number) will be
even.
B
B
273

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
Α
Β
Γ
Περισσὸς γὰρ ἀριθμὸς ὁ Α ἄρτιον τὸν Β πολλα- For let the odd number A make C (by) multiplying
πλασιάσαςτὸνΓποιείτω·λέγω,ὅτιὁΓἄρτιόςἐστιν. the even (number) B. I say that C is even.
᾿ΕπεὶγὰρὁΑτὸνΒπολλαπλασιάσαςτὸνΓπεποίηκεν, For since A has made C (by) multiplying B, C is thus
ὁΓἄρασύγκειταιἐκτοσούτωνἴσωντῷΒ,ὅσαιεἰσὶνἐν composed out of so many (magnitudes) equal to B, as
τῷΑμονάδες. καίἐστινὁΒἄρτιος·ὁΓἄρασύγκειται many as (there) are units in A [Def. 7.15]. And B is
ἐξἀρτίων.ἐὰνδὲἄρτιοιἀριθμοὶὁποσοιοῦνσυντεθῶσιν,ὁ even. Thus, C is composed out of even (numbers). And
ὅλοςἄρτιόςἐστιν.ἄρτιοςἄραἐστὶνὁΓ·ὅπερἔδειδεῖξαι. if any multitude whatsoever of even numbers is added
together then the whole is even [Prop. 9.21]. Thus, C is
even. (Which is) the very thing it was required to show.
. Proposition 29
᾿Εὰνπερισσὸςἀριθμὸςπερισσὸνἀριθμὸνπολλαπλασιάς- If an odd number makes some (number by) multiplyαςποιῇτινα,ὁγενόμενοςπερισσὸςἔσται.
ing an odd (number) then the created (number) will be
odd.
Α
Β
Γ
Περισσὸς γὰρ ἀριθμὸς ὁ Α περισσὸν τὸν Β πολλα- For let the odd number A make C (by) multiplying
πλασιάσαςτὸνΓποιείτω·λέγω,ὅτιὁΓπερισσόςἐστιν. the odd (number) B. I say that C is odd.
᾿ΕπεὶγὰρὁΑτὸνΒπολλαπλασιάσαςτὸνΓπεποίηκεν, For since A has made C (by) multiplying B, C is thus
ὁΓἄρασύγκειταιἐκτοσούτωνἴσωντῷΒ,ὅσαιεἰσὶνἐντῷ composed out of so many (magnitudes) equal to B, as
Αμονάδες.καίἐστινἑκάτεροςτῶνΑ,Βπερισσός·ὁΓἄρα many as (there) are units in A [Def. 7.15]. And each
σύγκειταιἐκπερισσῶνἀριθμῶν,ὧντὸπλῆθοςπερισσόν of A, B is odd. Thus, C is composed out of odd (num-
ἐστιν.ὥστεὁΓπερισσόςἐστιν·ὅπερἔδειδεῖξαι. bers), (and) the multitude of them is odd. Hence C is odd
[Prop. 9.23]. (Which is) the very thing it was required to
show.
Ð. Proposition 30
᾿Εὰνπερισσὸςἀριθμὸςἄρτιονἀριθμὸνμετρῇ,καὶτὸν If an odd number measures an even number then it
ἥμισυναὐτοῦμετρήσει.
will also measure (one) half of it.
Α
Β
Γ
ΠερισσὸςγὰρἀριθμὸςὁΑἄρτιοντὸνΒμετρείτω·λέγω, For let the odd number A measure the even (number)
ὅτικαὶτὸνἥμισυναὐτοῦμετρήσει.
B. I say that (A) will also measure (one) half of (B).
᾿ΕπεὶγὰρὁΑτὸνΒμετρεῖ,μετρείτωαὐτὸνκατὰτὸν For since A measures B, let it measure it according to
Γ·λέγω,ὅτιὁΓοὐκἔστιπερισσός.εἰγὰρδυνατόν,ἔστω. C. I say that C is not odd. For, if possible, let it be (odd).
καὶἐπεὶὁΑτὸνΒμετρεῖκατὰτὸνΓ,ὁΑἄρατὸνΓ And since A measures B according to C, A has thus made
πολλαπλασιάσαςτὸνΒπεποίηκεν. ὁΒἄρασύγκειταιἐκ B (by) multiplying C. Thus, B is composed out of odd
περισσῶνἀριθμῶν,ὧντὸπλῆθοςπερισσόνἐστιν.ὁΒἄρα numbers, (and) the multitude of them is odd. B is thus
A
B
C
A
B
C
A
B
C
274

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
περισσόςἐστιν· ὅπερἄτοπον·ὑπόκειταιγὰρἄρτιος. οὐκ odd [Prop. 9.23]. The very thing (is) absurd. For (B)
ἄραὁΓπερισσόςἐστιν·ἄρτιοςἄραἐστὶνὁΓ.ὥστεὁΑ was assumed (to be) even. Thus, C is not odd. Thus, C
τὸνΒμετρεῖἀρτιάκις. διὰδὴτοῦτοκαὶτὸνἥμισυναὐτοῦ is even. Hence, A measures B an even number of times.
μετρήσει·ὅπερἔδειδεῖξαι.
So, on account of this, (A) will also measure (one) half
of (B). (Which is) the very thing it was required to show.
Ð. Proposition 31
᾿Εὰνπερισσὸςἀριθμὸςπρόςτιναἀριθμὸνπρῶτοςᾖ,καὶ If an odd number is prime to some number then it will
πρὸςτὸνδιπλασίονααὐτοῦπρῶτοςἔσται.
also be prime to its double.
Α
Β
Γ
∆
ΠερισσὸςγὰρἀριθμὸςὁΑπρόςτιναἀριθμὸντὸνΒ For let the odd number A be prime to some number
πρῶτοςἔστω,τοῦδὲΒδιπλασίωνἔστωὁΓ·λέγω,ὅτιὁΑ B. And let C be double B. I say that A is [also] prime to
[καὶ]πρὸςτὸνΓπρῶτόςἐστιν. C.
Εἰγὰρμήεἰσιν[οἱΑ,Γ]πρῶτοι,μετρήσειτιςαὐτοὺς For if [A and C] are not prime (to one another) then
ἀριθμός.μετρείτω,καὶἔστωὁΔ.καίἐστινὁΑπερισσός· some number will measure them. Let it measure (them),
περισσὸςἄρακαὶὁΔ.καὶἐπεὶὁΔπερισσὸςὢντὸνΓ and let it be D. And A is odd. Thus, D (is) also odd.
μετρεῖ, καίἐστινὁΓἄρτιος, καὶτὸνἥμισυνἄρατοῦΓ And since D, which is odd, measures C, and C is even,
μετρήσει[ὁΔ].τοῦδὲΓἥμισύἐστινὁΒ·ὁΔἄρατὸν [D] will thus also measure half of C [Prop. 9.30]. And B
Βμετρεῖ. μετρεῖδὲκαὶτὸνΑ.ὁΔἄρατοὺςΑ,Βμετρεῖ is half of C. Thus, D measures B. And it also measures
πρώτουςὄνταςπρὸςἀλλήλους·ὅπερἐστὶνἀδύνατον. οὐκ A. Thus, D measures (both) A and B, (despite) them
ἄραὁΑπρὸςτὸνΓπρῶτοςοὔκἐστιν.οἱΑ,Γἄραπρῶτοι being prime to one another. The very thing is impossible.
πρὸςἀλλήλουςεἰσίν·ὅπερἔδειδεῖξαι.
Thus, A is not unprime to C. Thus, A and C are prime to
one another. (Which is) the very thing it was required to
show.
Ð. Proposition 32
Τῶν ἀπὸ δύαδος διπλασιαζομένων ἀριθμων ἕκαστος Each of the numbers (which is continually) doubled,
ἀρτιάκιςἄρτιόςἐστιμόνον.
(starting) from a dyad, is an even-times-even (number)
only.
Α
Β
Γ
∆
Ἀπὸ γὰρ δύαδος τῆς Α δεδιπλασιάσθωσαν ὁσοιδη- For let any multitude of numbers whatsoever, B, C,
ποτοῦνἀριθμοὶοἱΒ,Γ,Δ·λέγω,ὅτιοἱΒ,Γ,Δἀρτιάκις D, have been (continually) doubled, (starting) from the
ἄρτιοίεἰσιμόνον.
dyad A. I say that B, C, D are even-times-even (num-
῞Οτιμὲνοὖνἕκαστος[τῶνΒ,Γ,Δ]ἀρτιάκιςἄρτιός bers) only.
ἐστιν,φανερόν·ἀπὸγὰρδυάδοςἐστὶδιπλασιασθείς.λέγω, In fact, (it is) clear that each [of B, C, D] is an
ὅτικαὶμόνον.ἐκκείσθωγὰρμονάς.ἐπεὶοὖνἀπὸμονάδος even-times-even (number). For it is doubled from a dyad
ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογόν εἰσιν, ὁ δὲ μετὰ τὴν [Def. 7.8]. I also say that (they are even-times-even numμονάδαὁΑπρῶτόςἐστιν,ὁμέγιστοςτῶνΑ,Β,Γ,Δὁ
bers) only. For let a unit be laid down. Therefore, since
A
B
C
D
A
B
C
D
275

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
Δὑπ᾿οὐδενὸςἄλλουμετρηθήσεταιπαρὲξτῶνΑ,Β,Γ.καί any multitude of numbers whatsoever are continuously
ἐστινἕκαστοςτῶνΑ,Β,Γἄρτιος·ὁΔἄραἀρτιάκιςἄρτιός proportional, starting from a unit, and the (number) A af-
ἐστιμόνον.ὁμοίωςδὴδείξομεν,ὅτι[καὶ]ἑκάτεροςτῶνΒ, ter the unit is prime, the greatest of A, B, C, D, (namely)
Γἀρτιάκιςἄρτιόςἐστιμόνον·ὅπερἔδειδεῖξαι. D, will not be measured by any other (numbers) except
A, B, C [Prop. 9.13]. And each of A, B, C is even. Thus,
D is an even-time-even (number) only [Def. 7.8]. So,
similarly, we can show that each of B, C is [also] an eventime-even
(number) only. (Which is) the very thing it was
required to show.
Ð. Proposition 33
᾿Εὰν ἀριθμὸς τὸν ἥμισυν ἔχῃ περισσόν, ἀρτιάκις πε- If a number has an odd half then it is an even-timeρισσόςἐστιμόνον.
odd (number) only.
Α
ἈριθμὸςγὰρὁΑτὸνἥμισυνἐχέτωπερισσόν·λέγω,ὅτι For let the number A have an odd half. I say that A is
ὁΑἀρτιάκιςπερισσόςἐστιμόνον.
an even-times-odd (number) only.
῞Οτιμὲνοὖνἀρτιάκιςπερισσόςἐστιν,φανερόν·ὁγὰρ In fact, (it is) clear that (A) is an even-times-odd
ἥμισυςαὐτοῦπερισσὸςὢνμετρεῖαὐτὸνἀρτιάκις,λέγωδή, (number). For its half, being odd, measures it an even
ὅτικαὶμόνον. εἰγὰρἔσταιὁΑκαὶἀρτιάκιςἄρτιος,με- number of times [Def. 7.9]. So I also say that (it is
τρηθήσεταιὑπὸἀρτίουκατὰἄρτιονἀριθμόν· ὥστεκαὶὁ an even-times-odd number) only. For if A is also an
ἥμισυςαὐτοῦμετρηθήσεταιὑπὸἀρτίουἀριθμοῦπερισσὸς even-times-even (number) then it will be measured by an
ὤν· ὅπερἐστὶνἄτοπον. ὁΑἄραἀρτιάκιςπερισσόςἐστι even (number) according to an even number [Def. 7.8].
μόνον·ὅπερἔδειδεῖξαι.
Hence, its half will also be measured by an even number,
(despite) being odd. The very thing is absurd. Thus, A
is an even-times-odd (number) only. (Which is) the very
thing it was required to show.
Ð. Proposition 34
᾿Εὰνἀριθμὸςμήτετῶνἀπὸδυάδοςδιπλασιαζομένωνᾖ, If a number is neither (one) of the (numbers) doubled
μήτετὸνἥμισυνἔχῃπερισσόν,ἀρτιάκιςτεἄρτιόςἐστικαὶ from a dyad, nor has an odd half, then it is (both) an
ἀρτιάκιςπερισσός.
even-times-even and an even-times-odd (number).
Α
ἈριθμὸςγὰρὁΑμήτετῶνἀπὸδυάδοςδιπλασιαζομένων For let the number A neither be (one) of the (num-
ἔστω μήτε τὸν ἥμισυν ἐχέτω περισσόν· λέγω, ὅτι ὁ Α bers) doubled from a dyad, nor let it have an odd half.
ἀρτιάκιςτέἐστινἄρτιοςκαὶἀρτιάκιςπερισσός.
I say that A is (both) an even-times-even and an even-
῞ΟτιμὲνοὖνὁΑἀρτιάκιςἐστὶνἄρτιος,φανερόν·τὸν times-odd (number).
γὰρἥμισυνοὐκἔχειπερισσόν.λέγωδή,ὅτικαὶἀρτιάκιςπε- In fact, (it is) clear that A is an even-times-even (numρισσόςἐστιν.ἐὰνγὰρτὸνΑτέμνωμενδίχακαὶτὸνἥμισυν
ber) [Def. 7.8]. For it does not have an odd half. So I
αὐτοῦδίχακαὶτοῦτοἀεὶποιῶμεν,καταντήσομενεἴςτινα say that it is also an even-times-odd (number). For if we
ἀριθμὸνπερισσόν,ὃςμετρήσειτὸνΑκατὰἄρτιονἀριθμόν. cut A in half, and (then cut) its half in half, and we do
εἰγὰροὔ,καταντήσομενεἰςδυάδα,καὶἔσταιὁΑτῶνἀπὸ this continually, then we will arrive at some odd numδυάδοςδιπλασιαζομένων·ὅπεροὐχὑπόκειται.
ὥστεὁΑ ber which will measure A according to an even number.
ἀρτιάκιςπερισσόνἐστιν.ἐδείχθηδὲκαὶἀρτιάκιςἄρτιος.ὁ For if not, we will arrive at a dyad, and A will be (one)
Αἄραἀρτιάκιςτεἄρτιόςἐστικαὶἀρτιάκιςπερισσός·ὅπερ of the (numbers) doubled from a dyad. The very oppo-
ἔδειδεῖξαι.
site thing (was) assumed. Hence, A is an even-times-odd
(number) [Def. 7.9]. And it was also shown (to be) an
even-times-even (number). Thus, A is (both) an eventimes-even
and an even-times-odd (number). (Which is)
A
A
276

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
the very thing it was required to show.
Ð. Proposition 35 †
᾿Εὰνὦσινὁσοιδηποτοῦνἀριθμοὶἑξῆςἀνάλογον,ἀφαι- If there is any multitude whatsoever of continually
ρεθῶσιδὲἀπότετοῦδευτέρουκαὶτοῦἐσχάτουἴσοιτῷ proportional numbers, and (numbers) equal to the first
πρώτῳ,ἔσταιὡςἡτοῦδευτέρουὑπεροχὴπρὸςτὸνπρῶτον, are subtracted from (both) the second and the last, then
οὕτως ἡ τοῦ ἐσχάτου ὑπεροχὴ πρὸς τοὺς πρὸ ἑαυτοῦ as the excess of the second (number is) to the first, so the
πάντας.
excess of the last will be to (the sum of) all those (numbers)
before it.
Α
Β
Η
Γ
A
B
G
C
∆
Ε
Λ
Κ
Θ
Ζ
D
E
L
K
H
F
῎ΕστωσανὁποσοιδηποτοῦνἀριθμοὶἑξῆςἀνάλογονοἱΑ, Let A, BC, D, EF be any multitude whatsoever of
ΒΓ,Δ,ΕΖἀφχόμενοιἀπὸἐλαχίστουτοῦΑ,καὶἀφῃρήσθω continuously proportional numbers, beginning from the
ἀπὸτοῦΒΓκαὶτοῦΕΖτῲΑἴσοςἑκάτεροςτῶνΒΗ,ΖΘ· least A. And let BG and FH, each equal to A, have been
λέγω,ὅτιἐστὶνὡςὁΗΓπρὸςτὸνΑ,οὕτωςὁΕΘπρὸς subtracted from BC and EF (respectively). I say that as
τοὺςΑ,ΒΓ,Δ. GC is to A, so EH is to A, BC, D.
ΚείσθωγὰρτῷμὲνΒΓἴσοςὁΖΚ,τῷδὲΔἴσοςὁΖΛ. For let FK be made equal to BC, and FL to D. And
καὶἐπεὶὁΖΚτῷΒΓἴσοςἐστίν,ὧνὁΖΘτῷΒΗἴσοςἐστίν, since FK is equal to BC, of which FH is equal to BG,
λοιπὸςἄραὁΘΚλοιπῷτῷΗΓἐστινἴσος.καὶἐπείἐστινὡς the remainder HK is thus equal to the remainder GC.
ὁΕΖπρὸςτὸνΔ,οὕτωςὁΔπρὸςτὸνΒΓκαὶὁΒΓπρὸς And since as EF is to D, so D (is) to BC, and BC to
τὸνΑ,ἴσοςδὲὁμὲνΔτῷΖΛ,ὁδὲΒΓτῷΖΚ,ὁδὲΑτῷ A [Prop. 7.13], and D (is) equal to FL, and BC to FK,
ΖΘ,ἔστινἄραὡςὁΕΖπρὸςτὸνΖΛ,οὕτωςὁΛΖπρὸςτὸν and A to FH, thus as EF is to FL, so LF (is) to FK, and
ΖΚκαὶὁΖΚπρὸςτὸνΖΘ.διελόντι,ὡςὁΕΛπρὸςτὸνΛΖ, FK to FH. By separation, as EL (is) to LF , so LK (is)
οὕτωςὁΛΚπρὸςτὸνΖΚκαὶὁΚΘπρὸςτὸνΖΘ.ἔστινἄρα to FK, and KH to FH [Props. 7.11, 7.13]. And thus as
καὶὡςεἷςτῶνἡγουμένωνπρὸςἕνατῶνἑπομένων,οὕτως one of the leading (numbers) is to one of the following,
ἅπαντεςοἱἡγούμενοιπρὸςἅπανταςτοὺςἑπομένους·ἔστιν so (the sum of) all of the leading (numbers is) to (the
ἄραὡςὁΚΘπρὸςτὸνΖΘ,οὕτωςοἱΕΛ,ΛΚ,ΚΘπρὸς sum of) all of the following [Prop. 7.12]. Thus, as KH
τοὺςΛΖ,ΖΚ,ΘΖ.ἴσοςδὲὁμὲνΚΘτῷΓΗ,ὁδὲΖΘτῷ is to FH, so EL, LK, KH (are) to LF , FK, HF . And
Α,οἱδὲΛΖ,ΖΚ,ΘΖτοὶςΔ,ΒΓ,Α·ἔστινἄραὡςὁΓΗ KH (is) equal to CG, and FH to A, and LF , FK, HF
πρὸςτὸνΑ,οὕτωςὁΕΘπρὸςτοὺςΔ,ΒΓ,Α.ἔστινἄρα to D, BC, A. Thus, as CG is to A, so EH (is) to D,
ὡςἡτοῦδευτέρουὑπεροχὴπρὸςτὸνπρῶτον,οὕτωςἡτοῦ BC, A. Thus, as the excess of the second (number) is to
ἐσχάτουὑπεροχὴπρὸςτοὺςπρὸἑαυτοῦπάντας·ὅπερἔδει the first, so the excess of the last (is) to (the sum of) all
δεῖξαι.
those (numbers) before it. (Which is) the very thing it
was required to show.
† This proposition allows us to sum a geometric series of the form a, a r, a r 2 , a r 3 , · · · a r n−1 . According to Euclid, the sum S n satisfies
(a r − a)/a = (a r n − a)/S n. Hence, S n = a(r n − 1)/(r − 1).
Ц. Proposition 36 †
᾿Εὰνἀπὸμονάδοςὁποσοιοῦνἀριθμοὶἑξῆςἐκτεθῶσινἐν If any multitude whatsoever of numbers is set out conτῇδιπλασίονιἀναλογίᾳ,ἕωςοὗὁσύμπαςσυντεθεὶςπρῶτος
tinuously in a double proportion, (starting) from a unit,
γένηται,καὶὁσύμπαςἐπὶτὸνἔσχατονπολλαπλασιασθεὶς until the whole sum added together becomes prime, and
277

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
ποιῇτινα,ὁγενόμενοςτέλειοςἔσται.
the sum multiplied into the last (number) makes some
Ἀπὸ γὰρ μονάδος ἐκκείσθωσανὁσοιδηποτοῦνἀριθμ- (number), then the (number so) created will be perfect.
οὶἐντῇδιπλασίονιἀναλογίᾳ,ἕωςοὗὁσύμπαςσυντεθεὶς For let any multitude of numbers, A, B, C, D, be set
πρῶτοςγένηται,οἱΑ,Β,Γ,Δ,καὶτῷσύμπαντιἴσοςἔστω out (continuouly) in a double proportion, until the whole
ὁΕ,καὶὁΕτὸνΔπολλαπλασιάσαςτὸνΖΗποιείτω.λέγω, sum added together is made prime. And let E be equal to
ὅτιὁΖΗτέλειόςἐστιν.
the sum. And let E make FG (by) multiplying D. I say
that FG is a perfect (number).
Α
Β
Γ
∆
A
B
C
D
Ε
Θ
Ν Κ
E
H
N K
Λ
Μ
Ζ
Ξ
Η
L
M
F
O
G
Ο
P
Π
Q
῞ΟσοιγάρεἰσινοἱΑ,Β,Γ,Δτῷπλήθει,τοσοῦτοιἀπὸ For as many as is the multitude of A, B, C, D, let so
τοῦΕεἰλήφθωσανἐντῇδιπλασίονιἀναλογίᾳοἱΕ,ΘΚ,Λ, many (numbers), E, HK, L, M, have been taken in a
Μ·δι᾿ἴσουἄραἐστὶνὡςὁΑπρὸςτὸνΔ,οὕτωςὁΕπρὸς double proportion, (starting) from E. Thus, via equalτὸνΜ.ὁἄραἐκτῶνΕ,ΔἴσοςἐστὶτῷἐκτῶνΑ,Μ.καί
ity, as A is to D, so E (is) to M [Prop. 7.14]. Thus, the
ἐστινὁἐκτῶνΕ,ΔὁΖΗ·καὶὁἐκτῶνΑ,Μἄραἐστὶνὁ (number created) from (multiplying) E, D is equal to the
ΖΗ.ὁΑἄρατὸνΜπολλαπλασιάσαςτὸνΖΗπεποίηκεν·ὁ (number created) from (multiplying) A, M. And FG is
ΜἄρατὸνΖΗμετρεῖκατὰτὰςἐντῷΑμονάδας.καίἐστι the (number created) from (multiplying) E, D. Thus,
δυὰςὁΑ·διπλάσιοςἄραἐστὶνὁΖΗτοῦΜ.εἰσὶδὲκαὶοἱΜ, FG is also the (number created) from (multiplying) A,
Λ,ΘΚ,Εἑξῆςδιπλάσιοιἀλλήλων·οἱΕ,ΘΚ,Λ,Μ,ΖΗἄρα M [Prop. 7.19]. Thus, A has made FG (by) multiplying
ἑξῆςἀνάλογόνεἰσινἐντῇδιπλασίονιἀναλογίᾳ.ἀφῃρήσθω M. Thus, M measures FG according to the units in A.
δὴἀπὸτοῦδευτέρουτοῦΘΚκαὶτοῦἐσχάτουτοῦΖΗτῷ And A is a dyad. Thus, FG is double M. And M, L,
πρώτῳτῷΕἴσοςἑκάτεροςτῶνΘΝ,ΖΞ·ἔστινἄραὡςἡ HK, E are also continuously double one another. Thus,
τοῦδευτέρουἀριθμοῦὑπεροχὴπρὸςτὸνπρῶτον,οὕτωςἡ E, HK, L, M, FG are continuously proportional in a
τοῦἐσχάτουὑπεροχὴπρὸςτοὺςπρὸἑαυτοῦπάντας.ἔστιν double proportion. So let HN and FO, each equal to the
ἄραὡςὁΝΚπρὸςτὸνΕ,οὕτωςὁΞΗπρὸςτοὺςΜ,Λ, first (number) E, have been subtracted from the second
ΚΘ,Ε.καίἐστινὁΝΚἴσοςτῷΕ·καὶὁΞΗἄραἴσοςἐστὶ (number) HK and the last FG (respectively). Thus, as
τοῖςΜ,Λ,ΘΚ,Ε.ἔστιδὲκαὶὁΖΞτῷΕἴσος,ὁδὲΕ the excess of the second number is to the first, so the exτοῖςΑ,Β,Γ,Δκαὶτῇμονάδι.
ὅλοςἄραὁΖΗἴσοςἐστὶ cess of the last (is) to (the sum of) all those (numbers)
τοῖςτεΕ,ΘΚ,Λ,ΜκαὶτοῖςΑ,Β,Γ,Δκαὶτῇμονάδι· before it [Prop. 9.35]. Thus, as NK is to E, so OG (is)
καὶμετρεῖταιὑπ᾿αὐτῶν. λέγω,ὅτικαὶὁΖΗὐπ᾿οὐδενὸς to M, L, KH, E. And NK is equal to E. And thus OG
ἄλλουμετρηθήσεταιπαρὲξτῶνΑ,Β,Γ,Δ,Ε,ΘΚ,Λ,Μ is equal to M, L, HK, E. And FO is also equal to E,
καὶτῆςμονάδος. εἰγὰρδυνατόν,μετρείτωτιςτὸνΖΗὁ and E to A, B, C, D, and a unit. Thus, the whole of FG
Ο,καὶὁΟμηδενὶτῶνΑ,Β,Γ,Δ,Ε,ΘΚ,Λ,Μἔστωὁ is equal to E, HK, L, M, and A, B, C, D, and a unit.
αὐτός. καὶὁσάκιςὁΟτὸνΖΗμετρεῖ,τοσαῦταιμονάδες And it is measured by them. I also say that FG will be
278

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 9
ἔστωσανἐντῷΠ·ὁΠἄρατὸνΟπολλαπλασιάσαςτὸνΖΗ measured by no other (numbers) except A, B, C, D, E,
πεποίηκεν. ἀλλὰμὴνκαὶὁΕτὸνΔπολλαπλασιάσαςτὸν HK, L, M, and a unit. For, if possible, let some (num-
ΖΗπεποίηκεν·ἔστινἄραὡςὁΕπρὸςτὸνΠ,ὁΟπρὸςτὸν ber) P measure FG, and let P not be the same as any
Δ.καὶἐπεὶἀπὸμονάδοςἑξῆςἀνάλογόνεἰσινοἱΑ,Β,Γ, of A, B, C, D, E, HK, L, M. And as many times as P
Δ,ὁΔἄραὑπ᾿οὐδενὸςἄλλουἀριθμοῦμετρηθήσεταιπαρὲξ measures FG, so many units let there be in Q. Thus, Q
τῶνΑ,Β,Γ.καὶὑπόκειταιὁΟοὐδενὶτῶνΑ,Β,Γὁαὐτός· has made FG (by) multiplying P . But, in fact, E has also
οὐκἄραμετρήσειὁΟτὸνΔ.ἀλλ᾿ὡςὁΟπρὸςτὸνΔ,ὁ made FG (by) multiplying D. Thus, as E is to Q, so P
ΕπρὸςτὸνΠ·οὐδὲὁΕἄρατὸνΠμετρεῖ. καίἐστινὁΕ (is) to D [Prop. 7.19]. And since A, B, C, D are conπρῶτος·πᾶςδὲπρῶτοςἀριθμὸςπρὸςἅπαντα,ὃνμὴμετρεῖ,
tinually proportional, (starting) from a unit, D will thus
πρῶτός[ἐστιν].οἱΕ,Πἄραπρῶτοιπρὸςἀλλήλουςεἰσίν.οἱ not be measured by any other numbers except A, B, C
δὲπρῶτοικαὶἐλάχιστοι,οἱδὲἐλάχιστοιμετροῦσιτοὺςτὸν [Prop. 9.13]. And P was assumed not (to be) the same
αὐτὸνλόγονἔχονταςἰσάκιςὅτεἡγούμενοςτὸνἡγούμενον as any of A, B, C. Thus, P does not measure D. But,
καὶὁἑπόμενοςτὸνἑπόμενον·καίἐστινὡςὁΕπρὸςτὸνΠ, as P (is) to D, so E (is) to Q. Thus, E does not mea-
ὁΟπρὸςτὸνΔ.ἰσάκιςἄραὁΕτὸνΟμετρεῖκαὶὁΠτὸν sure Q either [Def. 7.20]. And E is a prime (number).
Δ.ὁδὲΔὑπ᾿οὐδενὸςἄλλουμετρεῖταιπαρὲξτῶνΑ,Β,Γ· And every prime number [is] prime to every (number)
ὁΠἄραἑνὶτῶνΑ,Β,Γἐστινὁαὐτός.ἔστωτῷΒὁαὐτός. which it does not measure [Prop. 7.29]. Thus, E and Q
καὶὅσοιεἰσὶνοἱΒ,Γ,Δτῷπλήθειτοσοῦτοιεἰλήφθωσαν are prime to one another. And (numbers) prime (to one
ἀπὸτοῦΕοἱΕ,ΘΚ,Λ.καίεἰσινοἱΕ,ΘΚ,ΛτοῖςΒ,Γ,Δ another are) also the least (of those numbers having the
ἐντῷαὐτῷλόγω·δι᾿ἴσουἄραἐστὶνὡςὁΒπρὸςτὸνΔ,ὁ same ratio as them) [Prop. 7.21], and the least (num-
ΕπρὸςτὸνΛ.ὁἄραἐκτῶνΒ,ΛἴσοςἐστὶτῷἐκτῶνΔ, bers) measure those (numbers) having the same ratio as
Ε·ἀλλ᾿ὁἐκτῶνΔ,ΕἴσοςἐστὶτῷἐκτῶνΠ,Ο·καὶὁἐκ them an equal number of times, the leading (measuring)
τῶνΠ,ΟἄραἴσοςἐστὶτῷἐκτῶνΒ,Λ.ἔστινἄραὡςὁΠ the leading, and the following the following [Prop. 7.20].
πρὸςτὸνΒ,ὁΛπρὸςτὸνΟ.καίἐστινὁΠτῷΒὁαὐτός· And as E is to Q, (so) P (is) to D. Thus, E measures P
καὶὁΛἄρατῳΟἐστινὁαὐτός·ὅπερἀδύνατον·ὁγὰρΟ the same number of times as Q (measures) D. And D
ὑπόκειταιμηδενὶτῶνἐκκειμένωνὁαὐτός·οὐκἄρατὸνΖΗ is not measured by any other (numbers) except A, B, C.
μετρήσειτιςἀριθμὸςπαρὲξτῶνΑ,Β,Γ,Δ,Ε,ΘΚ,Λ,Μ Thus, Q is the same as one of A, B, C. Let it be the same
καὶτῆςμονάδος.καὶἐδείχηὁΖΗτοῖςΑ,Β,Γ,Δ,Ε,ΘΚ, as B. And as many as is the multitude of B, C, D, let so
Λ,Μκαὶτῇμονάδιἴσος. τέλειοςδὲἀριθμόςἐστινὁτοῖς many (of the set out numbers) have been taken, (start-
ἑαυτοῦμέρεσινἴσοςὤν·τέλειοςἄραἐστὶνὁΖΗ·ὅπερἔδει ing) from E, (namely) E, HK, L. And E, HK, L are in
δεῖξαι.
the same ratio as B, C, D. Thus, via equality, as B (is)
to D, (so) E (is) to L [Prop. 7.14]. Thus, the (number
created) from (multiplying) B, L is equal to the (number
created) from multiplying D, E [Prop. 7.19]. But,
the (number created) from (multiplying) D, E is equal
to the (number created) from (multiplying) Q, P . Thus,
the (number created) from (multiplying) Q, P is equal
to the (number created) from (multiplying) B, L. Thus,
as Q is to B, (so) L (is) to P [Prop. 7.19]. And Q is the
same as B. Thus, L is also the same as P . The very thing
(is) impossible. For P was assumed not (to be) the same
as any of the (numbers) set out. Thus, FG cannot be
measured by any number except A, B, C, D, E, HK, L,
M, and a unit. And FG was shown (to be) equal to (the
sum of) A, B, C, D, E, HK, L, M, and a unit. And a
perfect number is one which is equal to (the sum of) its
own parts [Def. 7.22]. Thus, FG is a perfect (number).
(Which is) the very thing it was required to show.
† This proposition demonstrates that perfect numbers take the form 2 n−1 (2 n − 1) provided that 2 n − 1 is a prime number. The ancient Greeks
knew of four perfect numbers: 6, 28, 496, and 8128, which correspond to n = 2, 3, 5, and 7, respectively.
279

280

ELEMENTS BOOK 10
Incommensurable Magnitudes †
† The theory of incommensurable magntidues set out in this book is generally attributed to Theaetetus of Athens. In the footnotes throughout
this book, k, k ′ , etc. stand for distinct ratios of positive integers.
281

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ÎÇÖÓ.
αʹ.Σύμμετρα μεγέθη λέγεταιτὰ τῷ αὐτῷ μετρῳ με- 1. Those magnitudes measured by the same measure
Definitions
τρούμενα,ἀσύμμετραδέ,ὧνμηδὲνἐνδέχεταικοινὸνμέτρον are said (to be) commensurable, but (those) of which no
γενέσθαι.
(magnitude) admits to be a common measure (are said
βʹ.Εὐθεῖαιδυνάμεισύμμετροίεἰσιν,ὅταντὰἀπ᾿αὐτῶν to be) incommensurable. †
τετράγωνατῷαὐτῷχωρίῳμετρῆται,ἀσύμμετροιδέ,ὅταν 2. (Two) straight-lines are commensurable in square ‡
τοῖςἀπ᾿αὐτῶντετραγώνοιςμηδὲνἐνδέχηταιχωρίονκοινὸν when the squares on them are measured by the same
μέτρονγενέσθαι.
area, but (are) incommensurable (in square) when no
γʹ.Τούτωνὑποκειμένωνδείκνυται,ὅτιτῇπροτεθείσῃ area admits to be a common measure of the squares on
εὐθείᾳὑπάρχουσινεὐθεῖαιπλήθειἄπειροισύμμετροίτεκαὶ them. §
ἀσύμμετροιαἱμὲνμήκειμόνον,αἱδὲκαὶδυνάμει.καλείσθω 3. These things being assumed, it is proved that there
οὖνἡμὲνπροτεθεῖσαεὐθεῖαῥητή,καὶαἱταύτῃσύμμετροι exist an infinite multitude of straight-lines commensuεἴτεμήκεικαὶδυνάμειεἴτεδυνάμειμόνονῥηταί,αἱδὲταύτῃ
rable and incommensurable with an assigned straight-
ἀσύμμετροιἄλογοικαλείσθωσαν.
line—those (incommensurable) in length only, and those
δʹ.Καὶτὸμὲνἀπὸτῆςπροτεθείσηςεὐθείαςτετράγω- also (commensurable or incommensurable) in square.
νον ῥητόν, καὶ τὰ τούτῳ σύμμετρα ῥητά, τὰ δὲ τούτῳ Therefore, let the assigned straight-line be called ratio-
ἀσύμμετραἄλογακαλείσθω,καὶαἱδυνάμεναιαὐτὰἄλογοι, nal. And (let) the (straight-lines) commensurable with it,
εἰμὲντετράγωναεἴη, αὐταὶαἱπλευραί,εἰδὲἕτεράτινα either in length and square, or in square only, (also be
εὐθύγραμμα,αἱἴσααὐτοῖςτετράγωναἀναγράφουσαι.
called) rational. But let the (straight-lines) incommensurable
with it be called irrational. ∗
4. And let the square on the assigned straight-line be
called rational. And (let areas) commensurable with it
(also be called) rational. But (let areas) incommensurable
with it (be called) irrational, and (let) their squareroots
$ (also be called) irrational—the sides themselves, if
the (areas) are squares, and the (straight-lines) describing
squares equal to them, if the (areas) are some other
rectilinear (figure). ‖
† In other words, two magnitudes α and β are commensurable if α : β :: 1 : k, and incommensurable otherwise.
‡ Literally, “in power”.
§ In other words, two straight-lines of length α and β are commensurable in square if α : β :: 1 : k 1/2 , and incommensurable in square otherwise.
Likewise, the straight-lines are commensurable in length if α : β :: 1 : k, and incommensurable in length otherwise.
To be more exact, straight-lines can either be commensurable in square only, incommensurable in length only, or commenusrable/incommensurable
in both length and square, with an assigned straight-line.
∗ Let the length of the assigned straight-line be unity. Then rational straight-lines have lengths expressible as k or k 1/2 , depending on whether
the lengths are commensurable in length, or in square only, respectively, with unity. All other straight-lines are irrational.
$ The square-root of an area is the length of the side of an equal area square.
‖ The area of the square on the assigned straight-line is unity. Rational areas are expressible as k. All other areas are irrational. Thus, squares
whose sides are of rational length have rational areas, and vice versa.
. Proposition 1 †
Δύομεγεθῶνἀνίσωνἐκκειμένων,ἐὰνἀπὸτοῦμείζονος If, from the greater of two unequal magnitudes
ἀφαιρεθῇμεῖζονἢτὸἥμισυκαὶτοῦκαταλειπομένουμεῖζον (which are) laid out, (a part) greater than half is sub-
ἢτὸἥμισυ,καὶτοῦτοἀεὶγίγνηται,λειφθήσεταίτιμέγεθος, tracted, and (if from) the remainder (a part) greater than
ὃἔσταιἔλασσοντοῦἐκκειμένουἐλάσσονοςμεγέθους. half (is subtracted), and (if) this happens continually,
῎ΕστωδύομεγέθηἄνισατὰΑΒ,Γ,ὧνμεῖζοντὸΑΒ· then some magnitude will (eventually) be left which will
282

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
λέγω, ὅτι, ἐανἀπὸτοῦΑΒἀφαιρεθῇμεῖζονἢτὸἥμισυ be less than the lesser laid out magnitude.
καὶτοῦκαταλειπομένουμεῖζονἢτὸἥμισυ,καὶτοῦτοἀεὶ Let AB and C be two unequal magnitudes, of which
γίγνηται,λειφθήσεταίτιμέγεθος,ὃἔσταιἔλασσοντοῦΓ (let) AB (be) the greater. I say that if (a part) greater
μεγέθους.
than half is subtracted from AB, and (if a part) greater
than half (is subtracted) from the remainder, and (if) this
happens continually, then some magnitude will (eventually)
be left which will be less than the magnitude C.
Α Κ
Θ
Β
A
K
H
B
Γ
C
∆
Ζ
Η
Ε
Τὸ Γ γὰρ πολλαπλασιαζόμενον ἔσται ποτὲ τοῦ ΑΒ For C, when multiplied (by some number), will someμεῖζον.
πεπολλαπλασιάσθω,καὶἔστωτὸΔΕτοῦμὲνΓ times be greater than AB [Def. 5.4]. Let it have been
πολλαπλάσιον,τοῦδὲΑΒμεῖζον,καὶδιῃρήσθωτὸΔΕεἰς (so) multiplied. And let DE be (both) a multiple of C,
τὰτῷΓἴσατὰΔΖ,ΖΗ,ΗΕ,καὶἀφῃρήσθωἀπὸμὲντοῦ and greater than AB. And let DE have been divided into
ΑΒμεῖζονἢτὸἥμισυτὸΒΘ,ἀπὸδὲτοῦΑΘμεῖζονἢτὸ the (divisions) DF , FG, GE, equal to C. And let BH,
ἥμισυτὸΘΚ,καὶτοῦτοἀεὶγιγνέσθω,ἕωςἂναἱἐντῷΑΒ (which is) greater than half, have been subtracted from
διαιρέσειςἰσοπληθεῖςγένωνταιταῖςἐντῷΔΕδιαιρέσεσιν. AB. And (let) HK, (which is) greater than half, (have
῎ΕστωσανοὖναἱΑΚ,ΚΘ,ΘΒδιαιρέσειςἰσοπληθεῖς been subtracted) from AH. And let this happen continuοὖσαιταῖςΔΖ,ΖΗ,ΗΕ·καὶἐπεὶμεῖζόνἐστιτὸΔΕτοῦ
ally, until the divisions in AB become equal in number to
ΑΒ,καὶἀφῄρηταιἀπὸμὲντοῦΔΕἔλασσοντοῦἡμίσεωςτὸ the divisions in DE.
ΕΗ,ἀπὸδὲτοῦΑΒμεῖζονἢτὸἥμισυτὸΒΘ,λοιπὸνἄρα Therefore, let the divisions (in AB) be AK, KH, HB,
τὸΗΔλοιποῦτοῦΘΑμεῖζόνἐστιν.καὶἐπεὶμεῖζόνἐστιτὸ being equal in number to DF , FG, GE. And since DE is
ΗΔτοῦΘΑ,καὶἀφῄρηταιτοῦμὲνΗΔἥμισυτὸΗΖ,τοῦ greater than AB, and EG, (which is) less than half, has
δὲΘΑμεῖζονἢτὸἥμισυτὸΘΚ,λοιπὸνἄρατὸΔΖλοιποῦ been subtracted from DE, and BH, (which is) greater
τοῦΑΚμεῖζόνἐστιν. ἴσονδὲτὸΔΖτῷΓ·καὶτὸΓἄρα than half, from AB, the remainder GD is thus greater
τοῦΑΚμεῖζόνἐστιν.ἔλασσονἄρατὸΑΚτοῦΓ. than the remainder HA. And since GD is greater than
ΚαταλείπεταιἄραἀπὸτοῦΑΒμεγέθουςτὸΑΚμέγεθος HA, and the half GF has been subtracted from GD, and
ἔλασσονὂντοῦἐκκειμένουἐλάσσονοςμεγέθουςτοῦΓ· HK, (which is) greater than half, from HA, the remain-
ὅπερἔδειδεῖξαι. —ὁμοίωςδὲδειχθήσεται,κἂνἡμίσηᾖτὰ der DF is thus greater than the remainder AK. And DF
ἀφαιρούμενα.
(is) equal to C. C is thus also greater than AK. Thus,
AK (is) less than C.
Thus, the magnitude AK, which is less than the lesser
laid out magnitude C, is left over from the magnitude
AB. (Which is) the very thing it was required to show. —
(The theorem) can similarly be proved even if the (parts)
subtracted are halves.
† This theorem is the basis of the so-called method of exhaustion, and is generally attributed to Eudoxus of Cnidus.
. Proposition 2
᾿Εὰνδύομεγεθῶν[ἐκκειμένων]ἀνίσωνἀνθυφαιρουμένου If the remainder of two unequal magnitudes (which
ἀεὶτοῦἐλάσσονοςἀπὸτοῦμείζονοςτὸκαταλειπόμενον are) [laid out] never measures the (magnitude) before it,
μηδέποτεκαταμετρῇτὸπρὸἑαυτοῦ,ἀσύμμετραἔσταιτὰ (when) the lesser (magnitude is) continually subtracted
μεγέθη.
in turn from the greater, then the (original) magnitudes
Δύο γὰρ μεγεθῶν ὄντων ἀνίσων τῶν ΑΒ, ΓΔ καὶ will be incommensurable.
ἐλάσσονοςτοῦΑΒἀνθυφαιρουμένουἀεὶτοῦἐλάσσονος For, AB and CD being two unequal magnitudes, and
ἀπὸ τοῦ μείζονος τὸ περιλειπόμενον μηδέποτε καταμε- AB (being) the lesser, let the remainder never measure
D
F
G
E
283

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
τρείτωτὸπρὸἑαυτοῦ·λέγω,ὅτιἀσύμμετράἐστιτὰΑΒ, the (magnitude) before it, (when) the lesser (magnitude
ΓΔμεγέθη.
is) continually subtracted in turn from the greater. I say
that the magnitudes AB and CD are incommensurable.
Α Η
Β
A
G
B
Ε
E
Γ
Ζ
∆
Εἰγάρἐστισύμμετρα,μετρήσειτιαὐτὰμέγεθος. με- For if they are commensurable then some magnitude
τρείτω,εἰδυνατόν,καὶἔστωτὸΕ·καὶτὸμὲνΑΒτὸΖΔ will measure them (both). If possible, let it (so) measure
καταμετροῦνλειπέτωἑαυτοῦἔλασσοντὸΓΖ,τὸδὲΓΖτὸ (them), and let it be E. And let AB leave CF less than
ΒΗκαταμετροῦνλειπέτωἑαυτοῦἔλασσοντὸΑΗ,καὶτοῦτο itself (in) measuring FD, and let CF leave AG less than
ἀεὶγινέσθω,ἕωςοὗλειφθῇτιμέγεθος,ὅἐστινἔλασσοντοῦ itself (in) measuring BG, and let this happen continually,
Ε.γεγονέτω,καὶλελείφθωτὸΑΗἔλασσοντοῦΕ.ἐπεὶοὖν until some magnitude which is less than E is left. Let
τὸΕτὸΑΒμετρεῖ,ἀλλὰτὸΑΒτὸΔΖμετρεῖ,καὶτὸΕἄρα (this) have occurred, † and let AG, (which is) less than
τὸΖΔμετρήσει.μετρεῖδὲκαὶὅλοντὸΓΔ·καὶλοιπὸνἄρα E, have been left. Therefore, since E measures AB, but
τὸΓΖμετρήσει. ἀλλὰτὸΓΖτὸΒΗμετρεῖ·καὶτὸΕἄρα AB measures DF , E will thus also measure FD. And it
τὸΒΗμετρεῖ.μετρεῖδὲκαὶὅλοντὸΑΒ·καὶλοιπὸνἄρατὸ also measures the whole (of) CD. Thus, it will also mea-
ΑΗμετρήσει,τὸμεῖζοντὸἔλασσον·ὅπερἐστὶνἀδύνατον. sure the remainder CF . But, CF measures BG. Thus, E
οὐκἄρατὰΑΒ,ΓΔμεγέθημετρήσειτιμέγεθος·ἀσύμμετρα also measures BG. And it also measures the whole (of)
ἄραἐστὶτὰΑΒ,ΓΔμεγέθη.
᾿Εὰνἄραδύομεγεθῶνἀνίσων,καὶτὰἑξῆς.
† The fact that this will eventually occur is guaranteed by Prop. 10.1.
C
F
AB. Thus, it will also measure the remainder AG, the
greater (measuring) the lesser. The very thing is impossible.
Thus, some magnitude cannot measure (both) the
magnitudes AB and CD. Thus, the magnitudes AB and
CD are incommensurable [Def. 10.1].
Thus, if . . . of two unequal magnitudes, and so on . . . .
. Proposition 3
Δύομεγεθῶνσυμμέτρωνδοθέντωντὸμέγιστοναὐτῶν To find the greatest common measure of two given
κοινὸνμέτρονεὑρεῖν.
commensurable magnitudes.
Α
Ζ
Β
A F
B
D
Γ
Η
Ε
∆
῎ΕστωτὰδοθένταδύομεγέθησύμμετρατὰΑΒ,ΓΔ, Let AB and CD be the two given magnitudes, of
ὧνἔλασσοντὸΑΒ·δεῖδὴτῶνΑΒ,ΓΔτὸμέγιστονκοινὸν which (let) AB (be) the lesser. So, it is required to find
μέτρονεὑρεῖν.
the greatest common measure of AB and CD.
ΤὸΑΒγὰρμέγεθοςἤτοιμετρεῖτὸΓΔἢοὔ. εἰμὲν For the magnitude AB either measures, or (does) not
οὖνμετρεῖ,μετρεῖδὲκαὶἑαυτό,τὸΑΒἄρατῶνΑΒ,ΓΔ (measure), CD. Therefore, if it measures (CD), and
κοινὸνμέτρονἐστίν·καὶφανερόν,ὅτικαὶμέγιστον.μεῖζον (since) it also measures itself, AB is thus a common meaγὰρτοῦΑΒμεγέθουςτὸΑΒοὐμετρήσει.
sure of AB and CD. And (it is) clear that (it is) also (the)
ΜὴμετρείτωδὴτὸΑΒτὸΓΔ.καὶἀνθυφαιρουμένου greatest. For a (magnitude) greater than magnitude AB
ἀεὶτοῦἐλάσσονοςἀπὸτοῦμείζονος, τὸπεριλειπόμενον cannot measure AB.
μετρήσειποτὲτὸπρὸἑαυτοῦδιὰτὸμὴεἶναιἀσύμμετρατὰ So let AB not measure CD. And continually subtract-
ΑΒ,ΓΔ·καὶτὸμὲνΑΒτὸΕΔκαταμετροῦνλειπέτωἑαυτοῦ ing in turn the lesser (magnitude) from the greater, the
C
G
E
D
284

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ἔλασσον τὸ ΕΓ, τὸ δὲ ΕΓ τὸ ΖΒ καταμετροῦνλειπέτω remaining (magnitude) will (at) some time measure the
ἑαυτοῦἔλασσοντὸΑΖ,τὸδὲΑΖτὸΓΕμετρείτω. (magnitude) before it, on account of AB and CD not be-
᾿ΕπεὶοὖντὸΑΖτὸΓΕμετρεῖ,ἀλλὰτὸΓΕτὸΖΒμετρεῖ, ing incommensurable [Prop. 10.2]. And let AB leave EC
καὶτὸΑΖἄρατὸΖΒμετρήσει. μετρεῖδὲκαὶἑαυτό·καὶ less than itself (in) measuring ED, and let EC leave AF
ὅλονἄρατὸΑΒμετρήσειτὸΑΖ.ἀλλὰτὸΑΒτὸΔΕμετρεῖ· less than itself (in) measuring FB, and let AF measure
καὶτὸΑΖἄρατὸΕΔμετρήσει. μετρεῖδὲκαὶτὸΓΕ·καί CE.
ὅλονἄρατὸΓΔμετρεῖ· τὸΑΖἄρατῶνΑΒ,ΓΔκοινὸν Therefore, since AF measures CE, but CE measures
μέτρονἐστίν. λέγωδή,ὅτικαὶμέγιστον. εἰγὰρμή,ἔσται FB, AF will thus also measure FB. And it also meaτιμέγεθοςμεῖζοντοῦΑΖ,ὃμετρήσειτὰΑΒ,ΓΔ.ἔστωτὸ
sures itself. Thus, AF will also measure the whole (of)
Η.ἐπεὶοὖντὸΗτὸΑΒμετρεῖ,ἀλλὰτὸΑΒτὸΕΔμετρεῖ, AB. But, AB measures DE. Thus, AF will also meaκαὶτὸΗἄρατὸΕΔμετρήσει.
μετρεῖδὲκαὶὅλοντὸΓΔ· sure ED. And it also measures CE. Thus, it also meaκαὶλοιπὸνἄρατὸΓΕμετρήσειτὸΗ.ἀλλὰτὸΓΕτὸΖΒ
sures the whole of CD. Thus, AF is a common measure
μετρεῖ·καὶτὸΗἄρατὸΖΒμετρήσει. μετρεῖδὲκαὶὅλον of AB and CD. So I say that (it is) also (the) greatest
τὸΑΒ,καὶλοιπὸντὸΑΖμετρήσει,τὸμεῖζοντὸἔλασσον· (common measure). For, if not, there will be some mag-
ὅπερἐστὶνἀδύνατον. οὐκἄραμεῖζόντιμέγεθοςτοῦΑΖ nitude, greater than AF , which will measure (both) AB
τὰΑΒ,ΓΔμετρήσει·τὸΑΖἄρατῶνΑΒ,ΓΔτὸμέγιστον and CD. Let it be G. Therefore, since G measures AB,
κοινὸνμέτρονἐστίν.
but AB measures ED, G will thus also measure ED. And
ΔύοἄραμεγεθῶνσυμμέτρωνδοθέντωντῶνΑΒ,ΓΔ it also measures the whole of CD. Thus, G will also meaτὸμέγιστονκοινὸνμέτρονηὕρηται·ὅπερἔδειδεῖξαι.
sure the remainder CE. But CE measures FB. Thus, G
will also measure FB. And it also measures the whole
(of) AB. And (so) it will measure the remainder AF ,
the greater (measuring) the lesser. The very thing is impossible.
Thus, some magnitude greater than AF cannot
measure (both) AB and CD. Thus, AF is the greatest
common measure of AB and CD.
Thus, the greatest common measure of two given
commensurable magnitudes, AB and CD, has been
found. (Which is) the very thing it was required to show.
ÈÖ×Ñ.
Corollary
᾿Εκδὴτούτουφανερόν,ὅτι,ἐὰνμέγεθοςδύομεγέθη So (it is) clear, from this, that if a magnitude measures
μετρῇ,καὶτὸμέγιστοναὐτῶνκοινὸνμέτρονμετρήσει. two magnitudes then it will also measure their greatest
common measure.
. Proposition 4
Τριῶν μεγεθῶν συμμέτρων δοθέντων τὸ μέγιστον To find the greatest common measure of three given
αὐτῶνκοινὸνμέτρονεὑρεῖν.
commensurable magnitudes.
Α
Β
Γ
∆ Ε Ζ
A
B
C
D E F
῎ΕστωτὰδοθέντατρίαμεγέθησύμμετρατὰΑ,Β,Γ· Let A, B, C be the three given commensurable magδεῖδὴτῶνΑ,Β,Γτὸμέγιστονκοινὸνμέτρονεὑρεῖν.
nitudes. So it is required to find the greatest common
ΕἰλήφθωγὰρδύοτῶνΑ,Βτὸμέγιστονκοινὸνμέτρον, measure of A, B, C.
καὶἔστωτὸΔ·τὸδὴΔτὸΓἤτοιμετρεῖἢοὔ[μετρεῖ]. For let the greatest common measure of the two (magμετρείτωπρότερον.
ἐπεὶοὖντὸΔτὸΓμετρεῖ,μετρεῖδὲ nitudes) A and B have been taken [Prop. 10.3], and let it
285

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
καὶτὰΑ,Β,τὸΔἄρατὰΑ,Β,Γμετρεῖ·τὸΔἄρατῶνΑ, be D. So D either measures, or [does] not [measure], C.
Β,Γκοινὸνμέτρονἐστίν. καὶφανερόν,ὅτικαὶμέγιστον· Let it, first of all, measure (C). Therefore, since D meaμεῖζονγὰρτοῦΔμεγέθουςτὰΑ,Βοὐμετρεῖ.
sures C, and it also measures A and B, D thus measures
ΜὴμετρείτωδὴτὸΔτὸΓ.λέγωπρῶτον,ὅτισύμμετρά A, B, C. Thus, D is a common measure of A, B, C. And
ἐστιτὰΓ,Δ.ἐπεὶγὰρσύμμετράἐστιτὰΑ,Β,Γ,μετρήσει (it is) clear that (it is) also (the) greatest (common meaτιαὐτὰμέγεθος,
ὃδηλαδὴκαὶτὰΑ,Βμετρήσει· ὥστε sure). For no magnitude larger than D measures (both)
καὶτὸτῶνΑ,ΒμέγιστονκοινὸνμέτροντὸΔμετρήσει. A and B.
μετρεῖδὲκαὶτὸΓ·ὥστετὸεἰρημένονμέγεθοςμετρήσειτὰ So let D not measure C. I say, first, that C and D are
Γ,Δ·σύμμετραἄραἐστὶτὰΓ,Δ.εἰλήφθωοὖναὐτῶντὸ commensurable. For if A, B, C are commensurable then
μέγιστονκοινὸνμέτρον,καὶἔστωτὸΕ.ἐπεὶοὖντὸΕτὸ some magnitude will measure them which will clearly
Δμετρεῖ,ἀλλὰτὸΔτὰΑ,Βμετρεῖ,καὶτὸΕἄρατὰΑ,Β also measure A and B. Hence, it will also measure D, the
μετρήσει.μετρεῖδὲκαὶτὸΓ.τὸΕἄρατὰΑ,Β,Γμετρεῖ· greatest common measure of A and B [Prop. 10.3 corr.].
τὸΕἄρατῶνΑ,Β,Γκοινόνἐστιμέτρον.λέγωδή,ὅτικαὶ And it also measures C. Hence, the aforementioned magμέγιστον.
εἰγὰρδυνατόν,ἔστωτιτοῦΕμεῖζονμέγεθος nitude will measure (both) C and D. Thus, C and D are
τὸΖ,καὶμετρείτωτὰΑ,Β,Γ.καὶἐπεὶτὸΖτὰΑ,Β,Γ commensurable [Def. 10.1]. Therefore, let their greatest
μετρεῖ,καὶτὰΑ,ΒἄραμετρήσεικαὶτὸτῶνΑ,Βμέγιστον common measure have been taken [Prop. 10.3], and let
κοινὸνμέτρονμετρήσει. τὸδὲτῶνΑ,Βμέγιστονκοινὸν it be E. Therefore, since E measures D, but D measures
μέτρονἐστὶτὸΔ·τὸΖἄρατὸΔμετρεῖ.μετρεῖδὲκαὶτὸ (both) A and B, E will thus also measure A and B. And
Γ·τὸΖἄρατὰΓ,Δμετρεῖ·καὶτὸτῶνΓ,Δἄραμέγιστον it also measures C. Thus, E measures A, B, C. Thus, E
κοινὸνμέτρονμετρήσειτὸΖ.ἔστιδὲτὸΕ·τὸΖἄρατὸΕ is a common measure of A, B, C. So I say that (it is) also
μετρήσει,τὸμεῖζοντὸἔλασσον·ὅπερἐστὶνἀδύνατον.οὐκ (the) greatest (common measure). For, if possible, let F
ἄραμεῖζόντιτοῦΕμεγέθους[μέγεθος]τὰΑ,Β,Γμετρεῖ· be some magnitude greater than E, and let it measure A,
τὸΕἄρατῶνΑ,Β,Γτὸμέγιστονκοινὸνμέτρονἐστίν,ἐὰν B, C. And since F measures A, B, C, it will thus also
μὴμετρῇτὸΔτὸΓ,ἐὰνδὲμετρῇ,αὐτὸτὸΔ.
measure A and B, and will (thus) measure the greatest
Τριῶνἄραμεγεθῶνσυμμέτρωνδοθέντωντὸμέγιστον common measure of A and B [Prop. 10.3 corr.]. And D
κοινὸνμέτρονηὕρηται[ὅπερἔδειδεῖξαι].
is the greatest common measure of A and B. Thus, F
measures D. And it also measures C. Thus, F measures
(both) C and D. Thus, F will also measure the greatest
common measure of C and D [Prop. 10.3 corr.]. And it is
E. Thus, F will measure E, the greater (measuring) the
lesser. The very thing is impossible. Thus, some [magnitude]
greater than the magnitude E cannot measure A,
B, C. Thus, if D does not measure C then E is the greatest
common measure of A, B, C. And if it does measure
(C) then D itself (is the greatest common measure).
Thus, the greatest common measure of three given
commensurable magnitudes has been found. [(Which is)
the very thing it was required to show.]
ÈÖ×Ñ.
Corollary
᾿Εκδὴτούτουφανερόν,ὅτι,ἐὰνμέγεθοςτρίαμεγέθη So (it is) clear, from this, that if a magnitude measures
μετρῇ,καὶτὸμέγιστοναὐτῶνκοινὸνμέτρονμετρήσει. three magnitudes then it will also measure their greatest
῾Ομοίωςδὴκαὶἐπὶπλειόνωντὸμέγιστονκοινὸνμέτρον common measure.
ληφθήσεται,καὶτὸπόρισμαπροχωρήσει.ὅπερἔδειδεῖξαι. So, similarly, the greatest common measure of more
(magnitudes) can also be taken, and the (above) corollary
will go forward. (Which is) the very thing it was
required to show.
286

ËÌÇÁÉÁÏƺ ELEMENTS BOOK 10
. Proposition 5
Τὰ σύμμετρα μεγέθη πρὸς ἄλληλα λόγον ἔχει, ὃν Commensurable magnitudes have to one another the
ἀριθμὸςπρὸςἀριθμόν.
ratio which (some) number (has) to (some) number.
Α Β Γ
A B C
∆
Ε
῎ΕστωσύμμετραμεγέθητὰΑ,Β·λέγω,ὅτιτὸΑπρὸς Let A and B be commensurable magnitudes. I say
τὸΒλόγονἔχει,ὃνἀριθμὸςπρὸςἀριθμόν.
that A has to B the ratio which (some) number (has) to
᾿Επεὶ γὰρ σύμμετρά ἐστι τὰ Α, Β, μετρήσει τι αὐτὰ (some) number.
μέγεθος. μετρείτω, καὶἔστωτὸΓ.καὶὁσάκιςτὸΓτὸ For if A and B are commensurable (magnitudes) then
Αμετρεῖ,τοσαῦταιμονάδεςἔστωσανἐντῷΔ,ὁσάκιςδὲ some magnitude will measure them. Let it (so) measure
τὸΓτὸΒμετρεῖ,τοσαῦταιμονάδεςἔστωσανἐντῷΕ. (them), and let it be C. And as many times as C measures
᾿ΕπεὶοὖντὸΓτὸΑμετρεῖκατὰτὰςἐντῷΔμονάδας, A, so many units let there be in D. And as many times as
μετρεῖδὲκαὶἡμονὰςτὸνΔκατὰτὰςἐναὐτῷμονάδας, C measures B, so many units let there be in E.
ἰσάκιςἄραἡμονὰςτὸνΔμετρεῖἀριθμὸνκαὶτὸΓμέγεθος Therefore, since C measures A according to the units
τὸΑ·ἔστινἄραὡςτὸΓπρὸςτὸΑ,οὕτωςἡμονὰςπρὸς in D, and a unit also measures D according to the units
τὸνΔ·ἀνάπαλινἄρα,ὡςτὸΑπρὸςτὸΓ,οὕτωςὁΔπρὸς in it, a unit thus measures the number D as many times
τὴνμονάδα. πάλινἐπεὶτὸΓτὸΒμετρεῖκατὰτὰςἐντῷ as the magnitude C (measures) A. Thus, as C is to A,
Εμονάδας,μετρεῖδὲκαὶἡμονὰςτὸνΕκατὰτὰςἐναὐτῷ so a unit (is) to D [Def. 7.20]. † Thus, inversely, as A (is)
μονάδας,ἰσάκιςἄραἡμονὰςτὸνΕμετρεῖκαὶτὸΓτὸΒ· to C, so D (is) to a unit [Prop. 5.7 corr.]. Again, since
ἔστινἄραὡςτὸΓπρὸςτὸΒ,οὕτωςἡμονὰςπρὸςτὸνΕ. C measures B according to the units in E, and a unit
ἐδείχθηδὲκαὶὡςτὸΑπρὸςτὸΓ,ὁΔπρὸςτὴνμονάδα· also measures E according to the units in it, a unit thus
δι᾿ἴσουἄραἐστὶνὡςτὸΑπρὸςτὸΒ,οὕτωςὁΔἀριθμὸς measures E the same number of times that C (measures)
πρὸςτὸνΕ.
B. Thus, as C is to B, so a unit (is) to E [Def. 7.20]. And
ΤὰἄρασύμμετραμεγέθητὰΑ,Βπρὸςἄλληλαλόγον it was also shown that as A (is) to C, so D (is) to a unit.
ἔχει,ὃνἀριθμὸςὁΔπρὸςἀριθμὸντὸνΕ·ὅπερἔδειδεῖξαι. Thus, via equality, as A is to B, so the number D (is) to
the (number) E [Prop. 5.22].
Thus, the commensurable magnitudes A and B have
to one another the ratio which the number D (has) to the
number E. (Which is) the very thing it was required to
show.
† There is a slight logical gap here, since Def. 7.20 applies to four numbers, rather than two number and two magnitudes.
¦. Proposition 6
᾿Εὰνδύομεγέθηπρὸςἄλληλαλόγονἔχῃ,ὃνἀριθμὸς If two magnitudes have to one another the ratio which
πρὸςἀριθμόν,σύμμετραἔσταιτὰμεγέθη.
(some) number (has) to (some) number then the magnitudes
will be commensurable.
Α
∆
Γ
Β
Ε
Ζ
ΔύογὰρμεγέθητὰΑ,Βπρὸςἄλληλαλόγονἐχέτω,ὃν For let the two magnitudes A and B have to one an-
ἀριθμὸςὁΔπρὸςἀριθμὸντὸνΕ·λέγω,ὅτισύμμετράἐστι other the ratio which the number D (has) to the number
τὰΑ,Βμεγέθη.
E. I say that the magnitudes A and B are commensu-
῞Οσαι γάρ εἰσιν ἐν τῷ Δ μονάδες, εἰς τοσαῦτα ἴσα rable.
A
D
C
D
E
B
E
F
287

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
διῃρήσθωτὸΑ,καὶἑνὶαὐτῶνἴσονἔστωτὸΓ·ὅσαιδέ For, as many units as there are in D, let A have been
εἰσινἐντῷΕμονάδες,ἐκτοσούτωνμεγεθῶνἴσωντῷΓ divided into so many equal (divisions). And let C be
συγκείσθωτὸΖ.
equal to one of them. And as many units as there are
᾿Επεὶοὖν,ὅσαιεἰσὶνἐντῷΔμονάδες,τοσαῦτάεἰσικαὶ in E, let F be the sum of so many magnitudes equal to
ἐντῷΑμεγέθηἴσατῷΓ,ὃἄραμέροςἐστὶνἡμονὰςτοῦ C.
Δ,τὸαὐτὸμέροςἐστὶκαὶτὸΓτοῦΑ·ἔστινἄραὡςτὸΓ Therefore, since as many units as there are in D, so
πρὸςτὸΑ,οὕτωςἡμονὰςπρὸςτὸνΔ.μετρεῖδὲἡμονὰς many magnitudes equal to C are also in A, therefore
τὸνΔἀριθμόν·μετρεῖἄρακαὶτὸΓτὸΑ.καὶἐπείἐστιν whichever part a unit is of D, C is also the same part of
ὡςτὸΓπρὸςτὸΑ,οὕτωςἡμονὰςπρὸςτὸνΔ[ἀριθμόν], A. Thus, as C is to A, so a unit (is) to D [Def. 7.20]. And
ἀνάπαλινἄραὡςτὸΑπρὸςτὸΓ,οὕτωςὁΔἀριθμὸςπρὸς a unit measures the number D. Thus, C also measures
τὴνμονάδα.πάλινἐπεί,ὅσαιεἰσὶνἐντῷΕμονάδες,τοσαῦτά A. And since as C is to A, so a unit (is) to the [number]
εἰσικαὶἐντῷΖἴσατῷΓ,ἔστινἄραὡςτὸΓπρὸςτὸΖ, D, thus, inversely, as A (is) to C, so the number D (is)
οὕτωςἡμονὰςπρὸςτὸνΕ[ἀριθμόν]. ἐδείχθηδὲκαὶὡς to a unit [Prop. 5.7 corr.]. Again, since as many units as
τὸΑπρὸςτὸΓ,οὕτωςὁΔπρὸςτὴνμονάδα·δι᾿ἴσουἄρα there are in E, so many (magnitudes) equal to C are also
ἐστὶνὡςτὸΑπρὸςτὸΖ,οὕτωςὁΔπρὸςτὸνΕ.ἀλλ᾿ὡςὁ in F , thus as C is to F , so a unit (is) to the [number] E
ΔπρὸςτὸνΕ,οὕτωςἐστὶτὸΑπρὸςτὸΒ·καὶὡςἄρατὸΑ [Def. 7.20]. And it was also shown that as A (is) to C,
πρὸςτὸΒ,οὕτωςκαὶπρὸςτὸΖ.τὸΑἄραπρὸςἑκάτερον so D (is) to a unit. Thus, via equality, as A is to F , so D
τῶνΒ,Ζτὸναὐτὸνἔχειλόγον·ἴσονἄραἐστὶτὸΒτῷΖ. (is) to E [Prop. 5.22]. But, as D (is) to E, so A is to B.
μετρεῖδὲτὸΓτὸΖ·μετρεῖἄρακαὶτὸΒ.ἀλλὰμὴνκαὶτὸ And thus as A (is) to B, so (it) also is to F [Prop. 5.11].
Α·τὸΓἄρατὰΑ,Βμετρεῖ.σύμμετρονἄραἐστὶτὸΑτῷ Thus, A has the same ratio to each of B and F . Thus, B is
Β. equal to F [Prop. 5.9]. And C measures F . Thus, it also
᾿Εὰνἄραδύομεγέθηπρὸςἄλληλα,καὶτὰἑξῆς. measures B. But, in fact, (it) also (measures) A. Thus,
C measures (both) A and B. Thus, A is commensurable
with B [Def. 10.1].
Thus, if two magnitudes . . . to one another, and so on
. . . .
ÈÖ×Ñ.
Corollary
᾿Εκδὴτούτουφανερόν,ὅτι,ἐὰνὦσιδύοἀριθμοί,ὡς So it is clear, from this, that if there are two numbers,
οἱΔ,Ε,καὶεὐθεῖα,ὡςἡΑ,δύνατόνἐστιποιῆσαιὡςὁ like D and E, and a straight-line, like A, then it is possible
ΔἀριθμὸςπρὸςτὸνΕἀριθμόν,οὕτωςτὴνεὐθεῖανπρὸς to contrive that as the number D (is) to the number E,
εὐθεῖαν.ἐὰνδὲκαὶτῶνΑ,Ζμέσηἀνάλογονληφθῇ,ὡςἡ so the straight-line (is) to (another) straight-line (i.e., F ).
Β,ἔσταιὡςἡΑπρὸςτὴνΖ,οὕτωςτὸἀπὸτῆςΑπρὸςτὸ And if the mean proportion, (say) B, is taken of A and
ἀπὸτῆςΒ,τουτέστινὡςἡπρώτηπρὸςτὴντρίτην,οὕτως F , then as A is to F , so the (square) on A (will be) to the
τὸἀπὸτῆςπρώτηςπρὸςτὸἀπὸτῆςδευτέραςτὸὅμοιονκαὶ (square) on B. That is to say, as the first (is) to the third,
ὁμοίωςἀναγραφόμενον. ἀλλ᾿ὡςἡΑπρὸςτὴνΖ,οὕτως so the (figure) on the first (is) to the similar, and similarly
ἐστὶνὁΔἀριθμοςπρὸςτὸνΕἀριθμόν·γέγονενἄρακαὶ described, (figure) on the second [Prop. 6.19 corr.]. But,
ὡςὁΔἀριθμὸςπρὸςτὸνΕἀριθμόν,οὕτωςτὸἀπὸτῆςΑ as A (is) to F , so the number D is to the number E. Thus,
εὐθείαςπρὸςτὸἀπὸτῆςΒεὐθείας·ὅπερἔδειδεῖξαι. it has also been contrived that as the number D (is) to
the number E, so the (figure) on the straight-line A (is)
to the (similar figure) on the straight-line B. (Which is)
the very thing it was required to show.
Þ. Proposition 7
Τὰἀσύμμετραμεγέθηπρὸςἄλληλαλόγονοὐκἔχει,ὃν Incommensurable magnitudes do not have to one an-
ἀριθμὸςπρὸςἀριθμόν.
other the ratio which (some) number (has) to (some)
῎ΕστωἀσύμμετραμεγέθητὰΑ,Β·λέγω,ὅτιτὸΑπρὸς number.
τὸΒλόγονοὐκἔχει,ὃνἀριθμὸςπρὸςἀριθμόν.
Let A and B be incommensurable magnitudes. I say
that A does not have to B the ratio which (some) number
(has) to (some) number.
288

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
Α
Β
ΕἰγὰρἔχειτὸΑπρὸςτὸΒλόγον,ὃνἀριθμὸςπρὸς For if A has to B the ratio which (some) number (has)
ἀριθμόν,σύμμετρονἔσταιτὸΑτῷΒ.οὐκἔστιδέ·οὐκἄρα to (some) number then A will be commensurable with B
τὸΑπρὸςτὸΒλόγονἔχει,ὃνἀριθμὸςπρὸςἀριθμόν. [Prop. 10.6]. But it is not. Thus, A does not have to B
Τὰἄραἀσύμμετραμεγέθηπρὸςἄλληλαλόγονοὐκἔχει, the ratio which (some) number (has) to (some) number.
καὶτὰἑξῆς.
Thus, incommensurable numbers do not have to one
another, and so on . . . .
. Proposition 8
᾿Εὰνδύομεγέθηπρὸςἄλληλαλόγονμὴἔχῃ,ὃνἀριθμὸς If two magnitudes do not have to one another the raπρὸςἀριθμόν,ἀσύμμετραἔσταιτὰμεγέθη.
tio which (some) number (has) to (some) number then
the magnitudes will be incommensurable.
Α
Β
ΔύογὰρμεγέθητὰΑ,Βπρὸςἄλληλαλόγονμὴἐχέτω, For let the two magnitudes A and B not have to one
ὃνἀριθμὸςπρὸςἀριθμόν·λέγω,ὅτιἀσύμμετράἐστιτὰΑ, another the ratio which (some) number (has) to (some)
Βμεγέθη.
number. I say that the magnitudes A and B are incom-
Εἰγὰρἔσταισύμμετρα,τὸΑπρὸςτὸΒλόγονἕξει,ὃν mensurable.
ἀριθμὸςπρὸςἀριθμόν.οὐκἔχειδέ.ἀσύμμετραἄραἐστὶτὰ For if they are commensurable, A will have to B
Α,Βμεγέθη.
the ratio which (some) number (has) to (some) number
᾿Εὰνἄραδύομεγέθηπρὸςἄλληλα,καὶτὰἑξῆς. [Prop. 10.5]. But it does not have (such a ratio). Thus,
the magnitudes A and B are incommensurable.
Thus, if two magnitudes . . . to one another, and so on
. . . .
. Proposition 9
Τὰ ἀπὸ τῶν μήκει συμμέτρων εὐθειῶν τετράγωνα Squares on straight-lines (which are) commensurable
πρὸς ἄλληλα λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς in length have to one another the ratio which (some)
τετράγωνον ἀριθμόν· καὶτὰτετράγωνα τὰπρὸςἄλληλα square number (has) to (some) square number. And
λόγονἔχοντα, ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον squares having to one another the ratio which (some)
ἀριθμόν, καὶ τὰς πλευρὰς ἕξει μήκει συμμέτρους. τὰ square number (has) to (some) square number will also
δὲ ἀπὸ τῶν μήκει ἀσυμμέτρων εὐθειῶν τετράγωνα πρὸς have sides (which are) commensurable in length. But
ἄλληλαλόγονοὐκἔχει,ὅνπερτετράγωνοςἀριθμὸςπρὸς squares on straight-lines (which are) incommensurable
τετράγωνον ἀριθμόν· καὶτὰτετράγωνα τὰπρὸςἄλληλα in length do not have to one another the ratio which
λόγονμὴἔχοντα,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον (some) square number (has) to (some) square number.
ἀριθμόν,οὐδὲτὰςπλευρὰςἕξειμήκεισυμμέτρους. And squares not having to one another the ratio which
(some) square number (has) to (some) square number
will not have sides (which are) commensurable in length
either.
Α
Γ
Β
∆
῎ΕστωσανγὰραἱΑ,Βμήκεισύμμετροι·λέγω,ὅτιτὸ For let A and B be (straight-lines which are) commen-
ἀπὸτῆςΑτετράγωνον πρὸς τὸἀπὸτῆςΒτετράγωνον surable in length. I say that the square on A has to the
λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον square on B the ratio which (some) square number (has)
ἀριθμόν.
to (some) square number.
A
C
A
B
A
B
B
D
289

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
᾿ΕπεὶγὰρσύμμετρόςἐστινἡΑτῇΒμήκει,ἡΑἄραπρὸς For since A is commensurable in length with B, A
τὴνΒλόγονἔχει,ὃνἀριθμὸςπρὸςἀριθμόν. ἐχέτω,ὃνὁ thus has to B the ratio which (some) number (has) to
ΓπρὸςτὸνΔ.ἐπεὶοὖνἐστινὡςἡΑπρὸςτὴνΒ,οὕτωςὁ (some) number [Prop. 10.5]. Let it have (that) which
ΓπρὸςτὸνΔ,ἀλλὰτοῦμὲντῆςΑπρὸςτὴνΒλόγουδι- C (has) to D. Therefore, since as A is to B, so C (is)
πλασίωνἐστὶνὁτοῦἀπὸτῆςΑτετραγώνουπρὸςτὸἀπὸτῆς to D. But the (ratio) of the square on A to the square
Βτετράγωνον·τὰγὰρὅμοιασχήματαἐνδιπλασίονιλόγῳ on B is the square of the ratio of A to B. For similar
ἐστὶτῶνὁμολόγωνπλευρῶν·τοῦδὲτοῦΓ[ἀριθμοῦ]πρὸς figures are in the squared ratio of (their) corresponding
τὸνΔ[ἀριθμὸν]λόγουδιπλασίωνἐστὶνὁτοῦἀπὸτοῦΓ sides [Prop. 6.20 corr.]. And the (ratio) of the square
τετραγώνουπρὸςτὸνἀπὸτοῦΔτετράγωνον·δύογὰρτε- on C to the square on D is the square of the ratio of
τραγώνωνἀριθμῶνεἷςμέσοςἀνάλογόνἐστινἀριθμός,καί the [number] C to the [number] D. For there exits one
ὁτετράγωνοςπρὸςτὸντετράγωνον[ἀριθμὸν]διπλασίονα number in mean proportion to two square numbers, and
λόγονἔχει,ἤπερἡπλευρὰπρὸςτὴνπλευράν· ἔστινἄρα (one) square (number) has to the (other) square [numκαὶ
ὡς τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β ber] a squared ratio with respect to (that) the side (of the
τετράγωνον,οὕτωςὁἀπὸτοῦΓτετράγωνος[ἀριθμὸς]πρὸς former has) to the side (of the latter) [Prop. 8.11]. And,
τὸνἀπὸτοῦΔ[ἀριθμοῦ]τετράγωνον[ἀριθμόν]. thus, as the square on A is to the square on B, so the
ἈλλὰδὴἔστωὡςτὸἀπὸτῆςΑτετράγωνονπρὸςτὸ square [number] on the (number) C (is) to the square
ἀπὸτῆςΒ,οὕτωςὁἀπὸτοῦΓτετράγωνοςπρὸςτὸνἀπὸ [number] on the [number] D. †
τοῦΔ[τετράγωνον]·λέγω,ὅτισύμμετρόςἐστινἡΑτῇΒ And so let the square on A be to the (square) on B as
μήκει.
the square (number) on C (is) to the [square] (number)
᾿ΕπεὶγάρἐστινὡςτὸἀπὸτῆςΑτετράγωνονπρὸςτὸ on D. I say that A is commensurable in length with B.
ἀπὸτῆςΒ[τετράγωνον],οὕτωςὁἀπὸτοῦΓτετράγωνος For since as the square on A is to the [square] on B, so
πρὸςτὸνἀπὸτοῦΔ[τετράγωνον],ἀλλ᾿ὁμὲντοῦἀπὸτῆς the square (number) on C (is) to the [square] (number)
ΑτετραγώνουπρὸςτὸἀπὸτῆςΒ[τετράγωνον]λόγοςδι- on D. But, the ratio of the square on A to the (square)
πλασίωνἐστὶτοῦτῆςΑπρὸςτὴνΒλόγου,ὁδὲτοῦἀπὸ on B is the square of the (ratio) of A to B [Prop. 6.20
τοῦΓ[ἀριθμοῦ]τετραγώνου[ἀριθμοῦ]πρὸςτὸνἀπὸτοῦΔ corr.]. And the (ratio) of the square [number] on the
[ἀριθμοῦ]τετράγωνον[ἀριθμὸν]λόγοςδιπλασίωνἐστὶτοῦ [number] C to the square [number] on the [number] D is
τοῦΓ[ἀριθμοῦ]πρὸςτὸνΔ[ἀριθμὸν]λόγου,ἔστινἄρα the square of the ratio of the [number] C to the [number]
καὶὡςἡΑπρὸςτὴνΒ,οὕτωςὁΓ[ἀριθμὸς]πρὸςτὸνΔ D [Prop. 8.11]. Thus, as A is to B, so the [number] C
[ἀριθμόν].ἡΑἄραπρὸςτὴνΒλόγονἔχει,ὃνἀριθμὸςὁΓ also (is) to the [number] D. A, thus, has to B the ratio
πρὸςἀριθμὸντὸνΔ·σύμμετροςἄραἐστὶνἡΑτῇΒμήκει. which the number C has to the number D. Thus, A is
ἈλλὰδὴἀσύμμετροςἔστωἡΑτῇΒμήκει·λέγω,ὅτι commensurable in length with B [Prop. 10.6]. ‡
τὸἀπὸτῆςΑτετράγωνονπρὸςτὸἀπὸτῆςΒ[τετράγωνον] And so let A be incommensurable in length with B. I
λόγονοὐκἔχει,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον say that the square on A does not have to the [square] on
ἀριθμόν.
B the ratio which (some) square number (has) to (some)
Εἰ γὰρἔχειτὸ ἀπὸτῆςΑτετράγωνον πρὸς τὸἀπὸ square number.
τῆςΒ[τετράγωνον]λόγον,ὃντετράγωνοςἀριθμὸςπρὸς For if the square on A has to the [square] on B the raτετράγωνονἀριθμόν,σύμμετροςἔσταιἡΑτῇΒ.οὐκἔστι
tio which (some) square number (has) to (some) square
δέ· οὐκἄρατὸἀπὸτῆςΑτετράγωνονπρὸςτὸἀπὸτῆς number then A will be commensurable (in length) with
Β[τετράγωνον]λόγονἔχει,ὃντετράγωνοςἀριθμὸςπρὸς B. But it is not. Thus, the square on A does not have
τετράγωνονἀριθμόν.
to the [square] on the B the ratio which (some) square
ΠάλινδὴτὸἀπὸτῆςΑτετράγωνονπρὸςτὸἀπὸτῆς number (has) to (some) square number.
Β[τετράγωνον]λόγονμὴἐχέτω,ὃντετράγωνοςἀριθμὸς So, again, let the square on A not have to the [square]
πρὸςτετράγωνονἀριθμόν·λέγω,ὅτιἀσύμμετρόςἐστινἡΑ on B the ratio which (some) square number (has) to
τῇΒμήκει.
(some) square number. I say that A is incommensurable
ΕἰγάρἐστισύμμετροςἡΑτῇΒ,ἕξειτὸἀπὸτῆςΑ in length with B.
πρὸςτὸἀπὸτῆςΒλόγον,ὃντετράγωνοςἀριθμὸςπρὸς For if A is commensurable (in length) with B then
τετράγωνονἀριθμόν.οὐκἔχειδέ·οὐκἄρασύμμετρόςἐστιν the (square) on A will have to the (square) on B the ra-
ἡΑτῇΒμήκει.
Τὰἄραἀπὸτῶνμήκεισυμμέτρων,καὶτὰἑξῆς.
tio which (some) square number (has) to (some) square
number. But it does not have (such a ratio). Thus, A is
not commensurable in length with B.
Thus, (squares) on (straight-lines which are) com-
290

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
mensurable in length, and so on . . . .
Corollary
ÈÖ×Ñ.
Καὶφανερὸνἐκτῶν δεδειγμένωνἔσται, ὅτιαἱμήκει And it will be clear, from (what) has been demonσύμμετροιπάντωςκαὶδυνάμει,αἱδὲδυνάμειοὐπάντωςκαὶ
strated, that (straight-lines) commensurable in length
μήκει.
(are) always also (commensurable) in square, but (straightlines
commensurable) in square (are) not always also
(commensurable) in length.
† There is an unstated assumption here that if α : β :: γ : δ then α 2 : β 2 :: γ 2 : δ 2 .
‡ There is an unstated assumption here that if α 2 : β 2 :: γ 2 : δ 2 then α : β :: γ : δ.
. Proposition 10 †
Τῇπροτεθείσῃεὐθείᾳπροσευρεῖνδύοεὐθείαςἀσυμμέτ- To find two straight-lines incommensurable with a
ρους,τὴνμὲνμήκειμόνον,τὴνδὲκαὶδυνάμει.
given straight-line, the one (incommensurable) in length
only, the other also (incommensurable) in square.
Α
Ε
∆
Β
Γ
῎ΕστωἡπροτεθεῖσαεὐθεῖαἡΑ·δεῖδὴτῇΑπροσευρεῖν Let A be the given straight-line. So it is required to
δύοεὐθείαςἀσυμμέτρους,τὴνμὲνμήκειμόνον,τὴνδὲκαὶ find two straight-lines incommensurable with A, the one
δυνάμει.
(incommensurable) in length only, the other also (incom-
᾿ΕκκείσθωσανγὰρδύοαριθμοὶοἱΒ,Γπρὸςἀλλήλους mensurable) in square.
λόγονμὴἔχοντες,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον For let two numbers, B and C, not having to one
ἀριθμόν,τουτέστιμὴὅμοιοιἐπίπεδοι,καὶγεγονέτωὡςὁ another the ratio which (some) square number (has) to
ΒπρὸςτὸνΓ,οὕτωςτὸἀπὸτῆςΑτετράγωνονπρὸςτὸ (some) square number—that is to say, not (being) simi-
ἀπὸ τῆς Δ τετράγωνον· ἐμάθομεν γάρ· σύμμετρον ἄρα lar plane (numbers)—have been taken. And let it be conτὸἀπὸτῆςΑτῷἀπὸτῆςΔ.καὶἐπεὶὁΒπρὸςτὸνΓ
trived that as B (is) to C, so the square on A (is) to the
λόγονοὐκἔχει,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον square on D. For we learned (how to do this) [Prop. 10.6
ἀριθμόν,οὐδ᾿ἄρατὸἀπὸτῆςΑπρὸςτὸἀπὸτῆςΔλόγον corr.]. Thus, the (square) on A (is) commensurable with
ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· the (square) on D [Prop. 10.6]. And since B does not
ἀσύμμετροςἄραἐστὶνἡΑτῇΔμήκει.εἰλήφθωτῶνΑ,Δ have to C the ratio which (some) square number (has) to
μέσηἀνάλογονἡΕ·ἔστινἄραὡςἡΑπρὸςτὴνΔ,οὕτως (some) square number, the (square) on A thus does not
τὸἀπὸτῆςΑτετράγωνονπρὸςτὸἀπὸτῆςΕ.ἀσύμμετρος have to the (square) on D the ratio which (some) square
δέἐστινἡΑτῇΔμήκει·ἀσύμμετρονἄραἐστὶκαὶτὸἀπὸ number (has) to (some) square number either. Thus, A
τῆςΑτετράγωνοντῷἀπὸτῆςΕτετραγώνῳ·ἀσύμμετρος is incommensurable in length with D [Prop. 10.9]. Let
ἄραἐστὶνἡΑτῇΕδυνάμει.
the (straight-line) E (which is) in mean proportion to A
Τῂ ἄρα προτεθείσῃ εὐθείᾳ τῇ Α προσεύρηνται δύο and D have been taken [Prop. 6.13]. Thus, as A is to D,
εὐθεῖαιἀσύμμετροιαἱΔ,Ε,μήκειμὲνμόνονἡΔ,δυνάμει so the square on A (is) to the (square) on E [Def. 5.9].
δὲκαὶμήκειδηλαδὴἡΕ[ὅπερἔδειδεῖξαι].
And A is incommensurable in length with D. Thus, the
square on A is also incommensurble with the square on
E [Prop. 10.11]. Thus, A is incommensurable in square
with E.
A
E
D
B
C
291

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
† This whole proposition is regarded by Heiberg as an interpolation into the original text.
Thus, two straight-lines, D and E, (which are) incommensurable
with the given straight-line A, have been
found, the one, D, (incommensurable) in length only, the
other, E, (incommensurable) in square, and, clearly, also
in length. [(Which is) the very thing it was required to
show.]
. Proposition 11
᾿Εὰν τέσσαρα μεγέθηἀνάλογονᾖ, τὸδὲ πρῶτον τῷ If four magnitudes are proportional, and the first is
δευτέρῳσύμμετρονᾖ,καὶτὸτρίτοντῷτετάρτῳσύμμετρον commensurable with the second, then the third will also
ἔσται· κἂντὸπρῶτοντῷδευτέρῳἀσύμμετρονᾖ,καὶτὸ be commensurable with the fourth. And if the first is inτρίτοντῷτετάρτῳἀσύμμετρονἔσται.
commensurable with the second, then the third will also
be incommensurable with the fourth.
Α
Γ
Β
∆
῎ΕστωσαντέσσαραμεγέθηἀνάλογοντὰΑ,Β,Γ,Δ, Let A, B, C, D be four proportional magnitudes,
ὡςτὸΑπρὸςτὸΒ,οὕτωςτὸΓπρὸςτὸΔ,τὸΑδὲτῷ (such that) as A (is) to B, so C (is) to D. And let A
Βσύμμετρονἔστω·λέγω,ὅτικαὶτὸΓτῷΔσύμμετρον be commensurable with B. I say that C will also be com-
ἔσται. mensurable with D.
᾿ΕπεὶγὰρσύμμετρόνἐστιτὸΑτῷΒ,τὸΑἄραπρὸςτὸ For since A is commensurable with B, A thus has to B
Βλόγονἔχει,ὃνἀριθμὸςπρὸςἀριθμόν.καίἐστινὡςτὸΑ the ratio which (some) number (has) to (some) number
πρὸςτὸΒ,οὕτωςτὸΓπρὸςτὸΔ·καὶτὸΓἄραπρὸςτὸΔ [Prop. 10.5]. And as A is to B, so C (is) to D. Thus,
λόγονἔχει,ὃνἀριθμὸςπρὸςἀριθμόν·σύμμετρονἄραἐστὶ C also has to D the ratio which (some) number (has)
τὸΓτῷΔ.
to (some) number. Thus, C is commensurable with D
ἈλλὰδὴτὸΑτῷΒἀσύμμετρονἔστω·λέγω,ὅτικαὶτὸ [Prop. 10.6].
ΓτῷΔἀσύμμετρονἔσται.ἐπεὶγὰρἀσύμμετρόνἐστιτὸΑ And so let A be incommensurable with B. I say that
τῷΒ,τὸΑἄραπρὸςτὸΒλόγονοὐκἔχει,ὃνἀριθμὸςπρὸς C will also be incommensurable with D. For since A
ἀριθμόν.καίἐστινὡςτὸΑπρὸςτὸΒ,οὕτωςτὸΓπρὸςτὸ is incommensurable with B, A thus does not have to B
Δ·οὐδὲτὸΓἄραπρὸςτὸΔλόγονἔχει,ὃνἀριθμὸςπρὸς the ratio which (some) number (has) to (some) number
ἀριθμόν·ἀσύμμετρονἄραἐστὶτὸΓτῷΔ.
᾿Εὰνἄρατέσσαραμεγέθη,καὶτὰἑξῆς.
A
C
[Prop. 10.7]. And as A is to B, so C (is) to D. Thus, C
does not have to D the ratio which (some) number (has)
to (some) number either. Thus, C is incommensurable
with D [Prop. 10.8].
Thus, if four magnitudes, and so on . . . .
. Proposition 12
Τὰ τῷ αὐτῷ μεγέθει σύμμετρα καὶ ἀλλήλοις ἐστὶ (Magnitudes) commensurable with the same magniσύμμετρα.
tude are also commensurable with one another.
῾ΕκάτερονγὰρτῶνΑ,ΒτῷΓἔστωσύμμετρον.λέγω, For let A and B each be commensurable with C. I say
ὅτικαὶτὸΑτῷΒἐστισύμμετρον. that A is also commensurable with B.
᾿ΕπεὶγὰρσύμμετρόνἐστιτὸΑτῷΓ,τὸΑἄραπρὸς For since A is commensurable with C, A thus has
τὸΓλόγονἔχει,ὃνἀριθμὸςπρὸςἀριθμόν.ἐχέτω,ὃνὁΔ to C the ratio which (some) number (has) to (some)
πρὸςτὸνΕ.πάλιν,ἐπεὶσύμμετρόνἐστιτὸΓτῷΒ,τὸΓἄρα number [Prop. 10.5]. Let it have (the ratio) which D
πρὸςτὸΒλόγονἔχει,ὃνἀριθμὸςπρὸςἀριθμόν.ἐχέτω,ὃν (has) to E. Again, since C is commensurable with B,
ὁΖπρὸςτὸνΗ.καὶλόγωνδοθέντωνὁποσωνοῦντοῦτε, C thus has to B the ratio which (some) number (has)
ὃνἔχειὁΔπρὸςτὸνΕ,καὶὁΖπρὸςτὸνΗεἰλήφθωσαν to (some) number [Prop. 10.5]. Let it have (the ratio)
ἀριθμοὶἑξῆςἐντοῖςδοθεῖσιλόγοιςοἱΘ,Κ,Λ·ὥστεεἶναι which F (has) to G. And for any multitude whatsoever
B
D
292

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ὡςμὲντὸνΔπρὸςτὸνΕ,οὕτωςτὸνΘπρὸςτὸνΚ,ὡςδὲ of given ratios—(namely,) those which D has to E, and
τὸνΖπρὸςτὸνΗ,οὕτωςτὸνΚπρὸςτὸνΛ.
F to G—let the numbers H, K, L (which are) continuously
(proportional) in the(se) given ratios have been
taken [Prop. 8.4]. Hence, as D is to E, so H (is) to K,
and as F (is) to G, so K (is) to L.
Α Γ Β
A
C
B
∆
Ε
Ζ
Θ
Κ
Λ
D
E
F
H
K
L
Η
G
᾿ΕπεὶοὖνἐστινὡςτὸΑπρὸςτὸΓ,οὕτωςὁΔπρὸς Therefore, since as A is to C, so D (is) to E, but as
τὸνΕ,ἀλλ᾿ὡςὁΔπρὸςτὸνΕ,οὕτωςὁΘπρὸςτὸνΚ, D (is) to E, so H (is) to K, thus also as A is to C, so H
ἔστινἄρακαὶὡςτὸΑπρὸςτὸΓ,οὕτωςὁΘπρὸςτὸνΚ. (is) to K [Prop. 5.11]. Again, since as C is to B, so F
πάλιν,ἐπείἐστινὡςτὸΓπρὸςτὸΒ,οὕτωςὁΖπρὸςτὸν (is) to G, but as F (is) to G, [so] K (is) to L, thus also
Η,ἀλλ᾿ὡςὁΖπρὸςτὸνΗ,[οὕτως]ὁΚπρὸςτὸνΛ,καὶ as C (is) to B, so K (is) to L [Prop. 5.11]. And also as A
ὡςἄρατὸΓπρὸςτὸΒ,οὕτωςὁΚπρὸςτὸνΛ.ἔστιδὲκαὶ is to C, so H (is) to K. Thus, via equality, as A is to B,
ὡςτὸΑπρὸςτὸΓ,οὕτωςὁΘπρὸςτὸνΚ·δι᾿ἴσουἄρα so H (is) to L [Prop. 5.22]. Thus, A has to B the ratio
ἐστὶνὡςτὸΑπρὸςτὸΒ,οὕτωςὁΘπρὸςτὸνΛ.τὸΑἄρα which the number H (has) to the number L. Thus, A is
πρὸςτὸΒλόγονἔχει,ὃνἀριθμὸςὁΘπρὸςἀριθμὸντὸνΛ· commensurable with B [Prop. 10.6].
σύμμετρονἄραἐστὶτὸΑτῷΒ.
Thus, (magnitudes) commensurable with the same
Τὰἄρατῷαὐτῷμεγέθεισύμμετρακαὶἀλλήλοιςἐστὶ magnitude are also commensurable with one another.
σύμμετρα·ὅπερἔδειδεῖξαι.
(Which is) the very thing it was required to show.
. Proposition 13
᾿Εὰν ᾖ δύο μεγέθη σύμμετρα, τὸ δὲ ἕτερον αὐτῶν If two magnitudes are commensurable, and one of
μεγέθειτινὶἀσύμμετρονᾖ,καὶτὸλοιπὸντῷαὐτῷἀσύμμετρ- them is incommensurable with some magnitude, then
ονἔσται.
the remaining (magnitude) will also be incommensurable
with it.
Α
Γ
Β
῎Εστω δύο μεγέθησύμμετρα τὰ Α, Β, τὸ δὲ ἕτερον Let A and B be two commensurable magnitudes, and
αὐτῶντὸΑἄλλῳτινὶτῷΓἀσύμμετρονἔστω·λέγω,ὅτι let one of them, A, be incommensurable with some other
καὶτὸλοιπὸντὸΒτῷΓἀσύμμετρόνἐστιν.
(magnitude), C. I say that the remaining (magnitude),
ΕἰγάρἐστισύμμετροντὸΒτῷΓ,ἀλλὰκαὶτὸΑτῷ B, is also incommensurable with C.
Βσύμμετρόνἐστιν,καὶτὸΑἄρατῷΓσύμμετρόνἐστιν. For if B is commensurable with C, but A is also com-
ἀλλὰκαὶἀσύμμετρον·ὅπερἀδύνατον. οὐκἄρασύμμετρόν mensurable with B, A is thus also commensurable with
ἐστιτὸΒτῷΓ·ἀσύμμετρονἄρα.
C [Prop. 10.12]. But, (it is) also incommensurable (with
᾿Εὰνἄραᾖδύομεγέθησύμμετρα,καὶτὰἑξῆς. C). The very thing (is) impossible. Thus, B is not commensurable
with C. Thus, (it is) incommensurable.
Thus, if two magnitudes are commensurable, and so
on . . . .
ÄÑÑ.
Lemma
Δύο δοθεισῶν εὐθειῶν ἀνίσων εὑρεῖν, τίνι μεῖζον For two given unequal straight-lines, to find by (the
δύναταιἡμείζωντῆςἐλάσσονος.
square on) which (straight-line) the square on the greater
A
C
B
293

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
∆
Γ
(straight-line is) larger than (the square on) the lesser. †
D
C
Α
Β
῎ΕστωσαναἱδοθεῖσαιδύοἄνισοιεὐθεῖαιαἱΑΒ,Γ,ὧν Let AB and C be the two given unequal straight-lines,
μείζωνἔστωἡΑΒ·δεῖδὴεὑρεῖν,τίνιμεῖζονδύναταιἡΑΒ and let AB be the greater of them. So it is required to
τῆςΓ.
find by (the square on) which (straight-line) the square
ΓεγράφθωἐπὶτῆςΑΒἡμικύκλιοντὸΑΔΒ,καὶεἰςαὐτὸ on AB (is) greater than (the square on) C.
ἐνηρμόσθωτῇΓἴσηἡΑΔ,καὶἐπεζεύχθωἡΔΒ.φανερὸν Let the semi-circle ADB have been described on AB.
δή,ὅτιὀρθήἐστινἡὑπὸΑΔΒγωνία,καὶὅτιἡΑΒτῆς And let AD, equal to C, have been inserted into it
ΑΔ,τουτέστιτῆςΓ,μεῖζονδύναταιτῇΔΒ.
[Prop. 4.1]. And let DB have been joined. So (it is) clear
῾Ομοίωςδὲκαὶδύοδοθεισῶνεὐθειῶνἡδυναμένηαὐτὰς that the angle ADB is a right-angle [Prop. 3.31], and
εὑρίσκεταιοὕτως.
that the square on AB (is) greater than (the square on)
῎ΕστωσαναἱδοθεῖσαιδύοεὐθεῖαιαἱΑΔ,ΔΒ,καὶδέον AD—that is to say, (the square on) C—by (the square
ἔστωεὑρεῖντὴνδυναμένηναὐτάς. κείσθωσανγάρ,ὥστε on) DB [Prop. 1.47].
ὀρθὴνγωνίανπεριέχειντὴνὑπὸΑΔ,ΔΒ,καὶἐπεζεύχθω And, similarly, the square-root of (the sum of the
ἡΑΒ·φανερὸνπάλιν,ὅτιἡτὰςΑΔ,ΔΒδυναμένηἐστὶνἡ squares on) two given straight-lines is also found likeso.
ΑΒ·ὅπερἔδειδεῖξαι.
Let AD and DB be the two given straight-lines. And
let it be necessary to find the square-root of (the sum
of the squares on) them. For let them have been laid
down such as to encompass a right-angle—(namely), that
(angle encompassed) by AD and DB. And let AB have
been joined. (It is) again clear that AB is the square-root
of (the sum of the squares on) AD and DB [Prop. 1.47].
(Which is) the very thing it was required to show.
† That is, if α and β are the lengths of two given straight-lines, with α being greater than β, to find a straight-line of length γ such that
α 2 = β 2 + γ 2 . Similarly, we can also find γ such that γ 2 = α 2 + β 2 .
. Proposition 14
᾿Εὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, δύνηται δὲ ἡ If four straight-lines are proportional, and the square
πρώτη τῆς δευτέρας μεῖζον τῷ ἀπὸ συμμέτρου ἑαυτῇ on the first is greater than (the square on) the sec-
[μήκει],καὶἡτρίτητῆςτετάρτηςμεῖζονδυνήσεταιτῷἀπὸ ond by the (square) on (some straight-line) commenσυμμέτρουἑαυτῇ[μήκει].
καὶἐὰνἡπρώτητῆςδευτέρας surable [in length] with the first, then the square on
μεῖζον δύνηταιτῷ ἀπὸ ἀσυμμέτρουἑαυτῇ[μήκει], καὶἡ the third will also be greater than (the square on) the
τρίτητῆςτετάρτηςμεῖζονδυνήσεταιτῷἀπὸἀσυμμέτρου fourth by the (square) on (some straight-line) commen-
ἑαυτῇ[μήκει].
surable [in length] with the third. And if the square on
῎ΕστωσαντέσσαρεςεὐθεῖαιἀνάλογοναἱΑ,Β,Γ,Δ, the first is greater than (the square on) the second by
ὡςἡΑπρὸςτὴνΒ,οὕτωςἡΓπρὸςτὴνΔ,καὶἡΑμὲν the (square) on (some straight-line) incommensurable
τῆςΒμεῖζονδυνάσθωτῷἀπὸτῆςΕ,ἡδὲΓτῆςΔμεῖζον [in length] with the first, then the square on the third
δυνάσθωτῷἀπὸτῆςΖ·λέγω,ὅτι,εἴτεσύμμετρόςἐστιν will also be greater than (the square on) the fourth by
ἡΑτῇΕ,σύμμετρόςἐστικαὶἡΓτῇΖ,εἴτεἀσύμμετρός the (square) on (some straight-line) incommensurable
ἐστινἡΑτῇΕ,ἀσύμμετρόςἐστικαὶὁΓτῇΖ.
[in length] with the third.
Let A, B, C, D be four proportional straight-lines,
(such that) as A (is) to B, so C (is) to D. And let the
square on A be greater than (the square on) B by the
A
B
294

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
(square) on E, and let the square on C be greater than
(the square on) D by the (square) on F . I say that A
is either commensurable (in length) with E, and C is
also commensurable with F , or A is incommensurable
(in length) with E, and C is also incommensurable with
F .
Α Β Ε
Γ ∆
Ζ
A B E C
᾿ΕπεὶγάρἐστινὡςἡΑπρὸςτὴνΒ,οὕτωςἡΓπρὸςτὴν For since as A is to B, so C (is) to D, thus as the
Δ,ἔστινἄρακαὶὡςτὸἀπὸτῆςΑπρὸςτὸἀπὸτῆςΒ,οὕτως (square) on A is to the (square) on B, so the (square) on
τὸἀπὸτῆςΓπρὸςτὸἀπὸτῆςΔ.ἀλλὰτῷμὲνἀπὸτῆςΑ C (is) to the (square) on D [Prop. 6.22]. But the (sum
ἴσαἐστὶτὰἀπὸτῶνΕ,Β,τῷδὲἀπὸτῆςΓἴσαἐστὶτὰἀπὸ of the squares) on E and B is equal to the (square) on
τῶνΔ,Ζ.ἔστινἄραὡςτὰἀπὸτῶνΕ,Βπρὸςτὸἀπὸτῆς A, and the (sum of the squares) on D and F is equal
Β,οὕτωςτὰἀπὸτῶνΔ,ΖπρὸςτὸἀπὸτῆςΔ·διελόντιἄρα to the (square) on C. Thus, as the (sum of the squares)
ἐστὶνὡςτὸἀπὸτῆςΕπρὸςτὸἀπὸτῆςΒ,οὕτωςτὸἀπὸ on E and B is to the (square) on B, so the (sum of the
τῆςΖπρὸςτὸἀπὸτῆςΔ·ἔστινἄρακαὶὡςἡΕπρὸςτὴν squares) on D and F (is) to the (square) on D. Thus,
Β,οὕτωςἡΖπρὸςτὴνΔ·ἀνάπαλινἄραἐστὶνὡςἡΒπρὸς via separation, as the (square) on E is to the (square)
τὴνΕ,οὕτωςἡΔπρὸςτὴνΖ.ἔστιδὲκαὶὡςἡΑπρὸςτὴν on B, so the (square) on F (is) to the (square) on D
Β,οὕτωςἡΓπρὸςτὴνΔ·δι᾿ἴσουἄραἐστὶνὡςἡΑπρὸς [Prop. 5.17]. Thus, also, as E is to B, so F (is) to D
τὴνΕ,οὕτωςἡΓπρὸςτὴνΖ.εἴτεοὖνσύμμετρόςἐστινἡΑ [Prop. 6.22]. Thus, inversely, as B is to E, so D (is)
τῇΕ,συμμετρόςἐστικαὶἡΓτῇΖ,εἴτεἀσύμμετρόςἐστιν to F [Prop. 5.7 corr.]. But, as A is to B, so C also (is)
ἡΑτῇΕ,ἀσύμμετρόςἐστικαὶἡΓτῇΖ.
᾿Εὰνἄρα,καὶτὰἑξῆς.
to D. Thus, via equality, as A is to E, so C (is) to F
[Prop. 5.22]. Therefore, A is either commensurable (in
length) with E, and C is also commensurable with F , or
A is incommensurable (in length) with E, and C is also
incommensurable with F [Prop. 10.11].
Thus, if, and so on . . . .
. Proposition 15
᾿Εὰν δύο μεγέθη σύμμετρα συντεθῇ, καὶ τὸ ὅλον If two commensurable magnitudes are added together
ἑκατέρῳαὐτῶνσύμμετρονἔσται·κἂντὸὅλονἑνὶαὐτῶν then the whole will also be commensurable with each of
σύμμετρονᾖ,καὶτὰἐξἀρχῆςμεγέθησύμμετραἔσται. them. And if the whole is commensurable with one of
ΣυγκείσθωγὰρδύομεγέθησύμμετρατὰΑΒ,ΒΓ·λέγω, them then the original magnitudes will also be commen-
ὅτικαὶὅλοντὸΑΓἑκατέρῳτῶνΑΒ,ΒΓἐστισύμμετρον. surable (with one another).
For let the two commensurable magnitudes AB and
BC be laid down together. I say that the whole AC is
also commensurable with each of AB and BC.
D
F
295

ËÌÇÁÉÁÏƺ ELEMENTS
Α Β Γ
BOOK 10
A B C
∆
᾿ΕπεὶγὰρσύμμετράἐστιτὰΑΒ,ΒΓ,μετρήσειτιαὐτὰ For since AB and BC are commensurable, some magμέγεθος.
μετρείτω,καὶἔστωτὸΔ.ἐπεὶοὖντὸΔτὰΑΒ, nitude will measure them. Let it (so) measure (them),
ΒΓμετρεῖ,καὶὅλοντὸΑΓμετρήσει.μετρεῖδὲκαὶτὰΑΒ, and let it be D. Therefore, since D measures (both) AB
ΒΓ.τὸΔἄρατὰΑΒ,ΒΓ,ΑΓμετρεῖ·σύμμετρονἄραἐστὶ and BC, it will also measure the whole AC. And it also
τὸΑΓἑκατέρῳτῶνΑΒ,ΒΓ.
measures AB and BC. Thus, D measures AB, BC, and
ἈλλὰδὴτὸΑΓἔστωσύμμετροντῷΑΒ·λέγωδή,ὅτι AC. Thus, AC is commensurable with each of AB and
καὶτὰΑΒ,ΒΓσύμμετράἐστιν. BC [Def. 10.1].
᾿ΕπεὶγὰρσύμμετράἐστιτὰΑΓ,ΑΒ,μετρήσειτιαὐτὰ And so let AC be commensurable with AB. I say that
μέγεθος. μετρείτω,καὶἔστωτὸΔ.ἐπεὶοὖντὸΔτὰΓΑ, AB and BC are also commensurable.
ΑΒμετρεῖ,καὶλοιπὸνἄρατὸΒΓμετρήσει. μετρεῖδὲκαὶ For since AC and AB are commensurable, some magτὸΑΒ·τὸΔἄρατὰΑΒ,ΒΓμετρήσει·σύμμετραἄραἐστὶ
nitude will measure them. Let it (so) measure (them),
τὰΑΒ,ΒΓ.
and let it be D. Therefore, since D measures (both) CA
᾿Εὰνἄραδύομεγέθη,καὶτὰἑξῆς.
and AB, it will thus also measure the remainder BC.
And it also measures AB. Thus, D will measure (both)
AB and BC. Thus, AB and BC are commensurable
[Def. 10.1].
Thus, if two magnitudes, and so on . . . .
¦. Proposition 16
᾿Εὰν δύο μεγέθη ἀσύμμετρα συντεθῇ, καὶ τὸ ὅλον If two incommensurable magnitudes are added to-
ἑκατέρῳαὐτῶνἀσύμμετρονἔσται·κἂντὸὅλονἑνὶαὐτῶν gether then the whole will also be incommensurable with
ἀσύμμετρονᾖ,καὶτὰἐξἀρχῆςμεγέθηἀσύμμετραἔσται. each of them. And if the whole is incommensurable with
one of them then the original magnitudes will also be incommensurable
(with one another).
∆
Α Β Γ
D
D
A B C
Συγκείσθω γὰρ δύο μεγέθη ἀσύμμετρα τὰ ΑΒ, ΒΓ· For let the two incommensurable magnitudes AB and
λέγω,ὅτικαὶὅλοντὸΑΓἑκατέρῳτῶνΑΒ,ΒΓἀσύμμετρόν BC be laid down together. I say that that the whole AC
ἐστιν.
is also incommensurable with each of AB and BC.
Εἰ γὰρ μή ἐστιν ἀσύμμετρα τὰ ΓΑ, ΑΒ, μετρήσει τι For if CA and AB are not incommensurable then
[αὐτὰ]μέγεθος.μετρείτω,εἰδυνατόν,καὶἔστωτὸΔ.ἐπεὶ some magnitude will measure [them]. If possible, let it
οὖντὸΔτὰΓΑ,ΑΒμετρεῖ,καὶλοιπὸνἄρατὸΒΓμετρήσει. (so) measure (them), and let it be D. Therefore, since
μετρεῖδὲκαὶτὸΑΒ·τὸΔἄρατὰΑΒ,ΒΓμετρεῖ.σύμμετρα D measures (both) CA and AB, it will thus also mea-
ἄραἐστὶτὰΑΒ,ΒΓ·ὑπέκειντοδὲκαὶἀσύμμετρα· ὅπερ sure the remainder BC. And it also measures AB. Thus,
ἐστὶνἀδύνατον. οὐκἄρατὰΓΑ,ΑΒμετρήσειτιμέγεθος· D measures (both) AB and BC. Thus, AB and BC are
ἀσύμμετραἄραἐστὶτὰΓΑ,ΑΒ.ὁμοίωςδὴδείξομεν,ὅτικαὶ commensurable [Def. 10.1]. But they were also assumed
τὰΑΓ,ΓΒἀσύμμετράἐστιν.τὸΑΓἄραἑκατέρῳτῶνΑΒ, (to be) incommensurable. The very thing is impossible.
ΒΓἀσύμμετρόνἐστιν.
Thus, some magnitude cannot measure (both) CA and
Ἀλλὰδὴτὸ ΑΓἑνὶτῶν ΑΒ, ΒΓἀσύμμετρονἔστω. AB. Thus, CA and AB are incommensurable [Def. 10.1].
ἔστω δὴ πρότερον τῷ ΑΒ· λέγω, ὅτι καὶ τὰ ΑΒ, ΒΓ So, similarly, we can show that AC and CB are also
ἀσύμμετράἐστιν. εἰγὰρἔσταισύμμετρα,μετρήσειτιαὐτὰ incommensurable. Thus, AC is incommensurable with
μέγεθος. μετρείτω,καὶἔστωτὸΔ.ἐπεὶοὖντὸΔτὰΑΒ, each of AB and BC.
ΒΓμετρεῖ,καὶὅλονἄρατὸΑΓμετρήσει.μετρεῖδὲκαὶτὸ And so let AC be incommensurable with one of AB
ΑΒ·τὸΔἄρατὰΓΑ,ΑΒμετρεῖ. σύμμετραἄραἐστὶτὰ and BC. So let it, first of all, be incommensurable with
296

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ΓΑ,ΑΒ·ὑπέκειτοδὲκαὶἀσύμμετρα·ὅπερἐστὶνἀδύνατον. AB. I say that AB and BC are also incommensurable.
οὐκἄρατὰΑΒ,ΒΓμετρήσειτιμέγεθος·ἀσύμμετραἄρα For if they are commensurable then some magnitude will
ἐστὶτὰΑΒ,ΒΓ.
measure them. Let it (so) measure (them), and let it be
᾿Εὰνἄραδύομεγέθη,καὶτὰἑξῆς.
D. Therefore, since D measures (both) AB and BC, it
will thus also measure the whole AC. And it also measures
AB. Thus, D measures (both) CA and AB. Thus,
CA and AB are commensurable [Def. 10.1]. But they
were also assumed (to be) incommensurable. The very
thing is impossible. Thus, some magnitude cannot measure
(both) AB and BC. Thus, AB and BC are incommensurable
[Def. 10.1].
Thus, if two. . . magnitudes, and so on . . . .
ÄÑÑ.
Lemma
᾿Εὰνπαράτιναεὐθεῖανπαραβληθῇπαραλληλόγραμμον If a parallelogram, † falling short by a square figure, is
ἐλλεῖπονεἴδειτετραγώνῳ,τὸπαραβληθὲνἴσονἐστὶτῷὑπὸ applied to some straight-line then the applied (paralleloτῶνἐκτῆςπαραβολήςγενομένωντμημάτωντῆςεὐθείας.
gram) is equal (in area) to the (rectangle contained) by
the pieces of the straight-line created via the application
(of the parallelogram).
∆
D
Α
Γ
Β
Παρὰ γὰρ εὐθεῖαν τὴν ΑΒ παραβεβλήσθω παραλ- For let the parallelogram AD, falling short by the
ληλόγραμμον τὸ ΑΔ ἐλλεῖπονεἴδει τετραγώνῳ τῷ ΔΒ· square figure DB, have been applied to the straight-line
λέγω,ὅτιἴσονἐστὶτὸΑΔτῷὑπὸτῶνΑΓ,ΓΒ. AB. I say that AD is equal to the (rectangle contained)
Καίἐστιναὐτόθενφανερόν·ἐπεὶγὰρτετράγωνόνἐστι by AC and CB.
τὸΔΒ,ἴσηἐστὶνἡΔΓτῇΓΒ,καίἐστιτὸΑΔτὸὑπὸτῶν And it is immediately obvious. For since DB is a
ΑΓ,ΓΔ,τουτέστιτὸὑπὸτῶνΑΓ,ΓΒ.
square, DC is equal to CB. And AD is the (rectangle
᾿Εὰνἄραπαράτιναεὐθεῖαν,καὶτὰἑξῆς.
contained) by AC and CD—that is to say, by AC and
CB.
Thus, if . . . to some straight-line, and so on . . . .
† Note that this lemma only applies to rectangular parallelograms.
Þ. Proposition 17 †
᾿Εὰν ὦσι δύο εὐθεῖαι ἄνισοι, τῷ δὲ τετράτῳ μέρει If there are two unequal straight-lines, and a (rectτοῦἀπὸτῆςἐλάσσονοςἴσονπαρὰτὴνμείζοναπαραβληθῇ
angle) equal to the fourth part of the (square) on the
ἐλλεῖπονεἴδειτετραγώνῳκαὶεἰςσύμμετρααὐτὴνδιαιρῇ lesser, falling short by a square figure, is applied to the
μήκει,ἡμείζωντῆςἐλάσσονοςμεῖζονδυνήσεταιτῷἀπὸ greater, and divides it into (parts which are) commenσυμμέτουἑαυτῇ[μήκει].
καὶἐὰνἡμείζωντῆςἐλάσσονος surable in length, then the square on the greater will be
μεῖζον δύνηται τῷ ἀπὸ συμμέτρου ἑαυτῇ [μήκει], τῷ δὲ larger than (the square on) the lesser by the (square)
τετράρτῳτοῦἀπὸτῆςἐλάσσονοςἴσονπαρὰτὴνμείζονα on (some straight-line) commensurable [in length] with
παραβληθῇἐλλεῖπονεἴδειτετραγώνῳ,εἰςσύμμετρααὐτὴν the greater. And if the square on the greater is larger
διαιρεῖμήκει.
than (the square on) the lesser by the (square) on
῎ΕστωσανδύοεὐθεῖαιἄνισοιαἱΑ,ΒΓ,ὧνμείζωνἡ (some straight-line) commensurable [in length] with the
A
C
B
297

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ΒΓ, τῷ δὲ τετράρτῳ μέρει τοῦ ἀπὸ ἐλάσσονος τῆς Α, greater, and a (rectangle) equal to the fourth (part) of the
τουτέστιτῷἀπὸτῆςἡμισείαςτῆςΑ,ἴσονπαρὰτὴνΒΓ (square) on the lesser, falling short by a square figure, is
παραβεβλήσθωἐλλεῖπονεἴδειτετραγώνῳ,καὶἔστωτὸὑπὸ applied to the greater, then it divides it into (parts which
τῶνΒΔ,ΔΓ,σύμμετροςδὲἔστωἡΒΔτῇΔΓμήκει·λέγω, are) commensurable in length.
ὃτιἡΒΓτῆςΑμεῖζονδύναταιτῷἀπὸσυμμέτρουἑαυτῇ. Let A and BC be two unequal straight-lines, of which
(let) BC (be) the greater. And let a (rectangle) equal to
the fourth part of the (square) on the lesser, A—that is,
(equal) to the (square) on half of A—falling short by a
square figure, have been applied to BC. And let it be
the (rectangle contained) by BD and DC [see previous
lemma]. And let BD be commensurable in length with
DC. I say that that the square on BC is greater than
the (square on) A by (the square on some straight-line)
commensurable (in length) with (BC).
Α
A
Β Ζ
Ε
∆
Γ
ΤετμήσθωγὰρἡΒΓδίχακατὰτὸΕσημεῖον,καὶκείσθω For let BC have been cut in half at the point E [Prop.
τῇΔΕἴσηἡΕΖ.λοιπὴἄραἡΔΓἴσηἐστὶτῇΒΖ.καὶἐπεὶ 1.10]. And let EF be made equal to DE [Prop. 1.3].
εὐθεῖαἡΒΓτέτμηταιεἰςμὲνἴσακατὰτὸΕ,εἰςδὲἄνισα Thus, the remainder DC is equal to BF . And since the
κατὰτὸΔ,τὸἄραὑπὸΒΔ,ΔΓπερειχόμενονὀρθογώνιον straight-line BC has been cut into equal (pieces) at E,
μετὰτοῦἀπὸτῆςΕΔτετραγώνουἴσονἐστὶτῷἀπὸτῆς and into unequal (pieces) at D, the rectangle contained
ΕΓτετραγώνῳ·καὶτὰτετραπλάσια·τὸἄρατετράκιςὑπὸ by BD and DC, plus the square on ED, is thus equal to
τῶνΒΔ,ΔΓμετὰτοῦτετραπλασίουτοῦἀπὸτῆςΔΕἴσον the square on EC [Prop. 2.5]. (The same) also (for) the
ἐστὶτῷτετράκιςἀπὸτῆςΕΓτετραγώνῳ. ἀλλὰτῷμὲν quadruples. Thus, four times the (rectangle contained)
τετραπλασίῳτοῦὑπὸτῶνΒΔ,ΔΓἴσονἐστὶτὸἀπὸτῆς by BD and DC, plus the quadruple of the (square) on
Ατετράγωνον,τῷδὲτετραπλασίῳτοῦἀπὸτῆςΔΕἴσον DE, is equal to four times the square on EC. But, the
ἐστὶτὸἀπὸτῆςΔΖτετράγωνον·διπλασίωνγάρἐστινἡΔΖ square on A is equal to the quadruple of the (rectangle
τῆςΔΕ.τῷδὲτετραπλασίῳτοῦἀπὸτῆςΕΓἴσονἐστὶτὸ contained) by BD and DC, and the square on DF is
ἀπὸτῆςΒΓτετράγωνον·διπλασίωνγάρἐστιπάλινἡΒΓ equal to the quadruple of the (square) on DE. For DF
τῆςΓΕ.τὰἄραἀπὸτῶνΑ,ΔΖτετράγωναἴσαἐστὶτῷἀπὸ is double DE. And the square on BC is equal to the
τῆςΒΓτετράγωνῳ·ὥστετὸἀπὸτῆςΒΓτοῦἀπὸτῆςΑ quadruple of the (square) on EC. For, again, BC is douμεῖζόνἐστιτῷἀπὸτῆςΔΖ·ἡΒΓἄρατῆςΑμεῖζονδύναται
ble CE. Thus, the (sum of the) squares on A and DF is
τῇΔΖ.δεικτέον,ὅτικαὶσύμμετρόςἐστινἡΒΓτῇΔΖ. equal to the square on BC. Hence, the (square) on BC
ἐπεὶγὰρσύμμετρόςἐστινἡΒΔτῇΔΓμήκει,σύμμετρος is greater than the (square) on A by the (square) on DF .
ἄραἐστὶκαὶἡΒΓτῇΓΔμήκει. ἀλλὰἡΓΔταῖςΓΔ,ΒΖ Thus, BC is greater in square than A by DF . It must
ἐστισύμμετροςμήκει·ἴσηγάρἐστινἡΓΔτῇΒΖ.καὶἡΒΓ also be shown that BC is commensurable (in length)
ἄρασύμμετρόςἐστιταῖςΒΖ,ΓΔμήκει·ὥστεκαὶλοιπῇτῇ with DF . For since BD is commensurable in length
ΖΔσύμμετρόςἐστινἡΒΓμήκει·ἡΒΓἄρατῆςΑμεῖζον with DC, BC is thus also commensurable in length with
δύναταιτῷἀπὸσυμμέτρουἑαυτῇ.
CD [Prop. 10.15]. But, CD is commensurable in length
ἈλλὰδὴἡΒΓτῆςΑμεῖζονδυνάσθωτῷἀπὸσυμμέτρου with CD plus BF . For CD is equal to BF [Prop. 10.6].
ἑαυτῇ,τῷδὲτετράτρῳτοῦἀπὸτῆςΑἴσονπαρὰτὴνΒΓ Thus, BC is also commensurable in length with BF plus
παραβεβλήσθωἐλλεῖπονεἴδειτετραγώνῳ,καὶἔστωτὸὑπὸ CD [Prop. 10.12]. Hence, BC is also commensurable
τῶνΒΔ,ΔΓ.δεικτέον,ὅτισύμμετρόςἐστινἡΒΔτῇΔΓ in length with the remainder FD [Prop. 10.15]. Thus,
μήκει.
the square on BC is greater than (the square on) A by
Τῶνγὰραὐτῶνκατασκευασθέντωνὁμοίωςδείξομεν, the (square) on (some straight-line) commensurable (in
ὅτιἡΒΓτῆςΑμεῖζονδύναταιτῷἀπὸτῆςΖΔ.δύναταιδὲἡ length) with (BC).
B
F
E
D
C
298

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ΒΓτῆςΑμεῖζοντῷἀπὸσυμμέτρουἑαυτῇ.σύμμετροςἄρα And so let the square on BC be greater than the
ἐστὶνἡΒΓτῇΖΔμήκει·ὥστεκαὶλοιπῇσυναμφοτέρῳτῇ (square on) A by the (square) on (some straight-line)
ΒΖ,ΔΓσύμμετρόςἐστινἡΒΓμήκει.ἀλλὰσυναμφότερος commensurable (in length) with (BC). And let a (rect-
ἡΒΖ,ΔΓσύμμετρόςἐστιτῇΔΓ[μήκει]. ὥστεκαὶἡΒΓ angle) equal to the fourth (part) of the (square) on A,
τῇΓΔσύμμετρόςἐστιμήκει·καὶδιελόντιἄραἡΒΔτῇΔΓ falling short by a square figure, have been applied to BC.
ἐστισύμμετροςμήκει.
᾿Εὰνἄραὦσιδύοεὐθεῖαιἄνισοι,καὶτὰἑξῆς.
And let it be the (rectangle contained) by BD and DC. It
must be shown that BD is commensurable in length with
DC.
For, similarly, by the same construction, we can show
that the square on BC is greater than the (square on) A
by the (square) on FD. And the square on BC is greater
than the (square on) A by the (square) on (some straightline)
commensurable (in length) with (BC). Thus, BC
is commensurable in length with FD. Hence, BC is
also commensurable in length with the remaining sum
of BF and DC [Prop. 10.15]. But, the sum of BF and
DC is commensurable [in length] with DC [Prop. 10.6].
Hence, BC is also commensurable in length with CD
[Prop. 10.12]. Thus, via separation, BD is also commensurable
in length with DC [Prop. 10.15].
Thus, if there are two unequal straight-lines, and so
on . . . .
† This proposition states that if α x − x 2 = β 2 /4 (where α = BC, x = DC, and β = A) then α and p α 2 − β 2 are commensurable when α − x
are x are commensurable, and vice versa.
. Proposition 18 †
᾿Εὰν ὦσι δύο εὐθεῖαι ἄνισοι, τῷ δὲ τετάρτῳ μέρει If there are two unequal straight-lines, and a (rectτοῦἀπὸτῆςἐλάσσονοςἴσονπαρὰτὴνμείζοναπαραβληθῇ
angle) equal to the fourth part of the (square) on the
ἐλλεῖπονεἴδειτετραγώνῳ,καὶεἰςἀσυμμετρααὐτὴνδιαιρῇ lesser, falling short by a square figure, is applied to the
[μήκει],ἡμείζωντῆςἐλάσσονοςμεῖζονδυνήσεταιτῷἀπὸ greater, and divides it into (parts which are) incom-
ἀσυμμέτρουἑαυτῇ.καὶἐὰνἡμείζωντῆςἐλάσσονοςμεῖζον mensurable [in length], then the square on the greater
δύνηταιτῷἀπὸἀσυμμέτρουἑαυτῇ,τῷδὲτετράρτῳτοῦἀπὸ will be larger than the (square on the) lesser by the
τῆςἐλάσσονοςἴσονπαρὰτὴνμείζοναπαραβληθῇἐλλεῖπον (square) on (some straight-line) incommensurable (in
εἴδειτετραγώνῳ,εἰςἀσύμμετρααὐτὴνδιαιρεῖ[μήκει]. length) with the greater. And if the square on the
῎ΕστωσανδύοεὐθεῖαιἄνισοιαἱΑ,ΒΓ,ὧνμείζωνἡΒΓ, greater is larger than the (square on the) lesser by the
τῷδὲτετάρτῳ[μέρει]τοῦἀπὸτῆςἐλάσσονοςτῆςΑἴσον (square) on (some straight-line) incommensurable (in
παρὰτὴνΒΓπαραβεβλήσθωἐλλεῖπονεἴδειτετραγώνῳ,καὶ length) with the greater, and a (rectangle) equal to the
ἔστω τὸ ὑπὸ τῶν ΒΔΓ, ἀσύμμετρος δὲ ἔστω ἡ ΒΔ τῇ fourth (part) of the (square) on the lesser, falling short by
ΔΓμήκει·λέγω,ὅτιἡΒΓτῆςΑμεῖζονδύναταιτῷἀπὸ a square figure, is applied to the greater, then it divides it
ἀσυμμέτρουἑαυτῇ.
into (parts which are) incommensurable [in length].
Let A and BC be two unequal straight-lines, of which
(let) BC (be) the greater. And let a (rectangle) equal to
the fourth [part] of the (square) on the lesser, A, falling
short by a square figure, have been applied to BC. And
let it be the (rectangle contained) by BDC. And let BD
be incommensurable in length with DC. I say that that
the square on BC is greater than the (square on) A by
the (square) on (some straight-line) incommensurable
(in length) with (BC).
299

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
Α
A
Β
Ζ
Ε
∆ Γ
B F E
Τῶνγὰραὐτῶνκατασκευασθέντωντῷπρότερονὁμοίως For, similarly, by the same construction as before, we
δείξομεν,ὅτιἡΒΓτῆςΑμεῖζονδύναταιτῷἀπὸτῆςΖΔ. can show that the square on BC is greater than the
δεικτέον[οὖν],ὅτιἀσύμμετρόςἐστινἡΒΓτῇΔΖμήκει. (square on) A by the (square) on FD. [Therefore] it
ἐπεὶγὰρἀσύμμετρόςἐστινἡΒΔτῇΔΓμήκει,ἀσύμμετρος must be shown that BC is incommensurable in length
ἄραἐστὶκαὶἡΒΓτῇΓΔμήκει.ἀλλὰἡΔΓσύμμετρόςἐστι with DF . For since BD is incommensurable in length
συναμφοτέραις ταῖς ΒΖ, ΔΓ· καὶ ἡ ΒΓ ἄρα ἀσύμμετρός with DC, BC is thus also incommensurable in length
ἐστισυναμφοτέραιςταῖςΒΖ,ΔΓ.ὥστεκαὶλοιπῇτῇΖΔ with CD [Prop. 10.16]. But, DC is commensurable (in
ἀσύμμετρόςἔστιν ἡΒΓμήκει. καὶἡΒΓ τῆςΑμεῖζον length) with the sum of BF and DC [Prop. 10.6]. And,
δύναταιτῷἀπὸτῆςΖΔ·ἡΒΓἄρατῆςΑμεῖζονδύναταιτῷ thus, BC is incommensurable (in length) with the sum of
ἀπὸἀσυμμέτρουἑαυτῇ.
BF and DC [Prop. 10.13]. Hence, BC is also incommen-
ΔυνάσθωδὴπάλινἡΒΓτῆςΑμεῖζοντῷἀπὸἀσυμμέτρ- surable in length with the remainder FD [Prop. 10.16].
ουἑαυτῇ,τῷδὲτετάρτῳτοῦἀπὸτῆςΑἴσονπαρὰτὴνΒΓ And the square on BC is greater than the (square on)
παραβεβλήσθωἐλλεῖπονεἴδειτετραγώνῳ,καὶἔστωτὸὑπὸ A by the (square) on FD. Thus, the square on BC is
τῶνΒΔ,ΔΓ.δεικτέον,ὅτιἀσύμμετρόςἐστινἡΒΔτῇΔΓ greater than the (square on) A by the (square) on (some
μήκει.
straight-line) incommensurable (in length) with (BC).
Τῶνγὰραὐτῶνκατασκευασθέντωνὁμοίωςδείξομεν, So, again, let the square on BC be greater than the
ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ τῆς ΖΔ. ἀλλὰ (square on) A by the (square) on (some straight-line) in-
ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. commensurable (in length) with (BC). And let a (rect-
ἀσύμμετροςἄραἐστὶνἡΒΓτῇΖΔμήκει·ὥστεκαὶλοιπῇ angle) equal to the fourth [part] of the (square) on A,
συναμφοτέρῳτῇΒΖ,ΔΓἀσύμμετρόςἐστινἡΒΓ.ἀλλὰσυ- falling short by a square figure, have been applied to BC.
ναμφότεροςἡΒΖ,ΔΓτῇΔΓσύμμετρόςἐστιμήκει·καὶἡ And let it be the (rectangle contained) by BD and DC.
ΒΓἄρατῇΔΓἀσύμμετρόςἐστιμήκει·ὥστεκαὶδιελόντιἡ It must be shown that BD is incommensurable in length
ΒΔτῇΔΓἀσύμμετρόςἐστιμήκει.
᾿Εὰνἄραὦσιδύοεὐθεῖαι,καὶτὰἑξῆς.
with DC.
For, similarly, by the same construction, we can show
that the square on BC is greater than the (square) on
A by the (square) on FD. But, the square on BC is
greater than the (square) on A by the (square) on (some
straight-line) incommensurable (in length) with (BC).
Thus, BC is incommensurable in length with FD. Hence,
BC is also incommensurable (in length) with the remaining
sum of BF and DC [Prop. 10.16]. But, the
sum of BF and DC is commensurable in length with
DC [Prop. 10.6]. Thus, BC is also incommensurable
in length with DC [Prop. 10.13]. Hence, via separation,
BD is also incommensurable in length with DC
[Prop. 10.16].
Thus, if there are two . . . straight-lines, and so on . . . .
† This proposition states that if α x − x 2 = β 2 /4 (where α = BC, x = DC, and β = A) then α and p α 2 − β 2 are incommensurable when
α − x are x are incommensurable, and vice versa.
. Proposition 19
Τὸὑπὸῥητῶνμήκεισυμμέτρωνεὐθειῶνπεριεχόμενον The rectangle contained by rational straight-lines
ὀρθογώνιονῥητόνἐστιν.
(which are) commensurable in length is rational.
῾ΥπὸγὰρῥητῶνμήκεισυμμέτρωνεὐθειῶντῶνΑΒ,ΒΓ For let the rectangle AC have been enclosed by the
D
C
300

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ὀρθογώνιονπεριεχέσθωτὸΑΓ·λέγω,ὅτιῥητόνἐστιτὸ rational straight-lines AB and BC (which are) commen-
ΑΓ.
surable in length. I say that AC is rational.
∆
D
Γ
C
Α
Β
Ἀναγεγράφθω γὰρἀπὸτῆςΑΒτετράγωνοντὸΑΔ· For let the square AD have been described on AB.
ῥητὸνἄραἐστὶτὸΑΔ.καὶἐπεὶσύμμετρόςἐστινἡΑΒτῇ AD is thus rational [Def. 10.4]. And since AB is com-
ΒΓμήκει,ἴσηδέἐστινἡΑΒτῇΒΔ,σύμμετροςἄραἐστὶν mensurable in length with BC, and AB is equal to BD,
ἡΒΔτῇΒΓμήκει.καίἐστινὡςἡΒΔπρὸςτὴνΒΓ,οὕτως BD is thus commensurable in length with BC. And as
τὸΔΑπρὸςτὸΑΓ.σύμμετρονἄραἐστὶτὸΔΑτῷΑΓ. BD is to BC, so DA (is) to AC [Prop. 6.1]. Thus, DA
ῥητὸνδὲτὸΔΑ·ῥητὸνἄραἐστὶκαὶτὸΑΓ.
Τὸἄραὑπὸῥητῶνμήκεισυμμέτρων,καὶτὰἑξῆς.
A
is commensurable with AC [Prop. 10.11]. And DA (is)
rational. Thus, AC is also rational [Def. 10.4]. Thus,
the . . . by rational straight-lines . . . commensurable, and
so on . . . .
. Proposition 20
᾿Εὰνῥητὸνπαρὰῥητὴνπαραβληθῇ,πλάτοςποιεῖῥητὴν If a rational (area) is applied to a rational (straightκαὶσύμμετροντῇ,παρ᾿ἣνπαράκειται,μήκει.
line) then it produces as breadth a (straight-line which is)
rational, and commensurable in length with the (straightline)
to which it is applied.
∆
D
B
Β
Α
B
A
Γ
῾ΡητὸνγὰρτὸΑΓπαρὰῥητὴντὴνΑΒπαραβεβλήσθω For let the rational (area) AC have been applied to the
πλάτος ποιοῦν τὴν ΒΓ· λέγω, ὅτι ῥητή ἐστιν ἡ ΒΓ καὶ rational (straight-line) AB, producing the (straight-line)
σύμμετροςτῇΒΑμήκει.
BC as breadth. I say that BC is rational, and commen-
Ἀναγεγράφθω γὰρἀπὸτῆςΑΒτετράγωνοντὸΑΔ· surable in length with BA.
ῥητὸνἄραἐστὶτὸΑΔ.ῥητὸνδὲκαὶτὸΑΓ·σύμμετρονἄρα For let the square AD have been described on AB.
C
301

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ἐστὶτὸΔΑτῷΑΓ.καίἐστινὡςτὸΔΑπρὸςτὸΑΓ,οὕτως AD is thus rational [Def. 10.4]. And AC (is) also ratio-
ἡΔΒπρὸςτὴνΒΓ.σύμμετροςἄραἐστὶκαὶἡΔΒτῇΒΓ· nal. DA is thus commensurable with AC. And as DA
ἴσηδὲἡΔΒτῇΒΑ·σύμμετροςἄρακαὶἡΑΒτῇΒΓ.ῥητὴ is to AC, so DB (is) to BC [Prop. 6.1]. Thus, DB is
δέἐστινἡΑΒ·ῥητὴἄραἐστὶκαὶἡΒΓκαὶσύμμετροςτῇ also commensurable (in length) with BC [Prop. 10.11].
ΑΒμήκει.
And DB (is) equal to BA. Thus, AB (is) also commen-
᾿Εὰνἄραῥητὸνπαρὰῥητὴνπαραβληθῇ,καὶτὰἑξῆς. surable (in length) with BC. And AB is rational. Thus,
BC is also rational, and commensurable in length with
AB [Def. 10.3].
Thus, if a rational (area) is applied to a rational
(straight-line), and so on . . . .
. Proposition 21
Τὸὑπὸῥητῶνδυνάμειμόνονσυμμέτρωνεὐθειῶνπε- The rectangle contained by rational straight-lines
ριεχόμενονὀρθογώνιονἄλογόνἐστιν,καὶἡδυναμένηαὐτὸ (which are) commensurable in square only is irrational,
ἄλογόςἐστιν,καλείσθωδὲμέση. and its square-root is irrational—let it be called medial. †
∆
D
Β
Α
B
A
Γ
῾Υπὸγὰρῥητῶνδυνάμειμόνονσυμμέτρωνεὐθειῶντῶν For let the rectangle AC be contained by the rational
ΑΒ,ΒΓὀρθογώνιονπεριεχέσθωτὸΑΓ·λέγω,ὅτιἄλογόν straight-lines AB and BC (which are) commensurable in
ἐστιτὸΑΓ,καὶἡδυναμένηαὐτὸἄλογόςἐστιν,καλείσθω square only. I say that AC is irrational, and its squareδὲμέση.
root is irrational—let it be called medial.
Ἀναγεγράφθω γὰρἀπὸτῆςΑΒτετράγωνοντὸΑΔ· For let the square AD have been described on AB.
ῥητὸνἄραἐστὶτὸΑΔ.καὶἐπεὶἀσύμμετρόςἐστινἡΑΒ AD is thus rational [Def. 10.4]. And since AB is incomτῇΒΓμήκει·δυνάμειγὰρμόνονὑπόκεινταισύμμετροι·ἴση
mensurable in length with BC. For they were assumed
δὲἡΑΒτῇΒΔ,ἀσύμμετροςἄραἐστὶκαὶἡΔΒτῇΒΓ to be commensurable in square only. And AB (is) equal
μήκει. καίἐστινὡςἡΔΒπρὸςτὴνΒΓ,οὕτωςτὸΑΔ to BD. DB is thus also incommensurable in length with
πρὸςτὸΑΓ·ἀσύμμετρονἄρα[ἐστὶ]τὸΔΑτῷΑΓ.ῥητὸν BC. And as DB is to BC, so AD (is) to AC [Prop. 6.1].
δὲτὸΔΑ·ἄλογονἄραἐστὶτὸΑΓ·ὥστεκαὶἡδυναμένητὸ Thus, DA [is] incommensurable with AC [Prop. 10.11].
ΑΓ[τουτέστινἡἴσοναὐτῷτετράγωνονδυναμένη]ἄλογός And DA (is) rational. Thus, AC is irrational [Def. 10.4].
ἐστιν,καλείσθωδεμέση·ὅπερἔδειδεῖξαι.
Hence, its square-root [that is to say, the square-root of
the square equal to it] is also irrational [Def. 10.4]—let
it be called medial. (Which is) the very thing it was required
to show.
† Thus, a medial straight-line has a length expressible as k 1/4 .
C
302

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ÄÑÑ.
᾿Εὰν ὦσι δύο εὐθεῖαι, ἔστιν ὡς ἡ πρώτη πρὸς τὴν If there are two straight-lines then as the first is to the
Lemma
δευτέραν,οὕτωςτὸἀπὸτῆςπρώτηςπρὸςτὸὑπὸτῶνδύο second, so the (square) on the first (is) to the (rectangle
εὐθειῶν.
contained) by the two straight-lines.
Ζ Ε Η
F
E
G
∆
῎ΕστωσανδύοεὐθεῖαιαἱΖΕ,ΕΗ.λέγω,ὅτιἐστὶνὡςἡ Let FE and EG be two straight-lines. I say that as
ΖΕπρὸςτὴνΕΗ,οὕτωςτὸἀπὸτῆςΖΕπρὸςτὸὑπὸτῶν FE is to EG, so the (square) on FE (is) to the (rectangle
ΖΕ,ΕΗ.
contained) by FE and EG.
ἈναγεγράφθωγὰρἀπὸτῆςΖΕτετράγωνοντὸΔΖ,καὶ For let the square DF have been described on FE.
συμπεπληρώσθωτὸΗΔ.ἐπεὶοὖνἐστινὡςἡΖΕπρὸςτὴν And let GD have been completed. Therefore, since as
ΕΗ,οὕτωςτὸΖΔπρὸςτὸΔΗ,καίἐστιτὸμὲνΖΔτὸἀπὸ FE is to EG, so FD (is) to DG [Prop. 6.1], and FD is
τῆςΖΕ,τὸδὲΔΗτὸὑπὸτῶνΔΕ,ΕΗ,τουτέστιτὸὑπὸ the (square) on FE, and DG the (rectangle contained)
τῶνΖΕ,ΕΗ,ἔστινἄραὡςἡΖΕπρὸςτὴνΕΗ,οὕτωςτὸ by DE and EG—that is to say, the (rectangle contained)
ἀπὸτῆςΖΕπρὸςτὸὑπὸτῶνΖΕ,ΕΗ.ὁμοίωςδὲκαὶὡςτὸ by FE and EG—thus as FE is to EG, so the (square)
ὑπὸτῶνΗΕ,ΕΖπρὸςτὸἀπὸτῆςΕΖ,τουτέστινὡςτὸΗΔ on FE (is) to the (rectangle contained) by FE and EG.
πρὸςτὸΖΔ,οὕτωςἡΗΕπρὸςτὴνΕΖ·ὅπερἔδειδεῖξαι. And also, similarly, as the (rectangle contained) by GE
and EF is to the (square on) EF —that is to say, as GD
(is) to FD—so GE (is) to EF . (Which is) the very thing
it was required to show.
. Proposition 22
Τὸἀπὸμέσηςπαρὰῥητὴνπαραβαλλόμενονπλάτοςποιεῖ The square on a medial (straight-line), being ap-
ῥητὴνκαὶἀσύμμετροντῇ,παρ᾿ἣνπαράκειται,μήκει. plied to a rational (straight-line), produces as breadth a
(straight-line which is) rational, and incommensurable in
length with the (straight-line) to which it is applied.
Β
B
D
Η
G
Α Γ ∆ Ε Ζ
῎ΕστωμέσημὲνἡΑ,ῥητὴδὲἡΓΒ,καὶτῷἀπὸτῆς Let A be a medial (straight-line), and CB a rational
ΑἴσονπαρὰτὴνΒΓπαραβεβλήσθωχωρίονὀρθογώνιοντὸ (straight-line), and let the rectangular area BD, equal to
ΒΔπλάτοςποιοῦντὴνΓΔ·λέγω,ὅτιῥητήἐστινἡΓΔκαὶ the (square) on A, have been applied to BC, producing
ἀσύμμετροςτῇΓΒμήκει.
CD as breadth. I say that CD is rational, and incommen-
᾿ΕπεὶγὰρμέσηἐστὶνἡΑ,δύναταιχωρίονπεριεχόμενον surable in length with CB.
ὑπὸῥητῶνδυνάμειμόνονσυμμέτρων. δυνάσθωτὸΗΖ. For since A is medial, the square on it is equal to a
A
C
D
E
F
303

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
δύναταιδὲκαὶτὸΒΔ·ἴσονἄραἐστὶτὸΒΔτῷΗΖ.ἔστιδὲ (rectangular) area contained by rational (straight-lines
αὐτῷκαὶἰσογώνιον·τῶνδὲἴσωντεκαὶἰσογωνίωνπαραλ- which are) commensurable in square only [Prop. 10.21].
ληλογράμμωνἀντιπεπόνθασιναἱπλευραὶαἱπερὶτὰςἴσας Let the square on (A) be equal to GF . And the square on
γωνίας·ἀνάλογονἄραἐστὶνὡςἡΒΓπρὸςτὴνΕΗ,οὕτως (A) is also equal to BD. Thus, BD is equal to GF . And
ἡΕΖπρὸςτὴνΓΔ.ἔστινἄρακαὶὡςτὸἀπὸτῆςΒΓπρὸς (BD) is also equiangular with (GF ). And for equal and
τὸἀπὸτῆςΕΗ,οὕτωςτὸἀπὸτῆςΕΖπρὸςτὸἀπὸτῆςΓΔ. equiangular parallelograms, the sides about the equal anσύμμετρονδέἐστιτὸἀπὸτῆςΓΒτῷἀπὸτῆςΕΗ·ῥητὴγάρ
gles are reciprocally proportional [Prop. 6.14]. Thus, pro-
ἐστινἑκατέρααὐτῶν·σύμμετρονἄραἐστὶκαὶτὸἀπὸτῆς portionally, as BC is to EG, so EF (is) to CD. And, also,
ΕΖτῷἀπὸτῆςΓΔ.ῥητὸνδέἐστιτὸἀπὸτῆςΕΖ·ῥητὸν as the (square) on BC is to the (square) on EG, so the
ἄραἐστὶκαὶτὸἀπὸτῆςΓΔ·ῥητὴἄραἐστὶνἡΓΔ.καὶἐπεὶ (square) on EF (is) to the (square) on CD [Prop. 6.22].
ἀσύμμετρόςἐστινἡΕΖτῇΕΗμήκει·δυνάμειγὰρμόνον And the (square) on CB is commensurable with the
εἰσὶσύμμετροι·ὡςδὲἡΕΖπρὸςτὴνΕΗ,οὕτωςτὸἀπὸ (square) on EG. For they are each rational. Thus, the
τῆςΕΖπρὸςτὸὑπὸτῶνΖΕ,ΕΗ,ἀσύμμετρονἄρα[ἐστὶ] (square) on EF is also commensurable with the (square)
τὸἀπὸτῆςΕΖτῷὑπὸτῶνΖΕ,ΕΗ.ἀλλὰτῷμὲνἀπὸτῆς on CD [Prop. 10.11]. And the (square) on EF is ratio-
ΕΖσύμμετρόνἐστιτὸἀπὸτῆςΓΔ·ῥηταὶγάρεἰσιδυνάμει· nal. Thus, the (square) on CD is also rational [Def. 10.4].
τῷδὲὑπὸτῶνΖΕ,ΕΗσύμμετρόνἐστιτὸὑπὸτῶνΔΓ, Thus, CD is rational. And since EF is incommensurable
ΓΒ·ἴσαγάρἐστιτῷἀπὸτῆςΑ·ἀσύμμετρονἄραἐστὶκαὶ in length with EG. For they are commensurable in square
τὸἀπὸτῆςΓΔτῷὑπὸτῶνΔΓ,ΓΒ.ὡςδὲτὸἀπὸτῆςΓΔ only. And as EF (is) to EG, so the (square) on EF (is)
πρὸςτὸὑπὸτῶνΔΓ,ΓΒ,οὕτωςἐστὶνἡΔΓπρὸςτὴνΓΒ· to the (rectangle contained) by FE and EG [see previ-
ἀσύμμετροςἄραἐστὶνἡΔΓτῇΓΒμήκει.ῥητὴἄραἐστὶνἡ ous lemma]. The (square) on EF [is] thus incommen-
ΓΔκαὶἀσύμμετροςτῇΓΒμήκει·ὅπερἔδειδεῖξαι. surable with the (rectangle contained) by FE and EG
[Prop. 10.11]. But, the (square) on CD is commensurable
with the (square) on EF . For they are rational in
square. And the (rectangle contained) by DC and CB
is commensurable with the (rectangle contained) by FE
and EG. For they are (both) equal to the (square) on A.
Thus, the (square) on CD is also incommensurable with
the (rectangle contained) by DC and CB [Prop. 10.13].
And as the (square) on CD (is) to the (rectangle contained)
by DC and CB, so DC is to CB [see previous
lemma]. Thus, DC is incommensurable in length with
CB [Prop. 10.11]. Thus, CD is rational, and incommensurable
in length with CB. (Which is) the very thing it
was required to show.
† Literally, “rational”.
.
Proposition 23
῾Ητῇμέσῃσύμμετροςμέσηἐστίν.
A (straight-line) commensurable with a medial (straight-
῎ΕστωμέσηἡΑ,καὶτῇΑσύμμετροςἔστωἡΒ·λέγω, line) is medial.
ὅτικαὶἡΒμέσηἐστίν.
Let A be a medial (straight-line), and let B be com-
᾿Εκκείσθω γὰρ ῥητὴ ἡ ΓΔ, καὶ τῷ μὲν ἀπὸ τῆς Α mensurable with A. I say that B is also a medial (staight-
ἴσονπαρὰτὴνΓΔπαραβεβλήσθωχωρίονὀρθογώνιοντὸ line).
ΓΕ πλάτος ποιοῦν τὴν ΕΔ· ῥητὴ ἄρα ἐστὶν ἡ ΕΔ καὶ Let the rational (straight-line) CD be set out, and let
ἀσύμμετρος τῂ ΓΔ μήκει. τῷ δὲ ἀπὸ τῆς Β ἴσον παρὰ the rectangular area CE, equal to the (square) on A,
τὴνΓΔπαραβεβλήσθωχωρίονὀρθογώνιοντὸΓΖπλάτος have been applied to CD, producing ED as width. ED
ποιοῦν τὴν ΔΖ. ἐπεὶ οὖν σύμμετρός ἐστιν ἡ Α τῇ Β, is thus rational, and incommensurable in length with CD
σύμμετρόν ἐστι καὶ τὸ ἀπὸ τῆς Α τῷ ἀπὸ τῆς Β. ἀλλὰ [Prop. 10.22]. And let the rectangular area CF , equal
τῷ μὲν ἀπὸ τῆς Α ἴσον ἐστὶ τὸ ΕΓ, τῷ δὲ ἀπὸ τῆς Β to the (square) on B, have been applied to CD, produc-
ἴσονἐστὶτὸΓΖ·σύμμετρονἄραἐστὶτὸΕΓτῷΓΖ.καί ing DF as width. Therefore, since A is commensurable
ἐστινὡςτὸΕΓπρὸςτὸΓΖ,οὕτωςἡΕΔπρὸςτὴνΔΖ· with B, the (square) on A is also commensurable with
304

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
σύμμετροςἄραἐστὶνἡΕΔτῇΔΖμήκει. ῥητὴδέἐστιν the (square) on B. But, EC is equal to the (square) on A,
ἡΕΔκαὶἀσύμμετροςτῇΔΓμήκει·ῥητὴἄραἐστὶκαὶἡ and CF is equal to the (square) on B. Thus, EC is com-
ΔΖκαὶἀσύμμετροςτῇΔΓμήκει· αἱΓΔ,ΔΖἄραῥηταί mensurable with CF . And as EC is to CF , so ED (is) to
εἰσιδυνάμειμόνονσύμμετροι. ἡδὲτὸὑπὸῥητῶνδυνάμει DF [Prop. 6.1]. Thus, ED is commensurable in length
μόνονσυμμέτρωνδυναμένημέσηἐστίν. ἡἄρατὸὑπὸτῶν with DF [Prop. 10.11]. And ED is rational, and incom-
ΓΔ,ΔΖδυναμένημέσηἐστίν·καὶδύναταιτὸὑπὸτῶνΓΔ, mensurable in length with CD. DF is thus also ratio-
ΔΖἡΒ·μέσηἄραἐστὶνἡΒ.
nal [Def. 10.3], and incommensurable in length with DC
[Prop. 10.13]. Thus, CD and DF are rational, and commensurable
in square only. And the square-root of a (rectangle
contained) by rational (straight-lines which are)
commensurable in square only is medial [Prop. 10.21].
Thus, the square-root of the (rectangle contained) by CD
and DF is medial. And the square on B is equal to the
(rectangle contained) by CD and DF . Thus, B is a medial
(straight-line).
Α
Γ
Β
A
C
B
Ε ∆ Ζ E D F
ÈÖ×Ñ.
Corollary
᾿Εκδὴτούτουφανερόν,ὅτιτὸτῷμέσῳχωρίῳσύμμετρ- And (it is) clear, from this, that an (area) commensuονμέσονἐστίν.
rable with a medial area † is medial.
† A medial area is equal to the square on some medial straight-line. Hence, a medial area is expressible as k 1/2 .
. Proposition 24
Τὸὑπὸμέσωνμήκεισυμμέτρωνεὐθειῶνπεριεχόμενον A rectangle contained by medial straight-lines (which
ὀρθογώνιονμέσονἐστίν.
are) commensurable in length is medial.
῾ΥπὸγὰρμέσωνμήκεισυμμέτρωνεὐθειῶντῶνΑΒ,ΒΓ For let the rectangle AC be contained by the medial
περιεχέσθωὀρθογώνιοντὸΑΓ·λέγω, ὅτιτὸΑΓμέσον straight-lines AB and BC (which are) commensurable in
ἐστίν.
length. I say that AC is medial.
Ἀναγεγράφθω γὰρἀπὸτῆςΑΒτετράγωνοντὸΑΔ· For let the square AD have been described on AB.
μέσονἄραἐστὶτὸΑΔ.καὶἐπεὶσύμμετρόςἐστινἡΑΒτῇ AD is thus medial [see previous footnote]. And since
ΒΓμήκει,ἴσηδὲἡΑΒτῇΒΔ,σύμμετροςἄραἐστὶκαὶἡ AB is commensurable in length with BC, and AB (is)
ΔΒτῇΒΓμήκει·ὥστεκαὶτὸΔΑτῷΑΓσύμμετρόνἐστιν. equal to BD, DB is thus also commensurable in length
μέσονδὲτὸΔΑ·μέσονἄρακαὶτὸΑΓ·ὅπερἔδειδεῖξαι. with BC. Hence, DA is also commensurable with AC
[Props. 6.1, 10.11]. And DA (is) medial. Thus, AC (is)
also medial [Prop. 10.23 corr.]. (Which is) the very thing
it was required to show.
305

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
Γ
C
Α
Β
A
B
∆
D
. Proposition 25
Τὸὑπὸμέσωνδυνάμειμόνονσυμμέτρωνεὐθειῶνπε- The rectangle contained by medial straight-lines
ριεχόμενονὀρθογώνιονἤτοιῥητὸνἢμέσονἐστίν. (which are) commensurable in square only is either rational
or medial.
Α
A
Ζ Η
F G
∆
Β
Γ
Θ
Μ
D
B
C
H
M
Ξ
Ε
Κ
Λ
Ν
῾Υπὸγὰρμέσωνδυνάμειμόνονσυμμέτρωνεὐθειῶντῶν For let the rectangle AC be contained by the medial
ΑΒ,ΒΓὀρθογώνιονπεριεχέσθωτὸΑΓ·λέγω,ὅτιτὸΑΓ straight-lines AB and BC (which are) commensurable in
ἤτοιῥητὸνἢμέσονἐστίν.
square only. I say that AC is either rational or medial.
ἈναγεγράφθωγὰρἀπὸτῶνΑΒ,ΒΓτετράγωνατὰΑΔ, For let the squares AD and BE have been described
ΒΕ·μέσονἄραἐστὶνἑκάτεροντῶνΑΔ,ΒΕ.καὶἐκκείσθω on (the straight-lines) AB and BC (respectively). AD
ῥητὴ ἡ ΖΗ, καὶ τῷ μὲν ΑΔ ἴσον παρὰ τὴν ΖΗ παρα- and BE are thus each medial. And let the rational
βεβλήσθωὀρθογώνιονπαραλληλόγραμμοντὸΗΘπλάτος (straight-line) FG be laid out. And let the rectangular
ποιοῦντὴνΖΘ,τῷδὲΑΓἴσονπαρὰτὴνΘΜπαραβεβλήσθω parallelogram GH, equal to AD, have been applied to
ὀρθογώνιονπαραλληλόγραμμοντὸΜΚπλάτοςποιοῦντὴν FG, producing FH as breadth. And let the rectangular
ΘΚ, καὶ ἔτι τῷ ΒΕ ἴσον ὁμοίως παρὰ τὴν ΚΝ παρα- parallelogram MK, equal to AC, have been applied to
βεβλήσθωτὸΝΛπλάτοςποιοῦντὴνΚΛ·ἐπ᾿εὐθείαςἄρα HM, producing HK as breadth. And, finally, let NL,
εἰσὶναἱΖΘ,ΘΚ,ΚΛ.ἐπεὶοὖνμέσονἐστὶνἑκάτεροντῶν equal to BE, have similarly been applied to KN, pro-
ΑΔ,ΒΕ,καίἐστινἴσοντὸμὲνΑΔτῷΗΘ,τὸδὲΒΕτῷ ducing KL as breadth. Thus, FH, HK, and KL are in
ΝΛ,μέσονἄρακαὶἑκάτεροντῶνΗΘ,ΝΛ.καὶπαρὰῥητὴν a straight-line. Therefore, since AD and BE are each
τὴνΖΗπαράκειται·ῥητὴἄραἐστὶνἑκατέρατῶνΖΘ,ΚΛκαὶ medial, and AD is equal to GH, and BE to NL, GH
ἀσύμμετροςτῇΖΗμήκει. καὶἐπεὶσύμμετρόνἐστιτὸΑΔ and NL (are) thus each also medial. And they are apτῷΒΕ,σύμμετρονἄραἐστὶκαὶτὸΗΘτῷΝΛ.καίἐστινὡς
plied to the rational (straight-line) FG. FH and KL are
τὸΗΘπρὸςτὸΝΛ,οὕτωςἡΖΘπρὸςτὴνΚΛ·σύμμετρος thus each rational, and incommensurable in length with
ἄραἐστὶνἡΖΘτῇΚΛμήκει. αἱΖΘ,ΚΛἄραῥηταίεἰσι FG [Prop. 10.22]. And since AD is commensurable with
μήκεισύμμετροι·ῥητὸνἄραἐστὶτὸὑπὸτῶνΖΘ,ΚΛ.καὶ BE, GH is thus also commensurable with NL. And as
O
E
K
L
N
306

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ἐπεὶἴσηἐστὶνἡμὲνΔΒτῇΒΑ,ἡδὲΞΒτῇΒΓ,ἔστινἄρα GH is to NL, so FH (is) to KL [Prop. 6.1]. Thus, FH is
ὡςἡΔΒπρὸςτὴνΒΓ,οὕτωςἡΑΒπρὸςτὴνΒΞ.ἀλλ᾿ὡς commensurable in length with KL [Prop. 10.11]. Thus,
μὲνἡΔΒπρὸςτὴνΒΓ,οὕτωςτὸΔΑπρὸςτὸΑΓ·ὡςδὲἡ FH and KL are rational (straight-lines which are) com-
ΑΒπρὸςτὴνΒΞ,οὕτωςτὸΑΓπρὸςτὸΓΞ·ἔστινἄραὡς mensurable in length. Thus, the (rectangle contained) by
τὸΔΑπρὸςτὸΑΓ,οὕτωςτὸΑΓπρὸςτὸΓΞ.ἴσονδέἐστι FH and KL is rational [Prop. 10.19]. And since DB is
τὸμὲνΑΔτῷΗΘ,τὸδὲΑΓτῷΜΚ,τὸδὲΓΞτῷΝΛ· equal to BA, and OB to BC, thus as DB is to BC, so
ἔστινἄραὡςτὸΗΘπρὸςτὸΜΚ,οὕτωςτὸΜΚπρὸςτὸ AB (is) to BO. But, as DB (is) to BC, so DA (is) to
ΝΛ·ἔστινἄρακαὶὡςἡΖΘπρὸςτὴνΘΚ,οὕτωςἡΘΚπρὸς AC [Props. 6.1]. And as AB (is) to BO, so AC (is) to
τὴνΚΛ·τὸἄραὑπὸτῶνΖΘ,ΚΛἴσονἐστὶτῷἀπὸτῆςΘΚ. CO [Prop. 6.1]. Thus, as DA is to AC, so AC (is) to
ῥητὸνδὲτὸὑπὸτῶνΖΘ,ΚΛ·ῥητὸνἄραἐστὶκαὶτὸἀπὸ CO. And AD is equal to GH, and AC to MK, and CO
τῆςΘΚ·ῥητὴἄραἐστὶνἡΘΚ.καὶεἰμὲνσύμμετρόςἐστι to NL. Thus, as GH is to MK, so MK (is) to NL. Thus,
τῇΖΗμήκει,ῥητόνἐστιτὸΘΝ·εἰδὲἀσύμμετρόςἐστιτῇ also, as FH is to HK, so HK (is) to KL [Props. 6.1,
ΖΗμήκει,αἱΚΘ,ΘΜῥηταίεἰσιδυνάμειμόνονσύμμετροι· 5.11]. Thus, the (rectangle contained) by FH and KL
μέσονἄρατὸΘΝ.τὸΘΝἄραἤτοιῥητὸνἢμέσονἐστίν. is equal to the (square) on HK [Prop. 6.17]. And the
ἴσονδὲτὸΘΝτῷΑΓ·τὸΑΓἄραἤτοιῥητὸνἢμέσονἐστίν. (rectangle contained) by FH and KL (is) rational. Thus,
Τὸἄραὑπὸμέσωνδυνάμειμόνονσυμμέτρων,καὶτὰ the (square) on HK is also rational. Thus, HK is ratioεξῆς.
nal. And if it is commensurable in length with FG then
HN is rational [Prop. 10.19]. And if it is incommensurable
in length with FG then KH and HM are rational
(straight-lines which are) commensurable in square only:
thus, HN is medial [Prop. 10.21]. Thus, HN is either rational
or medial. And HN (is) equal to AC. Thus, AC is
either rational or medial.
Thus, the . . . by medial straight-lines (which are) commensurable
in square only, and so on . . . .
¦. Proposition
Μέσονμέσουοὐχὑπερέχειῥητῷ.
Α
Ζ
Ε
26
A medial (area) does not exceed a medial (area) by a
rational (area). †
A
F
E
∆
Γ
D
C
Β
Κ
Η
B
K
G
Θ
Εἰγὰρδυνατόν,μέσοντὸΑΒμέσουτοῦΑΓὑπερεχέτω For, if possible, let the medial (area) AB exceed the
ῥητῷτῷΔΒ,καὶἐκκείσθωῥητὴἡΕΖ,καὶτῷΑΒἴσονπαρὰ medial (area) AC by the rational (area) DB. And let
τὴνΕΖπαραβεβλήσθωπαραλληλόγραμμονὀρθογώνιοντὸ the rational (straight-line) EF be laid down. And let the
ΖΘπλάτοςποιοῦντὴνΕΘ,τῷδὲΑΓἴσονἀφῃρήσθωτὸ rectangular parallelogram FH, equal to AB, have been
ΖΗ·λοιπὸνἄρατὸΒΔλοιπῷτῷΚΘἐστινἴσον.ῥητὸνδέ applied to to EF , producing EH as breadth. And let FG,
ἐστιτὸΔΒ·ῥητὸνἄραἐστὶκαὶτὸΚΘ.ἐπεὶοὖνμέσονἐστὶν equal to AC, have been cut off (from FH). Thus, the
ἑκάτεροντῶνΑΒ,ΑΓ,καίἐστιτὸμὲνΑΒτῷΖΘἴσον,τὸ remainder BD is equal to the remainder KH. And DB
δὲΑΓτῷΖΗ,μέσονἄρακαὶἑκάτεροντῶνΖΘ,ΖΗ.καὶ is rational. Thus, KH is also rational. Therefore, since
παρὰῥητὴντὴνΕΖπαράκειται·ῥητὴἄραἐστὶνἑκατέρατῶν AB and AC are each medial, and AB is equal to FH,
ΘΕ,ΕΗκαὶἀσύμμετροςτῇΕΖμήκει.καὶἐπεὶῥητόνἐστι and AC to FG, FH and FG are thus each also medial.
H
307

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
τὸΔΒκαίἐστινἴσοντῷΚΘ,ῥητὸνἄραἐστὶκαὶτὸΚΘ. And they are applied to the rational (straight-line) EF .
καὶπαρὰῥητὴντὴνΕΖπαράκειται·ῥητὴἄραἐστὶνἡΗΘ Thus, HE and EG are each rational, and incommensuκαὶσύμμετροςτῇΕΖμήκει.
ἀλλάκαὶἡΕΗῥητήἐστικαὶ rable in length with EF [Prop. 10.22]. And since DB
ἀσύμμετροςτῇΕΖμήκει·ἀσύμμετροςἄραἐστὶνἡΕΗτῇ is rational, and is equal to KH, KH is thus also ratio-
ΗΘμήκει.καίἐστινὡςἡΕΗπρὸςτὴνΗΘ,οὕτωςτὸἀπὸ nal. And (KH) is applied to the rational (straight-line)
τῆςΕΗπρὸςτὸὑπὸτῶνΕΗ,ΗΘ·ἀσύμμετρονἄραἐστὶτὸ EF . GH is thus rational, and commensurable in length
ἀπὸτῆςΕΗτῷὑπὸτῶνΕΗ,ΗΘ.ἀλλὰτῷμὲνἀπὸτῆςΕΗ with EF [Prop. 10.20]. But, EG is also rational, and inσύμμετράἐστιτὰἀπὸτῶνΕΗ,ΗΘτετράγωνα·ῥητὰγὰρ
commensurable in length with EF . Thus, EG is incom-
ἀμφότερα·τῷδὲὑπὸτῶνΕΗ,ΗΘσύμμετρόνἐστιτὸδὶς mensurable in length with GH [Prop. 10.13]. And as
ὑπὸτῶνΕΗ,ΗΘ·διπλάσιονγάρἐστιναὐτοῦ·ἀσύμμετρα EG is to GH, so the (square) on EG (is) to the (rectan-
ἄραἐστὶτὰἀπὸτῶνΕΗ,ΗΘτῷδὶςὑπὸτῶνΕΗ,ΗΘ·καὶ gle contained) by EG and GH [Prop. 10.13 lem.]. Thus,
συναμφότεραἄρατάτεἀπὸτῶνΕΗ,ΗΘκαὶτὸδὶςὑπὸ the (square) on EG is incommensurable with the (rectτῶνΕΗ,ΗΘ,ὅπερἐστὶτὸἀπὸτῆςΕΘ,ἀσύμμετρόνἐστι
angle contained) by EG and GH [Prop. 10.11]. But, the
τοῖςἀπὸτῶνΕΗ,ΗΘ.ῥητὰδὲτὰἀπὸτῶνΕΗ,ΗΘ·ἄλογον (sum of the) squares on EG and GH is commensurable
ἄρατὸἀπὸτῆςΕΘ.ἄλογοςἄραἐστὶνἡΕΘ.ἀλλὰκαὶῥηρή· with the (square) on EG. For (EG and GH are) both
ὅπερἐστὶνἀδύνατον.
rational. And twice the (rectangle contained) by EG and
Μέσονἄραμέσουοὐχὑπερέχειῥητῷ·ὅπερἔδειδεῖξαι. GH is commensurable with the (rectangle contained) by
EG and GH [Prop. 10.6]. For (the former) is double
the latter. Thus, the (sum of the squares) on EG and
GH is incommensurable with twice the (rectangle contained)
by EG and GH [Prop. 10.13]. And thus the sum
of the (squares) on EG and GH plus twice the (rectangle
contained) by EG and GH, that is the (square) on
EH [Prop. 2.4], is incommensurable with the (sum of
the squares) on EG and GH [Prop. 10.16]. And the (sum
of the squares) on EG and GH (is) rational. Thus, the
(square) on EH is irrational [Def. 10.4]. Thus, EH is
irrational [Def. 10.4]. But, (it is) also rational. The very
thing is impossible.
Thus, a medial (area) does not exceed a medial (area)
by a rational (area). (Which is) the very thing it was
required to show.
† In other words, √ k − √ k ′ ≠ k ′′ .
Þ. Proposition 27
Μέσας εὑρεῖν δυνάμει μόνον συμμέτρους ῥητὸν πε- To find (two) medial (straight-lines), containing a raριεχούσας.
tional (area), (which are) commensurable in square only.
Α Β Γ
∆
A B
C
D
᾿ΕκκείσθωσανδύοῥηταὶδυνάμειμόνονσύμμετροιαἱΑ, Let the two rational (straight-lines) A and B, (which
Β,καὶεἰλήφθωτῶνΑ,ΒμέσηἀνάλογονἡΓ,καὶγεγονέτω are) commensurable in square only, be laid down. And let
ὡςἡΑπρὸςτὴνΒ,οὕτωςἡΓπρὸςτὴνΔ. C—the mean proportional (straight-line) to A and B—
308

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ΚαὶἐπεὶαἱΑ,Βῥηταίεἰσιδυνάμειμόνονσύμμετροι, have been taken [Prop. 6.13]. And let it be contrived
τὸἄραὑπὸτῶνΑ,Β,τουτέστιτὸἀπὸτῆςΓ,μέσονἐστίν. that as A (is) to B, so C (is) to D [Prop. 6.12].
μέσηἄραἡΓ.καὶἐπείἐστινὡςἡΑπρὸςτὴνΒ,[οὕτως]ἡΓ And since the rational (straight-lines) A and B
πρὸςτὴνΔ,αἱδὲΑ,Βδυνάμειμόνον[εἰσὶ]σύμμετροι,καὶ are commensurable in square only, the (rectangle conαἱΓ,Δἄραδυνάμειμόνονεἰσὶσύμμετροι.καίἐστιμέσηἡ
tained) by A and B—that is to say, the (square) on C
Γ·μέσηἄρακαὶἡΔ.αἱΓ,Δἄραμέσαιεἰσὶδυνάμειμόνον [Prop. 6.17]—is thus medial [Prop 10.21]. Thus, C is
σύμμετροι.λέγω,ὅτικαὶῥητὸνπεριέχουσιν.ἐπεὶγάρἐστιν medial [Prop. 10.21]. And since as A is to B, [so] C (is)
ὡςἡΑπρὸςτὴνΒ,οὕτωςἡΓπρὸςτὴνΔ,ἐναλλὰξἄρα to D, and A and B [are] commensurable in square only,
ἐστὶνὡςἡΑπρὸςτὴνΓ,ἡΒπρὸςτὴνΔ.ἀλλ᾿ὡςἡΑ C and D are thus also commensurable in square only
πρὸςτὴνΓ,ἡΓπρὸςτὴνΒ·καὶὡςἄραἡΓπρὸςτὴνΒ, [Prop. 10.11]. And C is medial. Thus, D is also medial
οὕτωςἡΒπρὸςτὴνΔ·τὸἄραὑπὸτῶνΓ,Δἴσονἐστὶτῷ [Prop. 10.23]. Thus, C and D are medial (straight-lines
ἀπὸτῆςΒ.ῥητὸνδὲτὸἀπὸτῆςΒ·ῥητὸνἄρα[ἐστὶ]καὶτὸ which are) commensurable in square only. I say that they
ὑπὸτῶνΓ,Δ.
also contain a rational (area). For since as A is to B, so
Εὕρηνται ἄρα μέσαι δυνάμει μόνον σύμμετροι ῥητὸν C (is) to D, thus, alternately, as A is to C, so B (is) to
περιέχουσαι·ὅπερἔδειδεῖξαι. D [Prop. 5.16]. But, as A (is) to C, (so) C (is) to B.
And thus as C (is) to B, so B (is) to D [Prop. 5.11].
Thus, the (rectangle contained) by C and D is equal to
the (square) on B [Prop. 6.17]. And the (square) on B
(is) rational. Thus, the (rectangle contained) by C and
D [is] also rational.
Thus, (two) medial (straight-lines, C and D), containing
a rational (area), (which are) commensurable in
square only, have been found. † (Which is) the very thing
it was required to show.
† C and D have lengths k 1/4 and k 3/4 times that of A, respectively, where the length of B is k 1/2 times that of A.
. Proposition 28
Μέσας εὑρεῖν δυνάμει μόνον συμμέτρους μέσον πει- To find (two) medial (straight-lines), containing a meριεχούσας.
dial (area), (which are) commensurable in square only.
Α ∆
A
D
Β Ε
B
E
Γ
C
᾿Εκκείσθωσαν[τρεῖς]ῥηταὶδυνάμειμόνονσύμμετροιαἱ Let the [three] rational (straight-lines) A, B, and C,
Α,Β,Γ,καὶεἰλήφθωτῶνΑ,ΒμέσηἀνάλογονἡΔ,καὶ (which are) commensurable in square only, be laid down.
γεγονέτωὡςἡΒπρὸςτὴνΓ,ἡΔπρὸςτὴνΕ.
And let, D, the mean proportional (straight-line) to A
᾿ΕπεὶαἱΑ,Βῥηταίεἰσιδυνάμειμόνονσύμμετροι,τὸἄρα and B, have been taken [Prop. 6.13]. And let it be con-
ὑπὸτῶνΑ,Β,τουτέστιτὸἀπὸτῆςΔ,μέσονἐστὶν. μέση trived that as B (is) to C, (so) D (is) to E [Prop. 6.12].
ἄραἡΔ.καὶἐπεὶαἱΒ,Γδυνάμειμόνονεἰσὶσύμμετροι,καί Since the rational (straight-lines) A and B are com-
ἐστινὡςἡΒπρὸςτὴνΓ,ἡΔπρὸςτὴνΕ,καὶαἱΔ,Εἄρα mensurable in square only, the (rectangle contained) by
δυνάμειμόνονεἰσὶσύμμετροι.μέσηδὲἡΔ·μέσηἄρακαὶἡ A and B—that is to say, the (square) on D [Prop. 6.17]—
Ε·αἱΔ,Εἄραμέσαιεἰσὶδυνάμειμόνονσύμμετροι. λέγω is medial [Prop. 10.21]. Thus, D (is) medial [Prop. 10.21].
δή,ὅτικαὶμέσονπεριέχουσιν.ἐπεὶγάρἐστινὡςἡΒπρὸς And since B and C are commensurable in square only,
τὴνΓ,ἡΔπρὸςτὴνΕ,ἐναλλὰξἄραὡςἡΒπρὸςτὴνΔ,ἡ and as B is to C, (so) D (is) to E, D and E are thus com-
ΓπρὸςτὴνΕ.ὡςδὲἡΒπρὸςτὴνΔ,ἡΔπρὸςτὴνΑ·καὶ mensurable in square only [Prop. 10.11]. And D (is) me-
ὡςἅραἡΔπρὸςτὴνΑ,ἡΓπρὸςτὴνΕ·τὸἄραὑπὸτῶν dial. E (is) thus also medial [Prop. 10.23]. Thus, D and
Α,ΓἴσονἐστὶτῷὑπὸτῶνΔ,Ε.μέσονδὲτὸὑπὸτῶνΑ, E are medial (straight-lines which are) commensurable
Γ·μέσονἄρακαὶτὸὑπὸτῶνΔ,Ε.
in square only. So, I say that they also enclose a medial
᾿Εὕρηνται ἄρα μέσαι δυνάμει μόνον σύμμετροι μέσον (area). For since as B is to C, (so) D (is) to E, thus,
309

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
περιέχουσαι·ὅπερἔδειδεῖξαι. alternately, as B (is) to D, (so) C (is) to E [Prop. 5.16].
And as B (is) to D, (so) D (is) to A. And thus as D (is) to
A, (so) C (is) to E. Thus, the (rectangle contained) by A
and C is equal to the (rectangle contained) by D and E
[Prop. 6.16]. And the (rectangle contained) by A and C
is medial [Prop. 10.21]. Thus, the (rectangle contained)
by D and E (is) also medial.
Thus, (two) medial (straight-lines, D and E), containing
a medial (area), (which are) commensurable in
ÄÑѺ
square only, have been found. (Which is) the very thing
it was required to show.
† D and E have lengths k 1/4 and k ′1/2 /k 1/4 times that of A, respectively, where the lengths of B and C are k 1/2 and k ′1/2 times that of A,
respectively.
Lemma I
Εὑρειν δύο τετραγώνους ἀριθμούς, ὥστε καὶ τὸν To find two square numbers such that the sum of them
συγκείμενονἐξαὐτῶνεἶναιτετράγωνον.
is also square.
Α ∆ Γ Β A D
᾿ΕκκείσθωσανδύοἀριθμοὶοἱΑΒ,ΒΓ,ἔστωσανδὲἤτοι Let the two numbers AB and BC be laid down. And
ἄρτιοιἢπεριττοί. καὶἐπεὶ,ἐάντεἀπὸἀρτίουἄρτιοςἀφαι- let them be either (both) even or (both) odd. And since, if
ρεθῇ,ἐάντεἀπὸπερισσοῦπερισσός,ὁλοιπὸςἄρτιόςἐστιν, an even (number) is subtracted from an even (number),
ὁλοιπὸςἄραὁΑΓἄρτιόςἐστιν.τετμήσθωὁΑΓδίχακατὰ or if an odd (number is subtracted) from an odd (numτὸΔ.ἔστωσανδὲκαὶοἱΑΒ,ΒΓἤτοιὅμοιοιἐπίπεδοιἢ
ber), then the remainder is even [Props. 9.24, 9.26], the
τετράγωνοι, οἳκαὶαὐτοὶὅμοιοίεἰσινἐπίπεδοι· ὁἄραἐκ remainder AC is thus even. Let AC have been cut in
τῶνΑΒ,ΒΓμετὰτοῦἀπὸ[τοῦ]ΓΔτετραγώνουἴσοςἐστὶ half at D. And let AB and BC also be either similar
τῷἀπὸτοῦΒΔτετραγώνῳ.καίἐστιτετράγωνοςὁἐκτῶν plane (numbers), or square (numbers)—which are them-
ΑΒ,ΒΓ,ἐπειδήπερἐδείχθη,ὅτι,ἐὰνδύοὅμοιοιἐπίπεδοι selves also similar plane (numbers). Thus, the (numπολλαπλασιάσαντες
ἀλλήλους ποιῶσι τινα, ὁ γενόμενος ber created) from (multiplying) AB and BC, plus the
τετράγωνόςἐστιν. εὕρηνταιἄραδύοτετράγωνοιἀριθμοὶ square on CD, is equal to the square on BD [Prop. 2.6].
ὅτεἐκτῶνΑΒ,ΒΓκαὶὁἀπὸτοῦΓΔ,οἳσυντεθέντες And the (number created) from (multiplying) AB and
ποιοῦσιτὸνἀπὸτοῦΒΔτετράγωνον.
BC is square—inasmuch as it was shown that if two
Καὶφανερόν,ὅτιεὕρηνταιπάλινδύοτετράγωνοιὅτε similar plane (numbers) make some (number) by mul-
ἀπὸτοῦΒΔκαὶὁἀπὸτοῦΓΔ,ὥστετὴνὑπεροχὴναὐτῶν tiplying one another then the (number so) created is
τὸνὑπὸΑΒ,ΒΓεἶναιτετράγωνον,ὅτανοἱΑΒ,ΒΓὅμοιοι square [Prop. 9.1]. Thus, two square numbers have
ὦσινἐπίπεδοι. ὅτανδὲμὴὦσινὅμοιοιἐπίπεδοι,εὕρηνται been found—(namely,) the (number created) from (mulδύοτετράγωνοιὅτεἀπὸτοῦΒΔκαὶὁἀπὸτοῦΔΓ,ὧν
tiplying) AB and BC, and the (square) on CD—which,
ἡὑπεροχὴὁὑπὸτῶνΑΒ,ΒΓοὐκἔστιτετράγωνος·ὅπερ (when) added (together), make the square on BD.
ἔδειδεῖξαι.
And (it is) clear that two square (numbers) have again
been found—(namely,) the (square) on BD, and the
(square) on CD—such that their difference—(namely,)
the (rectangle) contained by AB and BC—is square
whenever AB and BC are similar plane (numbers). But,
when they are not similar plane numbers, two square
(numbers) have been found—(namely,) the (square)
on BD, and the (square) on DC—between which the
difference—(namely,) the (rectangle) contained by AB
and BC—is not square. (Which is) the very thing it was
required to show.
C
B
310

ËÌÇÁÉÁÏƺ ÄÑѺ
ELEMENTS
Εὑρεῖνδύοτετραγώνουςἀριθμούς,ὥστετὸνἐξαὐτῶν To find two square numbers such that the sum of them
συγκείμενονμὴεἶναιτετράγωνον.
Α
Η
Θ
∆
Ε
Ζ
Γ
Β
is not square.
A
G
H
D
Lemma II
E
F
BOOK 10
῎ΕστωγὰρὁἐκτῶνΑΒ,ΒΓ,ὡςἔφαμεν,τετράγωνος, For let the (number created) from (multiplying) AB
καὶἄρτιοςὁΓΑ,καὶτετμήσθωὁΓΑδίχατῷΔ.φανερὸν and BC, as we said, be square. And (let) CA (be) even.
δή,ὅτιὁἐκτῶνΑΒ,ΒΓτετράγωνοςμετὰτοῦἀπὸ[τοῦ] And let CA have been cut in half at D. So it is clear
ΓΔτετραγώνουἴσοςἐστὶτῷἀπὸ[τοῦ]ΒΔτετραγώνῳ. that the square (number created) from (multiplying) AB
ἀφῃρήσθωμονὰςἡΔΕ·ὁἄραἐκτῶνΑΒ,ΒΓμετὰτοῦ and BC, plus the square on CD, is equal to the square
ἀπὸ[τοῦ]ΓΕἐλάσσωνἐστὶτοῦἀπὸ[τοῦ]ΒΔτετραγώνου. on BD [see previous lemma]. Let the unit DE have
λέγωοὖν,ὅτιὁἐκτῶνΑΒ,ΒΓτετράγωνοςμετὰτοῦἀπὸ been subtracted (from BD). Thus, the (number created)
[τοῦ]ΓΕοὐκἔσταιτετράγωνος.
from (multiplying) AB and BC, plus the (square) on
Εἰγὰρἔσταιτετράγωνος,ἤτοιἴσοςἐστὶτῷἀπὸ[τοῦ] CE, is less than the square on BD. I say, therefore, that
ΒΕἢἐλάσσωντοῦἀπὸ[τοῦ]ΒΕ,οὐκέτιδὲκαὶμείζων,ἵνα the square (number created) from (multiplying) AB and
μὴτμηθῇἡμονάς. ἔστω,εἰδυνατόν,πρότερονὁἐκτῶν BC, plus the (square) on CE, is not square.
ΑΒ,ΒΓμετὰτοῦἀπὸΓΕἴσοςτῷἀπὸΒΕ,καὶἔστωτῆς For if it is square, it is either equal to the (square)
ΔΕμονάδοςδιπλασίωνὁΗΑ.ἐπεὶοὖνὅλοςὁΑΓὅλου on BE, or less than the (square) on BE, but cannot any
τοῦΓΔἐστιδιπλασίων,ὧνὁΑΗτοῦΔΕἐστιδιπλασίων, more be greater (than the square on BE), lest the unit be
καὶλοιπὸςἄραὁΗΓλοιποῦτοῦΕΓἐστιδιπλασίων·δίχα divided. First of all, if possible, let the (number created)
ἄρατέτμηταιὁΗΓτῷΕ.ὁἄραἐκτῶνΗΒ,ΒΓμετὰτοῦ from (multiplying) AB and BC, plus the (square) on CE,
ἀπὸΓΕἴσοςἐστὶτῷἀπὸΒΕτετραγώνῳ. ἀλλὰκαὶὁἐκ be equal to the (square) on BE. And let GA be double
τῶνΑΒ,ΒΓμετὰτοῦἀπὸΓΕἴσοςὑπόκειταιτῷἀπὸ[τοῦ] the unit DE. Therefore, since the whole of AC is double
ΒΕτετραγώνῳ·ὁἄραἐκτῶνΗΒ,ΒΓμετὰτοῦἀπὸΓΕ the whole of CD, of which AG is double DE, the remain-
ἴσοςἐστὶτῷἐκτῶνΑΒ,ΒΓμετὰτοῦἀπὸΓΕ.καὶκοινοῦ der GC is thus double the remainder EC. Thus, GC has
ἀφαιρεθέντοςτοῦἀπὸΓΕσυνάγεταιὁΑΒἴσοςτῷΗΒ· been cut in half at E. Thus, the (number created) from
ὅπερἄτοπον.οὐκἄραὁἐκτῶνΑΒ,ΒΓμετὰτοῦἀπὸ[τοῦ] (multiplying) GB and BC, plus the (square) on CE, is
ΓΕἴσοςἐστὶτῷἀπὸΒΕ.λέγωδή,ὅτιοὐδὲἐλάσσωντοῦ equal to the square on BE [Prop. 2.6]. But, the (num-
ἀπὸΒΕ.εἰγὰρδυνατόν,ἔστωτῷἀπὸΒΖἴσος,καὶτοῦ ber created) from (multiplying) AB and BC, plus the
ΔΖδιπλασίωνὁΘΑ.καὶσυναχθήσεταιπάλινδιπλασίωνὁ (square) on CE, was also assumed (to be) equal to the
ΘΓτοῦΓΖ·ὥστεκαὶτὸνΓΘδίχατετμῆσθαικατὰτὸΖ, square on BE. Thus, the (number created) from (multiκαὶδιὰτοῦτοτὸνἐκτῶνΘΒ,ΒΓμετὰτοῦἀπὸΖΓἴσον
plying) GB and BC, plus the (square) on CE, is equal
γίνεσθαιτῷἀπὸΒΖ.ὑπόκειταιδὲκαὶὁἐκτῶνΑΒ,ΒΓ to the (number created) from (multiplying) AB and BC,
μετὰτοῦἀπὸΓΕἴσοςτῷἀπὸΒΖ.ὥστεκαὶὁἐκτῶνΘΒ, plus the (square) on CE. And subtracting the (square) on
ΒΓμετὰτοῦἀπὸΓΖἴσοςἔσταιτῷἐκτῶνΑΒ,ΒΓμετὰ CE from both, AB is inferred (to be) equal to GB. The
τοῦἀπὸΓΕ·ὅπερἄτοπον.οὐκἄραὁἐκτῶνΑΒ,ΒΓμετὰ very thing is absurd. Thus, the (number created) from
τοῦἀπὸΓΕἴσοςἐστὶ[τῷ]ἐλάσσονιτοῦἀπὸΒΕ.ἐδείχθη (multiplying) AB and BC, plus the (square) on CE, is
δέ,ὅτιοὐδὲ[αὐτῷ]τῷἀπὸΒΕ.οὐκἄραὁἐκτῶνΑΒ,ΒΓ not equal to the (square) on BE. So I say that (it is) not
μετὰτοῦἀπὸΓΕτετράγωνόςἐστιν.ὅπερἔδειδεῖξαι. less than the (square) on BE either. For, if possible, let it
be equal to the (square) on BF . And (let) HA (be) double
DF . And it can again be inferred that HC (is) double
CF . Hence, CH has also been cut in half at F . And, on
account of this, the (number created) from (multiplying)
HB and BC, plus the (square) on FC, becomes equal to
the (square) on BF [Prop. 2.6]. And the (number created)
from (multiplying) AB and BC, plus the (square)
on CE, was also assumed (to be) equal to the (square)
on BF . Hence, the (number created) from (multiplying)
HB and BC, plus the (square) on CF , will also be equal
to the (number created) from (multiplying) AB and BC,
C
B
311

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
plus the (square) on CE. The very thing is absurd. Thus,
the (number created) from (multiplying) AB and BC,
plus the (square) on CE, is not equal to less than the
(square) on BE. And it was shown that (is it) not equal
to the (square) on BE either. Thus, the (number created)
from (multiplying) AB and BC, plus the square on CE,
is not square. (Which is) the very thing it was required to
show.
. Proposition 29
Εὑρεῖνδύοῥητὰςδυνάμειμόνονσυμμέτρους,ὥστετὴν To find two rational (straight-lines which are) comμείζονατῆςἐλάσσονοςμεῖζονδύνασθαιτῷἀπὸσυμμέτρου
mensurable in square only, such that the square on the
ἑαυτῇμήκει.
greater is larger than the (square on the) lesser by the
(square) on (some straight-line which is) commensurable
in length with the greater.
Ζ
F
Α
Β
A
B
Γ Ε ∆
᾿Εκκείσθω γάρ τις ῥητὴ ἡ ΑΒ καὶ δύο τετράγωνοι For let some rational (straight-line) AB be laid down,
ἀριθμοὶοἱΓΔ,ΔΕ,ὥστετὴνὑπεροχὴναὐτῶντὸνΓΕμὴ and two square numbers, CD and DE, such that the difεἶναιτετράγωνον,καὶγεγράφθωἐπὶτῆςΑΒἡμικύκλιοντὸ
ference between them, CE, is not square [Prop. 10.28
ΑΖΒ,καὶπεποιήσθωὡςὁΔΓπρὸςτὸνΓΕ,οὕτωςτὸἀπὸ lem. I]. And let the semi-circle AFB have been drawn on
τῆςΒΑτετράγωνονπρὸςτὸἀπὸτῆςΑΖτετράγωνον,καὶ AB. And let it be contrived that as DC (is) to CE, so the
ἐπεζεύχθωἡΖΒ.
square on BA (is) to the square on AF [Prop. 10.6 corr.].
᾿Επεὶ[οὖν]ἐστινὡςτὸἀπὸτῆςΒΑπρὸςτὸἀπὸτῆςΑΖ, And let FB have been joined.
οὕτωςὁΔΓπρὸςτὸνΓΕ,τὸἀπὸτῆςΒΑἄραπρὸςτὸἀπὸ [Therefore,] since as the (square) on BA is to the
τῆςΑΖλόγονἔχει,ὅνἀριθμὸςὁΔΓπρὸςἀριθμὸντὸνΓΕ· (square) on AF , so DC (is) to CE, the (square) on
σύμμετρονἄραἐστὶτὸἀπὸτῆςΒΑτῷἀπὸτῆςΑΖ.ῥητὸν BA thus has to the (square) on AF the ratio which
δὲτὸἀπὸτῆςΑΒ·ῥητὸνἄρακαὶτὸἀπὸτῆςΑΖ·ῥητὴἄρα the number DC (has) to the number CE. Thus, the
καὶἡΑΖ.καὶἐπεὶὁΔΓπρὸςτὸνΓΕλόγονοὐκἔχει,ὃν (square) on BA is commensurable with the (square) on
τετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν,οὐδὲτὸἀπὸ AF [Prop. 10.6]. And the (square) on AB (is) rational
τῆςΒΑἄραπρὸςτὸἀπὸτῆςΑΖλόγονἔχει,ὃντετράγωνος [Def. 10.4]. Thus, the (square) on AF (is) also ratio-
ἀριθμὸςπρὸςτετράγωνονἀριθμόν·ἀσύμμετροςἄραἐστὶνἡ nal. Thus, AF (is) also rational. And since DC does
ΑΒτῇΑΖμήκει·αἱΒΑ,ΑΖἄραῥηταίεἰσιδυνάμειμόνον not have to CE the ratio which (some) square numσύμμετροι.
καὶἐπεί[ἐστιν]ὡςὁΔΓπρὸςτὸνΓΕ,οὕτως ber (has) to (some) square number, the (square) on
τὸἀπὸτῆςΒΑπρὸςτὸἀπὸτῆςΑΖ,ἀναστρέψαντιἄραὡς BA thus does not have to the (square) on AF the ra-
ὁΓΔπρὸςτὸνΔΕ,οὕτωςτὸἀπὸτῆςΑΒπρὸςτὸἀπὸ tio which (some) square number has to (some) square
τῆςΒΖ.ὁδὲΓΔπρὸςτὸνΔΕλόγονἔχει,ὃντετράγωνος number either. Thus, AB is incommensurable in length
ἀριθμὸςπρὸςτετράγωνονἀριθμόν·καὶτὸἀπὸτῆςΑΒἄρα with AF [Prop. 10.9]. Thus, the rational (straight-lines)
πρὸςτὸἀπὸτῆςΒΖλόγονἔχει,ὃντετράγωνοςἀριθμὸς BA and AF are commensurable in square only. And
πρὸςτετράγωνονἀριθμόν·σύμμετροςἄραἐστὶνἡΑΒτῇ since as DC [is] to CE, so the (square) on BA (is) to
ΒΖμήκει.καίἐστιτὸἀπὸτῆςΑΒἴσοντοῖςἀπὸτῶνΑΖ, the (square) on AF , thus, via conversion, as CD (is)
ΖΒ·ἡΑΒἄρατῆςΑΖμεῖζονδύναταιτῇΒΖσυμμέτρῳ to DE, so the (square) on AB (is) to the (square) on
C
E
D
312

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ἑαυτῇ.
BF [Props. 5.19 corr., 3.31, 1.47]. And CD has to DE
ΕὕρηνταιἄραδύοῥηταὶδυνάμειμόνονσύμμετροιαἱΒΑ, the ratio which (some) square number (has) to (some)
ΑΖ,ὥστετὴνμεῖζονατὴνΑΒτῆςἐλάσσονοςτῆςΑΖμεῖζον square number. Thus, the (square) on AB also has to the
δύνασθαιτῷἀπὸτῆςΒΖσυμμέτρουἑαυτῇμήκει·ὅπερἔδει (square) on BF the ratio which (some) square number
δεῖξαι.
has to (some) square number. AB is thus commensurable
in length with BF [Prop. 10.9]. And the (square)
on AB is equal to the (sum of the squares) on AF and
FB [Prop. 1.47]. Thus, the square on AB is greater than
(the square on) AF by (the square on) BF , (which is)
commensurable (in length) with (AB).
Thus, two rational (straight-lines), BA and AF , commensurable
in square only, have been found such that the
square on the greater, AB, is larger than (the square on)
the lesser, AF , by the (square) on BF , (which is) commensurable
in length with (AB). † (Which is) the very
thing it was required to show.
† BA and AF have lengths 1 and √ 1 − k 2 times that of AB, respectively, where k = p DE/CD.
Ð. Proposition 30
Εὑρεῖνδύοῥητὰςδυνάμειμόνονσυμμέτρους,ὥστετὴν To find two rational (straight-lines which are) comμείζονατῆςἐλάσσονοςμεῖζονδύνασθαιτῷἀπὸἀσυμμέτρου
mensurable in square only, such that the square on the
ἑαυτῇμήκει.
greater is larger than the (the square on) lesser by the
(square) on (some straight-line which is) incommensurable
in length with the greater.
Ζ
F
Α
Β
A
B
Γ
Ε
∆
᾿ΕκκείσθωῥητὴἡΑΒκαὶδύοτετράγωνοιἀριθμοὶοἱ Let the rational (straight-line) AB be laid out, and the
ΓΕ,ΕΔ,ὥστετὸνσυγκείμενονἐξαὐτῶντὸνΓΔμὴεἶναι two square numbers, CE and ED, such that the sum of
τετράγωνον,καὶγεγράφθωἐπὶτῆςΑΒἡμικύκλιοντὸΑΖΒ, them, CD, is not square [Prop. 10.28 lem. II]. And let the
καὶπεποιήσθωὡςὁΔΓπρὸςτὸνΓΕ,οὕτωςτὸἀπὸτῆς semi-circle AFB have been drawn on AB. And let it be
ΒΑπρὸςτὸἀπὸτῆςΑΖ,καὶἐπεζεύχθωἡΖΒ.
contrived that as DC (is) to CE, so the (square) on BA
῾Ομοίωςδὴδείξομεντῷ πρὸτούτου, ὅτιαἱΒΑ,ΑΖ (is) to the (square) on AF [Prop. 10.6 corr]. And let FB
ῥηταίεἰσιδυνάμειμόνονσύμμετροι. καὶἐπείἐστινὡςὁ have been joined.
ΔΓπρὸςτὸνΓΕ,οὕτωςτὸἀπὸτῆςΒΑπρὸςτὸἀπὸτῆς So, similarly to the (proposition) before this, we can
ΑΖ,ἀναστρέψαντιἄραὡςὁΓΔπρὸςτὸνΔΕ,οὕτωςτὸ show that BA and AF are rational (straight-lines which
ἀπὸτῆςΑΒπρὸςτὸἀπὸτῆςΒΖ.ὁδὲΓΔπρὸςτὸνΔΕ are) commensurable in square only. And since as DC is
λόγονοὐκἔχει,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον to CE, so the (square) on BA (is) to the (square) on AF ,
ἀριθμόν·οὐδ᾿ἄρατὸἀπὸτῆςΑΒπρὸςτὸἀπὸτῆςΒΖλόγον thus, via conversion, as CD (is) to DE, so the (square) on
ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· AB (is) to the (square) on BF [Props. 5.19 corr., 3.31,
ἀσύμμετροςἄραἐστὶνἡΑΒτῇΒΖμήκει. καὶδύναταιἡ 1.47]. And CD does not have to DE the ratio which
ΑΒτῆςΑΖμεῖζοντῷἀπὸτῆςΖΒἀσυμμέτρουἑαυτῇ. (some) square number (has) to (some) square number.
C
E
D
313

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ΑἱΑΒ,ΑΖἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι,καὶ Thus, the (square) on AB does not have to the (square)
ἡΑΒτῆςΑΖμεῖζονδύναταιτῷἀπὸτῆςΖΒἀσυμμέτρου on BF the ratio which (some) square number has to
ἑαυτῇμήκει·ὅπερἔδειδεῖξαι.
(some) square number either. Thus, AB is incommensurable
in length with BF [Prop. 10.9]. And the square on
AB is greater than the (square on) AF by the (square) on
FB [Prop. 1.47], (which is) incommensurable (in length)
with (AB).
Thus, AB and AF are rational (straight-lines which
are) commensurable in square only, and the square on
AB is greater than (the square on) AF by the (square) on
FB, (which is) incommensurable (in length) with (AB). †
(Which is) the very thing it was required to show.
† AB and AF have lengths 1 and 1/ √ 1 + k 2 times that of AB, respectively, where k = p DE/CE.
Ð. Proposition 31
Εὑρεῖν δύο μέσας δυνάμει μόνον συμμέτρους ῥητὸν To find two medial (straight-lines), commensurable in
περιεχούσας, ὥστε τὴν μείζονα τῆς ἐλάσσονος μεῖζον square only, (and) containing a rational (area), such that
δύνασθαιτῷἀπὸσυμμέτρουἑαυτῇμήκει.
the square on the greater is larger than the (square on
the) lesser by the (square) on (some straight-line) commensurable
in length with the greater.
Α
Β Γ ∆
A B C D
᾿ΕκκείσθωσανδύοῥηταὶδυνάμειμόνονσύμμετροιαἱΑ, Let two rational (straight-lines), A and B, commensu-
Β,ὥστετὴνΑμείζοναοὖσαντῆςἐλάσσονοςτῆςΒμεῖζον rable in square only, be laid out, such that the square on
δύνασθαιτῷἀπὸσυμμέτρουἑαυτῇμήκει.καὶτῷὑπὸτῶν the greater A is larger than the (square on the) lesser B
Α,ΒἴσονἔστωτὸἀπὸτῆςΓ.μέσονδὲτὸὑπὸτῶνΑ,Β· by the (square) on (some straight-line) commensurable
μέσονἄρακαὶτὸἀπὸτῆςΓ·μέσηἄρακαὶἡΓ.τῷδὲἀπὸ in length with (A) [Prop. 10.29]. And let the (square)
τῆςΒἴσονἔστωτὸὑπὸτῶνΓ,Δ·ῥητὸνδὲτὸἀπὸτῆςΒ· on C be equal to the (rectangle contained) by A and B.
ῥητὸνἄρακαὶτὸὑπὸτῶνΓ,Δ.καὶἐπείἐστινὡςἡΑπρὸς And the (rectangle contained by) A and B (is) medial
τὴνΒ,οὕτωςτὸὑπὸτῶνΑ,ΒπρὸςτὸἀπὸτῆςΒ,ἀλλὰτῷ [Prop. 10.21]. Thus, the (square) on C (is) also medial.
μὲνὑπὸτῶνΑ,ΒἴσονἐστὶτὸἀπὸτῆςΓ,τῷδὲἀπὸτῆς Thus, C (is) also medial [Prop. 10.21]. And let the (rect-
ΒἴσοντὸὑπὸτῶνΓ,Δ,ὡςἄραἡΑπρὸςτὴνΒ,οὕτως angle contained) by C and D be equal to the (square)
τὸἀπὸτῆςΓπρὸςτὸὑπὸτῶνΓ,Δ.ὡςδὲτὸἀπὸτῆςΓ on B. And the (square) on B (is) rational. Thus, the
πρὸςτὸὑπὸτῶνΓ,Δ,οὕτωςἡΓπρὸςτὴνΔ·καὶὡςἄρα (rectangle contained) by C and D (is) also rational. And
ἡΑπρὸςτὴνΒ,οὕτωςἡΓπρὸςτὴνΔ.σύμμετροςδὲἡΑ since as A is to B, so the (rectangle contained) by A and
τῇΒδυνάμειμόνον·σύμμετροςἄρακαὶἡΓτῇΔδυνάμει B (is) to the (square) on B [Prop. 10.21 lem.], but the
μόνον.καίἐστιμέσηἡΓ·μέσηἄρακαὶἡΔ.καὶἐπείἐστιν (square) on C is equal to the (rectangle contained) by
ὡςἡΑπρὸςτὴνΒ,ἡΓπρὸςτὴνΔ,ἡδὲΑτῆςΒμεῖζον A and B, and the (rectangle contained) by C and D to
δύναταιτῷἀπὸσυμμέτρουἑαυτῇ,καὶἡΓἄρατῆςΔμεῖζον the (square) on B, thus as A (is) to B, so the (square)
δύναταιτῷἀπὸσυμμέτρουἑαυτῇ.
on C (is) to the (rectangle contained) by C and D. And
ΕὕρηνταιἄραδύομέσαιδυνάμειμόνονσύμμετροιαἱΓ, as the (square) on C (is) to the (rectangle contained) by
314

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
Δῥητὸνπεριέχουσαι,καὶἡΓτῆςΔμεῖζονδυνάταιτῷἀπὸ C and D, so C (is) to D [Prop. 10.21 lem.]. And thus
συμμέτρουἑαυτῇμήκει.
as A (is) to B, so C (is) to D. And A is commensurable
῾Ομοίωςδὴδειχθήσεταικαὶτῷἀπὸἀσυμμέτρου,ὅτανἡ in square only with B. Thus, C (is) also commensurable
ΑτῆςΒμεῖζονδύνηταιτῷἀπὸἀσυμμέτρουἑαυτῇ. in square only with D [Prop. 10.11]. And C is medial.
Thus, D (is) also medial [Prop. 10.23]. And since as A is
to B, (so) C (is) to D, and the square on A is greater than
(the square on) B by the (square) on (some straight-line)
commensurable (in length) with (A), the square on C is
thus also greater than (the square on) D by the (square)
on (some straight-line) commensurable (in length) with
(C) [Prop. 10.14].
Thus, two medial (straight-lines), C and D, commensurable
in square only, (and) containing a rational (area),
have been found. And the square on C is greater than
(the square on) D by the (square) on (some straight-line)
commensurable in length with (C). †
So, similarly, (the proposition) can also be demonstrated
for (some straight-line) incommensurable (in
length with C), provided that the square on A is greater
than (the square on B) by the (square) on (some
straight-line) incommensurable (in length) with (A)
[Prop. 10.30]. ‡
† C and D have lengths (1 − k 2 ) 1/4 and (1 − k 2 ) 3/4 times that of A, respectively, where k is defined in the footnote to Prop. 10.29.
‡ C and D would have lengths 1/(1 + k 2 ) 1/4 and 1/(1 + k 2 ) 3/4 times that of A, respectively, where k is defined in the footnote to Prop. 10.30.
Ð. Proposition 32
Εὑρεῖν δύο μέσας δυνάμει μόνον συμμέτρους μέσον To find two medial (straight-lines), commensurable in
περιεχούσας, ὥστε τὴν μείζονα τῆς ἐλάσσονος μεῖζον square only, (and) containing a medial (area), such that
δύνασθαιτῷἀπὸσυμμέτρουἑαυτῇ.
the square on the greater is larger than the (square on
the) lesser by the (square) on (some straight-line) commensurable
(in length) with the greater.
Α
Β
Γ
∆
Ε
᾿Εκκείσθωσαντρεῖςῥηταὶδυνάμειμόνονσύμμετροιαἱ Let three rational (straight-lines), A, B and C, com-
Α, Β, Γ, ὥστε τὴν Α τῆς Γ μεῖζον δύνασθαι τῷ ἀπὸ mensurable in square only, be laid out such that the
συμμέτρουἑαυτῇ,καὶτῷμὲνὑπὸτῶνΑ,Βἴσονἔστωτὸ square on A is greater than (the square on C) by the
ἀπὸτὴςΔ.μέσονἄρατὸἀπὸτῆςΔ·καὶἡΔἄραμέση (square) on (some straight-line) commensurable (in
ἐστίν. τῷδὲὑπὸτῶνΒ,ΓἴσονἔστωτὸὑπὸτῶνΔ,Ε. length) with (A) [Prop. 10.29]. And let the (square)
καὶἐπείἐστινὡςτὸὑπὸτῶνΑ,ΒπρὸςτὸὑπὸτῶνΒ,Γ, on D be equal to the (rectangle contained) by A and B.
οὕτωςἡΑπρὸςτὴνΓ,ἀλλὰτῷμὲνὑπὸτῶνΑ,Βἴσον Thus, the (square) on D (is) medial. Thus, D is also me-
ἐστὶτὸἀπὸτῆςΔ,τῷδὲὑπὸτῶνΒ,Γἴσοντὸὑπὸτῶν dial [Prop. 10.21]. And let the (rectangle contained) by
Δ,Ε,ἔστινἄραὡςἡΑπρὸςτὴνΓ,οὕτωςτὸἀπὸτῆςΔ D and E be equal to the (rectangle contained) by B and
πρὸςτὸὑπὸτῶνΔ,Ε.ὡςδὲτὸἀπὸτῆςΔπρὸςτὸὑπὸ C. And since as the (rectangle contained) by A and B
τῶνΔ,Ε,οὕτωςἡΔπρὸςτὴνΕ·καὶὡςἄραἡΑπρὸςτὴν is to the (rectangle contained) by B and C, so A (is) to
Γ,οὕτωςἡΔπρὸςτὴνΕ.σύμμετροςδὲἡΑτῇΓδυνάμει C [Prop. 10.21 lem.], but the (square) on D is equal to
[μόνον].σύμμετροςἄρακαὶἡΔτῇΕδυνάμειμόνον.μέση the (rectangle contained) by A and B, and the (rectangle
A
B
C
D
E
315

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
δὲἡΔ·μέσηἄρακαὶἡΕ.καὶἐπείἐστινὡςἡΑπρὸςτὴν contained) by D and E to the (rectangle contained) by
Γ,ἡΔπρὸςτὴνΕ,ἡδὲΑτῆςΓμεῖζονδύναταιτῷἀπὸ B and C, thus as A is to C, so the (square) on D (is)
συμμέτρουἑαυτῇ,καὶἡΔἄρατῆςΕμεῖζονδυνήσεταιτῷ to the (rectangle contained) by D and E. And as the
ἀπὸσυμμέτρουἑαυτῇ.λέγωδή,ὅτικαὶμέσονἐστὶτὸὑπὸ (square) on D (is) to the (rectangle contained) by D and
τῶνΔ,Ε.ἐπεὶγὰρἴσονἐστὶτὸὑπὸτῶνΒ,Γτῷὑπὸτῶν E, so D (is) to E [Prop. 10.21 lem.]. And thus as A
Δ,Ε,μέσονδὲτὸὑπὸτῶνΒ,Γ[αἱγὰρΒ,Γῥηταίεἰσι (is) to C, so D (is) to E. And A (is) commensurable in
δυνάμειμόνονσύμμετροι],μέσονἄρακαὶτὸὑπὸτῶνΔ,Ε. square [only] with C. Thus, D (is) also commensurable
Εὕρηνται ἄρα δύο μέσαι δυνάμει μόνον σύμμετροι αἱ in square only with E [Prop. 10.11]. And D (is) me-
Δ,Εμέσονπεριέχουσαι,ὥστετὴνμείζονατῆςἐλάσσονος dial. Thus, E (is) also medial [Prop. 10.23]. And since
μεῖζονδύνασθαιτῷἀπὸσυμμέτρουἑαυτῇ.
as A is to C, (so) D (is) to E, and the square on A is
῾Ομοίωςδὴπάλινδιεχθήσεταικαὶτῷἀπὸἀσυμμέτρου, greater than (the square on) C by the (square) on (some
ὅτανἡΑτῆςΓμεῖζονδύνηταιτῷἀπὸἀσυμμέτρουἑαυτῃ. straight-line) commensurable (in length) with (A), the
square on D will thus also be greater than (the square
on) E by the (square) on (some straight-line) commensurable
(in length) with (D) [Prop. 10.14]. So, I also
say that the (rectangle contained) by D and E is medial.
For since the (rectangle contained) by B and C is equal
to the (rectangle contained) by D and E, and the (rectangle
contained) by B and C (is) medial [for B and C
are rational (straight-lines which are) commensurable in
square only] [Prop. 10.21], the (rectangle contained) by
D and E (is) thus also medial.
Thus, two medial (straight-lines), D and E, commensurable
in square only, (and) containing a medial (area),
have been found such that the square on the greater is
larger than the (square on the) lesser by the (square) on
(some straight-line) commensurable (in length) with the
greater. † .
So, similarly, (the proposition) can again also be
demonstrated for (some straight-line) incommensurable
(in length with the greater), provided that the square on
A is greater than (the square on) C by the (square) on
(some straight-line) incommensurable (in length) with
(A) [Prop. 10.30]. ‡
† D and E have lengths k ′1/4 and k ′1/4√ 1 − k 2 times that of A, respectively, where the length of B is k ′1/2 times that of A, and k is defined in
the footnote to Prop. 10.29.
‡ D and E would have lengths k ′1/4 and k ′1/4 / √ 1 + k 2 times that of A, respectively, where the length of B is k ′1/2 times that of A, and k is
defined in the footnote to Prop. 10.30.
ÄÑÑ.
Lemma
῎ΕστωτρίγωνονὀρθογώνιοντὸΑΒΓὀρθὴνἔχοντὴν Let ABC be a right-angled triangle having the (an-
Α,καὶἤχθωκάθετοςἡΑΔ·λέγω,ὅτιτὸμὲνὑπὸτῶνΓΒΔ gle) A a right-angle. And let the perpendicular AD have
ἴσονἐστὶτῷἀπὸτῆςΒΑ,τὸδὲὑπὸτῶνΒΓΑἴσοντῷἀπὸ been drawn. I say that the (rectangle contained) by CBD
τῆςΓΑ,καὶτὸὑπὸτῶνΒΔ,ΔΓἴσοντῷἀπὸτῆςΑΔ,καὶ is equal to the (square) on BA, and the (rectangle con-
ἔτιτὸὑπὸτῶνΒΓ,ΑΔἴσον[ἐστὶ]τῷὑπὸτῶνΒΑ,ΑΓ. tained) by BCD (is) equal to the (square) on CA, and
Καὶπρῶτον,ὅτιτὸὑπὸτῶνΓΒΔἴσον[ἐστὶ]τῷἀπὸ the (rectangle contained) by BD and DC (is) equal to the
τῆςΒΑ.
(square) on AD, and, further, the (rectangle contained)
by BC and AD [is] equal to the (rectangle contained) by
BA and AC.
316

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
Α
And, first of all, (let us prove) that the (rectangle contained)
by CBD [is] equal to the (square) on BA.
A
Β
∆
Γ
B
D
C
᾿Επεὶγὰρἐνὀρθογωνίῳτριγώνῳἀπὸτῆςὀρθῆςγωνίας For since AD has been drawn from the right-angle in
ἐπὶτὴνβάσινκάθετοςἦκταιἡΑΔ,τὰΑΒΔ,ΑΔΓἄρα a right-angled triangle, perpendicular to the base, ABD
τρίγωναὅμοιάἐστιτῷτεὅλῳτῷΑΒΓκαὶἀλλήλοις. καὶ and ADC are thus triangles (which are) similar to the
ἐπεὶὅμοιόνἐστιτὸΑΒΓτρίγωνοντῷΑΒΔτριγώνῳ,ἔστιν whole, ABC, and to one another [Prop. 6.8]. And since
ἄραὡςἡΓΒπρὸςτὴνΒΑ,οὕτωςἡΒΑπρὸςτὴνΒΔ·τὸ triangle ABC is similar to triangle ABD, thus as CB is
ἄραὑπὸτῶνΓΒΔἴσονἐστὶτῷἀπὸτῆςΑΒ.
to BA, so BA (is) to BD [Prop. 6.4]. Thus, the (rectan-
ΔιὰτὰαὐτὰδὴκαὶτὸὑπὸτῶνΒΓΔἴσονἐστὶτῷἀπὸ gle contained) by CBD is equal to the (square) on AB
τῆςΑΓ. [Prop. 6.17].
Καὶἐπεί, ἐὰνἐνὀρθογωνίῳτριγώνῳἀπὸτῆςὀρθῆς So, for the same (reasons), the (rectangle contained)
γωνίαςἐπὶτὴνβάσινκάθετοςἀχθῇ, ἡἀχθεῖσατῶντῆς by BCD is also equal to the (square) on AC.
βάσεωςτμημάτωνμέσηἀνάλογόνἐστιν,ἔστινἄραὡςἡΒΑ And since if a (straight-line) is drawn from the rightπρὸςτὴνΔΑ,οὕτωςἡΑΔπρὸςτὴνΔΓ·τὸἄραὑπὸτῶν
angle in a right-angled triangle, perpendicular to the
ΒΔ,ΔΓἴσονἐστὶτῷἀπὸτῆςΔΑ.
base, the (straight-line so) drawn is the mean propor-
Λέγω,ὅτικαὶτὸὑπὸτῶνΒΓ,ΑΔἴσονἐστὶτῷὑπὸ tional to the pieces of the base [Prop. 6.8 corr.], thus as
τῶνΒΑ,ΑΓ.ἐπεὶγὰρ,ὡςἔφαμεν,ὅμοιόνἐστιτὸΑΒΓτῷ BD is to DA, so AD (is) to DC. Thus, the (rectangle
ΑΒΔ,ἔστινἄραὡςἡΒΓπρὸςτὴνΓΑ,οὕτωςἡΒΑπρὸς contained) by BD and DC is equal to the (square) on
τὴνΑΔ.τὸἄραὑπὸτῶνΒΓ,ΑΔἴσονἐστὶτῷὑπὸτῶν DA [Prop. 6.17].
ΒΑ,ΑΓ·ὅπερἔδειδεῖξαι.
I also say that the (rectangle contained) by BC and
AD is equal to the (rectangle contained) by BA and AC.
For since, as we said, ABC is similar to ABD, thus as BC
is to CA, so BA (is) to AD [Prop. 6.4]. Thus, the (rectangle
contained) by BC and AD is equal to the (rectangle
contained) by BA and AC [Prop. 6.16]. (Which is) the
very thing it was required to show.
Ð. Proposition 33
Εὑρεῖνδύοεὐθείαςδυνάμειἀσυμμέτρουςποιούσαςτὸ To find two straight-lines (which are) incommensuμὲνσυγκείμενονἐκτῶνἀπ᾿αὐτῶντετραγώνωνῥητόν,τὸ
rable in square, making the sum of the squares on them
δ᾿ὑπ᾿αὐτῶνμέσον.
rational, and the (rectangle contained) by them medial.
᾿Εκκείσθωσαν δύο ῥηταὶδυνάμει μόνον σύμμετροι αἱ Let the two rational (straight-lines) AB and BC,
ΑΒ,ΒΓ,ὥστετὴνμείζονατὴνΑΒτῆςἐλάσσονοςτῆςΒΓ (which are) commensurable in square only, be laid out
μείζονδύνασθαιτῷἀπὸἀσυμμέτρουἑαυτῇ,καὶτετμήσθωἡ such that the square on the greater, AB, is larger than
ΒΓδίχακατὰτὸΔ,καὶτῷἀφ᾿ὁποτέραςτῶνΒΔ,ΔΓἴσον (the square on) the lesser, BC, by the (square) on (some
παρὰτὴνΑΒπαραβεβλήσθωπαραλληλόγραμμονἐλλεῖπον straight-line which is) incommensurable (in length) with
εἴδειτετραγώνῳ,καὶἔστωτὸὑπὸτῶνΑΕΒ,καὶγεγράφθω (AB) [Prop. 10.30]. And let BC have been cut in half at
ἐπὶτῆςΑΒημικύκλιοντὸΑΖΒ,καὶἤχθωτῇΑΒπρὸς D. And let a parallelogram equal to the (square) on ei-
317

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ὀρθὰςἡΕΖ,καὶἐπεζεύχθωσαναἱΑΖ,ΖΒ.
ther of BD or DC, (and) falling short by a square figure,
have been applied to AB [Prop. 6.28], and let it be the
(rectangle contained) by AEB. And let the semi-circle
AFB have been drawn on AB. And let EF have been
drawn at right-angles to AB. And let AF and FB have
been joined.
F
A E B D C
Καὶἐπεὶ[δύο]εὐθεῖαιἄνισοίεἰσιναἱΑΒ,ΒΓ,καὶἡ And since AB and BC are [two] unequal straight-
ΑΒ τῆς ΒΓ μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, lines, and the square on AB is greater than (the square
τῷ δὲ τετάρτῳ τοῦ ἀπὸ τῆς ΒΓ, τουτέστι τῷ ἀπὸ τῆς on) BC by the (square) on (some straight-line which is)
ἡμισείαςαὐτῆς, ἴσον παρὰτὴν ΑΒ παραβέβληταιπαραλ- incommensurable (in length) with (AB). And a paralληλόγραμμονἐλλεῖπονεἴδειτετραγώνῳκαὶποιεῖτὸὑπὸ
lelogram, equal to one quarter of the (square) on BC—
τῶνΑΕΒ,ἀσύμμετροςἄραἐστὶνἡΑΕτῇΕΒ.καίἐστινὡς that is to say, (equal) to the (square) on half of it—(and)
ἡΑΕπρὸςΕΒ,οὕτωςτὸὑπὸτῶνΒΑ,ΑΕπρὸςτὸὑπὸ falling short by a square figure, has been applied to AB,
τῶνΑΒ,ΒΕ,ἴσονδὲτὸμὲνὑπὸτῶνΒΑ,ΑΕτῷἀπὸτῆς and makes the (rectangle contained) by AEB. AE is thus
ΑΖ,τὸδὲὑπὸτῶνΑΒ,ΒΕτῷἁπὸτῆςΒΖ·ἀσύμμετρον incommensurable (in length) with EB [Prop. 10.18].
ἄραἐστὶτὸἀπὸτῆςΑΖτῷἀπὸτῆςΖΒ·αἱΑΖ,ΖΒἄρα And as AE is to EB, so the (rectangle contained) by BA
δυνάμειεἰσὶνἀσύμμετροι.καὶἐπεὶἡΑΒῥητήἐστιν,ῥητὸν and AE (is) to the (rectangle contained) by AB and BE.
ἄραἐστὶκαὶτὸἀπὸτῆςΑΒ·ὥστεκαὶτὸσυγκείμενονἐκ And the (rectangle contained) by BA and AE (is) equal
τῶνἀπὸτῶνΑΖ,ΖΒῥητόνἐστιν. καὶἐπεὶπάλιντὸὑπὸ to the (square) on AF , and the (rectangle contained) by
τῶνΑΕ,ΕΒἴσονἐστὶτῷἀπὸτῆςΕΖ,ὑπόκειταιδὲτὸὑπὸ AB and BE to the (square) on BF [Prop. 10.32 lem.].
τῶνΑΕ,ΕΒκαὶτῷἀπὸτῆςΒΔἴσον,ἴσηἄραἐστὶνἡΖΕ The (square) on AF is thus incommensurable with the
τῇΒΔ·διπλῆἄραἡΒΓτὴςΖΕ·ὥστεκαὶτὸὑπὸτῶνΑΒ, (square) on FB [Prop. 10.11]. Thus, AF and FB are in-
ΒΓσύμμετρόνἐστιτῷὑπὸτῶνΑΒ,ΕΖ.μέσονδὲτὸὑπὸ commensurable in square. And since AB is rational, the
τῶνΑΒ,ΒΓ·μέσονἄρακαὶτὸὑπὸτῶνΑΒ,ΕΖ.ἴσονδὲ (square) on AB is also rational. Hence, the sum of the
τὸὑπὸτῶνΑΒ,ΕΖτῷὑπὸτῶνΑΖ,ΖΒ·μέσονἄρακαὶτὸ (squares) on AF and FB is also rational [Prop. 1.47].
ὑπὸτῶνΑΖ,ΖΒ.ἐδείχθηδὲκαὶῥητὸντὸσυγκείμενονἐκ And, again, since the (rectangle contained) by AE and
τῶνἀπ᾿αὐτῶντετραγώνων.
EB is equal to the (square) on EF , and the (rectangle
ΕὕρηνταιἄραδύοεὐθεῖαιδυνάμειἀσύμμετροιαἱΑΖ,ΖΒ contained) by AE and EB was assumed (to be) equal to
ποιοῦσαιτὸμὲνσυγκείμενονἐκτῶνἀπ᾿αὐτῶντετραγώνων the (square) on BD, FE is thus equal to BD. Thus, BC
ῥητόν,τὸδὲὑπ᾿αὐτῶνμέσον·ὅπερἔδειδεῖξαι.
is double FE. And hence the (rectangle contained) by
AB and BC is commensurable with the (rectangle contained)
by AB and EF [Prop. 10.6]. And the (rectangle
contained) by AB and BC (is) medial [Prop. 10.21].
Thus, the (rectangle contained) by AB and EF (is) also
medial [Prop. 10.23 corr.]. And the (rectangle contained)
by AB and EF (is) equal to the (rectangle contained) by
AF and FB [Prop. 10.32 lem.]. Thus, the (rectangle contained)
by AF and FB (is) also medial. And the sum of
the squares on them was also shown (to be) rational.
Thus, the two straight-lines, AF and FB, (which are)
incommensurable in square, have been found, making
the sum of the squares on them rational, and the (rectangle
contained) by them medial. (Which is) the very thing
it was required to show.
318

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
† AF and FB have lengths
footnote to Prop. 10.30.
Ð.
q
q
[1 + k/(1 + k 2 ) 1/2 ]/2 and [1 − k/(1 + k 2 ) 1/2 ]/2 times that of AB, respectively, where k is defined in the
Proposition 34
Εὑρεῖνδύοεὐθείαςδυνάμειἀσυμμέτρουςποιούσαςτὸ To find two straight-lines (which are) incommensuμὲνσυγκείμενονἐκτῶνἀπ᾿αὐτῶντετραγώνωνμέσον,τὸ
rable in square, making the sum of the squares on them
δ᾿ὑπ᾿αὐτῶνῥητόν.
medial, and the (rectangle contained) by them rational.
∆
D
Α
Ζ Β Ε Γ A
F B E
᾿Εκκείσθωσανδύομέσαιδυνάμειμόνονσύμμετροιαἱ Let the two medial (straight-lines) AB and BC,
ΑΒ,ΒΓῥητὸνπεριέχουσαιτὸὑπ᾿αὐτῶν,ὥστετὴνΑΒ (which are) commensurable in square only, be laid out
τῆςΒΓμεῖζονδύνασθαιτῷἀπὸἀσυμμέτρουἑαυτῇ, καὶ having the (rectangle contained) by them rational, (and)
γεγράφθωἐπὶτῆςΑΒτὸΑΔΒἡμικύκλιον,καὶτετμήσθω such that the square on AB is greater than (the square
ἡΒΓδίχακατὰτὸΕ, καὶπαραβεβλήσθωπαρὰτὴνΑΒ on) BC by the (square) on (some straight-line) incomτῷἀπὸτῆςΒΕἴσονπαραλληλόγραμμονἐλλεῖπονεἴδειτε-
mensurable (in length) with (AB) [Prop. 10.31]. And
τραγώνῳτὸὑπὸτῶνΑΖΒ·ἀσύμμετροςἄρα[ἐστὶν]ἡΑΖ let the semi-circle ADB have been drawn on AB. And
τῇΖΒμήκει.καὶἤχθωἀπὸτοῦΖτῇΑΒπρὸςὀρθὰςἡΖΔ, let BC have been cut in half at E. And let a (rectanguκαὶἐπεζεύχθωσαναἱΑΔ,ΔΒ.
lar) parallelogram equal to the (square) on BE, (and)
᾿ΕπεὶἀσύμμετρόςἐστινἡΑΖτῇΖΒ,ἀσύμμετρονἄρα falling short by a square figure, have been applied to
ἐστὶκαὶτὸὑπὸτῶνΒΑ,ΑΖτῷὑπὸτῶνΑΒ,ΒΖ.ἴσονδὲ AB, (and let it be) the (rectangle contained by) AFB
τὸμὲνὑπὸτῶνΒΑ,ΑΖτῷἀπὸτῆςΑΔ,τὸδὲὑπὸτῶν [Prop. 6.28]. Thus, AF [is] incommensurable in length
ΑΒ,ΒΖτῷἀπὸτῆςΔΒ·ἀσύμμετρονἄραἐστὶκαὶτὸἀπὸ with FB [Prop. 10.18]. And let FD have been drawn
τῆςΑΔτῷἀπὸτῆςΔΒ.καὶἐπεὶμέσονἐστὶτὸἀπὸτῆς from F at right-angles to AB. And let AD and DB have
ΑΒ,μέσονἄρακαὶτὸσυγκείμενονἐκτῶνἀπὸτῶνΑΔ, been joined.
ΔΒ.καὶἐπεὶδιπλῆἐστινἡΒΓτῆςΔΖ,διπλάσιονἄρακαὶ Since AF is incommensurable (in length) with FB,
τὸὑπὸτῶνΑΒ,ΒΓτοῦὑπὸτῶνΑΒ,ΖΔ.ῥητὸνδὲτὸὑπὸ the (rectangle contained) by BA and AF is thus also
τῶνΑΒ,ΒΓ·ῥητὸνἄρακαὶτὸὑπὸτῶνΑΒ,ΖΔ.τὸδὲὑπὸ incommensurable with the (rectangle contained) by AB
τῶνΑΒ,ΖΔἴσοντῷὑπὸτῶνΑΔ,ΔΒ·ὥστεκαὶτὸὑπὸ and BF [Prop. 10.11]. And the (rectangle contained) by
τῶνΑΔ,ΔΒῥητόνἐστιν.
BA and AF (is) equal to the (square) on AD, and the
ΕὕρηνταιἄραδύοεὐθεῖαιδυνάμειἀσύμμετροιαἱΑΔ, (rectangle contained) by AB and BF to the (square) on
ΔΒποιοῦσαιτὸ[μὲν]συγκείμενονἐκτῶνἀπ᾿αὐτῶντε- DB [Prop. 10.32 lem.]. Thus, the (square) on AD is also
τραγώνωνμέσον,τὸδ᾿ὑπ᾿αὐτῶνῥητόν·ὅπερἔδειδεῖξαι. incommensurable with the (square) on DB. And since
the (square) on AB is medial, the sum of the (squares)
on AD and DB (is) thus also medial [Props. 3.31, 1.47].
And since BC is double DF [see previous proposition],
the (rectangle contained) by AB and BC (is) thus also
double the (rectangle contained) by AB and FD. And
the (rectangle contained) by AB and BC (is) rational.
Thus, the (rectangle contained) by AB and FD (is) also
rational [Prop. 10.6, Def. 10.4]. And the (rectangle contained)
by AB and FD (is) equal to the (rectangle contained)
by AD and DB [Prop. 10.32 lem.]. And hence
the (rectangle contained) by AD and DB is rational.
Thus, two straight-lines, AD and DB, (which are) incommensurable
in square, have been found, making the
sum of the squares on them medial, and the (rectangle
C
319

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
† AD and DB have lengths
defined in the footnote to Prop. 10.29.
contained) by them rational. † (Which is) the very thing it
was required to show.
q
q
[(1 + k 2 ) 1/2 + k]/[2(1 + k 2 )] and [(1 + k 2 ) 1/2 − k]/[2(1 + k 2 )] times that of AB, respectively, where k is
Ð. Proposition 35
Εὑρεῖνδύοεὐθείαςδυνάμειἀσυμμέτρουςποιούσαςτό To find two straight-lines (which are) incommensuτεσυγκείμενονἐκτῶνἀπ᾿αὐτῶντετραγώνωνμέσονκαὶ
rable in square, making the sum of the squares on them
τὸὑπ᾿αὐτῶνμέσονκαὶἔτιἀσύμμετροντῷσυγκειμένῳἐκ medial, and the (rectangle contained) by them medial,
τῶνἀπ᾿αὐτῶντετραγώνῳ.
and, moreover, incommensurable with the sum of the
squares on them.
∆
D
Α
Ζ Β Ε Γ A
F B E
᾿Εκκείσθωσανδύομέσαιδυνάμειμόνονσύμμετροιαἱ Let the two medial (straight-lines) AB and BC,
ΑΒ,ΒΓμέσονπεριέχουσαι,ὥστετὴνΑΒτῆςΒΓμεῖζον (which are) commensurable in square only, be laid out
δύνασθαιτῷἀπὸἀσυμμέτρουἑαυτῇ,καὶγεγράφθωἐπὶτῆς containing a medial (area), such that the square on AB
ΑΒἡμικύκλιοντὸΑΔΒ,καὶτὰλοιπὰγεγονέτωτοῖςἐπάνω is greater than (the square on) BC by the (square) on
ὁμοίως.
(some straight-line) incommensurable (in length) with
ΚαὶἐπεὶἀσύμμετρόςἐστινἡΑΖτῇΖΒμήκει,ἀσύμμετρ- (AB) [Prop. 10.32]. And let the semi-circle ADB have
όςἐστικαὶἡΑΔτῇΔΒδυνάμει. καὶἐπεὶμέσονἐστὶτὸ been drawn on AB. And let the remainder (of the figure)
ἀπὸτῆςΑΒ,μέσονἄρακαὶτὸσυγκείμενονἐκτῶνἀπὸ be generated similarly to the above (proposition).
τῶνΑΔ,ΔΒ.καὶἐπεὶτὸὑπὸτῶνΑΖ,ΖΒἴσονἐστὶτῷἀφ᾿ And since AF is incommensurable in length with FB
ἑκατέραςτῶνΒΕ,ΔΖ,ἴσηἄραἐστὶνἡΒΕτῇΔΖ·διπλῆ [Prop. 10.18], AD is also incommensurable in square
ἄραἡΒΓτῆςΖΔ·ὥστεκαὶτὸὑπὸτῶνΑΒ,ΒΓδιπλάσιόν with DB [Prop. 10.11]. And since the (square) on AB
ἐστιτοῦὑπὸτῶνΑΒ,ΖΔ.μέσονδὲτὸὑπὸτῶνΑΒ,ΒΓ· is medial, the sum of the (squares) on AD and DB (is)
μέσονἄρακαὶτὸὑπὸτῶνΑΒ,ΖΔ.καίἐστινἴσοντῷὑπὸ thus also medial [Props. 3.31, 1.47]. And since the (rectτῶνΑΔ,ΔΒ·μέσονἄρακαὶτὸὑπὸτῶνΑΔ,ΔΒ.καὶἐπεὶ
angle contained) by AF and FB is equal to the (square)
ἀσύμμετρόςἐστινἡΑΒτῇΒΓμήκει,σύμμετροςδὲἡΓΒ on each of BE and DF , BE is thus equal to DF . Thus,
τῇΒΕ,ἀσύμμετροςἄρακαὶἡΑΒτῇΒΕμήκει·ὥστεκαὶτὸ BC (is) double FD. And hence the (rectangle contained)
ἀπὸτῆςΑΒτῷὑπὸτῶνΑΒ,ΒΕἀσύμμετρόνἐστιν. ἀλλὰ by AB and BC is double the (rectangle) contained by
τῷμὲνἀπὸτῆςΑΒἴσαἐστὶτὰἀπὸτῶνΑΔ,ΔΒ,τῷδὲ AB and FD. And the (rectangle contained) by AB and
ὑπὸτῶνΑΒ,ΒΕἴσονἐστὶτὸὑπὸτῶνΑΒ,ΖΔ,τουτέστι BC (is) medial. Thus, the (rectangle contained) by AB
τὸὑπὸτῶνΑΔ,ΔΒ·ἀσύμμετρονἄραἐστὶτὸσυγκείμενον and FD (is) also medial. And it is equal to the (rectan-
ἐκτῶνἀπὸτῶνΑΔ,ΔΒτῷὑπὸτῶνΑΔ,ΔΒ. gle contained) by AD and DB [Prop. 10.32 lem.]. Thus,
ΕὕρηνταιἄραδύοεὐθεῖαιαἱΑΔ,ΔΒδυνάμειἀσύμμετροι the (rectangle contained) by AD and DB (is) also meποιοῦσαιτότεσυγκείμενονἐκτῶνἀπ᾿αὐτῶνμέσονκαὶ
dial. And since AB is incommensurable in length with
τὸὑπ᾿αὐτῶνμέσονκαὶἔτιἀσύμμετροντῷσυγκειμένῳἐκ BC, and CB (is) commensurable (in length) with BE,
τῶνἀπ᾿αὐτῶντετραγώνων·ὅπερἔδειδεῖξαι.
AB (is) thus also incommensurable in length with BE
[Prop. 10.13]. And hence the (square) on AB is also
incommensurable with the (rectangle contained) by AB
and BE [Prop. 10.11]. But the (sum of the squares) on
AD and DB is equal to the (square) on AB [Prop. 1.47].
And the (rectangle contained) by AB and FD—that is to
say, the (rectangle contained) by AD and DB—is equal
to the (rectangle contained) by AB and BE. Thus, the
C
320

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
sum of the (squares) on AD and DB is incommensurable
with the (rectangle contained) by AD and DB.
Thus, two straight-lines, AD and DB, (which are) incommensurable
in square, have been found, making the
sum of the (squares) on them medial, and the (rectangle
contained) by them medial, and, moreover, incommensurable
with the sum of the squares on them. † (Which is)
the very thing it was required to show.
† AD and DB have lengths k ′1/4 q
[1 + k/(1 + k 2 ) 1/2 ]/2 and k ′1/4 q[1 − k/(1 + k 2 ) 1/2 ]/2 times that of AB, respectively, where k and k ′ are
defined in the footnote to Prop. 10.32.
Ц. Proposition 36
᾿Εὰνδύοῥηταὶδυνάμειμόνονσύμμετροισυντεθῶσιν,ἡ If two rational (straight-lines which are) commensu-
ὅληἄλογόςἐστιν,καλείσθωδὲἐκδύοὀνομάτων. rable in square only are added together then the whole
(straight-line) is irrational—let it be called a binomial
(straight-line). †
Α Β Γ A B C
Συγκείσθωσανγὰρδύοῥηταὶδυνάμειμόνονσύμμετροι For let the two rational (straight-lines), AB and BC,
αἱΑΒ,ΒΓ·λέγω,ὅτιὅληἡΑΓἄλογόςἐστιν.
(which are) commensurable in square only, be laid down
᾿ΕπεὶγὰρἀσύμμετρόςἐστινἡΑΒτῇΒΓμήκει·δυνάμει together. I say that the whole (straight-line), AC, is irraγὰρμόνονεἰσὶσύμμετροι·ὡςδὲἡΑΒπρὸςτὴνΒΓ,οὕτως
tional. For since AB is incommensurable in length with
τὸὑπὸτῶνΑΒΓπρὸςτὸἀπὸτῆςΒΓ,ἀσύμμετρονἄραἐστὶ BC—for they are commensurable in square only—and as
τὸὑπὸτῶνΑΒ,ΒΓτῷἀπὸτῆςΒΓ.ἀλλὰτῷμὲνὑπὸτῶν AB (is) to BC, so the (rectangle contained) by ABC (is)
ΑΒ,ΒΓσύμμετρόνἐστιτὸδὶςὑπὸτῶνΑΒ,ΒΓ,τῷδὲἀπὸ to the (square) on BC, the (rectangle contained) by AB
τῆςΒΓσύμμετράἐστιτὰἀπὸτῶνΑΒ,ΒΓ·αἱγὰρΑΒ,ΒΓ and BC is thus incommensurable with the (square) on
ῥηταίεἰσιδυνάμειμόνονσύμμετροι·ἀσύμμετρονἄραἐστὶτὸ BC [Prop. 10.11]. But, twice the (rectangle contained)
δὶςὑπὸτῶνΑΒ,ΒΓτοῖςἀπὸτῶνΑΒ,ΒΓ.καὶσυνθέντιτὸ by AB and BC is commensurable with the (rectangle
δὶςὑπὸτῶνΑΒ,ΒΓμετὰτῶνἀπὸτῶνΑΒ,ΒΓ,τουτέστι contained) by AB and BC [Prop. 10.6]. And (the sum
τὸἀπὸτῆςΑΓ,ἀσύμμετρόνἐστιτῷσυγκειμένῳἐκτῶν of) the (squares) on AB and BC is commensurable with
ἀπὸτῶνΑΒ,ΒΓ·ῥητὸνδὲτὸσυγκείμενονἐκτῶνἀπὸ the (square) on BC—for the rational (straight-lines) AB
τῶνΑΒ,ΒΓ·ἄλογονἄρα[ἐστὶ]τὸἀπὸτῆςΑΓ·ὥστεκαὶἡ and BC are commensurable in square only [Prop. 10.15].
ΑΓἄλογόςἐστιν,καλείσθωδὲἐκδύοὀνομάτων·ὅπερἔδει Thus, twice the (rectangle contained) by AB and BC is
δεῖξαι.
incommensurable with (the sum of) the (squares) on AB
and BC [Prop. 10.13]. And, via composition, twice the
(rectangle contained) by AB and BC, plus (the sum of)
the (squares) on AB and BC—that is to say, the (square)
on AC [Prop. 2.4]—is incommensurable with the sum of
the (squares) on AB and BC [Prop. 10.16]. And the sum
of the (squares) on AB and BC (is) rational. Thus, the
(square) on AC [is] irrational [Def. 10.4]. Hence, AC
is also irrational [Def. 10.4]—let it be called a binomial
(straight-line). ‡ (Which is) the very thing it was required
to show.
† Literally, “from two names”.
‡ Thus, a binomial straight-line has a length expressible as 1 + k 1/2 [or, more generally, ρ (1 + k 1/2 ), where ρ is rational—the same proviso
applies to the definitions in the following propositions]. The binomial and the corresponding apotome, whose length is expressible as 1 − k 1/2
321

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
(see Prop. 10.73), are the positive roots of the quartic x 4 − 2(1 + k) x 2 + (1 − k) 2 = 0.
ÐÞ. Proposition 37
᾿Εὰν δύο μέσαι δυνάμει μόνον σύμμετροι συντεθῶσι If two medial (straight-lines), commensurable in
ῥητὸνπεριέχουσαι,ἡὅληἄλογόςἐστιν, καλείσθωδὲἐκ square only, which contain a rational (area), are added
δύομέσωνπρώτη.
together then the whole (straight-line) is irrational—let
it be called a first bimedial (straight-line). †
Α Β Γ A B C
Συγκείσθωσανγὰρδύομέσαιδυνάμειμόνονσύμμετροι For let the two medial (straight-lines), AB and BC,
αἱΑΒ,ΒΓῥητὸνπεριέχουσαι·λέγω,ὅτιὅληἡΑΓἄλογός commensurable in square only, (and) containing a ratio-
ἐστιν.
nal (area), be laid down together. I say that the whole
᾿ΕπεὶγὰρἀσύμμετρόςἐστινἡΑΒτῇΒΓμήκει,καὶτὰ (straight-line), AC, is irrational.
ἀπὸτῶνΑΒ,ΒΓἄραἀσύμμετράἐστιτῷδὶςὑπὸτῶνΑΒ, For since AB is incommensurable in length with BC,
ΒΓ·καὶσυνθέντιτὰἀπὸτῶνΑΒ,ΒΓμετὰτοῦδὶςὑπὸτῶν (the sum of) the (squares) on AB and BC is thus also in-
ΑΒ,ΒΓ,ὅπερἐστὶτὸἀπὸτῆςΑΓ,ἀσύμμετρόνἐστιτῷὑπὸ commensurable with twice the (rectangle contained) by
τῶνΑΒ,ΒΓ.ῥητὸνδὲτὸὑπὸτῶνΑΒ,ΒΓ·ὑπόκεινταιγὰρ AB and BC [see previous proposition]. And, via comαἱΑΒ,ΒΓῥητὸνπεριέχουσαι·ἄλογονἄρατὸἀπὸτῆςΑΓ·
position, (the sum of) the (squares) on AB and BC,
ἄλογοςἄραἡΑΓ,καλείσθωδὲἐκδύομέσωνπρώτη·ὅπερ plus twice the (rectangle contained) by AB and BC—
ἔδειδεῖξαι.
† Literally, “first from two medials”.
that is, the (square) on AC [Prop. 2.4]—is incommensurable
with the (rectangle contained) by AB and BC
[Prop. 10.16]. And the (rectangle contained) by AB and
BC (is) rational—for AB and BC were assumed to enclose
a rational (area). Thus, the (square) on AC (is)
irrational. Thus, AC (is) irrational [Def. 10.4]—let it be
called a first bimedial (straight-line). ‡ (Which is) the very
thing it was required to show.
‡ Thus, a first bimedial straight-line has a length expressible as k 1/4 + k 3/4 . The first bimedial and the corresponding first apotome of a medial,
whose length is expressible as k 1/4 − k 3/4 (see Prop. 10.74), are the positive roots of the quartic x 4 − 2 √ k (1 + k) x 2 + k (1 − k) 2 = 0.
Ð. Proposition 38
᾿Εὰν δύο μέσαι δυνάμει μόνον σύμμετροι συντεθῶσι If two medial (straight-lines), commensurable in square
μέσονπεριέχουσαι,ἡὅληἄλογόςἐστιν,καλείσθωδὲἐκ only, which contain a medial (area), are added together
δύομέσωνδυετέρα.
then the whole (straight-line) is irrational—let it be
called a second bimedial (straight-line).
Α Β Γ
A
B
C
∆
Θ
Η
D
H
G
Ε
Ζ
Συγκείσθωσανγὰρδύομέσαιδυνάμειμόνονσύμμετροι For let the two medial (straight-lines), AB and BC,
αἱΑΒ,ΒΓμέσονπεριέχουσαι·λέγω,ὅτιἄλογόςἐστινἡ commensurable in square only, (and) containing a medial
E
F
322

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ΑΓ.
(area), be laid down together [Prop. 10.28]. I say that
᾿ΕκκείσθωγὰρῥητὴἡΔΕ,καὶτῷἀπὸτῆςΑΓἴσονπαρὰ AC is irrational.
τὴνΔΕπαραβεβλήσθωτὸΔΖπλάτοςποιοῦντὴνΔΗ.καὶ For let the rational (straight-line) DE be laid down,
ἐπεὶτὸἀπὸτῆςΑΓἴσονἐστὶτοῖςτεἀπὸτῶνΑΒ,ΒΓκαὶτῷ and let (the rectangle) DF , equal to the (square) on
δὶςὑπὸτῶνΑΒ,ΒΓ,παραβεβλήσθωδὴτοῖςἀπὸτῶνΑΒ, AC, have been applied to DE, making DG as breadth
ΒΓπαρὰτὴνΔΕἴσοντὸΕΘ·λοιπὸνἄρατὸΘΖἴσονἐστὶ [Prop. 1.44]. And since the (square) on AC is equal to
τῷδὶςὑπὸτῶνΑΒ,ΒΓ.καὶἐπεὶμέσηἐστὶνἑκατέρατῶν (the sum of) the (squares) on AB and BC, plus twice
ΑΒ,ΒΓ,μέσαἄραἐστὶκαὶτὰἀπὸτῶνΑΒ,ΒΓ.μέσονδὲ the (rectangle contained) by AB and BC [Prop. 2.4], so
ὑπόκειταικαὶτὸδὶςὑπὸτῶνΑΒ,ΒΓ.καίἐστιτοῖςμὲνἀπὸ let (the rectangle) EH, equal to (the sum of) the squares
τῶνΑΒ,ΒΓἴσοντὸΕΘ,τῷδὲδὶςὑπὸτῶνΑΒ,ΒΓἴσον on AB and BC, have been applied to DE. The remainτὸΖΘ·μέσονἄραἑκάτεροντῶνΕΘ,ΘΖ.καὶπαρὰῥητὴν
der HF is thus equal to twice the (rectangle contained)
τὴνΔΕπαράκειται·ῥητὴἄραἐστὶνἑκατέρατῶνΔΘ,ΘΗ by AB and BC. And since AB and BC are each meκαὶἀσύμμετροςτῇΔΕμήκει.ἐπεὶοὖνἀσύμμετρόςἐστινἡ
dial, (the sum of) the squares on AB and BC is thus also
ΑΒτῇΒΓμήκει,καίἐστινὡςἡΑΒπρὸςτὴνΒΓ,οὕτωςτὸ medial. ‡ And twice the (rectangle contained) by AB and
ἀπὸτῆςΑΒπρὸςτὸὑπὸτῶνΑΒ,ΒΓ,ἀσύμμετρονἄραἐστὶ BC was also assumed (to be) medial. And EH is equal
τὸἀπὸτῆςΑΒτῷὑπὸτῶνΑΒ,ΒΓ.ἀλλὰτῷμὲνἀπὸτῆς to (the sum of) the squares on AB and BC, and FH (is)
ΑΒσύμμετρόνἐστιτὸσυγκείμενονἐκτῶνἀπὸτῶνΑΒ, equal to twice the (rectangle contained) by AB and BC.
ΒΓτετραγώνων,τῷδὲὑπὸτῶνΑΒ,ΒΓσύμμετρόνἐστιτὸ Thus, EH and HF (are) each medial. And they were apδὶςὑπὸτῶνΑΒ,ΒΓ.ἀσύμμετρονἄραἐστὶτὸσυγκείμενον
plied to the rational (straight-line) DE. Thus, DH and
ἐκτῶνἀπὸτῶνΑΒ,ΒΓτῷδὶςὑπὸτῶνΑΒ,ΒΓ.ἀλλὰ HG are each rational, and incommensurable in length
τοῖςμὲνἀπὸτῶνΑΒ,ΒΓἴσονἐστὶτὸΕΘ,τῷδὲδὶςὑπὸ with DE [Prop. 10.22]. Therefore, since AB is incomτῶνΑΒ,ΒΓἴσονἐστὶτὸΘΖ.ἀσύμμετρονἄραἐστὶτὸΕΘ
mensurable in length with BC, and as AB is to BC, so
τῷΘΖ·ὥστεκαὶἡΔΘτῇΘΗἐστινἀσύμμετροςμήκει.αἱ the (square) on AB (is) to the (rectangle contained) by
ΔΘ,ΘΗἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι. ὥστεἡ AB and BC [Prop. 10.21 lem.], the (square) on AB is
ΔΗἄλογόςἐστιν. ῥητὴδὲἡΔΕ·τὸδὲὑπὸἀλόγουκαὶ thus incommensurable with the (rectangle contained) by
ῥητῆςπεριεχόμενονὀρθογώνιονἄλογόνἐστιν·ἄλογονἄρα AB and BC [Prop. 10.11]. But, the sum of the squares
ἐστὶτὸΔΖχωρίον,καὶἡδυναμένη[αὐτὸ]ἄλογόςἐστιν. on AB and BC is commensurable with the (square) on
δύναταιδὲτὸΔΖἡΑΓ·ἄλογοςἄραἐστὶνἡΑΓ,καλείσθω AB [Prop. 10.15], and twice the (rectangle contained) by
δὲἐκδύομέσωνδευτέρα.ὅπερἔδειδεῖξαι.
AB and BC is commensurable with the (rectangle contained)
by AB and BC [Prop. 10.6]. Thus, the sum of the
(squares) on AB and BC is incommensurable with twice
the (rectangle contained) by AB and BC [Prop. 10.13].
But, EH is equal to (the sum of) the squares on AB and
BC, and HF is equal to twice the (rectangle) contained
by AB and BC. Thus, EH is incommensurable with
HF . Hence, DH is also incommensurable in length with
HG [Props. 6.1, 10.11]. Thus, DH and HG are rational
(straight-lines which are) commensurable in square
only. Hence, DG is irrational [Prop. 10.36]. And DE (is)
rational. And the rectangle contained by irrational and
rational (straight-lines) is irrational [Prop. 10.20]. The
area DF is thus irrational, and (so) the square-root [of
it] is irrational [Def. 10.4]. And AC is the square-root
of DF . AC is thus irrational—let it be called a second
bimedial (straight-line). § (Which is) the very thing it was
required to show.
† Literally, “second from two medials”.
‡ Since, by hypothesis, the squares on AB and BC are commensurable—see Props. 10.15, 10.23.
§ Thus, a second bimedial straight-line has a length expressible as k 1/4 +k ′1/2 /k 1/4 . The second bimedial and the corresponding second apotome
of a medial, whose length is expressible as k 1/4 − k ′1/2 /k 1/4 (see Prop. 10.75), are the positive roots of the quartic x 4 − 2[(k + k ′ )/ √ k ] x 2 +
323

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
[(k − k ′ ) 2 /k] = 0.
Ð. Proposition 39
᾿Εὰνδύοεὐθεῖαιδυνάμειἀσύμμετροισυντεθῶσιποιοῦς- If two straight-lines (which are) incommensurable in
αιτὸμὲνσυγκείμενονἐκτῶνἀπ᾿αὐτῶντετραγώνωνῥητόν, square, making the sum of the squares on them rational,
τὸδ᾿ὑπ᾿αὐτῶνμέσον,ἡὅληεὐθεῖαἄλογόςἐστιν,καλείσθω and the (rectangle contained) by them medial, are added
δὲμείζων.
together then the whole straight-line is irrational—let it
be called a major (straight-line).
Α Β Γ A B C
Συγκείσθωσανγὰρδύοεὐθεῖαιδυνάμειἀσύμμετροιαἱ For let the two straight-lines, AB and BC, incommen-
ΑΒ,ΒΓποιοῦσαιτὰπροκείμενα·λέγω,ὅτιἄλογόςἐστινἡ surable in square, and fulfilling the prescribed (condi-
ΑΓ.
tions), be laid down together [Prop. 10.33]. I say that
᾿ΕπεὶγὰρτὸὑπὸτῶνΑΒ,ΒΓμέσονἐστίν,καὶτὸδὶς AC is irrational.
[ἄρα]ὑπὸτῶνΑΒ,ΒΓμέσονἐστίν.τὸδὲσυγκείμενονἐκ For since the (rectangle contained) by AB and BC is
τῶνἀπὸτῶνΑΒ,ΒΓῥητόν·ἀσύμμετρονἄραἐστὶτὸδὶς medial, twice the (rectangle contained) by AB and BC
ὑπὸτῶνΑΒ,ΒΓτῷσυγκειμένῳἐκτῶνἀπὸτῶνΑΒ,ΒΓ· is [thus] also medial [Props. 10.6, 10.23 corr.]. And the
ὥστεκαὶτὰἀπὸτῶνΑΒ,ΒΓμετὰτοῦδὶςὑπὸτῶνΑΒ,ΒΓ, sum of the (squares) on AB and BC (is) rational. Thus,
ὅπερἐστὶτὸἀπὸτῆςΑΓ,ἀσύμμετρόνἐστιτῷσυγκειμένῳ twice the (rectangle contained) by AB and BC is incom-
ἐκτῶνἀπὸτῶνΑΒ,ΒΓ[ῥητὸνδὲτὸσυγμείμενονἐκτῶν mensurable with the sum of the (squares) on AB and
ἀπὸτῶνΑΒ,ΒΓ]·ἄλογονἄραἐστὶτὸἀπὸτῆςΑΓ.ὥστε BC [Def. 10.4]. Hence, (the sum of) the squares on AB
καὶἡΑΓἄλογόςἐστιν, καλείσθωδὲμείζων. ὅπερἔδει and BC, plus twice the (rectangle contained) by AB and
δεῖξαι.
q
† Thus, a major straight-line has a length expressible as
x 4 − 2 x 2 + k 2 /(1 + k 2 0.Ñ. ) =
Proposition
minor, whose length is expressible as
q
[1 + k/(1 + k 2 ) 1/2 ]/2 −
[1 + k/(1 + k 2 ) 1/2 ]/2 +
BC—that is, the (square) on AC [Prop. 2.4]—is also incommensurable
with the sum of the (squares) on AB and
BC [Prop. 10.16] [and the sum of the (squares) on AB
and BC (is) rational]. Thus, the (square) on AC is irrational.
Hence, AC is also irrational [Def. 10.4]—let it be
called a major (straight-line). † (Which is) the very thing
it was required to show.
q
[1 − k/(1 + k 2 ) 1/2 ]/2. The major and the corresponding
q
[1 − k/(1 + k 2 ) 1/2 ]/2 (see Prop. 10.76), are the positive roots of the quartic
40
᾿Εὰνδύοεὐθεῖαιδυνάμειἀσύμμετροισυντεθῶσιποιοῦς- If two straight-lines (which are) incommensurable
αιτὸμὲνσυγκείμενονἐκτῶνἀπ᾿αὐτῶντετραγώνωνμέσον, in square, making the sum of the squares on them
τὸδ᾿ὑπ᾿αὐτῶνῥητόν,ἡὅληεὐθεῖαἄλογόςἐστιν,καλείσθω medial, and the (rectangle contained) by them ratioδὲῥητὸνκαὶμέσονδυναμένη.
nal, are added together then the whole straight-line is
irrational—let it be called the square-root of a rational
plus a medial (area).
Α Β Γ A B C
Συγκείσθωσανγὰρδύοεὐθεῖαιδυνάμειἀσύμμετροιαἱ For let the two straight-lines, AB and BC, incommen-
ΑΒ,ΒΓποιοῦσαιτὰπροκείμενα·λέγω,ὅτιἄλογόςἐστινἡ surable in square, (and) fulfilling the prescribed (condi-
ΑΓ.
tions), be laid down together [Prop. 10.34]. I say that
᾿Επεὶ γὰρ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ AC is irrational.
μέσονἐστίν,τὸδὲδὶςὑπὸτῶνΑΒ,ΒΓῥητόν,ἀσύμμετρον For since the sum of the (squares) on AB and BC is
ἄραἐστὶτὸσυγκείμενονἐκτῶνἀπὸτῶνΑΒ,ΒΓτῷδὶς medial, and twice the (rectangle contained) by AB and
324

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ὑπὸτῶνΑΒ,ΒΓ·ὥστεκαὶτὸἁπὸτῆςΑΓἀσύμμετρόνἐστι BC (is) rational, the sum of the (squares) on AB and
τῷδὶςὑπὸτῶνΑΒ,ΒΓ.ῥητὸνδὲτὸδὶςὑπὸτῶνΑΒ,ΒΓ· BC is thus incommensurable with twice the (rectangle
ἄλογονἄρατὸἀπὸτῆςΑΓ.ἄλογοςἄραἡΑΓ,καλείσθωδὲ contained) by AB and BC. Hence, the (square) on AC
ῥητὸνκαὶμέσονδυναμένη.ὅπερἔδειδεῖξαι.
is also incommensurable with twice the (rectangle contained)
by AB and BC [Prop. 10.16]. And twice the
(rectangle contained) by AB and BC (is) rational. The
(square) on AC (is) thus irrational. Thus, AC (is) irrational
[Def. 10.4]—let it be called the square-root of a
rational plus a medial (area). † (Which is) the very thing
it was required to show.
q
q
† Thus, the square-root of a rational plus a medial (area) has a length expressible as [(1 + k 2 ) 1/2 + k]/[2(1 + k 2 )]+ [(1 + k 2 ) 1/2 − k]/[2(1 + k 2 )].
q
q
This and the corresponding irrational with a minus sign, whose length is expressible as [(1 + k 2 ) 1/2 + k]/[2(1 + k 2 )]− [(1 + k 2 ) 1/2 − k]/[2(1 + k 2 )]
(see Prop. 10.77), are the positive roots of the quartic x 4 − (2/ √ 1 + k 2 ) x 2 + k 2 /(1 + k 2 ) 2 = 0.
Ñ. Proposition 41
᾿Εὰνδύοεὐθεῖαιδυνάμειἀσύμμετροισυντεθῶσιποιοῦς- If two straight-lines (which are) incommensurable in
αιτότεσυγκείμενονἐκτῶνἀπ᾿αὐτῶντετραγώνωνμέσον square, making the sum of the squares on them meκαὶτὸὑπ᾿αὐτῶνμέσονκαὶἔτιἀσύμμετροντῷσυγκειμένῳ
dial, and the (rectangle contained) by them medial, and,
ἐκτῶνἀπ᾿αὐτῶντετραγώνων,ἡὅληεὐθεῖαἄλογόςἐστιν, moreover, incommensurable with the sum of the squares
καλείσθωδὲδύομέσαδυναμένη.
on them, are added together then the whole straight-line
is irrational—let it be called the square-root of (the sum
of) two medial (areas).
Κ Θ
K
H
Α
A
Β
Η
Ζ
B
G
F
Γ
∆
Ε
Συγκείσθωσανγὰρδύοεὐθεῖαιδυνάμειἀσύμμετροιαἱ For let the two straight-lines, AB and BC, incommen-
ΑΒ,ΒΓποιοῦσαιτὰπροκείμενα·λέγω,ὅτιἡΑΓἄλογός surable in square, (and) fulfilling the prescribed (condi-
ἐστιν.
tions), be laid down together [Prop. 10.35]. I say that
᾿ΕκκείσθωῥητὴἡΔΕ,καὶπαραβεβλήσθωπαρὰτὴνΔΕ AC is irrational.
τοῖςμὲνἀπὸτῶνΑΒ,ΒΓἴσοντὸΔΖ,τῷδὲδὶςὑπὸτῶν Let the rational (straight-line) DE be laid out, and let
ΑΒ,ΒΓἴσοντὸΗΘ·ὅλονἄρατὸΔΘἴσονἐστὶτῷἀπὸ (the rectangle) DF , equal to (the sum of) the (squares)
τῆςΑΓτετραγώνῳ. καὶἐπεὶμέσονἐστὶτὸσυγκείμενον on AB and BC, and (the rectangle) GH, equal to twice
ἐκτῶνἀπὸτῶνΑΒ,ΒΓ,καίἐστινἴσοντῷΔΖ,μέσονἄρα the (rectangle contained) by AB and BC, have been ap-
ἐστὶκαὶτὸΔΖ.καὶπαρὰῥητὴντὴνΔΕπαράκειται·ῥητὴ plied to DE. Thus, the whole of DH is equal to the
ἄραἐστὶνἡΔΗκαὶἀσύμμετροςτῇΔΕμήκει.διὰτὰαὐτὰ square on AC [Prop. 2.4]. And since the sum of the
δὴκαὶἡΗΚῥητήἐστικαὶἀσύμμετροςτῇΗΖ,τουτέστιτῇ (squares) on AB and BC is medial, and is equal to DF ,
ΔΕ,μήκει. καὶἐπεὶἀσύμμετράἐστιτὰἀπὸτῶνΑΒ,ΒΓ DF is thus also medial. And it is applied to the rational
τῷδὶςὑπὸτῶνΑΒ,ΒΓ,ἀσύμμετρόνἐστιτὸΔΖτῷΗΘ· (straight-line) DE. Thus, DG is rational, and incommen-
C
D
E
325

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ὥστεκαὶἡΔΗτῇΗΚἀσύμμετρόςἐστιν.καὶεἰσιῥηταί·αἱ surable in length with DE [Prop. 10.22]. So, for the same
ΔΗ,ΗΚἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι·ἄλογος (reasons), GK is also rational, and incommensurable in
ἄραἐστὶνἡΔΚἡκαλουμένηἐκδύοὀνομάτων.ῥητὴδὲἡ length with GF —that is to say, DE. And since (the sum
ΔΕ·ἄλογονἄραἐστὶτὸΔΘκαὶἡδυναμένηαὐτὸἄλογός of) the (squares) on AB and BC is incommensurable
ἐστιν. δύναταιδὲτὸΘΔἡΑΓ·ἄλογοςἄραἐστὶνἡΑΓ, with twice the (rectangle contained) by AB and BC,
καλείσθωδὲδύομέσαδυναμένη.ὅπερἔδειδεῖξαι. DF is incommensurable with GH. Hence, DG is also incommensurable
(in length) with GK [Props. 6.1, 10.11].
And they are rational. Thus, DG and GK are rational
(straight-lines which are) commensurable in square only.
Thus, DK is irrational, and that (straight-line which is)
called binomial [Prop. 10.36]. And DE (is) rational.
Thus, DH is irrational, and its square-root is irrational
[Def. 10.4]. And AC (is) the square-root of HD. Thus,
AC is irrational—let it be called the square-root of (the
sum of) two medial (areas). † (Which is) the very thing it
was required to show.
q
«
† Thus, the square-root of (the sum of) two medial (areas) has a length expressible as k
„q[1 ′1/4 + k/(1 + k 2 ) 1/2 ]/2 + [1 − k/(1 + k 2 ) 1/2 ]/2 .
q
«
This and the corresponding irrational with a minus sign, whose length is expressible as k
„q[1 ′1/4 + k/(1 + k 2 ) 1/2 ]/2 − [1 − k/(1 + k 2 ) 1/2 ]/2
(see Prop. 10.78), are the positive roots of the quartic x 4 − 2 k ′1/2 x 2 + k ′ k 2 /(1 + k 2 ) = 0.
ÄÑÑ.
Lemma
῞Οτιδὲαἱεἰρημέναιἄλογοιμοναχῶςδιαιροῦνταιεἰςτὰς We will now demonstrate that the aforementioned
εὐθείας, ἐξὧνσύγκεινταιποιουσῶντὰπροκείμεναεἴδη, irrational (straight-lines) are uniquely divided into the
δείξομενἤδηπροεκθέμενοιλημμάτιοντοιοῦτον·
straight-lines of which they are the sum, and which produce
the prescribed types, (after) setting forth the following
lemma.
Α ∆ Ε Γ Β
A D E C B
᾿ΕκκείσθωεὐθεῖαἡΑΒκαὶτετμήσθωἡὅληεἰςἄνισα Let the straight-line AB be laid out, and let the whole
καθ᾿ἑκάτεροντῶνΓ,Δ,ὑποκείσθωδὲμείζωνἡΑΓτῆς (straight-line) have been cut into unequal parts at each
ΔΒ·λέγω,ὅτιτὰἀπὸτῶνΑΓ,ΓΒμείζονάἐστιτῶνἀπὸ of the (points) C and D. And let AC be assumed (to be)
τῶνΑΔ,ΔΒ.
greater than DB. I say that (the sum of) the (squares) on
ΤετμήσθωγὰρἡΑΒδίχακατὰτὸΕ.καὶἐπεὶμείζων AC and CB is greater than (the sum of) the (squares) on
ἐστὶνἡΑΓτῆςΔΒ,κοινὴἀφῃρήσθωἡΔΓ·λοιπὴἄραἡΑΔ AD and DB.
λοιπῆςτῆςΓΒμείζωνἐστίν.ἴσηδὲἡΑΕτῇΕΒ·ἐλάττων For let AB have been cut in half at E. And since AC is
ἄραἡΔΕτῆςΕΓ·τὰΓ,Δἄρασημεῖαοὐκἴσονἀπέχουσι greater than DB, let DC have been subtracted from both.
τῆςδιχοτομίας.καὶἐπεὶτὸὑπὸτῶνΑΓ,ΓΒμετὰτοῦἀπὸ Thus, the remainder AD is greater than the remainder
τῆςΕΓἴσονἐστὶτῷἀπὸτῆςΕΒ,ἀλλὰμὴνκαὶτὸὑπὸτῶν CB. And AE (is) equal to EB. Thus, DE (is) less than
ΑΔ,ΔΒμετὰτοῦἀπὸΔΕἴσονἐστὶτῷἀπὸτῆςΕΒ,τὸ EC. Thus, points C and D are not equally far from the
ἄραὑπὸτῶνΑΓ,ΓΒμετὰτοῦἀπὸτῆςΕΓἴσονἐστὶτῷ point of bisection. And since the (rectangle contained)
ὑπὸτῶνΑΔ,ΔΒμετὰτοῦἀπὸτῆςΔΕ·ὧντὸἀπὸτῆς by AC and CB, plus the (square) on EC, is equal to the
ΔΕἔλασσόνἐστιτοῦἀπὸτῆςΕΓ·καὶλοιπὸνἄρατὸὑπὸ (square) on EB [Prop. 2.5], but, moreover, the (rectanτῶνΑΓ,ΓΒἔλασσόνἐστιτοῦὑπὸτῶνΑΔ,ΔΒ.ὥστεκαὶ
gle contained) by AD and DB, plus the (square) on DE,
τὸδὶςὑπὸτῶνΑΓ,ΓΒἔλασσόνἐστιτοῦδὶςὑπὸτῶνΑΔ, is also equal to the (square) on EB [Prop. 2.5], the (rect-
ΔΒ.καὶλοιπὸνἄρατὸσυγκείμενονἐκτῶνἀπὸτῶνΑΓ, angle contained) by AC and CB, plus the (square) on
ΓΒμεῖζόνἐστιτοῦσυγκειμένουἐκτῶνἀπὸτῶνΑΔ,ΔΒ. EC, is thus equal to the (rectangle contained) by AD and
ὅπερἔδειδεῖξαι.
DB, plus the (square) on DE. And, of these, the (square)
on DE is less than the (square) on EC. And, thus, the
326

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
† Since, AC 2 + CB 2 + 2 AC CB = AD 2 + DB 2 + 2 AD DB = AB 2 .
remaining (rectangle contained) by AC and CB is less
than the (rectangle contained) by AD and DB. And,
hence, twice the (rectangle contained) by AC and CB
is less than twice the (rectangle contained) by AD and
DB. And thus the remaining sum of the (squares) on AC
and CB is greater than the sum of the (squares) on AD
and DB. † (Which is) the very thing it was required to
show.
Ñ. Proposition 42
῾Ηἐκδύοὀνομάτωνκατὰἓνμόνονσημεῖονδιαιρεῖται A binomial (straight-line) can be divided into its (compoεἰςτὰὀνόματα.
nent) terms at one point only. †
Α ∆ Γ Β A D C B
῎ΕστωἐκδύοὀνομάτωνἡΑΒδιῃρημένηεἰςτὰὀνόματα Let AB be a binomial (straight-line) which has been
κατὰ τὸ Γ· αἱ ΑΓ, ΓΒ ἄρα ῥηταί εἰσι δυνάμει μόνον divided into its (component) terms at C. AC and CB are
σύμμετροι.λέγω,ὅτιἡΑΒκατ᾿ἄλλοσημεῖονοὐδιαιρεῖται thus rational (straight-lines which are) commensurable
εἰςδύοῥητὰςδυνάμειμόνονσυμμέτρους.
in square only [Prop. 10.36]. I say that AB cannot be
Εἰγὰρδυνατόν,διῃρήσθωκαὶκατὰτὸΔ,ὥστεκαὶτὰς divided at another point into two rational (straight-lines
ΑΔ,ΔΒῥητὰςεἶναιδυνάμειμόνονσυμμέτρους. φανερὸν which are) commensurable in square only.
δή,ὅτιἡΑΓτῇΔΒοὐκἔστινἡαὐτή. εἰγὰρδυνατόν, For, if possible, let it also have been divided at D, such
ἔστω.ἔσταιδὴκαὶἡΑΔτῇΓΒἡαὐτή·καὶἔσταιὡςἡΑΓ that AD and DB are also rational (straight-lines which
πρὸςτὴνΓΒ,οὕτωςἡΒΔπρὸςτὴνΔΑ,καὶἔσταιἡΑΒ are) commensurable in square only. So, (it is) clear that
κατὰτὸαὐτὸτῇκατὰτὸΓδιαιρέσειδιαιρεθεῖσακαὶκατὰ AC is not the same as DB. For, if possible, let it be (the
τὸΔ·ὅπεροὐχὑπόκειται. οὐκἄραἡΑΓτῇΔΒἐστινἡ same). So, AD will also be the same as CB. And as
αὐτή.διὰδὴτοῦτοκαὶτὰΓ,Δσημεῖαοὐκἴσονἀπέχουσι AC will be to CB, so BD (will be) to DA. And AB will
τῆςδιχοτομίας. ᾧἄραδιαφέρειτὰἀπὸτῶνΑΓ,ΓΒτῶν (thus) also be divided at D in the same (manner) as the
ἀπὸτῶνΑΔ,ΔΒ,τούτῳδιαφέρεικαὶτὸδὶςὑπὸτῶνΑΔ, division at C. The very opposite was assumed. Thus, AC
ΔΒτοῦδὶςὑπὸτῶνΑΓ,ΓΒδιὰτὸκαὶτὰἀπὸτῶνΑΓ, is not the same as DB. So, on account of this, points
ΓΒμετὰτοῦδὶςὑπὸτῶνΑΓ,ΓΒκαὶτὰἀπὸτῶνΑΔ,ΔΒ C and D are not equally far from the point of bisection.
μετὰτοῦδὶςὑπὸτῶνΑΔ,ΔΒἴσαεἶναιτῷἀπὸτῆςΑΒ. Thus, by whatever (amount the sum of) the (squares) on
ἀλλὰτὰἀπὸτῶνΑΓ,ΓΒτῶνἀπὸτῶνΑΔ,ΔΒδιαφέρει AC and CB differs from (the sum of) the (squares) on
ῥητῷ·ῥητὰγὰρἀμφότερα·καὶτὸδὶςἄραὑπὸτῶνΑΔ,ΔΒ AD and DB, twice the (rectangle contained) by AD and
τοῦδὶςὑπὸτῶνΑΓ,ΓΒδιαφέρειῥητῷμέσαὄντα·ὅπερ DB also differs from twice the (rectangle contained) by
ἄτοπον·μέσονγὰρμέσουοὐχὑπερέχειῥητῷ. AC and CB by this (same amount)—on account of both
Οὐχἄραἡἐκδύοὀνομάτωνκατ᾿ἄλλοκαὶἄλλοσημεῖον (the sum of) the (squares) on AC and CB, plus twice the
διαιρεῖται·καθ᾿ἓνἄραμόνον·ὅπερἔδειδεῖξαι.
(rectangle contained) by AC and CB, and (the sum of)
the (squares) on AD and DB, plus twice the (rectangle
contained) by AD and DB, being equal to the (square)
on AB [Prop. 2.4]. But, (the sum of) the (squares) on AC
and CB differs from (the sum of) the (squares) on AD
and DB by a rational (area). For (they are) both rational
(areas). Thus, twice the (rectangle contained) by AD
and DB also differs from twice the (rectangle contained)
by AC and CB by a rational (area, despite both) being
medial (areas) [Prop. 10.21]. The very thing is absurd.
For a medial (area) cannot exceed a medial (area) by a
rational (area) [Prop. 10.26].
327

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
Thus, a binomial (straight-line) cannot be divided
(into its component terms) at different points. Thus, (it
can be so divided) at one point only. (Which is) the very
thing it was required to show.
† In other words, k + k ′1/2 = k ′′ + k ′′′1/2 has only one solution: i.e., k ′′ = k and k ′′′ = k ′ . Likewise, k 1/2 + k ′1/2 = k ′′1/2 + k ′′′1/2 has only
one solution: i.e., k ′′ = k and k ′′′ = k ′ (or, equivalently, k ′′ = k ′ and k ′′′ = k).
Ñ. Proposition 43
῾Ηἐκδύομέσωνπρώτηκαθ᾿ἓνμόνονσημεῖονδιαιρεῖται. A first bimedial (straight-line) can be divided (into its
component terms) at one point only. †
Α ∆ Γ Β A D C B
῎ΕστωἐκδύομέσωνπρώτηἡΑΒδιῃρημένηκατὰτὸΓ, Let AB be a first bimedial (straight-line) which has
ὥστετὰςΑΓ,ΓΒμέσαςεἶναιδυνάμειμόνονσυμμέτρους been divided at C, such that AC and CB are medial
ῥητὸνπεριεχούσας·λέγω,ὅτιἡΑΒκατ᾿ἄλλοσημεῖονοὐ (straight-lines), commensurable in square only, (and)
διαιρεῖται.
containing a rational (area) [Prop. 10.37]. I say that AB
ΕἰγὰρδυνατόνδιῃρήσθωκαὶκατὰτὸΔ,ὥστεκαὶτὰς cannot be (so) divided at another point.
ΑΔ,ΔΒμέσαςεἶναιδυνάμειμόνονσυμμέτρουςῥητὸνπε- For, if possible, let it also have been divided at D,
ριεχούσας. ἐπεὶοὖν,ᾧδιαφέρειτὸδὶςὑπὸτῶνΑΔ,ΔΒ such that AD and DB are also medial (straight-lines),
τοῦδὶςὑπὸτῶνΑΓ,ΓΒ,τούτῳδιαφέρειτὰἀπὸτῶνΑΓ, commensurable in square only, (and) containing a ratio-
ΓΒτῶνἀπὸτῶνΑΔ,ΔΒ,ῥητῷδὲδιαφέρειτὸδὶςὑπὸ nal (area). Since, therefore, by whatever (amount) twice
τῶνΑΔ,ΔΒτοῦδὶςὑπὸτῶνΑΓ,ΓΒ·ῥητὰγὰρἀμφότερα· the (rectangle contained) by AD and DB differs from
ῥητῷἄραδιαφέρεικαὶτὰἀπὸτῶνΑΓ,ΓΒτῶνἀπὸτῶν twice the (rectangle contained) by AC and CB, (the sum
ΑΔ,ΔΒμέσαὄντα·ὅπερἄτοπον.
of) the (squares) on AC and CB differs from (the sum
Οὐκἄραἡἐκδύομέσωνπρώτηκατ᾿ἄλλοκαὶἄλλο of) the (squares) on AD and DB by this (same amount)
σημεῖονδιαιρεῖταιεἰςτὰὀνόματα·καθ᾿ἓνἄραμόνον·ὅπερ [Prop. 10.41 lem.]. And twice the (rectangle contained)
ἔδειδεῖξαι.
by AD and DB differs from twice the (rectangle contained)
by AC and CB by a rational (area). For (they
are) both rational (areas). (The sum of) the (squares) on
AC and CB thus differs from (the sum of) the (squares)
on AD and DB by a rational (area, despite both) being
medial (areas). The very thing is absurd [Prop. 10.26].
Thus, a first bimedial (straight-line) cannot be divided
into its (component) terms at different points. Thus, (it
can be so divided) at one point only. (Which is) the very
thing it was required to show.
† In other words, k 1/4 + k 3/4 = k ′1/4 + k ′3/4 has only one solution: i.e., k ′ = k.
Ñ. Proposition 44
῾Η ἐκ δύο μέσων δευτέρα καθ᾿ ἓν μόνον σημεῖον A second bimedial (straight-line) can be divided (into
διαιρεῖται. its component terms) at one point only. †
῎ΕστωἐκδύομέσωνδευτέραἡΑΒδιῃρημένηκατὰτὸ Let AB be a second bimedial (straight-line) which
Γ,ὥστετὰςΑΓ,ΓΒμέσαςεἶναιδυνάμειμόνονσυμμέτρους has been divided at C, so that AC and BC are medial
μέσονπεριεχούσας·φανερὸνδή,ὅτιτὸΓοὐκἔστικατὰ (straight-lines), commensurable in square only, (and)
τῆςδιχοτομίας,ὅτιοὐκεἰσὶμήκεισύμμετροι. λέγω,ὅτιἡ containing a medial (area) [Prop. 10.38]. So, (it is) clear
ΑΒκατ᾿ἄλλοσημεῖονοὐδιαιρεῖται.
that C is not (located) at the point of bisection, since (AC
and BC) are not commensurable in length. I say that AB
cannot be (so) divided at another point.
328

ËÌÇÁÉÁÏƺ ELEMENTS
Α ∆ Β
Γ
A D C B
BOOK 10
Ε Μ Θ Ν
E
M
H
N
Ζ
Λ
Η
Κ
Εἰγὰρδυνατόν, διῃρήσθωκαὶκατὰτὸΔ,ὥστετὴν For, if possible, let it also have been (so) divided at
ΑΓτῇΔΒμὴεἶναιτὴναὐτήν,ἀλλὰμείζονακαθ᾿ὑπόθεσιν D, so that AC is not the same as DB, but AC (is),
τὴνΑΓ·δῆλονδή,ὅτικαὶτὰἀπὸτῶνΑΔ,ΔΒ,ὡςἐπάνω by hypothesis, greater. So, (it is) clear that (the sum
ἐδείξαμεν,ἐλάσσονατῶνἀπὸτῶνΑΓ,ΓΒ·καὶτὰςΑΔ,ΔΒ of) the (squares) on AD and DB is also less than (the
μέσαςεἶναιδυνάμειμόνονσυμμέτρουςμέσονπεριεχούσας. sum of) the (squares) on AC and CB, as we showed
καὶἐκκείσθωῥητὴἡΕΖ,καὶτῷμὲνἀπὸτῆςΑΒἴσονπαρὰ above [Prop. 10.41 lem.]. And AD and DB are medial
τὴνΕΖπαραλληλόγραμμονὀρθογώνιονπαραβεβλήσθωτὸ (straight-lines), commensurable in square only, (and)
ΕΚ,τοῖςδὲἀπὸτῶνΑΓ,ΓΒἴσονἀφῃρήσθωτὸΕΗ·λοιπὸν containing a medial (area). And let the rational (straight-
ἄρατὸΘΚἴσονἐστὶτῷδὶςὑπὸτῶνΑΓ,ΓΒ.πάλινδὴτοῖς line) EF be laid down. And let the rectangular paral-
ἀπὸτῶνΑΔ,ΔΒ,ἅπερἐλάσσοναἐδείχθητῶνἀπὸτῶν lelogram EK, equal to the (square) on AB, have been
ΑΓ,ΓΒ,ἴσονἀφῃρήσθωτὸΕΛ·καὶλοιπὸνἄρατὸΜΚ applied to EF . And let EG, equal to (the sum of) the
ἴσοντῷδὶςὑπὸτῶνΑΔ,ΔΒ.καὶἐπεὶμέσαἐστὶτὰἀπὸ (squares) on AC and CB, have been cut off (from EK).
τῶνΑΓ,ΓΒ,μέσονἄρα[καὶ]τὸΕΗ.καὶπαρὰῥητὴντὴν Thus, the remainder, HK, is equal to twice the (rectan-
ΕΖπαράκειται·ῥητὴἄραἐστὶνἡΕΘκαὶἀσύμμετροςτῇΕΖ gle contained) by AC and CB [Prop. 2.4]. So, again,
μήκει.διὰτὰαὐτὰδὴκαὶἡΘΝῥητήἐστικαὶἀσύμμετρος let EL, equal to (the sum of) the (squares) on AD and
τῇΕΖμήκει.καὶἐπεὶαἱΑΓ,ΓΒμέσαιεἰσὶδυνάμειμόνον DB—which was shown (to be) less than (the sum of) the
σύμμετροι,ἀσύμμετροςἄραἐστὶνἡΑΓτῇΓΒμήκει.ὡςδὲ (squares) on AC and CB—have been cut off (from EK).
ἡΑΓπρὸςτὴνΓΒ,οὕτωςτὸἀπὸτῆςΑΓπρὸςτὸὑπὸτῶν And, thus, the remainder, MK, (is) equal to twice the
ΑΓ,ΓΒ·ἀσύμμετρονἄραἐστὶτὸἀπὸτῆςΑΓτῷὑπὸτῶν (rectangle contained) by AD and DB. And since (the
ΑΓ,ΓΒ.ἀλλὰτῷμὲνἀπὸτῆςΑΓσύμμετράἐστιτὰἀπὸτῶν sum of) the (squares) on AC and CB is medial, EG
ΑΓ,ΓΒ·δυνάμειγάρεἰσισύμμετροιαἱΑΓ,ΓΒ.τῷδὲὑπὸ (is) thus [also] medial. And it is applied to the ratioτῶνΑΓ,ΓΒσύμμετρόνἐστιτὸδὶςὑπὸτῶνΑΓ,ΓΒ.καὶτὰ
nal (straight-line) EF . Thus, EH is rational, and incom-
ἀπὸτῶνΑΓ,ΓΒἄραἀσύμμετράἐστιτῷδὶςὑπὸτῶνΑΓ, mensurable in length with EF [Prop. 10.22]. So, for the
ΓΒ.ἀλλὰτοῖςμὲνἀπὸτῶνΑΓ,ΓΒἴσονἐστὶτὸΕΗ,τῷδὲ same (reasons), HN is also rational, and incommensuδὶςὑπὸτῶνΑΓ,ΓΒἴσοντὸΘΚ·ἀσύμμετρονἄραἐστὶτὸ
rable in length with EF . And since AC and CB are me-
ΕΗτῷΘΚ·ὥστεκαὶἡΕΘτῇΘΝἀσύμμετρόςἐστιμήκει. dial (straight-lines which are) commensurable in square
καίεἰσιῥηταί· αἱΕΘ,ΘΝἄραῥηταίεἰσιδυνάμειμόνον only, AC is thus incommensurable in length with CB.
σύμμετροι.ἐὰνδὲδύοῥηταὶδυνάμειμόνονσύμμετροισυν- And as AC (is) to CB, so the (square) on AC (is) to the
τεθῶσιν,ἡὅληἄλογόςἐστινἡκαλουμένηἐκδύοὀνομάτων· (rectangle contained) by AC and CB [Prop. 10.21 lem.].
ἡΕΝἄραἐκδύοὀνομάτωνἐστὶδιῃρημένηκατὰτὸΘ.κατὰ Thus, the (square) on AC is incommensurable with the
τὰ αὐτὰδὴ δειχθήσονταικαὶ αἱΕΜ, ΜΝ ῥηταὶδυνάμει (rectangle contained) by AC and CB [Prop. 10.11]. But,
μόνονσύμμετροι· καὶἔσταιἡΕΝἐκδύοὀνομάτωνκατ᾿ (the sum of) the (squares) on AC and CB is commensu-
ἄλλοκαὶἄλλοδιῃρημένητότεΘκαὶτὸΜ,καὶοὐκἔστιν rable with the (square) on AC. For, AC and CB are com-
ἡΕΘτῇΜΝἡαὐτή,ὅτιτὰἀπὸτῶνΑΓ,ΓΒμείζονάἐστι mensurable in square [Prop. 10.15]. And twice the (rectτῶνἀπὸτῶνΑΔ,ΔΒ.ἀλλὰτὰἀπὸτῶνΑΔ,ΔΒμείζονά
angle contained) by AC and CB is commensurable with
ἐστιτοῦδὶςὑπὸΑΔ,ΔΒ·πολλῷἄρακαὶτὰἀπὸτῶνΑΓ, the (rectangle contained) by AC and CB [Prop. 10.6].
ΓΒ,τουτέστιτὸΕΗ,μεῖζόνἐστιτοῦδὶςὑπὸτῶνΑΔ,ΔΒ, And thus (the sum of) the squares on AC and CB is inτουτέστιτοῦΜΚ·ὥστεκαὶἡΕΘτῆςΜΝμείζωνἐστίν.ἡ
commensurable with twice the (rectangle contained) by
ἄραΕΘτῇΜΝοὐκἔστινἡαὐτή·ὅπερἔδειδεῖξαι. AC and CB [Prop. 10.13]. But, EG is equal to (the sum
of) the (squares) on AC and CB, and HK equal to twice
the (rectangle contained) by AC and CB. Thus, EG is
incommensurable with HK. Hence, EH is also incom-
F
L
G
K
329

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
mensurable in length with HN [Props. 6.1, 10.11]. And
(they are) rational (straight-lines). Thus, EH and HN
are rational (straight-lines which are) commensurable in
square only. And if two rational (straight-lines which
are) commensurable in square only are added together
then the whole (straight-line) is that irrational called binomial
[Prop. 10.36]. Thus, EN is a binomial (straightline)
which has been divided (into its component terms)
at H. So, according to the same (reasoning), EM and
MN can be shown (to be) rational (straight-lines which
are) commensurable in square only. And EN will (thus)
be a binomial (straight-line) which has been divided (into
its component terms) at the different (points) H and M
(which is absurd [Prop. 10.42]). And EH is not the same
as MN, since (the sum of) the (squares) on AC and CB
is greater than (the sum of) the (squares) on AD and
DB. But, (the sum of) the (squares) on AD and DB is
greater than twice the (rectangle contained) by AD and
DB [Prop. 10.59 lem.]. Thus, (the sum of) the (squares)
on AC and CB—that is to say, EG—is also much greater
than twice the (rectangle contained) by AD and DB—
that is to say, MK. Hence, EH is also greater than MN
[Prop. 6.1]. Thus, EH is not the same as MN. (Which
is) the very thing it was required to show.
† In other words, k 1/4 + k ′1/2 /k 1/4 = k ′′1/4 + k ′′′1/2 /k ′′1/4 has only one solution: i.e., k ′′ = k and k ′′′ = k ′ .
Ñ. Proposition
῾Ημείζωνκατὰτὸαὐτὸμόνονσημεῖονδιαιρεῖται.
45
A major (straight-line) can only be divided (into its
component terms) at the same point. †
Α ∆ Γ Β A D C B
῎ΕστωμείζωνἡΑΒδιῃρημένηκατὰτὸΓ,ὥστετὰςΑΓ, Let AB be a major (straight-line) which has been di-
ΓΒδυνάμειἀσυμμέτρουςεἶναιποιούσαςτὸμὲνσυγκείμενον vided at C, so that AC and CB are incommensurable in
ἐκ τῶν ἀπὸτῶν ΑΓ, ΓΒτετραγώνων ῥητόν, τὸ δ᾿ ὑπὸ square, making the sum of the squares on AC and CB
τῶνΑΓ,ΓΒμέσον· λέγω, ὅτιἡΑΒκατ᾿ἄλλοσημεῖον rational, and the (rectangle contained) by AC and CD
οὐδιαιρεῖται.
medial [Prop. 10.39]. I say that AB cannot be (so) di-
Εἰγὰρδυνατόν, διῃρήσθωκαὶκατὰτὸΔ,ὥστεκαὶ vided at another point.
τὰςΑΔ,ΔΒδυνάμειἀσυμμέτρουςεἶναιποιούσαςτὸμὲν For, if possible, let it also have been divided at D, such
συγκείμενονἐκτῶνἀπὸτῶνΑΔ,ΔΒῥητόν, τὸδ᾿ὑπ᾿ that AD and DB are also incommensurable in square,
αὐτῶνμέσον.καὶἐπεί,ᾧδιαφέρειτὰἀπὸτῶνΑΓ,ΓΒτῶν making the sum of the (squares) on AD and DB ratio-
ἀπὸτῶνΑΔ,ΔΒ,τούτῳδιαφέρεικαὶτὸδὶςὑπὸτῶνΑΔ, nal, and the (rectangle contained) by them medial. And
ΔΒτοῦδὶςὑπὸτῶνΑΓ,ΓΒ,ἀλλὰτὰἀπὸτῶνΑΓ,ΓΒ since, by whatever (amount the sum of) the (squares) on
τῶνἀπὸτῶνΑΔ,ΔΒὑπερέχειῥητῷ·ῥητὰγὰρἀμφότερα· AC and CB differs from (the sum of) the (squares) on
καὶτὸδὶςὑπὸτῶνΑΔ,ΔΒἄρατοῦδὶςὑπὸτῶνΑΓ,ΓΒ AD and DB, twice the (rectangle contained) by AD and
ὑπερέχειῥητῷμέσαὄντα·ὅπερἐστὶνἀδύνατον.οὐκἄραἡ DB also differs from twice the (rectangle contained) by
μείζωνκατ᾿ἄλλοκαὶἄλλοσημεῖονδιαιρεῖται·κατὰτὸαὐτὸ AC and CB by this (same amount). But, (the sum of)
ἄραμόνονδιαιρεῖται·ὅπερἔδειδεῖξαι.
the (squares) on AC and CB exceeds (the sum of) the
(squares) on AD and DB by a rational (area). For (they
are) both rational (areas). Thus, twice the (rectangle
330

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
† In other words,
contained) by AD and DB also exceeds twice the (rectangle
contained) by AC and CB by a rational (area),
(despite both) being medial (areas). The very thing is
impossible [Prop. 10.26]. Thus, a major (straight-line)
cannot be divided (into its component terms) at different
points. Thus, it can only be (so) divided at the same
(point). (Which is) the very thing it was required to show.
q
q
q
q
[1 + k/(1 + k 2 ) 1/2 ]/2 + [1 − k/(1 + k 2 ) 1/2 ]/2 = [1 + k ′ /(1 + k ′2 ) 1/2 ]/2 + [1 − k ′ /(1 + k ′2 ) 1/2 ]/2 has only one
solution: i.e., k ′ = k.
Ѧ. Proposition 46
῾Ηῥητὸνκαὶμέσονδυναμένηκαθ᾿ἓνμόνονσημεῖον The square-root of a rational plus a medial (area) can be
διαιρεῖται. divided (into its component terms) at one point only. †
Α ∆ Γ Β A D C B
῎ΕστωῥητὸνκαὶμέσονδυναμένηἡΑΒδιῃρημένηκατὰ Let AB be the square-root of a rational plus a medial
τὸΓ,ὥστετὰςΑΓ,ΓΒδυνάμειἀσυμμέτρουςεἶναιποιούσας (area) which has been divided at C, so that AC and CB
τὸμὲνσυγκείμενονἐκτῶνἀπὸτῶνΑΓ,ΓΒμέσον,τὸδὲ are incommensurable in square, making the sum of the
δὶςὑπὸτῶν ΑΓ,ΓΒῥητόν· λέγω, ὅτιἡΑΒκατ᾿ἄλλο (squares) on AC and CB medial, and twice the (rectanσημεῖονοὐδιαιρεῖται.
gle contained) by AC and CB rational [Prop. 10.40]. I
Εἰγὰρδυνατόν, διῃρήσθωκαὶκατὰτὸΔ,ὥστεκαὶ say that AB cannot be (so) divided at another point.
τὰςΑΔ,ΔΒδυνάμειἀσυμμέτρουςεἶναιποιούσαςτὸμὲν For, if possible, let it also have been divided at D, so
συγκείμενονἐκτῶνἀπὸτῶνΑΔ,ΔΒμέσον, τὸδὲδὶς that AD and DB are also incommensurable in square,
ὑπὸτῶνΑΔ,ΔΒῥητόν. ἐπεὶοὖν,ᾧδιαφέρειτὸδὶςὑπὸ making the sum of the (squares) on AD and DB medial,
τῶνΑΓ,ΓΒτοῦδὶςὑπὸτῶνΑΔ,ΔΒ,τούτῳδιαφέρεικαὶ and twice the (rectangle contained) by AD and DB raτὰἀπὸτῶνΑΔ,ΔΒτῶνἀπὸτῶνΑΓ,ΓΒ,τὸδὲδὶςὑπὸ
tional. Therefore, since by whatever (amount) twice the
τῶνΑΓ,ΓΒτοῦδὶςὑπὸτῶνΑΔ,ΔΒὑπερέχειῥητῷ,καὶ (rectangle contained) by AC and CB differs from twice
τὰἀπὸτῶνΑΔ,ΔΒἄρατῶνἀπὸτῶνΑΓ,ΓΒὑπερέχει the (rectangle contained) by AD and DB, (the sum of)
ῥητῷμέσαὄντα·ὅπερἐστὶνἀδύνατον. οὐκἄραἡῥητὸν the (squares) on AD and DB also differs from (the sum
καὶμέσονδυναμένηκατ᾿ἄλλοκαὶἄλλοσημεῖονδιαιρεῖται. of) the (squares) on AC and CB by this (same amount).
κατὰἓνἄρασημεῖονδιαιρεῖται·ὅπερἔδειδεῖξαι.
And twice the (rectangle contained) by AC and CB exceeds
twice the (rectangle contained) by AD and DB by
a rational (area). (The sum of) the (squares) on AD and
DB thus also exceeds (the sum of) the (squares) on AC
and CB by a rational (area), (despite both) being medial
(areas). The very thing is impossible [Prop. 10.26]. Thus,
the square-root of a rational plus a medial (area) cannot
be divided (into its component terms) at different points.
Thus, it can be (so) divided at one point (only). (Which
is) the very thing it was required to show.
q
† In other words,
q
+
[(1 + k 2 ) 1/2 + k]/[2(1 + k 2 )] +
[(1 + k ′2 ) 1/2 − k ′ ]/[2(1 + k ′2 )] has only one solution: i.e., k ′ = k.
q
q
[(1 + k 2 ) 1/2 − k]/[2(1 + k 2 )] = [(1 + k ′2 ) 1/2 + k ′ ]/[2(1 + k ′2 )]
ÑÞ. Proposition 47
῾Ηδύομέσαδυναμένηκαθ᾿ἓνμόνονσημεῖονδιαιρεῖται. The square-root of (the sum of) two medial (areas)
can be divided (into its component terms) at one point
only. †
331

ËÌÇÁÉÁÏƺ ELEMENTS
Α ∆ Β
Γ
BOOK 10
A D C B
Ε
Ζ
E
F
Μ
Θ
Λ
Η
M
H
L
G
Ν
Κ
῎Εστω[δύομέσαδυναμένη]ἡΑΒδιῃρημένηκατὰτὸ Let AB be [the square-root of (the sum of) two me-
Γ,ὥστετὰςΑΓ,ΓΒδυνάμειἀσυμμέτρουςεἶναιποιούσας dial (areas)] which has been divided at C, such that AC
τότεσυγκείμενονἐκτῶνἀπὸτῶνΑΓ,ΓΒμέσονκαὶτὸ and CB are incommensurable in square, making the sum
ὑπὸτῶνΑΓ,ΓΒμέσονκαὶἔτιἀσύμμετροντῷσυγκειμένῳ of the (squares) on AC and CB medial, and the (rect-
ἐκτῶνἀπ᾿αὐτῶν. λέγω,ὅτιἡΑΒκατ᾿ἄλλοσημεῖονοὐ angle contained) by AC and CB medial, and, moreover,
διαιρεῖταιποιοῦσατὰπροκείμενα.
incommensurable with the sum of the (squares) on (AC
Εἰγὰρδυνατόν,διῃρήσθωκατὰτὸΔ,ὥστεπάλινδη- and CB) [Prop. 10.41]. I say that AB cannot be divided
λονότιτὴνΑΓτῇΔΒμὴεἶναιτὴναὐτήν,ἀλλὰμείζονα at another point fulfilling the prescribed (conditions).
καθ᾿ὑπόθεσιντὴνΑΓ,καὶἐκκείσθωῥητὴἡΕΖ,καὶπαρα- For, if possible, let it have been divided at D, such that
βεβλήσθωπαρὰτὴνΕΖτοῖςμὲνἀπὸτῶνΑΓ,ΓΒἴσοντὸ AC is again manifestly not the same as DB, but AC (is),
ΕΗ,τῷδὲδὶςὑπὸτῶνΑΓ,ΓΒἴσοντὸΘΚ·ὅλονἄρατὸ by hypothesis, greater. And let the rational (straight-line)
ΕΚἴσονἐστὶτῷἀπὸτῆςΑΒτετραγώνῳ. πάλινδὴπαρα- EF be laid down. And let EG, equal to (the sum of) the
βεβλήσθωπαρὰτὴνΕΖτοῖςἀπὸτῶνΑΔ,ΔΒἴσοντὸΕΛ· (squares) on AC and CB, and HK, equal to twice the
λοιπὸνἄρατὸδὶςὑπὸτῶνΑΔ,ΔΒλοιπῷτῷΜΚἴσον (rectangle contained) by AC and CB, have been applied
ἐστίν.καὶἐπεὶμέσονὑπόκειταιτὸσυγκείμενονἐκτῶνἀπὸ to EF . Thus, the whole of EK is equal to the square on
τῶνΑΓ,ΓΒ,μέσονἄραἐστὶκαὶτὸΕΗ.καὶπαρὰῥητὴντὴν AB [Prop. 2.4]. So, again, let EL, equal to (the sum of)
ΕΖπαράκειται·ῥητὴἄραἐστὶνἡΘΕκαὶἀσύμμετροςτῇΕΖ the (squares) on AD and DB, have been applied to EF .
μήκει.διὰτὰαὐτὰδὴκαὶἡΘΝῥητήἐστικαὶἀσύμμετρος Thus, the remainder—twice the (rectangle contained) by
τῇΕΖμήκει. καὶἐπεὶἀσύμμετρόνἐστιτὸσυγκείμενον AD and DB—is equal to the remainder, MK. And since
ἐκτῶνἀπὸτῶνΑΓ,ΓΒτῷδὶςὑπὸτῶνΑΓ,ΓΒ,καὶτὸ the sum of the (squares) on AC and CB was assumed
ΕΗἄρατῷΗΝἀσύμμετρόνἐστιν·ὥστεκαὶἡΕΘτῇΘΝ (to be) medial, EG is also medial. And it is applied to
ἀσύμμετρόςἐστιν.καίεἰσιῥηταί·αἱΕΘ,ΘΝἄραῥηταίεἰσι the rational (straight-line) EF . HE is thus rational, and
δυνάμειμόνονσύμμετροι·ἡΕΝἄραἐκδύοὀνομάτωνἐστὶ incommensurable in length with EF [Prop. 10.22]. So,
διῃρημένηκατὰτὸΘ.ὁμοίωςδὴδείξομεν,ὅτικαὶκατὰτὸ for the same (reasons), HN is also rational, and incom-
Μδιῄρηται.καὶοὐκἔστινἡΕΘτῇΜΝἡαὐτή·ἡἄραἐκδύο mensurable in length with EF . And since the sum of
ὀνομάτωνκατ᾿ἄλλοκαὶἄλλοσημεῖονδιῄρηται·ὅπερἐστίν the (squares) on AC and CB is incommensurable with
ἄτοπον.οὐκἄραἡδύομέσαδυναμένηκατ᾿ἀλλοκαὶἄλλο twice the (rectangle contained) by AC and CB, EG is
σημεῖονδιαιρεῖται·καθ᾿ἓνἄραμόνον[σημεῖον]διαιρεῖται. thus also incommensurable with GN. Hence, EH is also
incommensurable with HN [Props. 6.1, 10.11]. And
they are (both) rational (straight-lines). Thus, EH and
HN are rational (straight-lines which are) commensurable
in square only. Thus, EN is a binomial (straightline)
which has been divided (into its component terms)
at H [Prop. 10.36]. So, similarly, we can show that it has
also been (so) divided at M. And EH is not the same as
MN. Thus, a binomial (straight-line) has been divided
(into its component terms) at different points. The very
thing is absurd [Prop. 10.42]. Thus, the square-root of
(the sum of) two medial (areas) cannot be divided (into
N
K
332

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
its component terms) at different points. Thus, it can be
(so) divided at one [point] only.
† In other words, k ′1/4 q
[1 + k/(1 + k 2 ) 1/2 ]/2 + k ′1/4 q
[1 − k/(1 + k 2 ) 1/2 ]/2 = k ′′′1/4 q[1 + k ′′ /(1 + k ′′2 ) 1/2 ]/2
+k
q[1 ′′′1/4 − k ′′ /(1 + k ′′2 ) 1/2 ]/2 has only one solution: i.e., k ′′ = k and k ′′′ = k ′ .
ÎÇÖÓØÖÓ.
Definitions II
εʹ. ῾Υποκειμένης ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων 5. Given a rational (straight-line), and a binomial
διῃρημένης εἰς τὰ ὀνόματα, ἧς τὸ μεῖζον ὄνομα τοῦ (straight-line) which has been divided into its (compo-
ἐλάσσονοςμεῖζονδύναταιτῷἀπὸσυμμέτρουἑαυτῇμήκει, nent) terms, of which the square on the greater term is
ἐὰνμὲντὸμεῖζονὄνομασύμμετρονᾖμήκειτῇἐκκειμένῃ larger than (the square on) the lesser by the (square)
ῥητῇ,καλείσθω[ἡὅλη]ἐκδύοὀνομάτωνπρώτη. on (some straight-line) commensurable in length with
ϛʹ.᾿Εὰνδὲτὸἐλάσσονὄνομασύμμετρονᾖμήκειτῇ (the greater) then, if the greater term is commensurable
ἐκκειμένῃῥητῇ,καλείσθωἐκδύοὀνομάτωνδευτέρα. in length with the rational (straight-line previously) laid
ζʹ.᾿Εὰνδὲμηδέτεροντῶνὀνομάτωνσύμμετρονᾖμήκει out, let [the whole] (straight-line) be called a first binoτῇἐκκειμένῃῥητῇ,καλείσθωἐκδύοὀνομάτωντρίτη.
mial (straight-line).
ηʹ.Πάλινδὴἐὰντὸμεῖζονὄνομα[τοῦἐλάσσονος]μεῖζον 6. And if the lesser term is commensurable in length
δύνηταιτῷἀπὸἀσυμμέτρουἑαυτῇμήκει,ἐὰνμὲντὸμεῖζον with the rational (straight-line previously) laid out then
ὄνομασύμμετρονᾖμήκειτῇἐκκειμένῃῥητῇ,καλείσθωἐκ let (the whole straight-line) be called a second binomial
δύοὀνομάτωντετάρτη.
(straight-line).
θʹ.᾿Εὰνδὲτὸἔλασσον,πέμπτη.
7. And if neither of the terms is commensurable in
ιʹ.᾿Εὰνδὲμηδέτερον,ἕκτη.
length with the rational (straight-line previously) laid out
then let (the whole straight-line) be called a third binomial
(straight-line).
8. So, again, if the square on the greater term is
larger than (the square on) [the lesser] by the (square)
on (some straight-line) incommensurable in length with
(the greater) then, if the greater term is commensurable
in length with the rational (straight-line previously) laid
out, let (the whole straight-line) be called a fourth binomial
(straight-line).
9. And if the lesser (term is commensurable), a fifth
(binomial straight-line).
10. And if neither (term is commensurable), a sixth
(binomial straight-line).
Ñ. Proposition 48
Εὑρεῖντὴνἐκδύοὀνομάτωνπρώτην.
To find a first binomial (straight-line).
᾿Εκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν Let two numbers AC and CB be laid down such that
συγκείμενονἐξ αὐτῶντὸνΑΒ πρὸςμὲντὸν ΒΓλόγον their sum AB has to BC the ratio which (some) square
ἔχειν,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν, number (has) to (some) square number, and does not
πρὸςδὲτὸνΓΑλόγονμὴἔχειν,ὃντετράγωνοςἀριθμὸς have to CA the ratio which (some) square number (has)
πρὸςτετράγωνονἀριθμόν,καὶἐκκείσθωτιςῥητὴἡΔ,καὶ to (some) square number [Prop. 10.28 lem. I]. And let
τῇΔσύμμετροςἔστωμήκειἡΕΖ.ῥητὴἄραἐστὶκαὶἡ some rational (straight-line) D be laid down. And let EF
ΕΖ.καὶγεγονέτωὡςὁΒΑἀριθμὸςπρὸςτὸνΑΓ,οὕτως be commensurable in length with D. EF is thus also raτὸἀπὸτῆςΕΖπρὸςτὸἀπὸτῆςΖΗ.ὁδὲΑΒπρὸςτὸν
tional [Def. 10.3]. And let it have been contrived that as
ΑΓλόγονἔχει,ὃνἀριθμὸςπρὸςἀριθμόν·καὶτὸἀπὸτῆς the number BA (is) to AC, so the (square) on EF (is)
ΕΖἄραπρὸςτὸἀπὸτῆςΖΗλόγονἔχει,ὃνἀριθμὸςπρὸς to the (square) on FG [Prop. 10.6 corr.]. And AB has to
ἀριθμόν·ὥστεσύμμετρόνἐστιτὸἀπὸτῆςΕΖτῷἀπὸτῆς AC the ratio which (some) number (has) to (some) num-
333

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ΖΗ.καὶἐστιῥητὴἡΕΖ·ῥητὴἄρακαὶἡΖΗ.καὶἐπεὶὁ ber. Thus, the (square) on EF also has to the (square)
ΒΑπρὸςτὸνΑΓλόγονοὐκἔχει,ὃντετράγωνοςἀριθμὸς on FG the ratio which (some) number (has) to (some)
πρὸςτετράγωνονἀριθμόν,οὐδὲτὸἀπὸτῆςΕΖἄραπρὸς number. Hence, the (square) on EF is commensurable
τὸἀπὸτῆςΖΗλόγονἔχει,ὃντετράγωνοςἀριθμὸςπρὸς with the (square) on FG [Prop. 10.6]. And EF is raτετράγωνονἀριθμόν·ἀσύμμετροςἄραἐστὶνἡΕΖτῇΖΗ
tional. Thus, FG (is) also rational. And since BA does
μήκει. αἱΕΖ,ΖΗἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι· not have to AC the ratio which (some) square number
ἐκδύοἄραὀνομάτωνἐστὶνἡΕΗ.λέγω,ὅτικαὶπρώτη. (has) to (some) square number, thus the (square) on EF
does not have to the (square) on FG the ratio which
(some) square number (has) to (some) square number
either. Thus, EF is incommensurable in length with FG
[Prop 10.9]. EF and FG are thus rational (straight-lines
which are) commensurable in square only. Thus, EG is
a binomial (straight-line) [Prop. 10.36]. I say that (it is)
also a first (binomial straight-line).
∆
Ε
Θ
Ζ
Η
D
E
H
F
G
Α Γ Β
A
C
B
᾿ΕπεὶγάρἐστινὡςὁΒΑἀριθμὸςπρὸςτὸνΑΓ,οὕτωςτὸ For since as the number BA is to AC, so the (square)
ἀπὸτῆςΕΖπρὸςτὸἀπὸτῆςΖΗ,μείζωνδὲὁΒΑτοῦΑΓ, on EF (is) to the (square) on FG, and BA (is) greater
μεῖζονἄρακαὶτὸἀπὸτῆςΕΖτοῦἀπὸτῆςΖΗ.ἔστωοὖντῷ than AC, the (square) on EF (is) thus also greater than
ἀπὸτῆςΕΖἴσατὰἀπὸτῶνΖΗ,Θ.καὶἐπείἐστινὡςὁΒΑ the (square) on FG [Prop. 5.14]. Therefore, let (the sum
πρὸςτὸνΑΓ,οὕτωςτὸἀπὸτῆςΕΖπρὸςτὸἀπὸτῆςΖΗ, of) the (squares) on FG and H be equal to the (square)
ἀναστρέψαντιἄραἐστὶνὡςὁΑΒπρὸςτὸνΒΓ,οὕτωςτὸ on EF . And since as BA is to AC, so the (square)
ἀπὸτῆςΕΖπρὸςτὸἀπὸτῆςΘ.ὁδὲΑΒπρὸςτὸνΒΓλόγον on EF (is) to the (square) on FG, thus, via conver-
ἔχει,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν.καὶ sion, as AB is to BC, so the (square) on EF (is) to the
τὸἀπὸτῆςΕΖἄραπρὸςτὸἀπὸτῆςΘλόγονἔχει, ὃν (square) on H [Prop. 5.19 corr.]. And AB has to BC
τετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν.σύμμετρος the ratio which (some) square number (has) to (some)
ἄραἐστὶνἡΕΖτῇΘμήκει·ἡΕΖἄρατῆςΖΗμεῖζονδύναται square number. Thus, the (square) on EF also has to
τῷἀπὸσυμμέτρουἑαυτῇ. καίεἰσιῥηταὶαἱΕΖ,ΖΗ,καὶ the (square) on H the ratio which (some) square number
σύμμετροςἡΕΖτῇΔμήκει.
(has) to (some) square number. Thus, EF is commensu-
᾿Η ΕΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ πρώτη· ὅπερ ἔδει rable in length with H [Prop. 10.9]. Thus, the square on
δεῖξαι.
EF is greater than (the square on) FG by the (square)
on (some straight-line) commensurable (in length) with
(EF ). And EF and FG are rational (straight-lines). And
EF (is) commensurable in length with D.
Thus, EG is a first binomial (straight-line) [Def. 10.5]. †
(Which is) the very thing it was required to show.
† If the rational straight-line has unit length then the length of a first binomial straight-line is k + k √ 1 − k ′ 2 . This, and the first apotome, whose
length is k − k √ 1 − k ′ 2 [Prop. 10.85], are the roots of x 2 − 2 k x + k 2 k ′2 = 0.
Ñ. Proposition
Εὑρεῖντὴνἐκδύοὀνομάτωνδευτέραν.
49
To find a second binomial (straight-line).
334

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
Θ
Ε
Ζ
∆
Η
H
E
F
D
G
Α
Γ
Β
A
C
B
᾿Εκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν Let the two numbers AC and CB be laid down such
συγκείμενονἐξ αὐτῶντὸνΑΒ πρὸςμὲντὸν ΒΓλόγον that their sum AB has to BC the ratio which (some)
ἔχειν,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν, square number (has) to (some) square number, and does
πρὸςδὲτὸνΑΓλόγονμὴἔχειν,ὃντετράγωνοςἀριθμὸς not have to AC the ratio which (some) square number
πρὸςτετράγωνονἀριθμόν,καὶἐκκείσθωῥητὴἡΔ,καὶτῇΔ (has) to (some) square number [Prop. 10.28 lem. I]. And
σύμμετροςἔστωἡΕΖμήκει·ῥητὴἄραἐστὶνἡΕΖ.γεγονέτω let the rational (straight-line) D be laid down. And let
δὴκαὶὡςὁΓΑἀριθμὸςπρὸςτὸνΑΒ,οὕτωςτὸἀπὸτῆςΕΖ EF be commensurable in length with D. EF is thus a
πρὸςτὸἀπὸτῆςΖΗ·σύμμετρονἄραἐστὶτὸἀπὸτῆςΕΖτῷ rational (straight-line). So, let it also have been contrived
ἀπὸτῆςΖΗ.ῥητὴἄραἐστὶκαὶἡΖΗ.καὶἐπεὶὁΓΑἀριθμὸς that as the number CA (is) to AB, so the (square) on EF
πρὸςτὸνΑΒλὸγονοὐκἔχει,ὃντετράγωνοςἀριθμὸςπρὸς (is) to the (square) on FG [Prop. 10.6 corr.]. Thus, the
τετράγωνονἀριθμόν,οὐδὲτὸἀπὸτῆςΕΖπρὸςτὸἀπὸτῆς (square) on EF is commensurable with the (square) on
ΖΗλόγονἔχει,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον FG [Prop. 10.6]. Thus, FG is also a rational (straight-
ἀριθμόν. ἀσύμμετροςἄραἐστὶνἡΕΖτῇΖΗμήκει·αἱΕΖ, line). And since the number CA does not have to AB
ΖΗἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι· ἐκδύοἄρα the ratio which (some) square number (has) to (some)
ὀνομάτωνἐστὶνἡΕΗ.δεικτέονδή,ὅτικαὶδευτέρα. square number, the (square) on EF does not have to the
᾿ΕπεὶγὰρἀνάπαλίνἐστινὡςὁΒΑἀριθμὸςπρὸςτὸνΑΓ, (square) on FG the ratio which (some) square number
οὕτωςτὸἀπὸτῆςΗΖπρὸςτὸἀπὸτῆςΖΕ,μείζωνδὲὁΒΑ (has) to (some) square number either. Thus, EF is inτοῦΑΓ,μεῖζονἄρα[καὶ]τὸἀπὸτῆςΗΖτοῦἀπὸτῆςΖΕ.
commensurable in length with FG [Prop. 10.9]. EF and
ἔστωτῷἀπὸτῆςΗΖἴσατὰἀπὸτῶνΕΖ,Θ·ἀναστρέψαντι FG are thus rational (straight-lines which are) commen-
ἄραἐστὶνὡςὁΑΒπρὸςτὸνΒΓ,οὕτωςτὸἀπὸτῆςΖΗ surable in square only. Thus, EG is a binomial (straightπρὸςτὸἀπὸτῆςΘ.ἀλλ᾿ὁΑΒπρὸςτὸνΒΓλόγονἔχει,ὃν
line) [Prop. 10.36]. So, we must show that (it is) also a
τετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν·καὶτὸἀπὸ second (binomial straight-line).
τῆςΖΗἄραπρὸςτὸἀπὸτῆςΘλόγονἔχει,ὃντετράγωνος For since, inversely, as the number BA is to AC, so
ἀριθμὸςπρὸςτετράγωνονἀριθμόν.σύμμετροςἄραἐστὶνἡ the (square) on GF (is) to the (square) on FE [Prop. 5.7
ΖΗτῇΘμήκει·ὥστεἡΖΗτῆςΖΕμεῖζονδύναταιτῷἀπὸ corr.], and BA (is) greater than AC, the (square) on
συμμέτρουἑαυτῇ. καίεἰσιῥηταὶαἱΖΗ,ΖΕδυνάμειμόνον GF (is) thus [also] greater than the (square) on FE
σύμμετροι,καὶτὸΕΖἔλασσονὄνοματῇἐκκειμένῃῥητῇ [Prop. 5.14]. Let (the sum of) the (squares) on EF and
σύμμετρόνἐστιτῇΔμήκει.
H be equal to the (square) on GF . Thus, via conver-
῾ΗΕΗἄραἐκδύοὀνομάτωνἐστὶδευτέρα·ὅπερἔδει sion, as AB is to BC, so the (square) on FG (is) to the
δεῖξαι.
(square) on H [Prop. 5.19 corr.]. But, AB has to BC
the ratio which (some) square number (has) to (some)
square number. Thus, the (square) on FG also has to
the (square) on H the ratio which (some) square number
(has) to (some) square number. Thus, FG is commensurable
in length with H [Prop. 10.9]. Hence, the square on
FG is greater than (the square on) FE by the (square) on
(some straight-line) commensurable in length with (FG).
And FG and FE are rational (straight-lines which are)
commensurable in square only. And the lesser term EF
is commensurable in length with the rational (straightline)
D (previously) laid down.
Thus, EG is a second binomial (straight-line) [Def.
10.6]. † (Which is) the very thing it was required to show.
† If the rational straight-line has unit length then the length of a second binomial straight-line is k/ √ 1 − k ′ 2 + k. This, and the second apotome,
335

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
whose length is k/ √ 1 − k ′ 2 − k [Prop. 10.86], are the roots of x 2 − (2 k/ √ 1 − k ′2 ) x + k 2 [k ′2 /(1 − k ′2 )] = 0.
Ò. Proposition
Εὑρεῖντὴνἐκδύοὀνομάτωντρίτην.
Α Γ Β
50
To find a third binomial (straight-line).
A
C
B
Ε
Ζ
Κ
Η
∆
Θ
E K D
F G H
᾿Εκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν Let the two numbers AC and CB be laid down such
συγκείμενονἐξ αὐτῶντὸνΑΒ πρὸςμὲντὸν ΒΓλόγον that their sum AB has to BC the ratio which (some)
ἔχειν,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν, square number (has) to (some) square number, and does
πρὸςδὲτὸνΑΓλόγονμὴἔχειν,ὃντετράγωνοςἀριθμὸς not have to AC the ratio which (some) square number
πρὸςτετράγωνονἀριθμόν. ἐκκείσθωδέτιςκαὶἄλλοςμὴ (has) to (some) square number. And let some other nonτετράγωνοςἀριθμὸςὁΔ,καὶπρὸςἑκάτεροντῶνΒΑ,ΑΓ
square number D also be laid down, and let it not have
λόγονμὴἐχέτω,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον to each of BA and AC the ratio which (some) square
ἀριθμόν·καὶἐκκείσθωτιςῥητὴεὐθεῖαἡΕ,καὶγεγονέτω number (has) to (some) square number. And let some ra-
ὡςὁΔπρὸςτὸνΑΒ,οὕτωςτὸἀπὸτῆςΕπρὸςτὸἀπὸ tional straight-line E be laid down, and let it have been
τῆςΖΗ·σύμμετρονἄραἐστὶτὸἀπὸτῆςΕτῷἀπὸτῆςΖΗ. contrived that as D (is) to AB, so the (square) on E
καίἐστιῥητὴἡΕ·ῥητὴἄραἐστὶκαὶἡΖΗ.καὶἐπεὶὁΔ (is) to the (square) on FG [Prop. 10.6 corr.]. Thus, the
πρὸςτὸνΑΒλόγονοὐκἔχει,ὃντετράγωνοςἀριθμὸςπρὸς (square) on E is commensurable with the (square) on
τετράγωνονἀριθμόν,οὐδὲτὸἀπὸτῆςΕπρὸςτὸἀπὸτῆς FG [Prop. 10.6]. And E is a rational (straight-line).
ΖΗλόγονἔχει,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον Thus, FG is also a rational (straight-line). And since
ἀριθμόν·ἀσύμμετροςἄραἐστὶνἡΕτῇΖΗμήκει.γεγονέτω D does not have to AB the ratio which (some) square
δὴπάλινὡςἡΒΑἀριθμὸςπρὸςτὸνΑΓ,οὕτωςτὸἀπὸτῆς number has to (some) square number, the (square) on
ΖΗπρὸςτὸἀπὸτῆςΗΘ·σύμμετρονἄραἐστὶτὸἀπὸτῆς E does not have to the (square) on FG the ratio which
ΖΗτῷἀπὸτῆςΗΘ.ῥητὴδὲἡΖΗ·ῥητὴἄρακαὶἡΗΘ.καὶ (some) square number (has) to (some) square number
ἐπεὶὁΒΑπρὸςτὸνΑΓλόγονοὐκἔχει,ὃντετράγωνος either. E is thus incommensurable in length with FG
ἀριθμὸςπρὸςτετράγωνονἀριθμόν, οὐδὲτὸἀπὸτῆςΖΗ [Prop. 10.9]. So, again, let it have been contrived that
πρὸςτὸἀπὸτῆςΘΗλόγονἔχει,ὃντετράγωνοςἀριθμὸς as the number BA (is) to AC, so the (square) on FG
πρὸς τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ ΖΗ (is) to the (square) on GH [Prop. 10.6 corr.]. Thus, the
τῇΗΘμήκει. αἱΖΗ,ΗΘἄραῥηταίεἰσιδυνάμειμόνον (square) on FG is commensurable with the (square) on
σύμμετροι·ἡΖΘἄραἐκδύοὀνομάτωνἐστίν.λέγωδή,ὅτι GH [Prop. 10.6]. And FG (is) a rational (straight-line).
καὶτρίτη.
Thus, GH (is) also a rational (straight-line). And since
᾿ΕπεὶγάρἐστινὡςὁΔπρὸςτὸνΑΒ,οὕτωςτὸἀπὸτῆς BA does not have to AC the ratio which (some) square
ΕπρὸςτὸἀπὸτῆςΖΗ,ὡςδὲὁΒΑπρὸςτὸνΑΓ,οὕτως number (has) to (some) square number, the (square) on
τὸἀπὸτῆςΖΗπρὸςτὸἀπὸτῆςΗΘ,δι᾿ἴσουἄραἐστὶνὡς FG does not have to the (square) on HG the ratio which
ὁΔπρὸςτὸνΑΓ,οὕτωςτὸἀπὸτῆςΕπρὸςτὸἀπὸτῆς (some) square number (has) to (some) square number
ΗΘ.ὁδὲΔπρὸςτὸνΑΓλόγονοὐκἔχει,ὃντετράγωνος either. Thus, FG is incommensurable in length with GH
ἀριθμὸςπρὸςτετράγωνονἀριθμόν·οὐδὲτὸἀπὸτῆςΕἄρα [Prop. 10.9]. FG and GH are thus rational (straightπρὸςτὸἀπὸτῆςΗΘλόγονἔχει,ὃντετράγωνοςἀριθμὸς
lines which are) commensurable in square only. Thus,
πρὸςτετράγωνονἀριθμόν· ἀσύμμετροςἄραἐστὶνἡΕτῇ FH is a binomial (straight-line) [Prop. 10.36]. So, I say
ΗΘμήκει. καὶἐπείἐστινὡςὁΒΑπρὸςτὸνΑΓ,οὕτως that (it is) also a third (binomial straight-line).
τὸἀπὸτῆςΖΗπρὸςτὸἀπὸτῆςΗΘ,μεῖζονἄρατὸἀπὸ For since as D is to AB, so the (square) on E (is)
τῆςΖΗτοῦἀπὸτῆςΗΘ.ἔστωοὖντῷἀπὸτῆςΖΗἴσατὰ to the (square) on FG, and as BA (is) to AC, so the
ἀπὸτῶνΗΘ,Κ·ἀναστρέψαντιἄρα[ἐστὶν]ὡςὁΑΒπρὸς (square) on FG (is) to the (square) on GH, thus, via
τὸνΒΓ,οὕτωςτὸἀπὸτῆςΖΗπρὸςτὸἀπὸτῆςΚ.ὁδὲ equality, as D (is) to AC, so the (square) on E (is)
ΑΒπρὸςτὸνΒΓλόγονἔχει,ὃντετράγωνοςἀριθμὸςπρὸς to the (square) on GH [Prop. 5.22]. And D does not
336

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
τετράγωνονἀριθμόν·καὶτὸἀπὸτῆςΖΗἄραπρὸςτὸἀπὸ have to AC the ratio which (some) square number (has)
τῆςΚλόγονἔχει,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον to (some) square number. Thus, the (square) on E
ἀριθμόν·σύμμετροςἄρα[ἐστὶν]ἡΖΗτῇΚμήκει.ἡΖΗἄρα does not have to the (square) on GH the ratio which
τῆςΗΘμεῖζονδύναταιτῷἀπὸσυμμέτρουἑαυτῇ.καίεἰσιν (some) square number (has) to (some) square number
αἱΖΗ,ΗΘῥηταὶδυνάμειμόνονσύμμετροι,καὶοὐδετέρα either. Thus, E is incommensurable in length with GH
αὐτῶνσύμμετρόςἐστιτῇΕμήκει.
[Prop. 10.9]. And since as BA is to AC, so the (square)
῾ΗΖΘἄραἐκδύοὀνομάτωνἐστὶτρίτη·ὅπερἔδειδεῖξαι. on FG (is) to the (square) on GH, the (square) on FG
(is) thus greater than the (square) on GH [Prop. 5.14].
Therefore, let (the sum of) the (squares) on GH and
K be equal to the (square) on FG. Thus, via conversion,
as AB [is] to BC, so the (square) on FG (is)
to the (square) on K [Prop. 5.19 corr.]. And AB has
to BC the ratio which (some) square number (has) to
(some) square number. Thus, the (square) on FG also
has to the (square) on K the ratio which (some) square
number (has) to (some) square number. Thus, FG [is]
commensurable in length with K [Prop. 10.9]. Thus,
the square on FG is greater than (the square on) GH
by the (square) on (some straight-line) commensurable
(in length) with (FG). And FG and GH are rational
(straight-lines which are) commensurable in square only,
and neither of them is commensurable in length with E.
Thus, FH is a third binomial (straight-line) [Def.
10.7]. † (Which is) the very thing it was required to show.
† If the rational straight-line has unit length then the length of a third binomial straight-line is k 1/2 (1 + √ 1 − k ′ 2 ). This, and the third apotome,
whose length is k 1/2 (1 − √ 1 − k ′2 ) [Prop. 10.87], are the roots of x 2 − 2 k 1/2 x + k k ′2 = 0.
Ò. Proposition
Εὑρεῖντὴνἐκδύοὀνομάτωντετάρτην.
Ε Ζ Η
51
To find a fourth binomial (straight-line).
E
F
G
∆
Θ
D
H
Α
Γ
Β
᾿ΕκκείσθωσανδύοἀριθμοὶοἱΑΓ,ΓΒ,ὥστετὸνΑΒ Let the two numbers AC and CB be laid down
πρὸς τὸν ΒΓ λόγον μὴ ἔχειν μήτε μὴν πρὸς τὸν ΑΓ, such that AB does not have to BC, or to AC either,
ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. καὶ the ratio which (some) square number (has) to (some)
ἐκκείσθωῥητὴἡΔ,καὶτῇΔσύμμετροςἔστωμήκειἡΕΖ· square number [Prop. 10.28 lem. I]. And let the rational
ῥητὴἄραἐστὶκαὶἡΕΖ.καὶγεγονέτωὡςὁΒΑἀριθμὸς (straight-line) D be laid down. And let EF be comπρὸςτὸνΑΓ,οὕτωςτὸἀπὸτῆςΕΖπρὸςτὸἀπὸτῆςΖΗ·
mensurable in length with D. Thus, EF is also a ratioσύμμετρονἄραἐστὶτὸἀπὸτῆςΕΖτῷἀπὸτῆςΖΗ·ῥητὴἄρα
nal (straight-line). And let it have been contrived that
ἐστὶκαὶἡΖΗ.καὶἐπεὶὁΒΑπρὸςτὸνΑΓλόγονοὐκἔχει, as the number BA (is) to AC, so the (square) on EF
ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν,οὐδὲτὸ (is) to the (square) on FG [Prop. 10.6 corr.]. Thus, the
ἀπὸτῆςΕΖπρὸςτὸἀπὸτῆςΖΗλόγονἔχει,ὃντετράγωνος (square) on EF is commensurable with the (square) on
ἀριθμὸςπρὸςτετράγωνονἀριθμόν·ἀσύμμετροςἄραἐστὶνἡ FG [Prop. 10.6]. Thus, FG is also a rational (straight-
ΕΖτῇΖΗμήκει. αἱΕΖ,ΖΗἄραῥηταίεἰσιδυνάμειμόνον line). And since BA does not have to AC the ratio which
σύμμετροι·ὥστεἡΕΗἐκδύοὀνομάτωνἐστίν. λέγωδή, (some) square number (has) to (some) square number,
A
C
B
337

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ὅτικαὶτετάρτη.
the (square) on EF does not have to the (square) on FG
᾿ΕπεὶγάρἐστινὡςὁΒΑπρὸςτὸνΑΓ,οὕτωςτὸἀπὸ the ratio which (some) square number (has) to (some)
τῆςΕΖπρὸςτὸἀπὸτῆςΖΗ[μείζωνδὲὁΒΑτοῦΑΓ], square number either. Thus, EF is incommensurable
μεῖζονἄρατὸἀπὸτῆςΕΖτοῦἀπὸτῆςΖΗ.ἔστωοὖντῷ in length with FG [Prop. 10.9]. Thus, EF and FG
ἀπὸτῆςΕΖἴσατὰἀπὸτῶνΖΗ,Θ·ἀναστρέψαντιἄραὡς are rational (straight-lines which are) commensurable
ὁΑΒἀριθμὸςπρὸςτὸνΒΓ,οὕτωςτὸἀπὸτῆςΕΖπρὸς in square only. Hence, EG is a binomial (straight-line)
τὸἀπὸτῆςΘ.ὁδὲΑΒπρὸςτὸνΒΓλόγονοὐκἔχει,ὃν [Prop. 10.36]. So, I say that (it is) also a fourth (binomial
τετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν·οὐδ᾿ἄρατὸ straight-line).
ἀπὸτῆςΕΖπρὸςτὸἀπὸτῆςΘλόγονἔχει,ὃντετράγωνος For since as BA is to AC, so the (square) on EF (is) to
ἀριθμὸςπρὸςτετράγωνονἀριθμόν. ἀσύμμετροςἄραἐστὶν the (square) on FG [and BA (is) greater than AC], the
ἡΕΖτῇΘμήκει·ἡΕΖἄρατῆςΗΖμεῖζονδύναταιτῷἀπὸ (square) on EF (is) thus greater than the (square) on FG
ἀσυμμέτρουἑαυτῇ.καίεἰσιναἱΕΖ,ΖΗῥηταὶδυνάμειμόνον [Prop. 5.14]. Therefore, let (the sum of) the squares on
σύμμετροι,καὶἡΕΖτῇΔσύμμετρόςἐστιμήκει. FG and H be equal to the (square) on EF . Thus, via con-
῾ΗΕΗἄραἐκδύοὀνομάτωνἐστὶτετάρτη·ὅπερἔδει version, as the number AB (is) to BC, so the (square) on
δεῖξαι.
EF (is) to the (square) on H [Prop. 5.19 corr.]. And AB
does not have to BC the ratio which (some) square number
(has) to (some) square number. Thus, the (square) on
EF does not have to the (square) on H the ratio which
(some) square number (has) to (some) square number
either. Thus, EF is incommensurable in length with H
[Prop. 10.9]. Thus, the square on EF is greater than (the
square on) GF by the (square) on (some straight-line) incommensurable
(in length) with (EF ). And EF and FG
are rational (straight-lines which are) commensurable in
square only. And EF is commensurable in length with D.
Thus, EG is a fourth binomial (straight-line) [Def.
10.8]. † (Which is) the very thing it was required to show.
† If the rational straight-line has unit length then the length of a fourth binomial straight-line is k (1 + 1/ √ 1 + k ′ ). This, and the fourth apotome,
whose length is k (1 − 1/ √ 1 + k ′ ) [Prop. 10.88], are the roots of x 2 − 2 k x + k 2 k ′ /(1 + k ′ ) = 0.
Ò. Proposition
Εὑρεῖντὴνἐκδύοὀνομάτωνπέμπτην.
Ε
Ζ
Η
52
To find a fifth binomial straight-line.
E
F
G
∆
Θ
D
H
Α
Γ
Β
᾿ΕκκείσθωσανδύοἀριθμοὶοἱΑΓ,ΓΒ,ὥστετὸνΑΒ Let the two numbers AC and CB be laid down such
πρὸς ἑκάτερον αὐτῶν λόγον μὴ ἔχειν, ὃν τετράγωνος that AB does not have to either of them the ratio which
ἀριθμὸςπρὸςτετράγωνονἀριθμόν,καὶἐκκείσθωῥητήτις (some) square number (has) to (some) square number
εὐθεῖαἡΔ,καὶτῇΔσύμμετροςἔστω[μήκει]ἡΕΖ·ῥητὴ [Prop. 10.38 lem.]. And let some rational straight-line
ἄραἡΕΖ.καὶγεγονέτωὡςὁΓΑπρὸςτὸνΑΒ,οὕτωςτὸ D be laid down. And let EF be commensurable [in
ἀπὸτῆςΕΖπρὸςτὸἀπὸτῆςΖΗ.ὁδὲΓΑπρὸςτὸνΑΒ length] with D. Thus, EF (is) a rational (straightλόγονοὐκἔχει,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον
line). And let it have been contrived that as CA (is) to
ἀριθμόν·οὑδὲτὸἀπὸτῆςΕΖἄραπρὸςτὸἀπὸτῆςΖΗλόγον AB, so the (square) on EF (is) to the (square) on FG
ἔχει,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν.αἱ [Prop. 10.6 corr.]. And CA does not have to AB the ra-
A
C
B
338

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ΕΖ,ΖΗἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι· ἐκδύο tio which (some) square number (has) to (some) square
ἄραὀνομάτωνἐστὶνἡΕΗ.λέγωδή,ὅτικαὶπέμπτη. number. Thus, the (square) on EF does not have to the
᾿ΕπεὶγάρἐστινὡςὁΓΑπρὸςτὸνΑΒ,οὕτωςτὸἀπὸ (square) on FG the ratio which (some) square number
τῆςΕΖπρὸςτὸἀπὸτῆςΖΗ,ἀνάπαλινὡςὁΒΑπρὸςτὸν (has) to (some) square number either. Thus, EF and
ΑΓ,οὕτωςτὸἀπὸτῆςΖΗπρὸςτὸἀπὸτῆςΖΕ·μεῖζον FG are rational (straight-lines which are) commensu-
ἄρατὸἀπὸτῆςΗΖτοῦἀπὸτῆςΖΕ.ἔστωοὖντῷἀπὸ rable in square only [Prop. 10.9]. Thus, EG is a binomial
τῆςΗΖἴσατὰἀπὸτῶνΕΖ,Θ·ἀναστρέψαντιἄραἐστὶνὡς (straight-line) [Prop. 10.36]. So, I say that (it is) also a
ὁΑΒἀριθμὸςπρὸςτὸνΒΓ,οὕτωςτὸἀπὸτῆςΗΖπρὸς fifth (binomial straight-line).
τὸἀπὸτῆςΘ.ὁδὲΑΒπρὸςτὸνΒΓλόγονοὐκἔχει,ὃν For since as CA is to AB, so the (square) on EF
τετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν·οὐδ᾿ἄρατὸ (is) to the (square) on FG, inversely, as BA (is) to
ἀπὸτῆςΖΗπρὸςτὸἀπὸτῆςΘλόγονἔχει,ὃντετράγωνος AC, so the (square) on FG (is) to the (square) on FE
ἀριθμὸςπρὸςτετράγωνονἀριθμόν. ἀσύμμετροςἄραἐστὶν [Prop. 5.7 corr.]. Thus, the (square) on GF (is) greater
ἡΖΗτῇΘμήκει·ὥστεἡΖΗτῆςΖΕμεῖζονδύναταιτῷ than the (square) on FE [Prop. 5.14]. Therefore, let
ἀπὸἀσυμμέτρουἑαυτῇ. καίεἰσιναἱΗΖ,ΖΕῥηταὶδυνάμει (the sum of) the (squares) on EF and H be equal to
μόνονσύμμετροι,καὶτὸΕΖἔλαττονὄνομασύμμετρόνἐστι the (square) on GF . Thus, via conversion, as the number
τῇἐκκειμένῃῥητῇτῇΔμήκει.
AB is to BC, so the (square) on GF (is) to the (square)
῾ΗΕΗἄραἐκδύοὀνομάτωνἐστὶπέμπτη· ὅπερἔδει on H [Prop. 5.19 corr.]. And AB does not have to BC
δεῖξαι.
the ratio which (some) square number (has) to (some)
square number. Thus, the (square) on FG does not have
to the (square) on H the ratio which (some) square number
(has) to (some) square number either. Thus, FG is
incommensurable in length with H [Prop. 10.9]. Hence,
the square on FG is greater than (the square on) FE
by the (square) on (some straight-line) incommensurable
(in length) with (FG). And GF and FE are rational
(straight-lines which are) commensurable in square only.
And the lesser term EF is commensurable in length with
the rational (straight-line previously) laid down, D.
Thus, EG is a fifth binomial (straight-line). † (Which
is) the very thing it was required to show.
† If the rational straight-line has unit length then the length of a fifth binomial straight-line is k ( √ 1 + k ′ +1). This, and the fifth apotome, whose
length is k ( √ 1 + k ′ − 1) [Prop. 10.89], are the roots of x 2 − 2 k √ 1 + k ′ x + k 2 k ′ = 0.
Ò. Proposition
Εὑρεῖντὴνἐκδύοὀνομάτωνἕκτην.
Ε
Κ
53
To find a sixth binomial (straight-line).
Ζ Η Θ
F G
∆
D
Α Γ Β
A C B
E
K
H
᾿ΕκκείσθωσανδύοἀριθμοὶοἱΑΓ,ΓΒ,ὥστετὸνΑΒ Let the two numbers AC and CB be laid down such
πρὸς ἑκάτερον αὐτῶν λόγον μὴ ἔχειν, ὃν τετράγωνος that AB does not have to each of them the ratio which
ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· ἔστω δὲ καὶ ἕτερος (some) square number (has) to (some) square number.
ἀριθμὸς ὁ Δ μὴ τετράγωνος ὢν μηδὲ πρὸς ἑκάτερον And let D also be another number, which is not square,
τῶν ΒΑ,ΑΓλόγονἔχων, ὃντετράγωνοςἀριθμὸςπρὸς and does not have to each of BA and AC the ratio which
τετράγωνονἀριθμόν· καὶἐκκείσθωτιςῥητὴεὐθεῖαἡΕ, (some) square number (has) to (some) square number eiκαὶγεγονέτωὡςὁΔπρὸςτὸνΑΒ,οὕτωςτὸἀπὸτῆςΕ
ther [Prop. 10.28 lem. I]. And let some rational straightπρὸςτὸἀπὸτῆςΖΗ·σύμμετρονἄρατὸἀπὸτῆςΕτῷἀπὸ
line E be laid down. And let it have been contrived that
339

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
τῆςΖΗ.καίἐστιῥητὴἡΕ·ῥητὴἄρακαὶἡΖΗ.καὶἐπεὶ as D (is) to AB, so the (square) on E (is) to the (square)
οὐκἔχειὁΔπρὸςτὸνΑΒλόγον,ὃντετράγωνοςἀριθμὸς on FG [Prop. 10.6 corr.]. Thus, the (square) on E (is)
πρὸςτετράγωνονἀριθμόν, οὐδὲτὸἀπὸτῆςΕἄραπρὸς commensurable with the (square) on FG [Prop. 10.6].
τὸἀπὸτῆςΖΗλόγονἔχει,ὃντετράγωνοςἀριθμὸςπρὸς And E is rational. Thus, FG (is) also rational. And since
τετράγωνονἀριθμόν· ἀσύμμετροςἄραἡΕτῇΖΗμήκει. D does not have to AB the ratio which (some) square
γεγονέτωδὴπάλινὡςὁΒΑπρὸςτὸνΑΓ,οὕτωςτὸἀπὸ number (has) to (some) square number, the (square) on
τῆςΖΗπρὸςτὸἀπὸτῆςΗΘ.σύμμετρονἄρατὸἀπὸτῆςΖΗ E thus does not have to the (square) on FG the raτῷἀπὸτῆςΘΗ.ῥητὸνἄρατὸἀπὸτῆςΘΗ·ῥητὴἄραἡΘΗ.
tio which (some) square number (has) to (some) square
καὶἐπεὶὁΒΑπρὸςτὸνΑΓλόγονοὐκἔχει,ὃντετράγωνος number either. Thus, E (is) incommensurable in length
ἀριθμὸςπρὸςτετράγωνονἀριθμόν, οὐδὲτὸἀπὸτῆςΖΗ with FG [Prop. 10.9]. So, again, let it have be contrived
πρὸςτὸἀπὸτῆςΗΘλόγονἔχει,ὃντετράγωνοςἀριθμὸς that as BA (is) to AC, so the (square) on FG (is) to
πρὸς τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ ΖΗ the (square) on GH [Prop. 10.6 corr.]. The (square) on
τῇΗΘμήκει. αἱΖΗ,ΗΘἄραῥηταίεἰσιδυνάμειμόνον FG (is) thus commensurable with the (square) on HG
σύμμετροι·ἐκδύοἄραὀνομάτωνἐστὶνἡΖΘ.δεικτέονδή, [Prop. 10.6]. The (square) on HG (is) thus rational.
ὅτικαὶἕκτη. Thus, HG (is) rational. And since BA does not have
᾿ΕπεὶγάρἐστινὡςὁΔπρὸςτὸνΑΒ,οὕτωςτὸἀπὸ to AC the ratio which (some) square number (has) to
τῆςΕπρὸςτὸἀπὸτῆςΖΗ,ἔστιδὲκαὶὡςὁΒΑπρὸς (some) square number, the (square) on FG does not have
τὸνΑΓ,οὕτωςτὸἀπὸτῆςΖΗπρὸςτὸἀπὸτῆςΗΘ,δι᾿ to the (square) on GH the ratio which (some) square
ἴσουἄραἐστὶνὡςὁΔπρὸςτὸνΑΓ,οὕτωςτὸἀπὸτῆςΕ number (has) to (some) square number either. Thus,
πρὸςτὸ ἀπὸτῆςΗΘ.ὁδὲ Δπρὸςτὸν ΑΓλόγονοὐκ FG is incommensurable in length with GH [Prop. 10.9].
ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· Thus, FG and GH are rational (straight-lines which are)
οὐδὲ τὸ ἀπὸ τῆς Ε ἄρα πρὸς τὸ ἀπὸ τῆς ΗΘ λόγον commensurable in square only. Thus, FH is a binomial
ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· (straight-line) [Prop. 10.36]. So, we must show that (it
ἀσύμμετροςἄραἐστὶνἡΕτῇΗΘμήκει. ἐδείχθηδὲκαὶ is) also a sixth (binomial straight-line).
τῇΖΗἀσύμμετρος·ἑκατέραἄρατῶνΖΗ,ΗΘἀσύμμετρός For since as D is to AB, so the (square) on E (is)
ἐστιτῇΕμήκει. καὶἐπείἐστινὡςὁΒΑπρὸςτὸνΑΓ, to the (square) on FG, and also as BA is to AC, so
οὕτωςτὸἀπὸτῆςΖΗπρὸςτὸἀπὸτῆςΗΘ,μεῖζονἄρα the (square) on FG (is) to the (square) on GH, thus,
τὸἀπὸτῆςΖΗτοῦἀπὸτῆςΗΘ.ἔστωοὖντῷἀπὸ[τῆς] via equality, as D is to AC, so the (square) on E (is)
ΖΗἴσατὰἀπὸτῶνΗΘ,Κ·ἀναστρέψαντιἄραὡςὁΑΒ to the (square) on GH [Prop. 5.22]. And D does not
πρὸςΒΓ,οὕτωςτὸἀπὸΖΗπρὸςτὸἀπὸτῆςΚ.ὁδὲΑΒ have to AC the ratio which (some) square number (has)
πρὸςτὸνΒΓλόγονοὐκἔχει,ὃντετράγωνοςἀριθμὸςπρὸς to (some) square number. Thus, the (square) on E
τετράγωνονἀριθμόν·ὥστεοὐδὲτὸἀπὸΖΗπρὸςτὸἀπὸ does not have to the (square) on GH the ratio which
τῆςΚλόγονἔχει,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον (some) square number (has) to (some) square number
ἀριθμόν.ἀσύμμετροςἄραἐστὶνἡΖΗτῇΚμήκει·ἡΖΗἄρα either. E is thus incommensurable in length with GH
τῆςΗΘμεῖζονδύναταιτῷἀπὸἀσυμμέτρουἑαυτῇ.καίεἰσιν [Prop. 10.9]. And (E) was also shown (to be) incomαἱΖΗ,ΗΘῥηταὶδυνάμειμόνονσύμμετροι,καὶοὐδετέρα
mensurable (in length) with FG. Thus, FG and GH
αὐτῶνσύμμετρόςἐστιμήκειτῇἐκκειμένῃῥητῃτῇΕ. are each incommensurable in length with E. And since
῾ΗΖΘἄραἐκδύοὀνομάτωνἐστὶνἕκτη·ὅπερἔδειδεῖξαι. as BA is to AC, so the (square) on FG (is) to the
(square) on GH, the (square) on FG (is) thus greater
than the (square) on GH [Prop. 5.14]. Therefore, let
(the sum of) the (squares) on GH and K be equal to
the (square) on FG. Thus, via conversion, as AB (is)
to BC, so the (square) on FG (is) to the (square) on K
[Prop. 5.19 corr.]. And AB does not have to BC the ratio
which (some) square number (has) to (some) square
number. Hence, the (square) on FG does not have to
the (square) on K the ratio which (some) square number
(has) to (some) square number either. Thus, FG is
incommensurable in length with K [Prop. 10.9]. The
square on FG is thus greater than (the square on) GH
by the (square) on (some straight-line which is) incom-
340

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
mensurable (in length) with (FG). And FG and GH
are rational (straight-lines which are) commensurable
in square only, and neither of them is commensurable
in length with the rational (straight-line) E (previously)
laid down.
Thus, FH is a sixth binomial (straight-line) [Def.
10.10]. † (Which is) the very thing it was required to
show.
† If the rational straight-line has unit length then the length of a sixth binomial straight-line is √ k + √ k ′ . This, and the sixth apotome, whose
length is √ k − √ k ′ [Prop. 10.90], are the roots of x 2 − 2 √ k x + (k − k ′ ) = 0.
ÄÑÑ.
Lemma
῎ΕστωδύοτετράγωνατὰΑΒ,ΒΓκαὶκείσθωσανὥστε Let AB and BC be two squares, and let them be laid
ἐπ᾿εὐθείαςεἶναιτὴνΔΒτῇΒΕ·ἐπ᾿εὐθείαςἄραἐστὶκαὶἡ down such that DB is straight-on to BE. FB is, thus,
ΖΒτῇΒΗ.καὶσυμπεπληρώσθωτὸΑΓπαραλληλόγραμμον· also straight-on to BG. And let the parallelogram AC
λέγω,ὅτιτετράγωνόνἐστιτὸΑΓ,καὶὅτιτῶνΑΒ,ΒΓ have been completed. I say that AC is a square, and
μέσονἀνάλογόνἐστιτὸΔΗ,καὶἔτιτῶνΑΓ,ΓΒμέσον that DG is the mean proportional to AB and BC, and,
ἀνάλογόνἐστιτὸΔΓ.
moreover, DC is the mean proportional to AC and CB.
Κ
Η
Γ
K
G
C
∆
Β
Ε
D
B
E
Α
Ζ
Θ
᾿ΕπεὶγὰρἴσηἐστὶνἡμὲνΔΒτῇΒΖ,ἡδὲΒΕτῇΒΗ, For since DB is equal to BF , and BE to BG, the
ὅληἄραἡΔΕὅλῃτῇΖΗἐστινἴση.ἀλλ᾿ἡμὲνΔΕἑκατέρᾳ whole of DE is thus equal to the whole of FG. But DE
τῶνΑΘ,ΚΓἐστινἴση,ἡδὲΖΗἑκατέρᾳτῶνΑΚ,ΘΓἐστιν is equal to each of AH and KC, and FG is equal to each
ἴση·καὶἑκατέραἄρατῶνΑΘ,ΚΓἑκατέρᾳτῶνΑΚ,ΘΓ of AK and HC [Prop. 1.34]. Thus, AH and KC are also
ἐστινἴση.ἰσόπλευρονἄραἐστὶτὸΑΓπαραλληλόγραμμον· equal to AK and HC, respectively. Thus, the parallel-
ἔστιδὲκαὶὀρθογώνιον·τετράγωνονἄραἐστὶτὸΑΓ. ogram AC is equilateral. And (it is) also right-angled.
ΚαὶἐπεὶἐστινὡςἡΖΒπρὸςτὴνΒΗ,οὕτωςἡΔΒπρὸς Thus, AC is a square.
τὴνΒΕ,ἀλλ᾿ὡςμὲνἡΖΒπρὸςτὴνΒΗ,οὕτωςτὸΑΒπρὸς And since as FB is to BG, so DB (is) to BE, but
τὸΔΗ,ὡςδὲἡΔΒπρὸςτὴνΒΕ,οὕτωςτὸΔΗπρὸςτὸ as FB (is) to BG, so AB (is) to DG, and as DB (is) to
ΒΓ,καὶὡςἄρατὸΑΒπρὸςτὸΔΗ,οὕτωςτὸΔΗπρὸςτὸ BE, so DG (is) to BC [Prop. 6.1], thus also as AB (is)
ΒΓ.τῶνΑΒ,ΒΓἄραμέσονἀνάλογόνἐστιτὸΔΗ. to DG, so DG (is) to BC [Prop. 5.11]. Thus, DG is the
Λέγωδή,ὅτικαὶτῶνΑΓ,ΓΒμέσονἀνάλογόν[ἐστι]τὸ mean proportional to AB and BC.
ΔΓ.
So I also say that DC [is] the mean proportional to
᾿ΕπεὶγάρἐστινὡςἡΑΔπρὸςτὴνΔΚ,οὕτωςἡΚΗ AC and CB.
πρὸςτὴνΗΓ·ἴσηγάρ[ἐστιν]ἑκατέραἑκατέρᾳ·καὶσυνθέντι For since as AD is to DK, so KG (is) to GC. For [they
ὡςἡΑΚπρὸςΚΔ,οὕτωςἡΚΓπρὸςΓΗ,ἀλλ᾿ὡςμὲνἡΑΚ are] respectively equal. And, via composition, as AK (is)
πρὸςΚΔ,οὕτωςτὸΑΓπρὸςτὸΓΔ,ὡςδὲἡΚΓπρὸςΓΗ, to KD, so KC (is) to CG [Prop. 5.18]. But as AK (is)
οὕτωςτὸΔΓπρὸςΓΒ,καὶὡςἄρατὸΑΓπρὸςΔΓ,οὕτως to KD, so AC (is) to CD, and as KC (is) to CG, so DC
τὸΔΓπρὸςτὸΒΓ.τῶνΑΓ,ΓΒἄραμέσονἀνάλογόνἐστι (is) to CB [Prop. 6.1]. Thus, also, as AC (is) to DC,
τὸΔΓ·ἃπροέκειτοδεῖξαι.
so DC (is) to BC [Prop. 5.11]. Thus, DC is the mean
proportional to AC and CB. Which (is the very thing) it
A
F
H
341

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
was prescribed to show.
Ò. Proposition 54
᾿Εὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο If an area is contained by a rational (straight-line)
ὀνομάτων πρώτης, ἡ τὸ χωρίον δυναμένη ἄλογός ἐστιν and a first binomial (straight-line) then the square-root
ἡκαλουμένηἐκδύοὀνομάτων.
of the area is the irrational (straight-line which is) called
binomial. †
Α Η Ε Ζ ∆ Ρ Π A G E F D
R Q
Μ
Ν
Ξ
M
N
O
Β
Θ
Κ
Λ
Γ
B
H
K
L
C
Σ Ο
S P
ΧωρίονγὰρτὸΑΓπεριεχέσθωὑπὸῥητῆςτῆςΑΒκαὶ For let the area AC be contained by the rational
τῆς ἐκ δύο ὀνομάτων πρώτης τῆς ΑΔ· λέγω, ὅτι ἡ τὸ (straight-line) AB and by the first binomial (straight-
ΑΓ χωρίον δυναμένη ἄλογόςἐστιν ἡ καλουμένηἐκδύο line) AD. I say that square-root of area AC is the ir-
ὀνομάτων.
rational (straight-line which is) called binomial.
᾿ΕπεὶγὰρἐκδύοὀνομάτωνἐστὶπρώτηἡΑΔ,διῃρήσθω For since AD is a first binomial (straight-line), let it
εἰς τὰ ὀνόματα κατὰ τὸ Ε, καὶ ἔστω τὸ μεῖζον ὄνομα have been divided into its (component) terms at E, and
τὸ ΑΕ. φανερὸν δή, ὅτι αἱ ΑΕ, ΕΔ ῥηταί εἰσι δυνάμει let AE be the greater term. So, (it is) clear that AE and
μόνονσύμμετροι,καὶἡΑΕτῆςΕΔμεῖζονδύναταιτῷἀπὸ ED are rational (straight-lines which are) commensuσυμμέτρουἑαυτῃ,καὶἡΑΕσύμμετρόςἐστιτῇἐκκειμένῃ
rable in square only, and that the square on AE is greater
ῥητῇτῇΑΒμήκει. τετμήσθωδὴἡΕΔδίχακατὰτὸΖ than (the square on) ED by the (square) on (some
σημεῖον. καὶἐπεὶἡΑΕτῆςΕΔμεῖζονδύναταιτῷἀπὸ straight-line) commensurable (in length) with (AE), and
συμμέτρουἑαυτῇ,ἐὰνἄρατῷτετάρτῳμέρειτοῦἀπὸτῆς that AE is commensurable (in length) with the rational
ἐλάσσονος,τουτέστιτῷἀπὸτῆςΕΖ,ἴσονπαρὰτὴνμείζονα (straight-line) AB (first) laid out [Def. 10.5]. So, let ED
τὴνΑΕπαραβληθῇἐλλεῖπονεἴδειτετραγώνῳ,εἰςσύμμετρα have been cut in half at point F . And since the square on
αὐτὴνδιαιρεῖ.παραβεβλήσθωοὖνπαρὰτὴνΑΕτῷἀπὸτῆς AE is greater than (the square on) ED by the (square)
ΕΖἴσοντὸὑπὸΑΗ,ΗΕ·σύμμετροςἄραἐστὶνἡΑΗτῇ on (some straight-line) commensurable (in length) with
ΕΗμήκει.καὶἤχθωσανἀπὸτῶνΗ,Ε,ΖὁποτέρᾳτῶνΑΒ, (AE), thus if a (rectangle) equal to the fourth part of the
ΓΔπαράλληλοιαἱΗΘ,ΕΚ,ΖΛ·καὶτῷμὲνΑΘπαραλλη- (square) on the lesser (term)—that is to say, the (square)
λογράμμῳἴσοντετράγωνονσυνεστάτωτὸΣΝ,τῷδὲΗΚ on EF —falling short by a square figure, is applied to the
ἴσοντὸΝΠ,καὶκείσθωὥστεἐπ᾿εὐθείαςεἶναιτὴνΜΝ greater (term) AE, then it divides it into (terms which
τῇΝΞ·ἐπ᾿εὐθείαςἄραἐστὶκαὶἡΡΝτῇΝΟ.καὶσυμ- are) commensurable (in length) [Prop 10.17]. Therefore,
πεπληρώσθωτὸΣΠπαραλληλόγραμμον·τετράγωνονἄρα let the (rectangle contained) by AG and GE, equal to the
ἐστὶτὸΣΠ.καὶἐπεὶτὸὑπὸτῶνΑΗ,ΗΕἴσονἐστὶτῷἀπὸ (square) on EF , have been applied to AE. AG is thus
τῆςΕΖ,ἔστινἄραὡςἡΑΗπρὸςΕΖ,οὕτωςἡΖΕπρὸς commensurable in length with EG. And let GH, EK,
ΕΗ·καὶὡςἄρατὸΑΘπρὸςΕΛ,τὸΕΛπρὸςΚΗ·τῶν and FL have been drawn from (points) G, E, and F (re-
ΑΘ,ΗΚἄραμέσονἀνάλογόνἐστιτὸΕΛ.ἀλλὰτὸμὲνΑΘ spectively), parallel to either of AB or CD. And let the
ἴσονἐστὶτῷΣΝ,τὸδὲΗΚἴσοντῷΝΠ·τῶνΣΝ,ΝΠἄρα square SN, equal to the parallelogram AH, have been
μέσονἀνάλογόνἐστιτὸΕΛ.ἔστιδὲτῶναὐτῶντῶνΣΝ, constructed, and (the square) NQ, equal to (the parallel-
ΝΠμέσονἀνάλογονκαὶτὸΜΡ·ἴσονἄραἐστὶτὸΕΛτῷ ogram) GK [Prop. 2.14]. And let MN be laid down so
ΜΡ·ὥστεκαὶτῷΟΞἴσονἐστίν. ἔστιδὲκαὶτὰΑΘ,ΗΚ as to be straight-on to NO. RN is thus also straight-on
τοῖςΣΝ,ΝΠἴσα·ὅλονἄρατὸΑΓἴσονἐστὶνὅλῳτῷΣΠ, to NP . And let the parallelogram SQ have been comτουτέστιτῷἀπὸτῆςΜΞτετραγώνῳ·τὸΑΓἄραδύναταιἡ
pleted. SQ is thus a square [Prop. 10.53 lem.]. And since
ΜΞ.λέγω,ὅτιἡΜΞἐκδύοὀνομάτωνἐστίν.
the (rectangle contained) by AG and GE is equal to the
᾿ΕπεὶγὰρσύμμετρόςἐστινἡΑΗτῇΗΕ,σύμμετρόςἐστι (square) on EF , thus as AG is to EF , so FE (is) to EG
καὶἡΑΕἑκατέρᾳτῶνΑΗ,ΗΕ.ὑπόκειταιδὲκαὶἡΑΕτῇ [Prop. 6.17]. And thus as AH (is) to EL, (so) EL (is)
342

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ΑΒσύμμετρος·καὶαἱΑΗ,ΗΕἄρατῇΑΒσύμμετροίεἰσιν. to KG [Prop. 6.1]. Thus, EL is the mean proportional to
καίἐστιῥητὴἡΑΒ·ῥητὴἄραἐστὶκαὶἑκατέρατῶνΑΗ,ΗΕ· AH and GK. But, AH is equal to SN, and GK (is) equal
ῥητὸνἄραἐστὶνἑκάτεροντῶνΑΘ,ΗΚ,καίἐστισύμμετρον to NQ. EL is thus the mean proportional to SN and NQ.
τὸΑΘτῷΗΚ.ἀλλὰτὸμὲνΑΘτῷΣΝἴσονἐστίν,τὸδὲ And MR is also the mean proportional to the same—
ΗΚτῷΝΠ·καὶτὰΣΝ,ΝΠἄρα,τουτέστιτὰἀπὸτῶνΜΝ, (namely), SN and NQ [Prop. 10.53 lem.]. EL is thus
ΝΞ,ῥητάἐστικαὶσύμμετρα. καὶἐπεὶἀσύμμετρόςἐστινἡ equal to MR. Hence, it is also equal to PO [Prop. 1.43].
ΑΕτῇΕΔμήκει,ἀλλ᾿ἡμὲνΑΕτῇΑΗἐστισύμμετρος, And AH plus GK is equal to SN plus NQ. Thus, the
ἡδὲΔΕτῇΕΖσύμμετρος,ἀσύμμετροςἄρακαὶἡΑΗτῇ whole of AC is equal to the whole of SQ—that is to say,
ΕΖ·ὥστεκαὶτὸΑΘτῷΕΛἀσύμμετρόνἐστιν.ἀλλὰτὸμὲν to the square on MO. Thus, MO (is) the square-root of
ΑΘτῷΣΝἐστινἴσον,τὸδὲΕΛτῷΜΡ·καὶτὸΣΝἄρα (area) AC. I say that MO is a binomial (straight-line).
τῷΜΡἀσύμμετρόνἐστιν.ἀλλ᾿ὡςτὸΣΝπρὸςΜΡ,ἡΟΝ For since AG is commensurable (in length) with GE,
πρὸςτὴνΝΡ·ἀσύμμετροςἄραἐστὶνἡΟΝτῇΝΡ.ἴσηδὲἡ AE is also commensurable (in length) with each of AG
μὲνΟΝτῇΜΝ,ἡδὲΝΡτῇΝΞ·ἀσύμμετροςἄραἐστὶνἡ and GE [Prop. 10.15]. And AE was also assumed (to
ΜΝτῇΝΞ.καίἐστιτὸἀπὸτῆςΜΝσύμμετροντῷἀπὸτῆς be) commensurable (in length) with AB. Thus, AG
ΝΞ,καὶῥητὸνἑκάτερον·αἱΜΝ,ΝΞἄραῥηταίεἰσιδυνάμει and GE are also commensurable (in length) with AB
μόνονσύμμετροι. [Prop. 10.12]. And AB is rational. AG and GE are
῾ΗΜΞἄραἐκδύοὀνομάτωνἐστὶκαὶδύναταιτὸΑΓ· thus each also rational.
ὅπερἔδειδεῖξαι.
Thus, AH and GK are each
rational (areas), and AH is commensurable with GK
[Prop. 10.19]. But, AH is equal to SN, and GK to NQ.
SN and NQ—that is to say, the (squares) on MN and
NO (respectively)—are thus also rational and commensurable.
And since AE is incommensurable in length
with ED, but AE is commensurable (in length) with
AG, and DE (is) commensurable (in length) with EF ,
AG (is) thus also incommensurable (in length) with EF
[Prop. 10.13]. Hence, AH is also incommensurable with
EL [Props. 6.1, 10.11]. But, AH is equal to SN, and
EL to MR. Thus, SN is also incommensurable with
MR. But, as SN (is) to MR, (so) PN (is) to NR
[Prop. 6.1]. PN is thus incommensurable (in length)
with NR [Prop. 10.11]. And PN (is) equal to MN, and
NR to NO. Thus, MN is incommensurable (in length)
with NO. And the (square) on MN is commensurable
with the (square) on NO, and each (is) rational. MN
and NO are thus rational (straight-lines which are) commensurable
in square only.
Thus, MO is (both) a binomial (straight-line) [Prop.
10.36], and the square-root of AC. (Which is) the very
thing it was required to show.
† If the rational straight-line has unit length then this proposition states that the square-root of a first binomial straight-line is a binomial straightline:
i.e., a first binomial straight-line has a length k + k √ 1 − k ′2 whose square-root can be written ρ (1 + √ k ′′ ), where ρ = p k (1 + k ′ )/2 and
k ′′ = (1 − k ′ )/(1 + k ′ ). This is the length of a binomial straight-line (see Prop. 10.36), since ρ is rational.
Ò. Proposition 55
᾿Εὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο If an area is contained by a rational (straight-line) and
ὁνομάτωνδευτέρας,ἡτὸχωρίονδυναμένηἄλογόςἐστιν a second binomial (straight-line) then the square-root of
ἡκαλουμένηἐκδύομέσωνπρώτη.
the area is the irrational (straight-line which is) called
first bimedial. †
343

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
Α
Η
Ε
Ζ
∆
Ρ
Π
A
G
E
F
D
R
Q
Μ
Ν
Ξ
M
N
O
Β
Θ
Κ
Λ
Γ
B
H
K
L
C
Σ Ο
S P
ΠεριεχέσθωγὰρχωρίοντὸΑΒΓΔὑπὸῥητῆςτῆςΑΒ For let the area ABCD be contained by the rational
καὶτῆςἐκδύοὀνομάτωνδυετέραςτῆςΑΔ·λέγω,ὅτιἡτὸ (straight-line) AB and by the second binomial (straight-
ΑΓχωρίονδυναμένηἐκδύομέσωνπρώτηἐστίν. line) AD. I say that the square-root of area AC is a first
᾿Επεὶ γὰρ ἐκ δύο ὀνομάτων δευτέρα ἐστὶν ἡ ΑΔ, bimedial (straight-line).
διῃρήσθω εἰς τὰ ὀνόματα κατὰ τὸ Ε, ὥστε τὸ μεῖζον For since AD is a second binomial (straight-line), let it
ὄνομα εἶναι τὸ ΑΕ· αἱ ΑΕ, ΕΔ ἄρα ῥηταί εἰσι δυνάμει have been divided into its (component) terms at E, such
μόνονσύμμετροι,καὶἡΑΕτῆςΕΔμεῖζονδύναταιτῷἀπὸ that AE is the greater term. Thus, AE and ED are ratioσυμμέτρουἑαυτῇ,καὶτὸἔλαττονὄνομαἡΕΔσύμμετρόν
nal (straight-lines which are) commensurable in square
ἐστιτῇΑΒμήκει. τετμήσθωἡΕΔδίχακατὰτὸΖ,καὶ only, and the square on AE is greater than (the square
τῷἀπὸτῆςΕΖἴσονπαρὰτὴνΑΕπαραβεβλήσθωἐλλεῖπον on) ED by the (square) on (some straight-line) commenεἴδειτετραγώνῳτὸὑπὸτῶνΑΗΕ·σύμμετροςἄραἡΑΗ
surable (in length) with (AE), and the lesser term ED
τῇΗΕμήκει. καὶδιὰτῶνΗ,Ε,Ζπαράλληλοιἤχθωσαν is commensurable in length with AB [Def. 10.6]. Let
ταῖςΑΒ,ΓΔαἱΗΘ,ΕΚ,ΖΛ,καὶτῷμὲνΑΘπαραλλη- ED have been cut in half at F . And let the (rectanλογράμμῳἴσοντετράγωνονσυνεστάτωτὸΣΝ,τῷδὲΗΚ
gle contained) by AGE, equal to the (square) on EF ,
ἴσοντετράγωνοντὸΝΠ,καὶκείσθωὥστεἐπ᾿εὐθείαςεἶναι have been applied to AE, falling short by a square figτὴνΜΝτῇΝΞ·ἐπ᾿εὐθείαςἄρα[ἐστὶ]καὶἡΡΝτῆΝΟ.καὶ
ure. AG (is) thus commensurable in length with GE
συμπεπληρώσθωτὸΣΠτετράγωνον· φανερὸνδὴἐκτοῦ [Prop. 10.17]. And let GH, EK, and FL have been
προδεδειγμένου,ὅτιτὸΜΡμέσονἀνάλογόνἐστιτῶνΣΝ, drawn through (points) G, E, and F (respectively), par-
ΝΠ,καὶἴσοντῷΕΛ,καὶὅτιτὸΑΓχωρίονδύναταιἡΜΞ. allel to AB and CD. And let the square SN, equal to
δεικτέονδή,ὅτιἡΜΞἐκδύομέσωνἐστὶπρώτη. the parallelogram AH, have been constructed, and the
᾿ΕπεὶἀσύμμετρόςἐστινἡΑΕτῇΕΔμήκει,σύμμετρος square NQ, equal to GK. And let MN be laid down so
δὲἡΕΔτῇΑΒ,ἀσύμμετροςἄραἡΑΕτῇΑΒ.καὶἐπεὶ as to be straight-on to NO. Thus, RN [is] also straight-on
σύμμετρόςἐστινἡΑΗτῇΕΗ,σύμμετρόςἐστικαὶἡΑΕ to NP . And let the square SQ have been completed. So,
ἑκατέρᾳτῶνΑΗ,ΗΕ.ἀλλὰἡΑΕἀσύμμετροςτῇΑΒμήκει· (it is) clear from what has been previously demonstrated
καὶαἱΑΗ,ΗΕἄραἀσύμμετροίεἰσιτῇΑΒ.αἱΒΑ,ΑΗ, [Prop. 10.53 lem.] that MR is the mean proportional to
ΗΕἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι·ὥστεμέσον SN and NQ, and (is) equal to EL, and that MO is the
ἐστὶνἑκάτεροντῶνΑΘ,ΗΚ.ὥστεκαὶἑκάτεροντῶνΣΝ, square-root of the area AC. So, we must show that MO
ΝΠ μέσονἐστίν. καὶαἱΜΝ,ΝΞ ἄραμέσαιεἰσίν. καὶ is a first bimedial (straight-line).
ἐπεὶσύμμετροςἡΑΗτῇΗΕμήκει, σύμμετρόνἐστικαὶ Since AE is incommensurable in length with ED,
τὸΑΘτῷΗΚ,τουτέστιτὸΣΝτῷΝΠ,τουτέστιτὸἀπὸ and ED (is) commensurable (in length) with AB,
τῆςΜΝτῷἀπὸτῆςΝΞ[ὥστεδυνάμειεἰσὶσύμμετροιαἱ AE (is) thus incommensurable (in length) with AB
ΜΝ,ΝΞ].καὶἐπεὶἀσύμμετρόςἐστινἡΑΕτῇΕΔμήκει, [Prop. 10.13]. And since AG is commensurable (in
ἀλλ᾿ἡμὲνΑΕσύμμετρόςἐστιτῇΑΗ,ἡδὲΕΔτῇΕΖ length) with EG, AE is also commensurable (in length)
σύμμετρος, ἀσύμμετρος ἄρα ἡ ΑΗ τῇ ΕΖ· ὥστε καὶ τὸ with each of AG and GE [Prop. 10.15]. But, AE is in-
ΑΘτῷΕΛἀσύμμετρόνἐστιν, τουτέστιτὸΣΝτῷΜΡ, commensurable in length with AB. Thus, AG and GE
τουτέστινὁΟΝτῇΝΡ,τουτέστινἡΜΝτῇΝΞἀσύμμετρός are also (both) incommensurable (in length) with AB
ἐστιμήκει.ἐδείχθησανδὲαἱΜΝ,ΝΞκαὶμέσαιοὖσαικαὶ [Prop. 10.13]. Thus, BA, AG, and (BA, and) GE are
δυνάμεισύμμετροι·αἱΜΝ,ΝΞἄραμέσαιεἰσὶδυνάμειμόνον (pairs of) rational (straight-lines which are) commensuσύμμετροι.λέγωδή,ὅτικαὶῥητὸνπεριέχουσιν.ἐπεὶγὰρἡ
rable in square only. And, hence, each of AH and GK
ΔΕὑπόκειταιἑκατέρᾳτῶνΑΒ,ΕΖσύμμετρος,σύμμετρος is a medial (area) [Prop. 10.21]. Hence, each of SN
ἄρακαὶἡΕΖτῇΕΚ.καὶῥητὴἑκατέρααὐτῶν·ῥητὸνἄρα and NQ is also a medial (area). Thus, MN and NO
τὸΕΛ,τουτέστιτὸΜΡ·τὸδὲΜΡἐστιτὸὑπὸτῶνΜΝΞ. are medial (straight-lines). And since AG (is) commen-
ἐὰνδὲδύομέσαιδυνάμειμόνονσύμμετροισυντεθῶσιῥητὸν surable in length with GE, AH is also commensurable
344

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
περιέχουσαι,ἡὅληἄλογόςἐστιν,καλεῖταιδὲἐκδύομέσων with GK—that is to say, SN with NQ—that is to say,
πρώτη.
the (square) on MN with the (square) on NO [hence,
῾ΗἄραΜΞἐκδύομέσωνἐστὶπρώτη·ὅπερἔδειδεῖξαι. MN and NO are commensurable in square] [Props. 6.1,
10.11]. And since AE is incommensurable in length with
ED, but AE is commensurable (in length) with AG, and
ED commensurable (in length) with EF , AG (is) thus
incommensurable (in length) with EF [Prop. 10.13].
Hence, AH is also incommensurable with EL—that is
to say, SN with MR—that is to say, PN with NR—that
is to say, MN is incommensurable in length with NO
[Props. 6.1, 10.11]. But MN and NO have also been
shown to be medial (straight-lines) which are commensurable
in square. Thus, MN and NO are medial (straightlines
which are) commensurable in square only. So, I say
that they also contain a rational (area). For since DE was
assumed (to be) commensurable (in length) with each of
AB and EF , EF (is) thus also commensurable with EK
[Prop. 10.12]. And they (are) each rational. Thus, EL—
that is to say, MR—(is) rational [Prop. 10.19]. And MR
is the (rectangle contained) by MNO. And if two medial
(straight-lines), commensurable in square only, which
contain a rational (area), are added together, then the
whole is (that) irrational (straight-line which is) called
first bimedial [Prop. 10.37].
Thus, MO is a first bimedial (straight-line). (Which
is) the very thing it was required to show.
† If the rational straight-line has unit length then this proposition states that the square-root of a second binomial straight-line is a first bimedial
straight-line: i.e., a second binomial straight-line has a length k/ √ 1 − k ′ 2 + k whose square-root can be written ρ (k ′′1/4 + k ′′3/4 ), where
ρ = p (k/2) (1 + k ′ )/(1 − k ′ ) and k ′′ = (1 − k ′ )/(1 + k ′ ). This is the length of a first bimedial straight-line (see Prop. 10.37), since ρ is rational.
Ò¦. Proposition 56
᾿Εὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο If an area is contained by a rational (straight-line) and
ὀνομάτωντρίτης, ἡτὸχωρίονδυναμένηἄλογόςἐστινἡ a third binomial (straight-line) then the square-root of
καλουμένηἐκδύομέσωνδευτέρα.
the area is the irrational (straight-line which is) called
second bimedial. †
Α
Η
Ε
Ζ
∆
Ρ
Π
A
G
E
F
D
R
Q
Μ
Ν
Ξ
M
N
O
Β
Θ
Κ
Λ
Γ
B
H
K
L
C
Σ Ο
S P
ΧωρίονγὰρτὸΑΒΓΔπεριεχέσθωὑπὸῥητῆςτῆςΑΒ For let the area ABCD be contained by the rational
καὶτῆςἐκδύοὀνομάτωντρίτηςτῆςΑΔδιῃρημένηςεἰςτὰ (straight-line) AB and by the third binomial (straight-
ὀνόματακατὰτὸΕ,ὧνμεῖζόνἐστιτὸΑΕ·λέγω,ὅτιἡ line) AD, which has been divided into its (component)
τὸΑΓχωρίονδυναμένηἄλογόςἐστινἡκαλουμένηἐκδύο terms at E, of which AE is the greater. I say that the
μέσωνδευτέρα.
square-root of area AC is the irrational (straight-line
Κατεσκευάσθωγὰρτὰαὐτὰτοῖςπρότερον. καὶἐπεὶ which is) called second bimedial.
345

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ἐκδύοὀνομάτωνἐστὶτρίτηἡΑΔ,αἱΑΕ,ΕΔἄραῥηταί For let the same construction be made as previously.
εἰσιδυνάμειμόνονσύμμετροι, καὶἡΑΕτῆςΕΔμεῖζον And since AD is a third binomial (straight-line), AE and
δύναταιτῷἀπὸσυμμέτρουἑαυτῇ,καὶοὐδετέρατῶνΑΕ, ED are thus rational (straight-lines which are) commen-
ΕΔσύμμετρός[ἐστι]τῇΑΒμήκει.ὁμοίωςδὴτοῖςπροδε- surable in square only, and the square on AE is greater
δειγμένοιςδείξομεν,ὅτιἡΜΞἐστινἡτὸΑΓχωρίονδυ- than (the square on) ED by the (square) on (some
ναμένη,καὶαἱΜΝ,ΝΞμέσαιεἰσὶδυνάμειμόνονσύμμετροι· straight-line) commensurable (in length) with (AE), and
ὥστε ἡ ΜΞ ἐκ δύο μέσων ἐστίν. δεικτέον δή, ὅτι καὶ neither of AE and ED [is] commensurable in length with
δευτέρα. AB [Def. 10.7]. So, similarly to that which has been
[Καὶ] ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΔΕ τῇ ΑΒ μήκει, previously demonstrated, we can show that MO is the
τουτέστιτῇΕΚ,σύμμετροςδὲἡΔΕτῇΕΖ,ἀσύμμετρος square-root of area AC, and MN and NO are medial
ἄραἐστὶνἡΕΖτῇΕΚμήκει.καίεἰσιῥηταί·αἱΖΕ,ΕΚἄρα (straight-lines which are) commensurable in square only.
ῥηταίεἰσιδυνάμειμόνονσύμμετροι. μέσονἄρα[ἐστὶ]τὸ Hence, MO is bimedial. So, we must show that (it is)
ΕΛ,τουτέστιτὸΜΡ·καὶπεριέχεταιὑπὸτῶνΜΝΞ·μέσον also second (bimedial).
ἄραἐστὶτὸὑπὸτῶνΜΝΞ.
[And] since DE is incommensurable in length with
῾ΗΜΞἄραἐκδύομέσωνἐστὶδευτέρα·ὅπερἔδειδεῖξαι. AB—that is to say, with EK—and DE (is) commensurable
(in length) with EF , EF is thus incommensurable
in length with EK [Prop. 10.13]. And they are (both)
rational (straight-lines). Thus, FE and EK are rational
(straight-lines which are) commensurable in square only.
EL—that is to say, MR—[is] thus medial [Prop. 10.21].
And it is contained by MNO. Thus, the (rectangle contained)
by MNO is medial.
Thus, MO is a second bimedial (straight-line) [Prop.
10.38]. (Which is) the very thing it was required to show.
† If the rational straight-line has unit length then this proposition states that the square-root of a third binomial straight-line is a second bimedial
straight-line: i.e., a third binomial straight-line has a length k 1/2 (1+ √ 1 − k ′2 ) whose square-root can be written ρ (k 1/4 + k ′′1/2 /k 1/4 ), where
ρ = p (1 + k ′ )/2 and k ′′ = k (1 − k ′ )/(1 + k ′ ). This is the length of a second bimedial straight-line (see Prop. 10.38), since ρ is rational.
ÒÞ. Proposition 57
᾿Εὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο If an area is contained by a rational (straight-line) and
ὀνομάτωντετάρτης,ἡτὸχωρίονδυναμένηἄλογόςἐστιν a fourth binomial (straight-line) then the square-root of
ἡκαλουμένημείζων.
the area is the irrational (straight-line which is) called
major. †
Α
Η
Ε
Ζ
∆
Ρ
Π
A
G
E
F
D
R
Q
Μ
Ν
Ξ
M
N
O
Β
Θ
Κ
Λ
Γ
B
H
K
L
C
Σ Ο
S P
ΧωρίονγὰρτὸΑΓπεριεχέσθωὑπὸῥητῆςτῆςΑΒκαὶ For let the area AC be contained by the rational
τῆςἐκδύοὀνομάτωντετάρτηςτῆςΑΔδιῃρημένηςεἰςτὰ (straight-line) AB and the fourth binomial (straight-line)
ὀνόματακατὰτὸΕ,ὧνμεῖζονἔστωτὸΑΕ·λέγω,ὅτιἡτὸ AD, which has been divided into its (component) terms
ΑΓχωρίονδυναμένηἄλογόςἐστινἡκαλουμένημείζων. at E, of which let AE be the greater. I say that the square-
᾿ΕπεὶγὰρἡΑΔἐκδύοὀνομάτωνἐστὶτετάρτη,αἱΑΕ, root of AC is the irrational (straight-line which is) called
ΕΔἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι,καὶἡΑΕτῆς major.
ΕΔμεῖζονδύναταιτῷἀπὸἀσυμμέτρουἑαυτῇ,καὶἡΑΕ For since AD is a fourth binomial (straight-line), AE
τῇΑΒσύμμετρός[ἐστι]μήκει.τετμήσθωἡΔΕδίχακατὰ and ED are thus rational (straight-lines which are) com-
346

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
τὸΖ,καὶτῷἀπὸτῆςΕΖἴσονπαρὰτὴνΑΕπαραβεβλήσθω mensurable in square only, and the square on AE is
παραλληλόγραμμοντὸὑπὸΑΗ,ΗΕ·ἀσύμμετροςἄραἐστὶν greater than (the square on) ED by the (square) on
ἡΑΗτῇΗΕμήκει. ἤχθωσανπαράλληλοιτῇΑΒαἱΗΘ, (some straight-line) incommensurable (in length) with
ΕΚ,ΖΛ,καὶτὰλοιπὰτὰαὐτὰτοῖςπρὸτούτουγεγονέτω· (AE), and AE [is] commensurable in length with AB
φανερὸνδή, ὅτιἡτὸΑΓχωρίονδυναμένηἐστὶνἡΜΞ. [Def. 10.8]. Let DE have been cut in half at F , and let
δεικτέονδή,ὅτιἡΜΞἄλογόςἐστινἡκαλουμένημείζων. the parallelogram (contained by) AG and GE, equal to
᾿ΕπεὶἀσύμμετρόςἐστινἡΑΗτῇΕΗμήκει,ἀσύμμετρόν the (square) on EF , (and falling short by a square figure)
ἐστικαὶτὸΑΘτῷΗΚ,τουτέστιτὸΣΝτῷΝΠ·αἱΜΝ, have been applied to AE. AG is thus incommensurable
ΝΞἄραδυνάμειεἰσὶνἀσύμμετροι.καὶἐπεὶσύμμετρόςἐστιν in length with GE [Prop. 10.18]. Let GH, EK, and FL
ἡΑΕτῇΑΒμήκει,ῥητόνἐστιτὸΑΚ·καίἐστινἴσοντοῖς have been drawn parallel to AB, and let the rest (of the
ἀπὸτῶνΜΝ,ΝΞ·ῥητὸνἄρα[ἐστὶ]καὶτὸσυγκείμενονἐκ construction) have been made the same as the (proposiτῶνἀπὸτῶνΜΝ,ΝΞ.καὶἐπεὶἀσύμμετρός[ἐστιν]ἡΔΕ
tion) before this. So, it is clear that MO is the square-root
τῇΑΒμήκει,τουτέστιτῇΕΚ,ἀλλὰἡΔΕσύμμετρόςἐστι of area AC. So, we must show that MO is the irrational
τῇΕΖ,ἀσύμμετροςἄραἡΕΖτῇΕΚμήκει. αἱΕΚ,ΕΖ (straight-line which is) called major.
ἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι·μέσονἄρατὸΛΕ, Since AG is incommensurable in length with EG, AH
τουτέστιτὸΜΡ.καὶπεριέχεταιὑπὸτῶνΜΝ,ΝΞ·μέσον is also incommensurable with GK—that is to say, SN
ἄραἐστὶτὸὑπὸτῶνΜΝ,ΝΞ.καὶῥητὸντὸ[συγκείμενον] with NQ [Props. 6.1, 10.11]. Thus, MN and NO are
ἐκτῶνἀπὸτῶνΜΝ,ΝΞ,καίεἰσινἀσύμμετροιαἱΜΝ,ΝΞ incommensurable in square. And since AE is commensuδυνάμει.ἐὰνδὲδύοεὐθεῖαιδυνάμειἀσύμμετροισυντεθῶσι
rable in length with AB, AK is rational [Prop. 10.19].
ποιοῦσαιτὸμὲνσυγκείμενονἐκτῶνἀπ᾿αὐτῶντετραγώνων And it is equal to the (sum of the squares) on MN
ῥητόν,τὸδ᾿ὑπ᾿αὐτῶνμέσον,ἡὅληἄλογόςἐστιν,καλεῖται and NO. Thus, the sum of the (squares) on MN and
δὲμείζων.
NO [is] also rational. And since DE [is] incommensu-
῾Η ΜΞ ἄρα ἄλογός ἐστιν ἡ καλουμένη μείζων, καὶ rable in length with AB [Prop. 10.13]—that is to say,
δύναταιτὸΑΓχωρίον·ὅπερἔδειδεῖξαι.
with EK—but DE is commensurable (in length) with
EF , EF (is) thus incommensurable in length with EK
[Prop. 10.13]. Thus, EK and EF are rational (straightlines
which are) commensurable in square only. LE—
that is to say, MR—(is) thus medial [Prop. 10.21]. And it
is contained by MN and NO. The (rectangle contained)
by MN and NO is thus medial. And the [sum] of the
(squares) on MN and NO (is) rational, and MN and
NO are incommensurable in square. And if two straightlines
(which are) incommensurable in square, making the
sum of the squares on them rational, and the (rectangle
contained) by them medial, are added together, then the
whole is the irrational (straight-line which is) called major
[Prop. 10.39].
Thus, MO is the irrational (straight-line which is)
called major. And (it is) the square-root of area AC.
(Which is) the very thing it was required to show.
† If the rational straight-line has unit length then this proposition states that the square-root of a fourth binomial straight-line is a major straightline:
i.e., a fourth binomial straight-line has a length k (1 + 1/ √ q
1 + k ′ ) whose square-root can be written ρ [1 + k ′′ /(1 + k ′′2 ) 1/2 ]/2 +
q
ρ [1 − k ′′ /(1 + k ′′ 2 ) 1/2 ]/2, where ρ = √ k and k ′′ 2 = k ′ . This is the length of a major straight-line (see Prop. 10.39), since ρ is rational.
Ò. Proposition 58
᾿Εὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο If an area is contained by a rational (straight-line) and
ὀνομάτων πέμπτης, ἡ τὸ χωρίον δυναμένη ἄλογόςἐστιν a fifth binomial (straight-line) then the square-root of the
ἡκαλουμένηῥητὸνκαὶμέσονδυναμένη.
area is the irrational (straight-line which is) called the
ΧωρίονγὰρτὸΑΓπεριεχέσθωὑπὸῥητῆςτῆςΑΒκαὶ square-root of a rational plus a medial (area). †
347

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
τῆςἐκδύοὀνομάτωνπέμπτηςτῆςΑΔδιῃρημένηςεἰςτὰ For let the area AC be contained by the rational
ὀνόματακατὰτὸΕ,ὥστετὸμεῖζονὄνομαεἶναιτὸΑΕ· (straight-line) AB and the fifth binomial (straight-line)
λέγω[δή],ὅτιἡτὸΑΓχωρίονδυναμένηἄλογόςἐστινἡ AD, which has been divided into its (component) terms
καλουμένηῥητὸνκαὶμέσονδυναμένη.
at E, such that AE is the greater term. [So] I say that
the square-root of area AC is the irrational (straight-line
which is) called the square-root of a rational plus a medial
(area).
Α Η Ε Ζ ∆ Ρ Π A G E F D
R Q
Μ
Ν
Ξ
M
N
O
Β
Θ
Κ
Λ
Γ
B
H
K
L
C
Σ Ο
S P
Κατεσκευάσθωγὰρτὰαὐτὰτοῖςπρότερονδεδειγμένοις· For let the same construction be made as that shown
φανερὸνδή, ὅτιἡτὸΑΓχωρίονδυναμένηἐστὶνἡΜΞ. previously. So, (it is) clear that MO is the square-root of
δεικτέονδή,ὅτιἡΜΞἐστινἡῥητὸνκαὶμέσονδυναμένη. area AC. So, we must show that MO is the square-root
᾿ΕπεὶγὰρἀσύμμετρόςἐστινἡΑΗτῇΗΕ,ἀσύμμετρον of a rational plus a medial (area).
ἄραἐστὶκαὶτὸΑΘτῷΘΕ,τουτέστιτὸἀπὸτῆςΜΝτῷἀπὸ For since AG is incommensurable (in length) with
τῆςΝΞ·αἱΜΝ,ΝΞἄραδυνάμειεἰσὶνἀσύμμετροι.καὶἐπεὶ GE [Prop. 10.18], AH is thus also incommensurable
ἡΑΔἐκδύοὀνομάτωνἐστὶπέμπτη,καί[ἐστιν]ἔλασσον with HE—that is to say, the (square) on MN with the
αὐτῆςτμῆματὸΕΔ,σύμμετροςἄραἡΕΔτῇΑΒμήκει. (square) on NO [Props. 6.1, 10.11]. Thus, MN and
ἀλλὰἡΑΕτῇΕΔἐστινἀσύμμετρος· καὶἡΑΒἄρατῇ NO are incommensurable in square. And since AD is
ΑΕἐστινἀσύμμετροςμήκει[αἱΒΑ,ΑΕῥηταίεἰσιδυνάμει a fifth binomial (straight-line), and ED [is] its lesser segμόνον
σύμμετροι]· μέσον ἄρα ἐστὶ τὸ ΑΚ, τουτέστι τὸ ment, ED (is) thus commensurable in length with AB
συγκείμενονἐκτῶνἀπὸτῶνΜΝ,ΝΞ.καὶἐπεὶσύμμετρός [Def. 10.9]. But, AE is incommensurable (in length)
ἐστινἡΔΕτῇΑΒμήκει,τουτέστιτῇΕΚ,ἀλλὰἡΔΕτῇ with ED. Thus, AB is also incommensurable in length
ΕΖσύμμετρόςἐστιν,καὶἡΕΖἄρατῇΕΚσύμμετρόςἐστιν. with AE [BA and AE are rational (straight-lines which
καὶῥητὴἡΕΚ·ῥητὸνἄρακαὶτὸΕΛ,τουτέστιτὸΜΡ, are) commensurable in square only] [Prop. 10.13]. Thus,
τουτέστιτὸὑπὸΜΝΞ·αἱΜΝ,ΝΞἄραδυνάμειἀσύμμετροί AK—that is to say, the sum of the (squares) on MN
εἰσιποιοῦσαιτὸμὲνσυγκείμενονἐκτῶν ἀπ᾿αὐτῶντε- and NO—is medial [Prop. 10.21]. And since DE is
τραγώνωνμέσον,τὸδ᾿ὑπ᾿αὐτῶνῥητόν.
commensurable in length with AB—that is to say, with
῾ΗΜΞἄραῥητὸνκαὶμέσονδυναμένηἐστὶκαὶδύναται EK—but, DE is commensurable (in length) with EF ,
τὸΑΓχωρίον·ὅπερἔδειδεῖξαι.
EF is thus also commensurable (in length) with EK
[Prop. 10.12]. And EK (is) rational. Thus, EL—that
is to say, MR—that is to say, the (rectangle contained)
by MNO—(is) also rational [Prop. 10.19]. MN and NO
are thus (straight-lines which are) incommensurable in
square, making the sum of the squares on them medial,
and the (rectangle contained) by them rational.
Thus, MO is the square-root of a rational plus a medial
(area) [Prop. 10.40]. And (it is) the square-root of
area AC. (Which is) the very thing it was required to
show.
† If the rational straight-line has unit length then this proposition states that the square-root of a fifth binomial straight-line is the square root of
a rational plus a medial area: i.e., a fifth binomial straight-line has a length k ( √ q
q
1 + k ′ + 1) whose square-root can be written
ρ [(1 + k ′′2 ) 1/2 + k ′′ ]/[2(1 + k ′′2 )] + ρ [(1 + k ′′ 2 ) 1/2 − k ′′ ]/[2(1 + k ′′2 )], where ρ = p k (1 + k ′′ 2 ) and k ′′ 2 = k ′ . This is the length of
348

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
the square root of a rational plus a medial area (see Prop. 10.40), since ρ is rational.
Ò. Proposition 59
᾿Εὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ τῆς ἐκ δύο If an area is contained by a rational (straight-line) and
ὀνομάτων ἕκτης, ἡ τὸ χωρίον δυναμένηἄλογόςἐστιν ἡ a sixth binomial (straight-line) then the square-root of
καλουμένηδύομέσαδυναμένη.
the area is the irrational (straight-line which is) called
the square-root of (the sum of) two medial (areas). †
Α
Η Ε Ζ ∆ Ρ Π A G E F D
R Q
Μ
Ν
Ξ
M
N
O
Β
Θ
Κ
Λ
Γ
B
H
K
L
C
Σ Ο
S P
ΧωρίονγὰρτὸΑΒΓΔπεριεχέσθωὑπὸῥητῆςτῆςΑΒ For let the area ABCD be contained by the rational
καὶτῆςἐκδύοὀνομάτωνἕκτηςτῆςΑΔδιῃρημένηςεἰςτὰ (straight-line) AB and the sixth binomial (straight-line)
ὀνόματακατὰτὸΕ,ὥστετὸμεῖζονὄνομαεἶναιτὸΑΕ· AD, which has been divided into its (component) terms
λέγω,ὅτιἡτὸΑΓδυναμένηἡδύομέσαδυναμένηἐστίν. at E, such that AE is the greater term. So, I say that the
Κατεσκευάσθω [γὰρ] τὰ αὐτὰ τοῖς προδεδειγμένοις. square-root of AC is the square-root of (the sum of) two
φανερὸν δή, ὅτι [ἡ] τὸ ΑΓ δυναμένη ἐστὶν ἡ ΜΞ, καὶ medial (areas).
ὅτι ἀσύμμετρός ἐστιν ἡ ΜΝ τῇ ΝΞ δυνάμει. καὶ ἐπεὶ [For] let the same construction be made as that shown
ἀσύμμετρός ἐστιν ἡ ΕΑ τῇ ΑΒ μήκει, αἱ ΕΑ, ΑΒ ἄρα previously. So, (it is) clear that MO is the square-root of
ῥηταίεἰσιδυνάμειμόνονσύμμετροι·μέσονἄραἐστὶτὸΑΚ, AC, and that MN is incommensurable in square with
τουτέστιτὸσυγκείμενονἐκτῶνἀπὸτῶνΜΝ,ΝΞ.πάλιν, NO. And since EA is incommensurable in length with
ἐπεὶἀσύμμετρόςἐστινἡΕΔτῇΑΒμήκει,ἀσύμμετροςἄρα AB [Def. 10.10], EA and AB are thus rational (straight-
ἐστὶκαὶἡΖΕτῇΕΚ·αἱΖΕ,ΕΚἄραῥηταίεἰσιδυνάμει lines which are) commensurable in square only. Thus,
μόνονσύμμετροι·μέσονἄραἐστὶτὸΕΛ,τουτέστιτὸΜΡ, AK—that is to say, the sum of the (squares) on MN
τουτέστιτὸὑπὸτῶνΜΝΞ.καὶἐπεὶἀσύμμετροςἡΑΕτῇ and NO—is medial [Prop. 10.21]. Again, since ED
ΕΖ,καὶτὸΑΚτῷΕΛἀσύμμετρόνἐστιν. ἀλλὰτὸμὲν is incommensurable in length with AB [Def. 10.10],
ΑΚ ἐστιτὸ συγκείμενονἐκτῶν ἀπὸτῶν ΜΝ, ΝΞ, τὸ FE is thus also incommensurable (in length) with EK
δὲ ΕΛἐστιτὸὑπὸτῶν ΜΝΞ· ἀσύμμετρονἄραἐστὶτὸ [Prop. 10.13]. Thus, FE and EK are rational (straightσυγκείμενονἐκτῶνἀπὸτῶνΜΝΞτῷὑπὸτῶνΜΝΞ.καί
lines which are) commensurable in square only. Thus,
ἐστιμέσονἑκάτεροναὐτῶν,καὶαἱΜΝ,ΝΞδυνάμειεἰσὶν EL—that is to say, MR—that is to say, the (rectangle
ἀσύμμετροι.
contained) by MNO—is medial [Prop. 10.21]. And since
῾ΗΜΞἄραδύομέσαδυναμένηἐστὶκαὶδύναταιτὸΑΓ· AE is incommensurable (in length) with EF , AK is also
ὅπερἔδειδεῖξαι.
incommensurable with EL [Props. 6.1, 10.11]. But, AK
is the sum of the (squares) on MN and NO, and EL is
the (rectangle contained) by MNO. Thus, the sum of the
(squares) on MNO is incommensurable with the (rectangle
contained) by MNO. And each of them is medial.
And MN and NO are incommensurable in square.
Thus, MO is the square-root of (the sum of) two medial
(areas) [Prop. 10.41]. And (it is) the square-root of
AC. (Which is) the very thing it was required to show.
† If the rational straight-line has unit length then this proposition states that the square-root of a sixth binomial straight-line is the square root of
the sum of two medial areas: i.e., a sixth binomial straight-line has a length √ k + √ k ′ whose square-root can be written
q
«
k
„q[1 1/4 + k ′′ /(1 + k ′′2 ) 1/2 ]/2 + [1 − k ′′ /(1 + k ′′2 ) 1/2 ]/2 , where k ′′ 2 = (k − k ′ )/k ′ . This is the length of the square-root of the sum of
349

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
two medial areas (see Prop. 10.41).
ÄÑÑ.
᾿Εὰνεὐθεῖαγραμμὴτμηθῇεἰςἄνισα,τὰἀπὸτῶνἀνίσων If a straight-line is cut unequally then (the sum of) the
Lemma
τετράγωνα μείζονά ἐστι τοῦ δὶς ὑπὸ τῶν ἀνίσων περιε- squares on the unequal (parts) is greater than twice the
χομένουὀρθογωνίου.
rectangle contained by the unequal (parts).
Α ∆ Γ Β A D C
῎ΕστωεὐθεῖαἡΑΒκαὶτετμήσθωεἰςἄνισακατὰτὸ Let AB be a straight-line, and let it have been cut
Γ,καὶἔστωμείζωνἡΑΓ·λέγω,ὅτιτὰἀπὸτῶνΑΓ,ΓΒ unequally at C, and let AC be greater (than CB). I say
μείζονάἐστιτοῦδὶςὑπὸτῶνΑΓ,ΓΒ.
that (the sum of) the (squares) on AC and CB is greater
ΤετμήσθωγὰρἡΑΒδίχακατὰτὸΔ.ἐπεὶοὖνεὐθεῖα than twice the (rectangle contained) by AC and CB.
γραμμὴτέτμηταιεἰςμὲνἴσακατὰτὸΔ,εἰςδὲἄνισακατὰ For let AB have been cut in half at D. Therefore,
τὸΓ,τὸἄραὑπὸτῶνΑΓ,ΓΒμετὰτοῦἀπὸΓΔἴσονἐστὶ since a straight-line has been cut into equal (parts) at D,
τῷἀπὸΑΔ·ὥστετὸὑπὸτῶνΑΓ,ΓΒἔλαττόνἐστιτοῦ and into unequal (parts) at C, the (rectangle contained)
ἀπὸΑΔ·τὸἄραδὶςὑπὸτῶνΑΓ,ΓΒἔλαττονἢδιπλάσιόν by AC and CB, plus the (square) on CD, is thus equal
ἐστιτοῦἀπὸΑΔ.ἀλλὰτὰἀπὸτῶνΑΓ,ΓΒδιπλάσιά[ἐστι] to the (square) on AD [Prop. 2.5]. Hence, the (rectangle
τῶνἀπὸτῶνΑΔ,ΔΓ·τὰἄραἀπὸτῶνΑΓ,ΓΒμείζονά contained) by AC and CB is less than the (square) on
ἐστιτοῦδὶςὑπὸτῶνΑΓ,ΓΒ·ὅπερἔδειδεῖξαι.
AD. Thus, twice the (rectangle contained) by AC and
CB is less than double the (square) on AD. But, (the
sum of) the (squares) on AC and CB [is] double (the
sum of) the (squares) on AD and DC [Prop. 2.9]. Thus,
(the sum of) the (squares) on AC and CB is greater than
twice the (rectangle contained) by AC and CB. (Which
is) the very thing it was required to show.
Ü. Proposition 60
Τὸ ἀπὸ τῆς ἐκ δύο ὀνομάτων παρὰ ῥητὴν παρα- The square on a binomial (straight-line) applied to a
βαλλόμενονπλάτοςποιεῖτὴνἐκδύοὀνομάτωνπρώτην. rational (straight-line) produces as breadth a first binomial
(straight-line). †
∆ Κ Μ Ν Η
D K M N G
B
Ε
Θ
Λ
Ξ
Ζ
E
H
L
O
F
Α Γ Β
A C B
῎ΕστωἐκδύοὀνομάτωνἡΑΒδιῃρημένηεἰςτὰὀνόματα Let AB be a binomial (straight-line), having been diκατὰτὸΓ,ὥστετὸμεῖζονὄνομαεἶναιτὸΑΓ,καὶἐκκείσθω
vided into its (component) terms at C, such that AC is
ῥητὴἡΔΕ,καὶτῷἀπὸτῆςΑΒἴσονπαρὰτὴνΔΕπαρα- the greater term. And let the rational (straight-line) DE
βεβλήσθωτὸΔΕΖΗπλάτοςποιοῦντὴνΔΗ·λέγω,ὅτιἡ be laid down. And let the (rectangle) DEFG, equal to
ΔΗἐκδύοὀνομάτωνἐστὶπρώτη.
the (square) on AB, have been applied to DE, producing
ΠαραβεβλήσθωγὰρπαρὰτὴνΔΕτῷμὲνἀπὸτῆςΑΓ DG as breadth. I say that DG is a first binomial (straight-
ἴσοντὸΔΘ,τῷδὲἀπὸτῆςΒΓἴσοντὸΚΛ·λοιπὸνἄρα line).
τὸ δὶςὑπὸτῶνΑΓ, ΓΒἴσονἐστὶτῷ ΜΖ.τετμήσθωἡ For let DH, equal to the (square) on AC, and KL,
ΜΗδίχακατὰτὸΝ,καὶπαράλληλοςἤχθωἡΝΞ[ἑκατέρᾳ equal to the (square) on BC, have been applied to DE.
350

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
τῶνΜΛ,ΗΖ].ἑκάτερονἄρατῶνΜΞ,ΝΖἴσονἐστὶτῷ Thus, the remaining twice the (rectangle contained) by
ἅπαξὑπὸτῶνΑΓΒ.καὶἐπεὶἐκδύοὀνομάτωνἐστὶνἡΑΒ AC and CB is equal to MF [Prop. 2.4]. Let MG have
διῃρημένηεἰςτὰὀνόματακατὰτὸΓ,αἱΑΓ,ΓΒἄραῥηταί been cut in half at N, and let NO have been drawn parεἰσιδυνάμειμόνονσύμμετροι·τὰἄραἀπὸτῶνΑΓ,ΓΒῥητά
allel [to each of ML and GF ]. MO and NF are thus
ἐστικαὶσύμμετραἀλλήλοις·ὥστεκαὶτὸσυγκείμενονἐκ each equal to once the (rectangle contained) by ACB.
τῶνἀπὸτῶνΑΓ,ΓΒ.καίἐστινἴσοντῷΔΛ·ῥητὸνἄρα And since AB is a binomial (straight-line), having been
ἐστὶτὸΔΛ.καὶπαρὰῥητὴντὴνΔΕπαράκειται·ῥητὴἄρα divided into its (component) terms at C, AC and CB are
ἐστὶνἡΔΜκαὶσύμμετροςτῇΔΕμήκει.πάλιν,ἐπεὶαἱΑΓ, thus rational (straight-lines which are) commensurable
ΓΒῥηταίεἰσιδυνάμειμόνον σύμμετροι, μέσον ἄραἐστὶ in square only [Prop. 10.36]. Thus, the (squares) on AC
τὸδὶςὑπὸτῶνΑΓ,ΓΒ,τουτέστιτὸΜΖ.καὶπαρὰῥητὴν and CB are rational, and commensurable with one anτὴνΜΛπαράκειται·ῥητὴἄρακαὶἡΜΗκαὶἀσύμμετρος
other. And hence the sum of the (squares) on AC and
τῇΜΛ,τουτέστιτῇΔΕ,μήκει. ἔστιδὲκαὶἡΜΔῥητὴ CB (is rational) [Prop. 10.15], and is equal to DL. Thus,
καὶτῇΔΕμήκεισύμμετρος·ἀσύμμετροςἄραἐστὶνἡΔΜ DL is rational. And it is applied to the rational (straightτῇΜΗμήκει.
καίεἰσιῥηταί·αἱΔΜ,ΜΗἄραῥηταίεἰσι line) DE. DM is thus rational, and commensurable in
δυνάμειμόνονσύμμετροι· ἐκδύοἄραὀνομάτωνἐστὶνἡ length with DE [Prop. 10.20]. Again, since AC and CB
ΔΗ.δεικτέονδή,ὅτικαὶπρώτη.
are rational (straight-lines which are) commensurable in
᾿ΕπεὶτῶνἀπὸτῶνΑΓ,ΓΒμέσονἀνάλογόνἐστιτὸ square only, twice the (rectangle contained) by AC and
ὑπὸτῶνΑΓΒ,καὶτῶνΔΘ,ΚΛἄραμέσονἀνάλογόνἐστι CB—that is to say, MF —is thus medial [Prop. 10.21].
τὸΜΞ.ἔστινἄραὡςτὸΔΘπρὸςτὸΜΞ,οὕτωςτὸΜΞ And it is applied to the rational (straight-line) ML. MG
πρὸςτὸΚΛ,τουτέστινὡςἡΔΚπρὸςτὴνΜΝ,ἡΜΝπρὸς is thus also rational, and incommensurable in length
τὴνΜΚ·τὸἄραὑπὸτῶνΔΚ,ΚΜἴσονἐστὶτῷἀπὸτῆς with ML—that is to say, with DE [Prop. 10.22]. And
ΜΝ.καὶἐπεὶσύμμετρόνἐστιτὸἀπὸτῆςΑΓτῷἀπὸτῆς MD is also rational, and commensurable in length with
ΓΒ,σύμμετρόνἐστικαὶτὸΔΘτῷΚΛ·ὥστεκαὶἡΔΚ DE. Thus, DM is incommensurable in length with MG
τῇΚΜσύμμετρόςἐστιν.καὶἐπεὶμείζονάἐστιτὰἀπὸτῶν [Prop. 10.13]. And they are rational. DM and MG are
ΑΓ,ΓΒτοῦδὶςὑπὸτῶνΑΓ,ΓΒ,μεῖζονἄρακαὶτὸΔΛ thus rational (straight-lines which are) commensurable
τοῦΜΖ·ὥστεκαὶἡΔΜτῆςΜΗμείζωνἐστίν. καίἐστιν in square only. Thus, DG is a binomial (straight-line)
ἴσοντὸὑπὸτῶνΔΚ,ΚΜτῷἀπὸτῆςΜΝ,τουτέστιτῷ [Prop. 10.36]. So, we must show that (it is) also a first
τετάρτῳτοῦἀπὸτῆςΜΗ,καὶσύμμετροςἡΔΚτῇΚΜ. (binomial straight-line).
ἐὰνδὲὦσιδύοεὐθεῖαιἄνισοι,τῷδὲτετάρτῳμέρειτοῦἀπὸ Since the (rectangle contained) by ACB is the
τῆςἐλάσσονοςἴσονπαρὰτὴνμείζοναπαραβληθῇἐλλεῖπον mean proportional to the squares on AC and CB
εἴδειτετραγώνῳκαὶεἰςσύμμετρααὐτὴνδιαιρῇ,ἡμείζων [Prop. 10.53 lem.], MO is thus also the mean proporτῆςἐλάσσονοςμεῖζονδύναταιτῷἀπὸσυμμέτρουἑαυτῇ·ἡ
tional to DH and KL. Thus, as DH is to MO, so
ΔΜἄρατῆςΜΗμεῖζονδύναταιτῷἀπὸσυμμέτρουἑαυτῇ. MO (is) to KL—that is to say, as DK (is) to MN,
καίεἰσιῥηταὶαἱΔΜ,ΜΗ,καὶἡΔΜμεῖζονὄνομαοὖσα (so) MN (is) to MK [Prop. 6.1]. Thus, the (rectanσύμμετρόςἐστιτῇἐκκειμένῃῥητῇτῇΔΕμήκει.
gle contained) by DK and KM is equal to the (square)
῾Η ΔΗ ἄραἐκδύοὀνομάτων ἐστὶπρώτη· ὅπερἔδει on MN [Prop. 6.17]. And since the (square) on AC is
δεῖξαι.
commensurable with the (square) on CB, DH is also
commensurable with KL. Hence, DK is also commensurable
with KM [Props. 6.1, 10.11]. And since (the sum
of) the squares on AC and CB is greater than twice the
(rectangle contained) by AC and CB [Prop. 10.59 lem.],
DL (is) thus also greater than MF . Hence, DM is also
greater than MG [Props. 6.1, 5.14]. And the (rectangle
contained) by DK and KM is equal to the (square)
on MN—that is to say, to one quarter the (square) on
MG. And DK (is) commensurable (in length) with KM.
And if there are two unequal straight-lines, and a (rectangle)
equal to the fourth part of the (square) on the
lesser, falling short by a square figure, is applied to the
greater, and divides it into (parts which are) commensurable
(in length), then the square on the greater is larger
351

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
† In other words, the square of a binomial is a first binomial. See Prop. 10.54.
than (the square on) the lesser by the (square) on (some
straight-line) commensurable (in length) with the greater
[Prop. 10.17]. Thus, the square on DM is greater than
(the square on) MG by the (square) on (some straightline)
commensurable (in length) with (DM). And DM
and MG are rational. And DM, which is the greater
term, is commensurable in length with the (previously)
laid down rational (straight-line) DE.
Thus, DG is a first binomial (straight-line) [Def.
10.5]. (Which is) the very thing it was required to show.
Ü. Proposition 61
Τὸἀπὸτῆςἐκδύομέσωνπρώτηςπαρὰῥητὴνπαρα- The square on a first bimedial (straight-line) applied
βαλλόμενονπλάτοςποιεῖτὴνἐκδύοὀνομάτωνδευτέραν. to a rational (straight-line) produces as breadth a second
binomial (straight-line). †
∆ Κ Μ Ν Η
D K M N G
Ε
Θ
Λ
Ξ
Ζ
E
H
L
O
F
Α
Γ
Β
῎ΕστωἐκδύομέσωνπρώτηἡΑΒδιῃρημένηεἰςτὰς Let AB be a first bimedial (straight-line) having been
μέσαςκατὰτὸΓ,ὧνμείζωνἡΑΓ,καὶἐκκείσθωῥητὴἡ divided into its (component) medial (straight-lines) at
ΔΕ,καὶπαραβεβλήσθωπαρὰτὴνΔΕτῷἀπὸτῆςΑΒἴσον C, of which AC (is) the greater. And let the rational
παραλληλόγραμμοντὸΔΖπλάτοςποιοῦντὴνΔΗ·λέγω, (straight-line) DE be laid down. And let the parallelo-
ὅτιἡΔΗἐκδύοὀνομάτωνἐστὶδευτέρα.
gram DF , equal to the (square) on AB, have been ap-
Κατεσκευάσθω γὰρ τὰ αὐτὰ τοῖς πρὸ τούτου. καὶ plied to DE, producing DG as breadth. I say that DG is
ἐπεὶἡΑΒἐκδύομέσωνἐστὶπρώτηδιῃρημένηκατὰτὸ a second binomial (straight-line).
Γ, αἱ ΑΓ, ΓΒ ἄρα μέσαι εἰσὶ δυνάμει μόνον σύμμετροι For let the same construction have been made as
ῥητὸν περιέχουσαι· ὥστε καὶτὰἀπὸτῶν ΑΓ, ΓΒμέσα in the (proposition) before this. And since AB is a
ἐστίν.μέσονἄραἐστὶτὸΔΛ.καὶπαρὰῥητὴντὴνΔΕπα- first bimedial (straight-line), having been divided at C,
ραβέβληται·ῥητὴἄραἐστίνἡΜΔκαὶἀσύμμετροςτῇΔΕ AC and CB are thus medial (straight-lines) commenμήκει.πάλιν,ἐπεὶῥητόνἐστιτὸδὶςὑπὸτῶνΑΓ,ΓΒ,ῥητόν
surable in square only, and containing a rational (area)
ἐστικαὶτὸΜΖ.καὶπαρὰῥητὴντὴνΜΛπαράκειται·ῥητὴ [Prop. 10.37]. Hence, the (squares) on AC and CB
ἄρα[ἐστὶ]καὶἡΜΗκαὶμήκεισύμμετροςτῇΜΛ,τουτέστι are also medial [Prop. 10.21]. Thus, DL is medial
τῇΔΕ·ἀσύμμετροςἄραἐστὶνἡΔΜτῇΜΗμήκει.καίεἰσι [Props. 10.15, 10.23 corr.]. And it has been applied to the
ῥηταί·αἱΔΜ,ΜΗἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι· rational (straight-line) DE. MD is thus rational, and in-
ἐκδύοἄραὀνομάτωνἐστὶν ἡΔΗ. δεικτέονδή, ὅτικαὶ commensurable in length with DE [Prop. 10.22]. Again,
δευτέρα.
since twice the (rectangle contained) by AC and CB is
᾿ΕπεὶγὰρτὰἀπὸτῶνΑΓ,ΓΒμείζονάἐστιτοῦδὶςὑπὸ rational, MF is also rational. And it is applied to the
τῶνΑΓ,ΓΒ,μεῖζονἄρακαὶτὸΔΛτοῦΜΖ·ὥστεκαὶἡ rational (straight-line) ML. Thus, MG [is] also ratio-
ΔΜτῆςΜΗ.καὶἐπεὶσύμμετρόνἐστιτὸἀπὸτῆςΑΓτῷ nal, and commensurable in length with ML—that is to
ἀπὸτῆςΓΒ,σύμμετρόνἐστικαὶτὸΔΘτῷΚΛ·ὥστεκαὶ say, with DE [Prop. 10.20]. DM is thus incommensu-
ἡΔΚτῇΚΜσύμμετρόςἐστιν.καίἐστιτὸὑπὸτῶνΔΚΜ rable in length with MG [Prop. 10.13]. And they are
ἴσοντῷἀπὸτῆςΜΝ·ἡΔΜἄρατῆςΜΗμεῖζονδύναταιτῷ rational. DM and MG are thus rational, and commensu-
A
C
B
352

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ἀπὸσυμμέτρουἑαυτῇ. καίἐστινἡΜΗσύμμετροςτῇΔΕ rable in square only. DG is thus a binomial (straight-line)
μήκει.
[Prop. 10.36]. So, we must show that (it is) also a second
῾ΗΔΗἄραἐκδύοὀνομάτωνἐστὶδευτέρα.
(binomial straight-line).
For since (the sum of) the squares on AC and CB is
greater than twice the (rectangle contained) by AC and
CB [Prop. 10.59], DL (is) thus also greater than MF .
Hence, DM (is) also (greater) than MG [Prop. 6.1].
And since the (square) on AC is commensurable with
the (square) on CB, DH is also commensurable with
KL. Hence, DK is also commensurable (in length) with
KM [Props. 6.1, 10.11]. And the (rectangle contained)
by DKM is equal to the (square) on MN. Thus, the
square on DM is greater than (the square on) MG by
the (square) on (some straight-line) commensurable (in
length) with (DM) [Prop. 10.17]. And MG is commensurable
in length with DE.
Thus, DG is a second binomial (straight-line) [Def.
10.6].
† In other words, the square of a first bimedial is a second binomial. See Prop. 10.55.
Ü. Proposition 62
Τὸἀπὸτῆςἐκδύομέσωνδευτέραςπαρὰῥητὴνπαρα- The square on a second bimedial (straight-line) apβαλλόμενονπλάτοςποιεῖτὴνἐκδύοὀνομάτωντρίτην.
plied to a rational (straight-line) produces as breadth a
third binomial (straight-line). †
∆ Κ Μ Ν Η
D K M N G
Ε
Θ
Λ
Ξ
Ζ
E
H
L
O
F
Α Γ Β
A C B
῎ΕστωἐκδύομέσωνδευτέραἡΑΒδιῃρημένηεἰςτὰς Let AB be a second bimedial (straight-line) having
μέσαςκατὰτὸΓ,ὥστετὸμεῖζοντμῆμαεἶναιτὸΑΓ,ῥητὴ been divided into its (component) medial (straight-lines)
δέτιςἔστωἡΔΕ,καὶπαρὰτὴνΔΕτῷἀπὸτῆςΑΒἴσον at C, such that AC is the greater segment. And let DE be
παραλληλόγραμμονπαραβεβλήσθωτὸΔΖπλάτοςποιοῦν some rational (straight-line). And let the parallelogram
τὴνΔΗ·λέγω,ὅτιἡΔΗἐκδύοὀνομάτωνἐστὶτρίτη. DF , equal to the (square) on AB, have been applied to
Κατεσκευάσθωτὰαὐτὰτοῖςπροδεδειγμένοις. καὶἐπεὶ DE, producing DG as breadth. I say that DG is a third
ἐκδύομέσωνδευτέραἐστὶνἡΑΒδιῃρημένηκατὰτὸΓ, binomial (straight-line).
αἱΑΓ,ΓΒἄραμέσαιεἰσὶδυνάμειμόνονσύμμετροιμέσον Let the same construction be made as that shown preπεριέχουσαι·ὥστεκαὶτὸσυγκείμενονἐκτῶνἀπὸτῶνΑΓ,
viously. And since AB is a second bimedial (straight-
ΓΒμέσονἐστίν. καίἐστινἴσοντῷΔΛ·μέσονἄρακαὶτὸ line), having been divided at C, AC and CB are thus
ΔΛ.καὶπαράκειταιπαρὰῥητὴντὴνΔΕ·ῥητὴἄραἐστὶκαὶἡ medial (straight-lines) commensurable in square only,
ΜΔκαὶἀσύμμετροςτῇΔΕμήκει.διὰτὰαὐτὰδὴκαὶἡΜΗ and containing a medial (area) [Prop. 10.38]. Hence,
ῥητήἐστικαὶἀσύμμετροςτῇΜΛ,τουτέστιτῇΔΕ,μήκει· the sum of the (squares) on AC and CB is also medial
ῥητὴἄραἐστὶνἑκατέρατῶνΔΜ,ΜΗκαὶἀσύμμετροςτῇ [Props. 10.15, 10.23 corr.]. And it is equal to DL. Thus,
ΔΕμήκει. καὶἐπεὶἀσύμμετρόςἐστινἡΑΓτῇΓΒμήκει, DL (is) also medial. And it is applied to the rational
ὡςδὲἡΑΓπρὸςτὴνΓΒ,οὕτωςτὸἀπὸτῆςΑΓπρὸςτὸ (straight-line) DE. MD is thus also rational, and in-
353

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ὑπὸτῶνΑΓΒ,ἀσύμμετρονἄρακαὶτὸἀπὸτῆςΑΓτῷὑπὸ commensurable in length with DE [Prop. 10.22]. So,
τῶνΑΓΒ.ὥστεκαὶτὸσυγκείμενονἐκτῶνἀπὸτῶνΑΓ, for the same (reasons), MG is also rational, and incom-
ΓΒτῷδὶςὑπὸτῶνΑΓΒἀσύμμετρόνἐστιν,τουτέστιτὸ mensurable in length with ML—that is to say, with DE.
ΔΛτῷΜΖ·ὥστεκαὶἡΔΜτῷΜΗἀσύμμετρόςἐστιν.καί Thus, DM and MG are each rational, and incommenεἰσιῥηταί·ἐκδύοἄραὀνομάτωνἐστὶνἡΔΗ.δεικτέον[δή],
surable in length with DE. And since AC is incommen-
ὅτικαὶτρίτη.
surable in length with CB, and as AC (is) to CB, so
῾Ομοίωςδὴτοῖςπροτέροιςἐπιλογιούμεθα,ὅτιμείζων the (square) on AC (is) to the (rectangle contained) by
ἐστὶνἡΔΜτῆςΜΗ,καὶσύμμετροςἡΔΚτῇΚΜ.καίἐστι ACB [Prop. 10.21 lem.], the (square) on AC (is) also inτὸὑπὸτῶνΔΚΜἴσοντῷἀπὸτῆςΜΝ·ἡΔΜἄρατῆςΜΗ
commensurable with the (rectangle contained) by ACB
μεῖζονδύναταιτῷἀπὸσυμμέτρουἑαυτῇ.καὶοὐδετέρατῶν [Prop. 10.11]. And hence the sum of the (squares) on
ΔΜ,ΜΗσύμμετρόςἐστιτῇΔΕμήκει.
AC and CB is incommensurable with twice the (rect-
῾ΗΔΗἄραἐκδύοὀνομάτωνἐστὶτρίτη·ὅπερἔδειδεῖξαι. angle contained) by ACB—that is to say, DL with MF
[Props. 10.12, 10.13]. Hence, DM is also incommensurable
(in length) with MG [Props. 6.1, 10.11]. And
they are rational. DG is thus a binomial (straight-line)
[Prop. 10.36]. [So] we must show that (it is) also a third
(binomial straight-line).
So, similarly to the previous (propositions), we can
conclude that DM is greater than MG, and DK (is) commensurable
(in length) with KM. And the (rectangle
contained) by DKM is equal to the (square) on MN.
Thus, the square on DM is greater than (the square on)
MG by the (square) on (some straight-line) commensurable
(in length) with (DM) [Prop. 10.17]. And neither
of DM and MG is commensurable in length with DE.
Thus, DG is a third binomial (straight-line) [Def.
10.7]. (Which is) the very thing it was required to show.
† In other words, the square of a second bimedial is a third binomial. See Prop. 10.56.
Ü. Proposition 63
Τὸ ἀπὸ τῆς μείζονος παρὰ ῥητὴν παραβαλλόμενον The square on a major (straight-line) applied to a raπλάτοςποιεῖτὴνἐκδύοὀνομάτωντετάρτην.
tional (straight-line) produces as breadth a fourth binomial
(straight-line). †
∆ Κ Μ Ν Η
D K M N G
Ε
Θ
Λ
Ξ
Ζ
E
H
L
O
F
Α
Γ
Β
῎ΕστωμείζωνἡΑΒδιῃρημένηκατὰτὸΓ,ὥστεμείζονα Let AB be a major (straight-line) having been divided
εἶναιτὴνΑΓτῆςΓΒ,ῥητὴδὲἡΔΕ,καὶτῷἀπὸτῆςΑΒἴσον at C, such that AC is greater than CB, and (let) DE
παρὰ τὴν ΔΕ παραβεβλήσθω τὸ ΔΖ παραλληλόγραμμον (be) a rational (straight-line). And let the parallelogram
πλάτοςποιοῦντὴνΔΗ·λέγω,ὅτιἡΔΗἐκδύοὀνομάτων DF , equal to the (square) on AB, have been applied to
ἐστὶτετάρτη.
DE, producing DG as breadth. I say that DG is a fourth
Κατεσκευάσθωτὰαὐτὰτοῖςπροδεδειγμένοις. καὶἐπεὶ binomial (straight-line).
μείζωνἐστὶνἡΑΒδιῃρημένηκατὰτὸΓ,αἱΑΓ,ΓΒδυνάμει Let the same construction be made as that shown pre-
A
C
B
354

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
εἰσὶνἀσύμμετροιποιοῦσαιτὸμὲνσυγκείμενονἐκτῶνἀπ᾿ viously. And since AB is a major (straight-line), havαὐτῶντετραγώνωνῥητόν,τὸδὲὑπ᾿αὐτῶνμέσον.ἐπεὶοὖν
ing been divided at C, AC and CB are incommensu-
ῥητόνἐστιτὸσυγκείμενονἐκτῶνἀπὸτῶνΑΓ,ΓΒ,ῥητὸν rable in square, making the sum of the squares on them
ἄραἐστὶτὸΔΛ·ῥητὴἄρακαὶἡΔΜκαὶσύμμετροςτῇ rational, and the (rectangle contained) by them medial
ΔΕμήκει.πάλιν,ἐπεὶμέσονἐστὶτὸδὶςὑπὸτῶνΑΓ,ΓΒ, [Prop. 10.39]. Therefore, since the sum of the (squares)
τουτέστιτὸΜΖ,καὶπαρὰῥητήνἐστιτὴνΜΛ,ῥητὴἄρα on AC and CB is rational, DL is thus rational. Thus,
ἐστὶκαὶἡΜΗκαὶἀσύμμετροςτῇΔΕμήκει·ἀσύμμετρος DM (is) also rational, and commensurable in length with
ἄραἐστὶκαὶἡΔΜτῇΜΗμήκει. αἱΔΜ,ΜΗἄραῥηταί DE [Prop. 10.20]. Again, since twice the (rectangle conεἰσιδυνάμειμόνονσύμμετροι·ἐκδύοἄραὀνομάτωνἐστὶν
tained) by AC and CB—that is to say, MF —is medial,
ἡΔΗ.δεικτέον[δή],ὅτικαὶτετάρτη.
and is (applied to) the rational (straight-line) ML, MG
῾Ομοίως δὴδείξομεντοῖςπρότερον, ὅτιμείζων ἐστὶν is thus also rational, and incommensurable in length with
ἡΔΜτῆςΜΗ,καὶὅτιτὸὑπὸΔΚΜἴσονἐστὶτῷἀπὸ DE [Prop. 10.22]. DM is thus also incommensurable
τῆς ΜΝ. ἐπεὶ οὖν ἀσύμμετρόν ἐστι τὸ ἀπὸ τῆς ΑΓ τῷ in length with MG [Prop. 10.13]. DM and MG are
ἀπὸ τῆς ΓΒ, ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ΔΘ τῷ ΚΛ· thus rational (straight-lines which are) commensurable
ὥστε ἀσύμμετροςκαὶἡΔΚ τῇΚΜ ἐστιν. ἐὰν δὲ ὦσι in square only. Thus, DG is a binomial (straight-line)
δύο εὐθεῖαι ἄνισοι, τῷ δὲ τετάρτῳ μέρει τοῦ ἀπὸ τῆς [Prop. 10.36]. [So] we must show that (it is) also a fourth
ἐλάσσονοςἴσονπαραλληλόγραμμονπαρὰτὴνμείζοναπαρα- (binomial straight-line).
βληθῇἐλλεῖπονεἴδειτετραγώνῳκαὶεἰςἀσύμμετρααὐτὴν So, similarly to the previous (propositions), we can
διαιρῇ,ἡμείζωντῆςἐλάσσονοςμεῖζονδυνήσεταιτῷἀπὸ show that DM is greater than MG, and that the (rectan-
ἀσύμμέτρουἑαυτῇμήκει·ἡΔΜἄρατῆςΜΗμεῖζονδύναται gle contained) by DKM is equal to the (square) on MN.
τῷἀπὸἀσυμμέτρουἑαυτῇ. καίεἰσιναἱΔΜ,ΜΗῥηταὶ Therefore, since the (square) on AC is incommensurable
δυνάμειμόνονσύμμετροι, καὶἡΔΜσύμμετρόςἐστιτῇ with the (square) on CB, DH is also incommensurable
ἐκκειμένῃῥητῇτῇΔΕ. with KL. Hence, DK is also incommensurable with
῾ΗΔΗἄραἐκδύοὀνομάτωνἐστὶτετάρτη·ὅπερἔδει KM [Props. 6.1, 10.11]. And if there are two unequal
δεῖξαι.
straight-lines, and a parallelogram equal to the fourth
part of the (square) on the lesser, falling short by a square
figure, is applied to the greater, and divides it into (parts
which are) incommensurable (in length), then the square
on the greater will be larger than (the square on) the
lesser by the (square) on (some straight-line) incommensurable
in length with the greater [Prop. 10.18]. Thus,
the square on DM is greater than (the square on) MG
by the (square) on (some straight-line) incommensurable
(in length) with (DM). And DM and MG are rational
(straight-lines which are) commensurable in square only.
And DM is commensurable (in length) with the (previously)
laid down rational (straight-line) DE.
Thus, DG is a fourth binomial (straight-line) [Def.
10.8]. (Which is) the very thing it was required to show.
† In other words, the square of a major is a fourth binomial. See Prop. 10.57.
Ü. Proposition 64
Τὸἀπὸτῆςῥητὸνκαὶμέσονδυναμένηςπαρὰῥητὴνπα- The square on the square-root of a rational plus a meραβαλλόμενονπλάτοςποιεῖτὴνἐκδύοὀνομάτωνπέμπτην.
dial (area) applied to a rational (straight-line) produces
῎ΕστωῥητὸνκαὶμέσονδυναμένηἡΑΒδιῃρημένηεἰς as breadth a fifth binomial (straight-line). †
τὰςεὐθείαςκατὰτὸΓ,ὥστεμείζοναεἶναιτὴνΑΓ,καὶ Let AB be the square-root of a rational plus a medial
ἐκκείσθωῥητὴἡΔΕ,καὶτῷἀπὸτῆςΑΒἴσονπαρὰτὴν (area) having been divided into its (component) straight-
ΔΕπαραβεβλήσθωτὸΔΖπλάτοςποιοῦντὴνΔΗ·λέγω, lines at C, such that AC is greater. And let the rational
ὅτιἡΔΗἐκδύοὀνομάτωνἐστὶπέμπτη.
(straight-line) DE be laid down. And let the (parallelogram)
DF , equal to the (square) on AB, have been ap-
355

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
∆
Κ
Μ
Ν
Η
plied to DE, producing DG as breadth. I say that DG is
a fifth binomial straight-line.
D K M N G
Ε
Θ
Λ
Ξ
Ζ
E
H
L
O
F
Α Γ Β
A C B
Κατεσκευάσθω τὰ αὐτα τοῖς πρὸ τούτου. ἐπεὶ οὖν Let the same construction be made as in the (proposi-
ῥητὸνκαὶμέσονδυναμένηἐστὶνἡΑΒδιῃρημένηκατὰτὸ tions) before this. Therefore, since AB is the square-root
Γ,αἱΑΓ,ΓΒἄραδυνάμειεἰσὶνἀσύμμετροιποιοῦσαιτὸμὲν of a rational plus a medial (area), having been divided
συγκείμενονἐκτῶνἀπ᾿αὐτῶντετραγώνωνμέσον,τὸδ᾿ὑπ᾿ at C, AC and CB are thus incommensurable in square,
αὐτῶνῥητόν. ἐπεὶοὖνμέσονἐστὶτὸσυγκείμενονἐκτῶν making the sum of the squares on them medial, and the
ἀπὸτῶνΑΓ,ΓΒ,μέσονἄραἐστὶτὸΔΛ·ὥστεῥητήἐστιν (rectangle contained) by them rational [Prop. 10.40].
ἡΔΜκαὶμήκειἀσύμμετροςτῇΔΕ.πάλιν,ἐπεὶῥητόνἐστι Therefore, since the sum of the (squares) on AC and
τὸδὶςὑπὸτῶνΑΓΒ,τουτέστιτὸΜΖ,ῥητὴἄραἡΜΗκαὶ CB is medial, DL is thus medial. Hence, DM is rational
σύμμετροςτῇΔΕ.ἀσύμμετροςἄραἡΔΜτῇΜΗ·αἱΔΜ, and incommensurable in length with DE [Prop. 10.22].
ΜΗἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι·ἐκδύοἄρα Again, since twice the (rectangle contained) by ACB—
ὀνομάτωνἐστὶνἡΔΗ.λέγωδή,ὅτικαὶπέμπτη. that is to say, MF —is rational, MG (is) thus rational
῾Ομοίωςγὰρδιεχθήσεται,ὅτιτὸὑπὸτῶνΔΚΜἴσον and commensurable (in length) with DE [Prop. 10.20].
ἐστὶτῷἀπὸτῆςΜΝ,καὶἀσύμμετροςἡΔΚτῇΚΜμήκει·ἡ DM (is) thus incommensurable (in length) with MG
ΔΜἄρατῆςΜΗμεῖζονδύναταιτῷἀπὸἀσυμμέτρουἑαυτῇ. [Prop. 10.13]. Thus, DM and MG are rational (straightκαίεἰσιναἱΔΜ,ΜΗ[ῥηταὶ]δυνάμειμόνονσύμμετροι,καὶ
lines which are) commensurable in square only. Thus,
ἡἐλάσσωνἡΜΗσύμμετροςτῇΔΕμήκει.
DG is a binomial (straight-line) [Prop. 10.36]. So, I say
῾ΗΔΗἄραἐκδύοὀνομάτωνἐστὶπέμπτη· ὅπερἔδει that (it is) also a fifth (binomial straight-line).
δεῖξαι.
For, similarly (to the previous propositions), it can
be shown that the (rectangle contained) by DKM is
equal to the (square) on MN, and DK (is) incommensurable
in length with KM. Thus, the square on DM
is greater than (the square on) MG by the (square) on
(some straight-line) incommensurable (in length) with
(DM) [Prop. 10.18]. And DM and MG are [rational]
(straight-lines which are) commensurable in square only,
and the lesser MG is commensurable in length with DE.
Thus, DG is a fifth binomial (straight-line) [Def. 10.9].
(Which is) the very thing it was required to show.
† In other words, the square of the square-root of a rational plus medial is a fifth binomial. See Prop. 10.58.
Ü. Proposition 65
Τὸ ἀπὸ τῆς δύο μέσα δυναμένης παρὰ ῥητὴν παρα- The square on the square-root of (the sum of) two meβαλλόμενονπλάτοςποιεῖτὴνἐκδύοὀνομάτωνἕκτην.
dial (areas) applied to a rational (straight-line) produces
῎ΕστωδύομέσαδυναμένηἡΑΒδιῃρημένηκατὰτὸΓ, as breadth a sixth binomial (straight-line). †
ῥητὴδὲἔστωἡΔΕ,καὶπαρὰτὴνΔΕτῷἀπὸτῆςΑΒἴσον Let AB be the square-root of (the sum of) two meπαραβεβλήσθωτὸΔΖπλάτοςποιοῦντὴνΔΗ·λέγω,ὅτιἡ
dial (areas), having been divided at C. And let DE be a
ΔΗἐκδύοὀνομάτωνἐστὶνἕκτη. rational (straight-line). And let the (parallelogram) DF ,
equal to the (square) on AB, have been applied to DE,
356

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
∆
Κ
Μ
Ν
Η
producing DG as breadth. I say that DG is a sixth binomial
(straight-line).
D
K
M
N
G
Ε
Θ
Λ
Ξ
Ζ
E
H
L
O
F
Α
Γ
Β
Κατεσκευάσθωγὰρτὰαὐτὰτοῖςπρότερον. καὶἐπεὶἡ For let the same construction be made as in the pre-
ΑΒδύομέσαδυναμένηἐστὶδιῃρημένηκατὰτὸΓ,αἱΑΓ,ΓΒ vious (propositions). And since AB is the square-root
ἄραδυνάμειεἰσὶνἀσύμμετροιποιοῦσαιτότεσυγκείμενον of (the sum of) two medial (areas), having been divided
ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων μέσον καὶ τὸ ὑπ᾿ αὐτῶν at C, AC and CB are thus incommensurable in square,
μέσονκαὶἔτιἀσύμμετροντὸἐκτῶνἀπ᾿αὐτῶντετραγώνων making the sum of the squares on them medial, and the
συγκείμενοντῷὑπ᾿αὐτῶν· ὥστεκατὰτὰπροδεδειγμένα (rectangle contained) by them medial, and, moreover,
μέσονἐστὶνἑκάτεροντῶνΔΛ,ΜΖ.καὶπαρὰῥητὴντὴν the sum of the squares on them incommensurable with
ΔΕπαράκειται·ῥητὴἄραἐστὶνἑκατέρατῶνΔΜ,ΜΗκαὶ the (rectangle contained) by them [Prop. 10.41]. Hence,
ἀσύμμετροςτῇΔΕ μήκει. καὶἐπεὶἀσύμμετρόνἐστιτὸ according to what has been previously demonstrated, DL
συγκείμενονἐκτῶνἀπὸτῶνΑΓ,ΓΒτῷδὶςὑπὸτῶνΑΓ, and MF are each medial. And they are applied to the
ΓΒ,ἀσύμμετρονἄραἐστὶτὸΔΛτῷΜΖ.ἀσύμμετροςἄρα rational (straight-line) DE. Thus, DM and MG are
καὶἡΔΜτῇΜΗ·αἱΔΜ,ΜΗἄραῥηταίεἰσιδυνάμειμόνον each rational, and incommensurable in length with DE
σύμμετροι·ἐκδύοἄραὀνομάτωνἐστὶνἡΔΗ.λέγωδή,ὅτι [Prop. 10.22]. And since the sum of the (squares) on
καὶἕκτη.
AC and CB is incommensurable with twice the (rectan-
῾Ομοίωςδὴπάλινδεῖξομεν,ὅτιτὸὑπὸτῶνΔΚΜἴσον gle contained) by AC and CB, DL is thus incommensu-
ἐστὶτῷἀπὸτῆςΜΝ,καὶὅτιἡΔΚτῇΚΜμήκειἐστὶν rable with MF . Thus, DM (is) also incommensurable (in
ἀσύμμετρος· καὶδιὰ τὰ αὐτὰδὴ ἡ ΔΜ τῆς ΜΗ μεῖζον length) with MG [Props. 6.1, 10.11]. DM and MG are
δύναταιτῷἀπὸἀσυμμέτρουἑαυτῇμήκει.καὶοὐδετέρατῶν thus rational (straight-lines which are) commensurable
ΔΜ,ΜΗσύμμετρόςἐστιτῇἐκκειμένῃῥητῇτῇΔΕμήκει. in square only. Thus, DG is a binomial (straight-line)
῾Η ΔΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶν ἕκτη· ὅπερ ἔδει [Prop. 10.36]. So, I say that (it is) also a sixth (binomial
δεῖξαι.
straight-line).
So, similarly (to the previous propositions), we can
again show that the (rectangle contained) by DKM is
equal to the (square) on MN, and that DK is incommensurable
in length with KM. And so, for the same
(reasons), the square on DM is greater than (the square
on) MG by the (square) on (some straight-line) incommensurable
in length with (DM) [Prop. 10.18]. And neither
of DM and MG is commensurable in length with the
(previously) laid down rational (straight-line) DE.
Thus, DG is a sixth binomial (straight-line) [Def.
10.10]. (Which is) the very thing it was required to show.
† In other words, the square of the square-root of two medials is a sixth binomial. See Prop. 10.59.
ܦ. Proposition 66
῾Ητῇἐκδύοὀνομάτωνμήκεισύμμετροςκαὶαὐτὴἐκ A (straight-line) commensurable in length with a biδύοὀνομάτωνἐστὶκαὶτῇτάξειἡαὐτή.
nomial (straight-line) is itself also binomial, and the same
῎Εστω ἐκ δύο ὀνομάτων ἡ ΑΒ, καὶ τῇ ΑΒ μήκει in order.
A
C
B
357

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
σύμμετροςἔστωἡΓΔ·λέγω,ὅτιἡΓΔἐκδύοὀνομάτων Let AB be a binomial (straight-line), and let CD be
ἐστὶκαὶτῇτάξειἡαὐτὴτῇΑΒ.
commensurable in length with AB. I say that CD is a binomial
(straight-line), and (is) the same in order as AB.
Α
Ε
Β
A E B
Γ Ζ ∆
C
F
D
᾿ΕπεὶγὰρἐκδύοὀνομάτωνἐστὶνἡΑΒ,διῃρήσθωεἰςτὰ For since AB is a binomial (straight-line), let it have
ὀνόματακατὰτὸΕ,καὶἔστωμεῖζονὄνοματὸΑΕ·αἱΑΕ, been divided into its (component) terms at E, and let
ΕΒἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι. γεγονέτωὡς AE be the greater term. AE and EB are thus rational
ἡΑΒπρὸςτὴνΓΔ,οὕτωςἡΑΕπρὸςτὴνΓΖ·καὶλοιπὴ (straight-lines which are) commensurable in square only
ἄραἡΕΒπρὸςλοιπὴντὴνΖΔἐστιν,ὡςἡΑΒπρὸςτὴν [Prop. 10.36]. Let it have been contrived that as AB (is)
ΓΔ.σύμμετροςδὲἡΑΒτῇΓΔμήκει·σύμμετροςἄραἐστὶ to CD, so AE (is) to CF [Prop. 6.12]. Thus, the remainκαὶἡμὲνΑΕτῇΓΖ,ἡδὲΕΒτῇΖΔ.καίεἰσιῥηταὶαἱΑΕ,
der EB is also to the remainder FD, as AB (is) to CD
ΕΒ·ῥηταὶἄραεἰσὶκαὶαἱΓΖ,ΖΔ.καὶἐστινὡςἡΑΕπρὸς [Props. 6.16, 5.19 corr.]. And AB (is) commensurable
ΓΖ,ἡΕΒπρὸςΖΔ.ἐναλλὰξἄραἐστὶνὡςἡΑΕπρὸςΕΒ,ἡ in length with CD. Thus, AE is also commensurable
ΓΖπρὸςΖΔ.αἱδὲΑΕ,ΕΒδυνάμειμόνον[εἰσὶ]σύμμετροι· (in length) with CF , and EB with FD [Prop. 10.11].
καὶαἱΓΖ,ΖΔἄραδυνάμειμόνονεἰσὶσύμμετροι. καίεἰσι And AE and EB are rational. Thus, CF and FD are
ῥηταί·ἐκδύοἄραὀνομάτωνἐστὶνἡΓΔ.λέγωδή,ὅτιτῇ also rational. And as AE is to CF , (so) EB (is) to FD
τάξειἐστὶνἡαὐτὴτῇΑΒ. [Prop. 5.11]. Thus, alternately, as AE is to EB, (so)
῾Η γὰρ ΑΕ τῆς ΕΒ μεῖζον δύναται ἤτοι τῷ ἀπὸ CF (is) to FD [Prop. 5.16]. And AE and EB [are]
συμμέτρου ἑαυτῇ ἢ τῷ ἀπὸ ἀσυμμέτρου. εἰ μὲν οὖν ἡ commensurable in square only. Thus, CF and FD are
ΑΕτῆςΕΒμεῖζονδύναταιτῷἀπὸσυμμέτρουἑαυτῇ,καὶ also commensurable in square only [Prop. 10.11]. And
ἡΓΖτῆςΖΔμεῖζονδυνήσεταιτῷἀπὸσυμμέτρουἑαυτῇ. they are rational. CD is thus a binomial (straight-line)
καὶεἰμὲνσύμμετρόςἐστινἡΑΕτῇἐκκειμένῃῥητῇ,καὶ [Prop. 10.36]. So, I say that it is the same in order as
ἡΓΖσύμμετροςαὐτῇἔσται, καὶδιὰτοῦτοἑκατέρατῶν AB.
ΑΒ,ΓΔἐκδύοὀνομάτωνἐστὶπρώτη,τουτέστιτῇτάξει For the square on AE is greater than (the square on)
ἡαὐτή. εἰδὲἡΕΒσύμμετρόςἐστιτῇἐκκειμένῃῥητῇ, EB by the (square) on (some straight-line) either comκαὶἡΖΔσύμμετρόςἐστιναὐτῇ,καὶδιὰτοῦτοπάλιντῇ
mensurable or incommensurable (in length) with (AE).
τάξειἡαὐτὴἔσταιτῇΑΒ·ἑκατέραγὰραὐτῶνἔσταιἐκδύο Therefore, if the square on AE is greater than (the square
ὀνομάτωνδευτέρα.εἰδὲοὐδετέρατῶνΑΕ,ΕΒσύμμετρός on) EB by the (square) on (some straight-line) com-
ἐστιτῇἐκκειμένῃῥητῇ,οὐδετέρατῶνΓΖ,ΖΔσύμμετρος mensurable (in length) with (AE) then the square on
αὐτῇἔσται,καίἐστινἑκατέρατρίτη. εἰδὲἡΑΕτῆςΕΒ CF will also be greater than (the square on) FD by
μεῖζονδύναταιτῷἀπὸἀσυμμέτρουἑαυτῇ,καὶἡΓΖτὴςΖΔ the (square) on (some straight-line) commensurable (in
μεῖζονδύναταιτῷἀπὸἀσυμμέτρουἑαυτῇ.καὶεἰμὲνἡΑΕ length) with (CF ) [Prop. 10.14]. And if AE is comσύμμετρόςἐστιτῇἐκκειμένῃῥητῃ,
καὶἡΓΖσύμμετρός mensurable (in length) with (some previously) laid down
ἐστιναὐτῇ,καὶἐστινἑκατέρατετάρτη. εἰδὲἡΕΒ,καὶ rational (straight-line) then CF will also be commensu-
ἡΖΔ,καὶἔσταιἑκατέραπέμπτη. εἰδὲοὐδετέρατῶνΑΕ, rable (in length) with it [Prop. 10.12]. And, on account
ΕΒ,καὶτῶνΓΖ,ΖΔοὐδετέρασύμμετρόςἐστιτῇἐκκειμένῃ of this, AB and CD are each first binomial (straight-
ῥητῇ,καὶἔσταιἑκατέραἕκτη.
lines) [Def. 10.5]—that is to say, the same in order. And if
῞Ωστεἡτῇἐκδύοὀνομάτωνμήκεισύμμετροςἐκδύο EB is commensurable (in length) with the (previously)
ὀνομάτωνἐστὶκαὶτῇτάξειἡαὐτή·ὅπερἔδειδεῖξαι. laid down rational (straight-line) then FD is also commensurable
(in length) with it [Prop. 10.12], and, again,
on account of this, (CD) will be the same in order as
AB. For each of them will be second binomial (straightlines)
[Def. 10.6]. And if neither of AE and EB is commensurable
(in length) with the (previously) laid down
rational (straight-line) then neither of CF and FD will
be commensurable (in length) with it [Prop. 10.13], and
each (of AB and CD) is a third (binomial straight-line)
358

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
[Def. 10.7]. And if the square on AE is greater than
(the square on) EB by the (square) on (some straightline)
incommensurable (in length) with (AE) then the
square on CF is also greater than (the square on) FD
by the (square) on (some straight-line) incommensurable
(in length) with (CF ) [Prop. 10.14]. And if AE is commensurable
(in length) with the (previously) laid down
rational (straight-line) then CF is also commensurable
(in length) with it [Prop. 10.12], and each (of AB and
CD) is a fourth (binomial straight-line) [Def. 10.8]. And
if EB (is commensurable in length with the previously
laid down rational straight-line) then FD (is) also (commensurable
in length with it), and each (of AB and CD)
will be a fifth (binomial straight-line) [Def. 10.9]. And
if neither of AE and EB (is commensurable in length
with the previously laid down rational straight-line) then
also neither of CF and FD is commensurable (in length)
with the laid down rational (straight-line), and each (of
AB and CD) will be a sixth (binomial straight-line)
[Def. 10.10].
Hence, a (straight-line) commensurable in length
with a binomial (straight-line) is a binomial (straightline),
and the same in order. (Which is) the very thing it
was required to show.
ÜÞ. Proposition 67
῾Ητῇἐκδύομέσωνμήκεισύμμετροςκαὶαὐτὴἐκδύο A (straight-line) commensurable in length with a biμέσωνἐστὶκαὶτῇτάξειἡαὐτή.
medial (straight-line) is itself also bimedial, and the same
in order.
Α
Ε
Β
A E B
Γ Ζ ∆
C
F
D
῎ΕστωἐκδύομέσωνἡΑΒ,καὶτῇΑΒσύμμετροςἔστω Let AB be a bimedial (straight-line), and let CD be
μήκειἡΓΔ·λέγω,ὅτιἡΓΔἐκδύομέσωνἐστὶκαὶτῇτάξει commensurable in length with AB. I say that CD is bi-
ἡαὐτὴτῇΑΒ.
medial, and the same in order as AB.
᾿ΕπεὶγὰρἐκδύομέσωνἐστὶνἡΑΒ,διῃρήσθωεἰςτὰς For since AB is a bimedial (straight-line), let it have
μέσαςκατὰτὸΕ·αἱΑΕ,ΕΒἄραμέσαιεἰσὶδυνάμειμόνον been divided into its (component) medial (straight-lines)
σύμμετροι. καὶγεγονέτωὡςἡΑΒπρὸςΓΔ,ἡΑΕπρὸς at E. Thus, AE and EB are medial (straight-lines
ΓΖ·καὶλοιπὴἄραἡΕΒπρὸςλοιπὴντὴνΖΔἐστιν,ὡςἡ which are) commensurable in square only [Props. 10.37,
ΑΒπρὸςΓΔ.σύμμετροςδὲἡΑΒτῇΓΔμήκει·σύμμετρος 10.38]. And let it have been contrived that as AB (is) to
ἄρακαὶἑκατέρατῶνΑΕ,ΕΒἑκατέρᾳτῶνΓΖ,ΖΔ.μέσαι CD, (so) AE (is) to CF [Prop. 6.12]. And thus as the
δὲαἱΑΕ,ΕΒ·μέσαιἄρακαὶαἱΓΖ,ΖΔ.καὶἐπείἐστιν remainder EB is to the remainder FD, so AB (is) to CD
ὡςἡΑΕπρὸςΕΒ,ἡΓΖπρὸςΖΔ,αἱδὲΑΕ,ΕΒδυνάμει [Props. 5.19 corr., 6.16]. And AB (is) commensurable
μόνονσύμμετροίεἰσιν,καὶαἱΓΖ,ΖΔ[ἄρα]δυνάμειμόνον in length with CD. Thus, AE and EB are also comσύμμετροίεἰσιν,ἐδείχθησανδὲκαὶμέσαι·ἡΓΔἄραἐκδύο
mensurable (in length) with CF and FD, respectively
μέσωνἐστίν.λέγωδή,ὅτικαὶτῇτάξειἡαὐτήἐστιτῇΑΒ. [Prop. 10.11]. And AE and EB (are) medial. Thus, CF
᾿ΕπεὶγάρἐστινὡςἡΑΕπρὸςΕΒ,ἡΓΖπρὸςΖΔ,καὶ and FD (are) also medial [Prop. 10.23]. And since as
ὡςἄρατὸἀπὸτῆςΑΕπρὸςτὸὑπὸτῶνΑΕΒ,οὕτωςτὸ AE is to EB, (so) CF (is) to FD, and AE and EB are
ἀπὸτῆςΓΖπρὸςτὸὑπὸτῶνΓΖΔ·ἐναλλὰξὡςτὸἀπὸτῆς commensurable in square only, CF and FD are [thus]
359

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ΑΕπρὸςτὸἀπὸτῆςΓΖ,οὕτωςτὸὑπὸτῶνΑΕΒπρὸςτὸ also commensurable in square only [Prop. 10.11]. And
ὑπὸτῶνΓΖΔ.σύμμετρονδὲτὸἀπὸτῆςΑΕτῷἀπὸτῆς they were also shown (to be) medial. Thus, CD is a bi-
ΓΖ·σύμμετρονἄρακαὶτὸὑπὸτῶνΑΕΒτῷὑπὸτῶνΓΖΔ. medial (straight-line). So, I say that it is also the same in
εἴτεοὖνῥητόνἐστιτὸὑπὸτῶνΑΕΒ,καὶτὸὑπὸτῶνΓΖΔ order as AB.
ῥητόνἐστιν[καὶδιὰτοῦτόἐστινἐκδύομέσωνπρώτη].εἴτε For since as AE is to EB, (so) CF (is) to FD, thus
μέσον,μέσον,καίἐστινἑκατέραδευτέρα.
also as the (square) on AE (is) to the (rectangle con-
ΚαὶδιὰτοῦτοἔσταιἡΓΔτῇΑΒτῇτάξειἡαὐτή·ὅπερ tained) by AEB, so the (square) on CF (is) to the (rect-
ἔδειδεῖξαι. angle contained) by CFD [Prop. 10.21 lem.]. Alternately,
as the (square) on AE (is) to the (square) on
CF , so the (rectangle contained) by AEB (is) to the
(rectangle contained) by CFD [Prop. 5.16]. And the
(square) on AE (is) commensurable with the (square)
on CF . Thus, the (rectangle contained) by AEB (is)
also commensurable with the (rectangle contained) by
CFD [Prop. 10.11]. Therefore, either the (rectangle
contained) by AEB is rational, and the (rectangle contained)
by CFD is rational [and, on account of this,
(AE and CD) are first bimedial (straight-lines)], or (the
rectangle contained by AEB is) medial, and (the rectangle
contained by CFD is) medial, and (AB and CD)
are each second (bimedial straight-lines) [Props. 10.23,
10.37, 10.38].
And, on account of this, CD will be the same in order
as AB. (Which is) the very thing it was required to show.
Ü. Proposition
῾Ητῇμείζονισύμμετροςκαὶαὐτὴμείζωνἐστίν.
Α
Ε
Β
68
A (straight-line) commensurable (in length) with a
major (straight-line) is itself also major.
A E B
Γ Ζ ∆
C
F
D
῎ΕστωμείζωνἡΑΒ,καὶτῇΑΒσύμμετροςἔστωἡΓΔ· Let AB be a major (straight-line), and let CD be comλέγω,ὅτιἡΓΔμείζωνἐστίν.
mensurable (in length) with AB. I say that CD is a major
ΔιῃρήσθωἡΑΒκατὰτὸΕ·αἱΑΕ,ΕΒἄραδυνάμειεἰσὶν (straight-line).
ἀσύμμετροιποιοῦσαιτὸμὲνσυγκείμενονἐκτῶνἀπ᾿αὐτῶν Let AB have been divided (into its component terms)
τετραγώνωνῥητόν,τὸδ᾿ὑπ᾿αὐτῶνμέσον·καὶγεγονέτω at E. AE and EB are thus incommensurable in square,
τὰαὐτὰτοῖςπρότερον. καὶἐπείἐστινὡςἡΑΒπρὸςτὴν making the sum of the squares on them rational, and the
ΓΔ,οὕτωςἥτεΑΕπρὸςτὴνΓΖκαὶἡΕΒπρὸςτὴνΖΔ, (rectangle contained) by them medial [Prop. 10.39]. And
καὶὡςἄραἡΑΕπρὸςτὴνΓΖ,οὕτωςἡΕΒπρὸςτὴνΖΔ. let (the) same (things) have been contrived as in the preσύμμετροςδὲἡΑΒτῇΓΔ·σύμμετροςἄρακαὶἑκατέρατῶν
vious (propositions). And since as AB is to CD, so AE
ΑΕ,ΕΒἑκατέρᾳτῶνΓΖ,ΖΔ.καὶἐπείἐστινὡςἡΑΕπρὸς (is) to CF and EB to FD, thus also as AE (is) to CF ,
τὴνΓΖ,οὕτωςἡΕΒπρὸςτὴνΖΔ,καὶἐναλλὰξὡςἡΑΕ so EB (is) to FD [Prop. 5.11]. And AB (is) commenπρὸςΕΒ,οὕτωςἡΓΖπρὸςΖΔ,καὶσυνθέντιἄραἑστὶνὡς
surable (in length) with CD. Thus, AE and EB (are)
ἡΑΒπρὸςτὴνΒΕ,οὕτωςἡΓΔπρὸςτὴνΔΖ·καὶὡςἄρα also commensurable (in length) with CF and FD, reτὸἀπὸτῆςΑΒπρὸςτὸἀπὸτῆςΒΕ,οὕτωςτὸἀπὸτῆςΓΔ
spectively [Prop. 10.11]. And since as AE is to CF , so
πρὸςτὸἀπὸτῆςΔΖ.ὁμοίωςδὴδείξομεν,ὅτικαὶὡςτὸἀπὸ EB (is) to FD, also, alternately, as AE (is) to EB, so
τῆςΑΒπρὸςτὸἀπὸτῆςΑΕ,οὕτωςτὸἀπὸτῆςΓΔπρὸς CF (is) to FD [Prop. 5.16], and thus, via composition,
τὸἀπὸτῆςΓΖ.καὶὡςἄρατὸἀπὸτῆςΑΒπρὸςτὰἀπὸτῶν as AB is to BE, so CD (is) to DF [Prop. 5.18]. And thus
ΑΕ,ΕΒ,οὕτωςτὸἀπὸτῆςΓΔπρὸςτὰἀπὸτῶνΓΖ,ΖΔ· as the (square) on AB (is) to the (square) on BE, so the
360

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
καὶἐναλλὰξἄραἐστὶνὡςτὸἀπὸτῆςΑΒπρὸςτὸἀπὸτῆς (square) on CD (is) to the (square) on DF [Prop. 6.20].
ΓΔ,οὕτωςτὰἀπὸτῶνΑΕ,ΕΒπρὸςτὰἀπὸτῶνΓΖ,ΖΔ. So, similarly, we can also show that as the (square) on
σύμμετρονδὲτὸἀπὸτῆςΑΒτῷἀπὸτῆςΓΔ·σύμμετρα AB (is) to the (square) on AE, so the (square) on CD
ἄρακαὶτὰἀπὸτῶνΑΕ,ΕΒτοῖςἀπὸτῶνΓΖ,ΖΔ.καίἐστι (is) to the (square) on CF . And thus as the (square) on
τὰἀπὸτῶνΑΕ,ΕΒἅμαῥητόν,καὶτὰἀπὸτῶνΓΖ,ΖΔ AB (is) to (the sum of) the (squares) on AE and EB, so
ἅμαῥητόνἐστιν. ὁμοίωςδὲκαὶτὸδὶςὑπὸτῶνΑΕ,ΕΒ the (square) on CD (is) to (the sum of) the (squares) on
σύμμετρόνἐστιτῷδὶςὑπὸτῶνΓΖ,ΖΔ.καίἐστιμέσοντὸ CF and FD. And thus, alternately, as the (square) on AB
δὶςὑπὸτῶνΑΕ,ΕΒ·μέσονἄρακαὶτὸδὶςὑπὸτῶνΓΖ, is to the (square) on CD, so (the sum of) the (squares) on
ΖΔ.αἱΓΖ,ΖΔἄραδυνάμειἀσύμμετροίεἰσιποιοῦσαιτὸ AE and EB (is) to (the sum of) the (squares) on CF and
μὲνσυγκείμενονἐκτῶνἀπ᾿αὐτῶντετραγώνωνἅμαῥητόν, FD [Prop. 5.16]. And the (square) on AB (is) commenτὸδὲδὶςὑπ᾿αὐτῶνμέσον·ὅληἄραἡΓΔἄλογόςἐστινἡ
surable with the (square) on CD. Thus, (the sum of) the
καλουμένημείζων.
(squares) on AE and EB (is) also commensurable with
῾Ηἄρατῇμείζονισύμμετροςμείζωνἐστίν· ὅπερἔδει (the sum of) the (squares) on CF and FD [Prop. 10.11].
δεῖξαι.
And the (squares) on AE and EB (added) together are
rational. The (squares) on CF and FD (added) together
(are) thus also rational. So, similarly, twice the (rectangle
contained) by AE and EB is also commensurable
with twice the (rectangle contained) by CF and FD. And
twice the (rectangle contained) by AE and EB is medial.
Therefore, twice the (rectangle contained) by CF
and FD (is) also medial [Prop. 10.23 corr.]. CF and FD
are thus (straight-lines which are) incommensurable in
square [Prop 10.13], simultaneously making the sum of
the squares on them rational, and twice the (rectangle
contained) by them medial. The whole, CD, is thus that
irrational (straight-line) called major [Prop. 10.39].
Thus, a (straight-line) commensurable (in length)
with a major (straight-line) is major. (Which is) the very
thing it was required to show.
Ü. Proposition 69
῾Ητῇῥητὸνκαὶμέσονδυναμένῃσύμμετρος[καὶαὐτὴ] A (straight-line) commensurable (in length) with the
ῥητὸνκαὶμέσονδυναμένηἐστίν.
square-root of a rational plus a medial (area) is [itself
also] the square-root of a rational plus a medial (area).
Α
Ε
Β
A E B
Γ Ζ ∆
C
F
D
῎Εστω ῥητὸν καὶ μέσον δυναμένη ἡ ΑΒ, καὶ τῇ ΑΒ Let AB be the square-root of a rational plus a medial
σύμμετροςἔστωἡΓΔ·δεικτέον,ὅτικαὶἡΓΔῥητὸνκαὶ (area), and let CD be commensurable (in length) with
μέσονδυναμένηἐστίν.
AB. We must show that CD is also the square-root of a
ΔιῃρήσθωἡΑΒεἰςτὰςεὐθείαςκατὰτὸΕ·αἱΑΕ,ΕΒ rational plus a medial (area).
ἄραδυνάμειεἰσὶνἀσύμμετροιποιοῦσαιτὸμὲνσυγκείμενον Let AB have been divided into its (component)
ἐκ τῶν ἀπ᾿ αὐτῶν τετραγώνων μέσον, τὸ δ᾿ ὑπ᾿ αὐτῶν straight-lines at E. AE and EB are thus incommensu-
ῥητόν·καὶτὰαὐτὰκατεσκευάσθωτοῖςπρότερον. ὁμοίως rable in square, making the sum of the squares on them
δὴδείξομεν,ὅτικαὶαἱΓΖ,ΖΔδυνάμειεἰσὶνἀσύμμετροι, medial, and the (rectangle contained) by them rational
καὶσύμμετροντὸμὲνσυγκείμενονἐκτῶνἀπὸτῶνΑΕ,ΕΒ [Prop. 10.40]. And let the same construction have been
τῷσυγκειμένῳἐκτῶνἀπὸτῶνΓΖ,ΖΔ,τὸδὲὑπὸΑΕ,ΕΒ made as in the previous (propositions). So, similarly, we
τῷὑπὸΓΖ,ΖΔ·ὥστεκαὶτὸ[μὲν]συγκείμενονἐκτῶνἀπὸ can show that CF and FD are also incommensurable
τῶνΓΖ,ΖΔτετραγώνωνἐστὶμέσον,τὸδ᾿ὑπὸτῶνΓΖ, in square, and that the sum of the (squares) on AE and
361

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ΖΔῥητόν.
EB (is) commensurable with the sum of the (squares)
῾ΡητὸνἄρακαὶμέσονδυναμένηἐστὶνἡΓΔ·ὅπερἔδει on CF and FD, and the (rectangle contained) by AE
δεῖξαι.
and EB with the (rectangle contained) by CF and FD.
And hence the sum of the squares on CF and FD is medial,
and the (rectangle contained) by CF and FD (is)
rational.
Thus, CD is the square-root of a rational plus a medial
(area) [Prop. 10.40]. (Which is) the very thing it was
required to show.
Ó. Proposition 70
῾Ητῇδύομέσαδυναμένῃσύμμετροςδύομέσαδυναμένη A (straight-line) commensurable (in length) with the
ἐστίν.
square-root of (the sum of) two medial (areas) is (itself
also) the square-root of (the sum of) two medial (areas).
Α
Ε
Β
A E B
Γ Ζ ∆
C
F
D
῎ΕστωδύομέσαδυναμένηἡΑΒ,καὶτῇΑΒσύμμετρος Let AB be the square-root of (the sum of) two medial
ἡΓΔ·δεικτέον,ὅτικαὶἡΓΔδύομέσαδυναμένηἐστίν. (areas), and (let) CD (be) commensurable (in length)
᾿ΕπεὶγὰρδύομέσαδυναμένηἐστὶνἡΑΒ,διῃρήσθω with AB. We must show that CD is also the square-root
εἰς τὰς εὐθείας κατὰ τὸ Ε· αἱ ΑΕ, ΕΒ ἄρα δυνάμει of (the sum of) two medial (areas).
εἰσὶνἀσύμμετροιποιοῦσαιτότεσυγκείμενονἐκτῶνἀπ᾿ For since AB is the square-root of (the sum of) two
αὐτῶν [τετραγώνων] μέσον καὶτὸ ὑπ᾿ αὐτῶν μέσον καὶ medial (areas), let it have been divided into its (compo-
ἔτιἀσύμμετροντὸσυγκείμενονἐκτῶνἀπὸτῶνΑΕ,ΕΒ nent) straight-lines at E. Thus, AE and EB are incomτετραγώνωντῷὑπὸτῶνΑΕ,ΕΒ·καὶκατεσκευάσθωτὰ
mensurable in square, making the sum of the [squares]
αὐτὰτοῖςπρότερον.ὁμοίωςδὴδείξομεν,ὅτικαὶαἱΓΖ,ΖΔ on them medial, and the (rectangle contained) by them
δυνάμειεἰσὶνἀσύμμετροικαὶσύμμετροντὸμὲνσυγκείμενον medial, and, moreover, the sum of the (squares) on AE
ἐκτῶνἀπὸτῶνΑΕ,ΕΒτῷσυγκειμένῳἐκτῶνἀπὸτῶνΓΖ, and EB incommensurable with the (rectangle) contained
ΖΔ,τὸδὲὑπὸτῶνΑΕ,ΕΒτῷὑπὸτῶνΓΖ,ΖΔ·ὥστεκαὶ by AE and EB [Prop. 10.41]. And let the same construcτὸσυγκείμενονἐκτῶνἀπὸτῶνΓΖ,ΖΔτετραγώνωνμέσον
tion have been made as in the previous (propositions).
ἐστὶκαὶτὸὑπὸτῶνΓΖ,ΖΔμέσονκαὶἔτιἀσύμμετροντὸ So, similarly, we can show that CF and FD are also
συγκείμενονἐκτῶνἀπὸτῶνΓΖ,ΖΔτετραγώνωντῷὑπὸ incommensurable in square, and (that) the sum of the
τῶνΓΖ,ΖΔ.
(squares) on AE and EB (is) commensurable with the
῾ΗἄραΓΔδύομέσαδυναμένηἐστίν·ὅπερἔδειδεῖξαι. sum of the (squares) on CF and FD, and the (rectangle
contained) by AE and EB with the (rectangle contained)
by CF and FD. Hence, the sum of the squares on CF
and FD is also medial, and the (rectangle contained) by
CF and FD (is) medial, and, moreover, the sum of the
squares on CF and FD (is) incommensurable with the
(rectangle contained) by CF and FD.
Thus, CD is the square-root of (the sum of) two medial
(areas) [Prop. 10.41]. (Which is) the very thing it
was required to show.
Ó. Proposition 71
῾Ρητοῦκαὶμέσουσυντιθεμένουτέσσαρεςἄλογοιγίγνον- When a rational and a medial (area) are added toταιἤτοιἐκδύοὀνομάτωνἢἐκδύομέσωνπρώτηἢμείζων
gether, four irrational (straight-lines) arise (as the square-
ἢῥητὸνκαὶμέσονδυναμένη.
roots of the total area)—either a binomial, or a first bi-
362

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
῎ΕστωῥητὸνμὲντὸΑΒ,μέσονδὲτὸΓΔ·λέγω,ὅτιἡ medial, or a major, or the square-root of a rational plus a
τὸΑΔχωρίονδυναμένηἤτοιἐκδύοὀνομάτωνἐστὶνἢἐκ medial (area).
δύομέσωνπρώτηἢμείζωνἢῥητὸνκαὶμέσονδυναμένη. Let AB be a rational (area), and CD a medial (area).
I say that the square-root of area AD is either binomial,
or first bimedial, or major, or the square-root of a rational
plus a medial (area).
Α Γ Ε Θ Κ A C E H K
Ζ Η Ι
F
G
I
Β ∆
Τὸ γὰρ ΑΒ τοῦ ΓΔ ἤτοι μεῖζόν ἐστιν ἢ ἔλασσον. For AB is either greater or less than CD. Let it, first
ἔστωπρότερονμεῖζον·καὶἐκκείσθωῥητὴἡΕΖ,καὶπαρα- of all, be greater. And let the rational (straight-line) EF
βεβλήσθωπαρὰτὴνΕΖτῷΑΒἴσοντὸΕΗπλάτοςποιοῦν be laid down. And let (the rectangle) EG, equal to AB,
τὴνΕΘ·τῷδὲΔΓἴσονπαρὰτὴνΕΖπαραβεβλήσθωτὸ have been applied to EF , producing EH as breadth. And
ΘΙπλάτοςποιοῦντὴνΘΚ.καὶἐπεὶῥητόνἐστιτὸΑΒκαί let (the recatangle) HI, equal to DC, have been ap-
ἐστινἴσοντῷΕΗ,ῥητὸνἄρακαὶτὸΕΗ.καὶπαρὰ[ῥητὴν] plied to EF , producing HK as breadth. And since AB
τὴνΕΖπαραβέβληταιπλάτοςποιοῦντὴνΕΘ·ἡΕΘἄρα is rational, and is equal to EG, EG is thus also rational.
ῥητήἐστικαὶσύμμετροςτῇΕΖμήκει. πάλιν,ἐπεὶμέσον And it has been applied to the [rational] (straight-line)
ἐστὶτὸΓΔκαίἐστινἴσοντῷΘΙ,μέσονἄραἐστὶκαὶτὸ EF , producing EH as breadth. EH is thus rational, and
ΘΙ.καὶπαρὰῥητὴντὴνΕΖπαράκειταιπλάτοςποιοῦντὴν commensurable in length with EF [Prop. 10.20]. Again,
ΘΚ·ῥητὴἄραἐστὶνἡΘΚκαὶἀσύμμετροςτῇΕΖμήκει.καὶ since CD is medial, and is equal to HI, HI is thus also
ἐπεὶμέσονἐστὶτὸΓΔ,ῥητὸνδὲτὸΑΒ,ἀσύμμετρονἄρα medial. And it is applied to the rational (straight-line)
ἐστὶτὸΑΒτῷΓΔ·ὥστεκαὶτὸΕΗἀσύμμετρόνἐστιτῷ EF , producing HK as breadth. HK is thus rational,
ΘΙ.ὡςδὲτὸΕΗπρὸςτὸΘΙ,οὕτωςἐστὶνἡΕΘπρὸςτὴν and incommensurable in length with EF [Prop. 10.22].
ΘΚ·ἀσύμμετροςἄραἐστὶκαὶἡΕΘτῇΘΚμήκει.καίεἰσιν And since CD is medial, and AB rational, AB is thus
ἀμφότεραιῥηταί·αἱΕΘ,ΘΚἄραῥηταίεἰσιδυνάμειμόνον incommensurable with CD. Hence, EG is also incomσύμμετροι·
ἐκδύοἄραὀνομάτωνἐστὶνἡΕΚ διῃρημένη mensurable with HI. And as EG (is) to HI, so EH is
κατὰτὸΘ.καὶἐπεὶμεῖζόνἐστιτὸΑΒτοῦΓΔ,ἴσονδὲτὸ to HK [Prop. 6.1]. Thus, EH is also incommensurable
μὲνΑΒτῷΕΗ,τὸδὲΓΔτῷΘΙ,μεῖζονἄρακαὶτὸΕΗ in length with HK [Prop. 10.11]. And they are both raτοῦΘΙ·καὶἡΕΘἄραμείζωνἐστὶτῆςΘΚ.ἤτοιοὖνἡΕΘ
tional. Thus, EH and HK are rational (straight-lines
τῆςΘΚμεῖζονδύναταιτῷἀπὸσυμμέτρουἑαυτῇμήκειἢ which are) commensurable in square only. EK is thus
τῷἀπὸἀσυμμέτρου.δυνάσθωπρότεροντῷἀπὸσυμμέτρου a binomial (straight-line), having been divided (into its
ἑαυτῇ·καίἐστινἡμείζωνἡΘΕσύμμετροςτῇἐκκειμένῃ component terms) at H [Prop. 10.36]. And since AB
ῥητῃτῇΕΖ·ἡἄραΕΚἐκδύοὀνομάτωνἐστὶπρώτη.ῥητὴ is greater than CD, and AB (is) equal to EG, and CD
δὲἡΕΖ·ἐὰνδὲχωρίονπεριέχηταιὑπὸῥητῆςκαὶτῆςἐκδύο to HI, EG (is) thus also greater than HI. Thus, EH is
ὀνομάτωνπρώτης,ἡτὸχωρίονδυναμένηἐκδύοὀνομάτων also greater than HK [Prop. 5.14]. Therefore, the square
ἐστίν.ἡἄρατὸΕΙδυναμένηἐκδύοὀνομάτωνἐστίν·ὥστε on EH is greater than (the square on) HK either by
καὶἡτὸΑΔδυναμένηἐκδύοὀνομάτωνἐστίν. ἀλλὰδὴ the (square) on (some straight-line) commensurable in
δυνάσθωἡΕΘτῆςΘΚμεῖζοντῷἀπὸἀσυμμέτρουἑαυτῇ· length with (EH), or by the (square) on (some straightκαίἐστινἡμείζωνἡΕΘσύμμετροςτῇἐκκειμένῃῥητῇτῇ
line) incommensurable (in length with EH). Let it, first
ΕΖμήκει·ἡἄραΕΚἐκδύοὀνομάτωνἐστὶτετάρτη. ῥητὴ of all, be greater by the (square) on (some straight-line)
δὲἡΕΖ·ἐὰνδὲχωρίονπεριέχηταιὑπὸῥητῆςκαὶτῆςἐκδύο commensurable (in length with EH). And the greater
B
D
363

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ὀνομάτωντετάρτης,ἡτὸχωρίονδυναμένηἄλογόςἐστινἡ (of the two components of EK) HE is commensurable
καλουμένημείζων. ἡἄρατὸΕΙχωρίονδυναμένημείζων (in length) with the (previously) laid down (straight-
ἐστίν·ὥστεκαὶἡτὸΑΔδυναμένημείζωνἐστίν. line) EF . EK is thus a first binomial (straight-line)
Ἀλλὰ δὴ ἔστω ἔλασσον τὸ ΑΒ τοῦ ΓΔ· καὶ τὸ ΕΗ [Def. 10.5]. And EF (is) rational. And if an area is con-
ἄραἔλασσόνἐστιτοῦΘΙ·ὥστεκαὶἡΕΘἐλάσσωνἐστὶ tained by a rational (straight-line) and a first binomial
τῆς ΘΚ. ἤτοιδὲ ἡΘΚ τῆςΕΘ μεῖζον δύναταιτῷ ἀπὸ (straight-line) then the square-root of the area is a binoσυμμέτρουἑαυτῇἢτῷἀπὸἀσυμμέτρου.δυνάσθωπρότερον
mial (straight-line) [Prop. 10.54]. Thus, the square-root
τῷἀπὸσυμμέτρουἑαυτῇμήκει·καίἐστινἡἐλάσσωνἡΕΘ of EI is a binomial (straight-line). Hence the squareσύμμετροςτῇἐκκειμένῃῥητῇτῇΕΖμήκει·ἡἄραΕΚἐκ
root of AD is also a binomial (straight-line). And, so, let
δύοὀνομάτωνἐστὶδευτέρα. ῥητὴδὲἡΕΖ·ἐὰνδὲχωρίον the square on EH be greater than (the square on) HK
περιέχηταιὑπὸῥητῆςκαὶτῆςἐκδύοὀνομάτωνδευτέρας,ἡ by the (square) on (some straight-line) incommensurable
τὸχωρίονδυναμένηἐκδύομέσωνἐστὶπρώτη.ἡἄρατὸΕΙ (in length) with (EH). And the greater (of the two comχωρίονδυναμένηἐκδύομέσωνἐστὶπρώτη·ὥστεκαὶἡτὸ
ponents of EK) EH is commensurable in length with the
ΑΔδυναμένηἐκδύομέσωνἐστὶπρώτη.ἀλλὰδὴἡΘΚτῆς (previously) laid down rational (straight-line) EF . Thus,
ΘΕμεῖζονδυνάσθωτῷἀπὸἀσυμμέτρουἑαυτῇ. καίἐστιν EK is a fourth binomial (straight-line) [Def. 10.8]. And
ἡἐλάσσωνἡΕΘσύμμετροςτῇἐκκειμένῃῥητῇτῇΕΖ·ἡ EF (is) rational. And if an area is contained by a rational
ἄραΕΚἐκδύοὀνομάτωνἐστὶπέμπτη. ῥητὴδὲἡΕΖ·ἐὰν (straight-line) and a fourth binomial (straight-line) then
δὲχωρίονπεριέχηταιὑπὸῥητῆςκαὶτῆςἐκδύοὀνομάτων the square-root of the area is the irrational (straight-line)
πέμπτης,ἡτὸχωρίονδυναμένηῥητὸνκαὶμέσονδυναμένη called major [Prop. 10.57]. Thus, the square-root of area
ἐστίν. ἡἄρατὸΕΙχωρίονδυναμένηῥητὸνκαὶμέσονδυ- EI is a major (straight-line). Hence, the square-root of
ναμένηἐστίν·ὥστεκαὶἡτὸΑΔχωρίονδυναμένηῥητὸν AD is also major.
καὶμέσονδυναμένηἐστίν.
And so, let AB be less than CD. Thus, EG is also less
῾Ρητοῦἄρακαὶμέσουσυντιθεμένουτέσσαρεςἄλογοι than HI. Hence, EH is also less than HK [Props. 6.1,
γίγνονταιἤτοιἐκδύοὀνομάτωνἢἐκδύομέσωνπρώτηἢ 5.14].
μείζωνἢῥητὸνκαὶμέσονδυναμένη·ὅπερἔδειδεῖξαι.
And the square on HK is greater than (the
square on) EH either by the (square) on (some straightline)
commensurable (in length) with (HK), or by the
(square) on (some straight-line) incommensurable (in
length) with (HK). Let it, first of all, be greater by the
square on (some straight-line) commensurable in length
with (HK). And the lesser (of the two components of
EK) EH is commensurable in length with the (previously)
laid down rational (straight-line) EF . Thus, EK
is a second binomial (straight-line) [Def. 10.6]. And EF
(is) rational. And if an area is contained by a rational
(straight-line) and a second binomial (straight-line) then
the square-root of the area is a first bimedial (straightline)
[Prop. 10.55]. Thus, the square-root of area EI is
a first bimedial (straight-line). Hence, the square-root of
AD is also a first bimedial (straight-line). And so, let
the square on HK be greater than (the square on) HE
by the (square) on (some straight-line) incommensurable
(in length) with (HK). And the lesser (of the two components
of EK) EH is commensurable (in length) with the
(previously) laid down rational (straight-line) EF . Thus,
EK is a fifth binomial (straight-line) [Def. 10.9]. And
EF (is) rational. And if an area is contained by a rational
(straight-line) and a fifth binomial (straight-line) then
the square-root of the area is the square-root of a rational
plus a medial (area) [Prop. 10.58]. Thus, the square-root
of area EI is the square-root of a rational plus a medial
(area). Hence, the square-root of area AD is also the
364

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
square-root of a rational plus a medial (area).
Thus, when a rational and a medial area are added together,
four irrational (straight-lines) arise (as the squareroots
of the total area)—either a binomial, or a first bimedial,
or a major, or the square-root of a rational plus a
medial (area). (Which is) the very thing it was required
to show.
Ü. Proposition 72
Δύο μέσων ἀσυμμέτρων ἀλλήλοις συντιθεμένων αἱ When two medial (areas which are) incommensuλοιπαὶδύοἄλογοιγίγνονταιἤτοιἐκδύομέσωνδευτέρα
rable with one another are added together, the remaining
ἢ[ἡ]δύομέσαδυναμένη.
two irrational (straight-lines) arise (as the square-roots of
the total area)—either a second bimedial, or the squareroot
of (the sum of) two medial (areas).
Α Γ Ε Θ Κ A C E H K
Ζ Η Ι
F
G
I
Β ∆
ΣυγκείσθωγὰρδύομέσαἀσύμμετραἀλλήλοιςτὰΑΒ, For let the two medial (areas) AB and CD, (which
ΓΔ·λέγω,ὅτιἡτὸΑΔχωρίονδυναμένηἤτοιἐκδύομέσων are) incommensurable with one another, have been
ἐστὶδευτέραἢδύομέσαδυναμένη.
added together. I say that the square-root of area AD
ΤὸγὰρΑΒτοῦΓΔἤτοιμεῖζόνἐστινἢἔλασσον.ἔστω, is either a second bimedial, or the square-root of (the
εἰτύχον,πρότερονμεῖζοντὸΑΒτοῦΓΔ·καὶἐκκείσθω sum of) two medial (areas).
ῥητὴἡΕΖ,καὶτῷμὲνΑΒἴσονπαρὰτὴνΕΖπαραβεβλήσθω For AB is either greater than or less than CD. By
τὸΕΗπλάτοςποιοῦντὴνΕΘ,τῷδὲΓΔἴσοντὸΘΙπλάτος chance, let AB, first of all, be greater than CD. And
ποιοῦντὴνΘΚ.καὶἐπεὶμέσονἐστὶνἑκάτεροντῶνΑΒ,ΓΔ, let the rational (straight-line) EF be laid down. And let
μέσονἄρακαὶἐκάτεροντῶνΕΗ,ΘΙ.καὶπαρὰῥητὴντὴν EG, equal to AB, have been applied to EF , producing
ΖΕπαράκειταιπλάτοςποιοῦντὰςΕΘ,ΘΚ·ἑκατέραἄρατῶν EH as breadth, and HI, equal to CD, producing HK
ΕΘ,ΘΚῥητήἐστικαὶἀσύμμετροςτῇΕΖμήκει. καὶἐπεὶ as breadth. And since AB and CD are each medial, EG
ἀσύμμετρόνἐστιτὸΑΒτῷΓΔ,καίἐστινἴσοντὸμὲνΑΒ and HI (are) thus also each medial. And they are apτῷΕΗ,τὸδὲΓΔτῷΘΙ,ἀσύμμετρονἄραἐστὶκαὶτὸΕΗτῷ
plied to the rational straight-line FE, producing EH and
ΘΙ.ὡςδὲτὸΕΗπρὸςτὸΘΙ,οὕτωςἐστὶνἡΕΘπρὸςΘΚ· HK (respectively) as breadth. Thus, EH and HK are
ἀσύμμετροςἄραἐστὶνἡΕΘτῇΘΚμήκει.αἱΕΘ,ΘΚἄρα each rational (straight-lines which are) incommensurable
ῥηταίεἰσιδυνάμειμόνονσύμμετροι·ἐκδύοἄραὀνομάτων in length with EF [Prop. 10.22]. And since AB is incom-
ἐστὶνἡΕΚ.ἤτοιδὲἡΕΘτῆςΘΚμεῖζονδύναταιτῷἀπὸ mensurable with CD, and AB is equal to EG, and CD
συμμέτρουἑαυτῇἢτῷἀπὸἀσυμμέτρου.δυνάσθωπρότερον to HI, EG is thus also incommensurable with HI. And
τῷἀπὸσυμμέτρουἑαυτῇμήκει·καὶοὐδετέρατῶνΕΘ,ΘΚ as EG (is) to HI, so EH is to HK [Prop. 6.1]. EH is
σύμμετρόςἐστιτῇἐκκειμένῃῥητῇτῇΕΖμήκει·ἡΕΚἄρα thus incommensurable in length with HK [Prop. 10.11].
ἐκδύοὀνομάτωνἐστὶτρίτη.ῥητὴδὲἡΕΖ·ἐὰνδὲχωρίον Thus, EH and HK are rational (straight-lines which are)
περιέχηταιὑπὸῥητῆςκαὶτῆςἐκδύοὀνομάτωντρίτης,ἡ commensurable in square only. EK is thus a binomial
τὸχωρίονδυναμένηἐκδύομέσωνἐστὶδευτέρα·ἡἄρατὸ (straight-line) [Prop. 10.36]. And the square on EH is
ΕΙ,τουτέστιτὸΑΔ,δυναμένηἐκδύομέσωνἐστὶδευτέρα. greater than (the square on) HK either by the (square)
B
D
365

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ἀλλαδὴἡΕΘτῆςΘΚμεῖζονδυνάσθωτῷἀπὸἀσυμμέτρου on (some straight-line) commensurable (in length) with
ἑαυτῇμήκει·καὶἀσύμμετρόςἐστινἑκατέρατῶνΕΘ,ΘΚ (EH), or by the (square) on (some straight-line) incomτῇΕΖμήκει·ἡἄραΕΚἐκδύοὀνομάτωνἐστὶνἕκτη.
ἐὰν mensurable (in length with EH). Let it, first of all, be
δὲχωρίονπεριέχηταιὑπὸῥητῆςκαὶτῆςἐκδύοὀνομάτων greater by the square on (some straight-line) commensu-
ἕκτης,ἡτὸχωρίονδυναμένηἡδύομέσαδυναμένηἐστίν· rable in length with (EH). And neither of EH or HK is
ὥστεκαὶἡτὸΑΔχωρίονδυναμένηἡδύομέσαδυναμένη commensurable in length with the (previously) laid down
ἐστίν.
rational (straight-line) EF . Thus, EK is a third binomial
[῾Ομοίωςδὴδείξομεν,ὅτικἂνἔλαττονᾖτὸΑΒτοῦΓΔ, (straight-line) [Def. 10.7]. And EF (is) rational. And if
ἡτὸΑΔχωρίονδυναμένηἢἐκδύομέσωνδευτέραἐστὶν an area is contained by a rational (straight-line) and a
ἤτοιδύομέσαδυναμένη].
third binomial (straight-line) then the square-root of the
Δύοἄραμέσωνἀσυμμέτρωνἀλλήλοιςσυντιθεμένωναἱ area is a second bimedial (straight-line) [Prop. 10.56].
λοιπαὶδύοἄλογοιγίγνονταιἤτοιἐκδύομέσωνδευτέραἢ Thus, the square-root of EI—that is to say, of AD—
δύομέσαδυναμένη. is a second bimedial. And so, let the square on EH
be greater than (the square) on HK by the (square)
᾿Η ἐκ δύο ὀνομάτων καὶ αἱ μετ᾿ αὐτὴν ἄλογοι οὔτε on (some straight-line) incommensurable in length with
τῇ μέσῃ οὔτε ἀλλήλαιςεἰσὶν αἱ αὐταί. τὸ μὲν γὰρ ἀπὸ (EH). And EH and HK are each incommensurable in
μέσηςπαρὰῥητὴνπαραβαλλόμενονπλάτοςποιεῖῥητὴνκαὶ length with EF . Thus, EK is a sixth binomial (straight-
ἀσύμμετροντῇπαρ᾿ἣνπαράκειταιμήκει. τὸδὲἀπὸτῆς line) [Def. 10.10]. And if an area is contained by a ra-
ἐκδύοὀνομάτωνπαρὰῥητὴνπαραβαλλόμενονπλάτοςποιεῖ tional (straight-line) and a sixth binomial (straight-line)
τὴν ἐκ δύο ὀνομάτων πρώτην. τὸ δὲ ἀπὸ τῆς ἐκ δύο then the square-root of the area is the square-root of (the
μέσωνπρώτηςπαρὰῥητὴνπαραβαλλόμενονπλάτοςποιεῖ sum of) two medial (areas) [Prop. 10.59]. Hence, the
τὴν ἐκ δύοὀνομάτων δευτέραν. τὸ δὲ ἀπὸτῆςἐκδύο square-root of area AD is also the square-root of (the
μέσωνδευτέραςπαρὰῥητὴνπαραβαλλόμενονπλάτοςποιεῖ sum of) two medial (areas).
τὴνἐκδύοὀνομάτωντρίτην.τὸδὲἀπὸτῆςμείζονοςπαρὰ [So, similarly, we can show that, even if AB is less
ῥητὴνπαραβαλλόμενονπλάτοςποιεῖτὴνἐκδύοὀνομάτων than CD, the square-root of area AD is either a second
τετάρτην. τὸδὲἀπὸτῆςῥητὸνκαὶμέσονδυναμένηςπαρὰ bimedial or the square-root of (the sum of) two medial
ῥητὴνπαραβαλλόμενονπλάτοςποιεῖτὴνἐκδύοὀνομάτων (areas).]
πέμπτην. τὸδὲἀπὸτῆςδύομέσαδυναμένηςπαρὰῥητὴν Thus, when two medial (areas which are) incommenπαραβαλλόμενονπλάτοςποιεῖτὴνἐκδύοὀνομάτωνἕκτην.
surable with one another are added together, the remainτὰδ᾿εἰρημέναπλάτηδιαφέρειτοῦτεπρώτουκαὶἀλλήλων,
ing two irrational (straight-lines) arise (as the squareτοῦμὲνπρώτου,ὅτιῥητήἐστιν,ἀλλήλωνδέ,ὅτιτῇτάξει
roots of the total area)—either a second bimedial, or the
οὐκεἰσὶναἱαὐταί·ὥστεκαὶαὐταὶαἱἄλογοιδιαφέρουσιν square-root of (the sum of) two medial (areas).
ἀλλήλων.
A binomial (straight-line), and the (other) irrational
(straight-lines) after it, are neither the same as a medial
(straight-line) nor (the same) as one another. For the
(square) on a medial (straight-line), applied to a rational
(straight-line), produces as breadth a rational (straightline
which is) also incommensurable in length with (the
straight-line) to which it is applied [Prop. 10.22]. And
the (square) on a binomial (straight-line), applied to a
rational (straight-line), produces as breadth a first binomial
[Prop. 10.60]. And the (square) on a first bimedial
(straight-line), applied to a rational (straight-line), produces
as breadth a second binomial [Prop. 10.61]. And
the (square) on a second bimedial (straight-line), applied
to a rational (straight-line), produces as breadth a third
binomial [Prop. 10.62]. And the (square) on a major
(straight-line), applied to a rational (straight-line), produces
as breadth a fourth binomial [Prop. 10.63]. And
the (square) on the square-root of a rational plus a medial
366

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
(area), applied to a rational (straight-line), produces as
breadth a fifth binomial [Prop. 10.64]. And the (square)
on the square-root of (the sum of) two medial (areas),
applied to a rational (straight-line), produces as breadth
a sixth binomial [Prop. 10.65]. And the aforementioned
breadths differ from the first (breadth), and from one
another—from the first, because it is rational—and from
one another, because they are not the same in order.
Hence, the (previously mentioned) irrational (straightlines)
themselves also differ from one another.
Ó. Proposition 73
᾿Εὰνἀπὸῥητῆςῥητὴἀφαιρεθῇδυνάμειμόνονσύμμετρος If a rational (straight-line), which is commensuοὖσατῇὅλῃ,ἡλοιπὴἄλογόςἐστιν·καλείσθωδὲἀποτομή.
rable in square only with the whole, is subtracted from
a(nother) rational (straight-line) then the remainder is
an irrational (straight-line). Let it be called an apotome.
Α
Γ
Β
ἈπὸγὰρῥητῆςτῆςΑΒῥητὴἀφῃρήσθωἡΒΓδυνάμει For let the rational (straight-line) BC, which comμόνονσύμμετροςοὖσατῇὅλῃ·
λέγω,ὅτιἡλοιπὴἡΑΓ mensurable in square only with the whole, have been
ἄλογόςἐστινἡκαλουμένηἀποτομή.
subtracted from the rational (straight-line) AB. I say that
᾿ΕπεὶγὰρἀσύμμετρόςἐστινἡΑΒτῇΒΓμήκει,καίἐστιν the remainder AC is that irrational (straight-line) called
ὡςἡΑΒπρὸςτὴνΒΓ,οὕτωςτὸἀπὸτῆςΑΒπρὸςτὸὑπὸ an apotome.
τῶνΑΒ,ΒΓ,ἀσύμμετρονἄραἐστὶτὸἀπὸτῆςΑΒτῷὑπὸ For since AB is incommensurable in length with BC,
τῶνΑΒ,ΒΓ.ἀλλὰτῷμὲνἀπὸτῆςΑΒσύμμετράἐστιτὰἀπὸ and as AB is to BC, so the (square) on AB (is) to the
τῶνΑΒ,ΒΓτετράγωνα,τῷδὲὑπὸτῶνΑΒ,ΒΓσύμμετρόν (rectangle contained) by AB and BC [Prop. 10.21 lem.],
ἐστιτὸδὶςὑπὸτῶνΑΒ,ΒΓ.καὶἐπειδήπερτὰἀπὸτῶνΑΒ, the (square) on AB is thus incommensurable with the
ΒΓἴσαἐστὶτῷδὶςὑπὸτῶνΑΒ,ΒΓμετὰτοῦἀπὸΓΑ,καὶ (rectangle contained) by AB and BC [Prop. 10.11]. But,
λοιπῷἄρατῷἀπὸτῆςΑΓἀσύμμετράἐστιτὰἀπὸτῶνΑΒ, the (sum of the) squares on AB and BC is commen-
ΒΓ.ῥητὰδὲτὰἀπὸτῶνΑΒ,ΒΓ·ἄλογοςἄραἐστὶνἡΑΓ· surable with the (square) on AB [Prop. 10.15], and
καλείσθωδὲἀποτομή.ὅπερἔδειδεῖξαι.
twice the (rectangle contained) by AB and BC is commensurable
with the (rectangle contained) by AB and
BC [Prop. 10.6]. And, inasmuch as the (sum of the
squares) on AB and BC is equal to twice the (rectangle
contained) by AB and BC plus the (square) on CA
[Prop. 2.7], the (sum of the squares) on AB and BC is
thus also incommensurable with the remaining (square)
on AC [Props. 10.13, 10.16]. And the (sum of the
squares) on AB and BC is rational. AC is thus an irrational
(straight-line) [Def. 10.4]. And let it be called
an apotome. † (Which is) the very thing it was required to
show.
A
C
B
† See footnote to Prop. 10.36.Ó. Proposition 74
᾿Εὰνἀπὸμέσηςμέσηἀφαιρεθῇδυνάμειμόνονσύμμετρος If a medial (straight-line), which is commensurable in
οὖσατῇὅλῃ,μετὰδὲτῆςὅληςῥητὸνπεριέχουσα,ἡλοιπὴ square only with the whole, and which contains a ratio-
ἄλογόςἐστιν·καλείσθωδὲμέσηςἀποτομὴπρώτη. nal (area) with the whole, is subtracted from a(nother)
medial (straight-line) then the remainder is an irrational
367

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
Α
Γ
Β
(straight-line). Let it be called a first apotome of a medial
(straight-line).
A
C
B
ἈπὸγὰρμέσηςτῆςΑΒμέσηἀφῃρήσθωἡΒΓδυνάμει For let the medial (straight-line) BC, which is comμόνον
σύμμετρος οὖσα τῇ ΑΒ, μετὰ δὲ τῆς ΑΒ ῥητὸν mensurable in square only with AB, and which makes
ποιοῦσατὸ ὑπὸτῶν ΑΒ, ΒΓ· λέγω, ὅτι ἡλοιπὴἡΑΓ with AB the rational (rectangle contained) by AB and
ἄλογόςἐστιν·καλείσθωδὲμέσηςἀποτομὴπρώτη. BC, have been subtracted from the medial (straight-line)
᾿ΕπεὶγὰραἱΑΒ,ΒΓμέσαιεἰσίν,μέσαἐστὶκαὶτὰἀπὸ AB [Prop. 10.27]. I say that the remainder AC is an irτῶνΑΒ,ΒΓ.ῥητὸνδὲτὸδὶςὑπὸτῶνΑΒ,ΒΓ·ἀσύμμετρα
rational (straight-line). Let it be called the first apotome
ἄρατὰἀπὸτῶνΑΒ,ΒΓτῷδὶςὑπὸτῶνΑΒ,ΒΓ·καὶλοιπῷ of a medial (straight-line).
ἄρατῷἀπὸτῆςΑΓἀσύμμετρόνἐστιτὸδὶςὑπὸτῶνΑΒ, For since AB and BC are medial (straight-lines), the
ΒΓ,ἐπεὶκἂντὸὅλονἑνὶαὐτῶνἀσύμμετρονᾖ,καὶτὰἐξ (sum of the squares) on AB and BC is also medial. And
ἀρχῆςμεγέθηἀσύμμετραἔσται. ῥητὸνδὲτὸδὶςὑπὸτῶν twice the (rectangle contained) by AB and BC (is) ratio-
ΑΒ,ΒΓ·ἄλογονἄρατὸἀπὸτῆςΑΓ·ἄλογοςἄραἐστὶνἡ nal. The (sum of the squares) on AB and BC (is) thus in-
ΑΓ·καλείσθωδὲμέσηςἀποτομὴπρώτη.
commensurable with twice the (rectangle contained) by
AB and BC. Thus, twice the (rectangle contained) by
AB and BC is also incommensurable with the remaining
(square) on AC [Prop. 2.7], since if the whole is incommensurable
with one of the (constituent magnitudes)
then the original magnitudes will also be incommensurable
(with one another) [Prop. 10.16]. And twice the
(rectangle contained) by AB and BC (is) rational. Thus,
the (square) on AC is irrational. Thus, AC is an irrational
(straight-line) [Def. 10.4]. Let it be called a first
apotome of a medial (straight-line). †
† See footnote to Prop. 10.37.Ó. Proposition 75
᾿Εὰνἀπὸμέσηςμέσηἀφαιρεθῇδυνάμειμόνονσύμμετρος If a medial (straight-line), which is commensurable in
οὖσατῇὅλη,μετὰδὲτῆςὅληςμέσονπεριέχουσα,ἡλοιπὴ square only with the whole, and which contains a me-
ἄλογόςἐστιν·καλείσθωδὲμέσηςἀποτομὴδευτέρα. dial (area) with the whole, is subtracted from a(nother)
ἈπὸγὰρμέσηςτῆςΑΒμέσηἀφῃρήσθωἡΓΒδυνάμει medial (straight-line) then the remainder is an irrational
μόνονσύμμετροςοὖσατῇὅλῃτῇΑΒ,μετὰδὲτῆςὅλης (straight-line). Let it be called a second apotome of a
τῆςΑΒμέσονπεριέχουσατὸὑπὸτῶνΑΒ,ΒΓ·λέγω,ὅτι medial (straight-line).
ἡλοιπὴἡΑΓἄλογόςἐστιν·καλείσθωδὲμέσηςἀποτομὴ For let the medial (straight-line) CB, which is comδευτέρα.
mensurable in square only with the whole, AB, and
which contains with the whole, AB, the medial (rectangle
contained) by AB and BC, have been subtracted
from the medial (straight-line) AB [Prop. 10.28]. I say
that the remainder AC is an irrational (straight-line). Let
it be called a second apotome of a medial (straight-line).
368

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
Α
Γ
Β
A
C
B
∆
Ζ
Η
D
F
G
Ι
Θ
Ε
᾿ΕκκείσθωγὰρῥητὴἡΔΙ,καὶτοῖςμὲνἀπὸτῶνΑΒ, For let the rational (straight-line) DI be laid down.
ΒΓἴσονπαρὰτὴνΔΙπαραβεβλήσθωτὸΔΕπλάτοςποιοῦν And let DE, equal to the (sum of the squares) on AB
τὴνΔΗ,τῷδὲδὶςὑπὸτῶνΑΒ,ΒΓἴσονπαρὰτὴνΔΙ and BC, have been applied to DI, producing DG as
παραβεβλήσθωτὸΔΘπλάτοςποιοῦντὴνΔΖ·λοιπὸνἄρα breadth. And let DH, equal to twice the (rectangle conτὸΖΕἴσονἐστὶτῷἀπὸτῆςΑΓ.καὶἐπεὶμέσακαὶσύμμετρά
tained) by AB and BC, have been applied to DI, produc-
ἐστιτὰἀπὸτῶνΑΒ,ΒΓ,μέσονἄρακαὶτὸΔΕ.καὶπαρὰ ing DF as breadth. The remainder FE is thus equal to
ῥητὴντὴνΔΙπαράκειταιπλάτοςποιοῦντὴνΔΗ·ῥητὴἄρα the (square) on AC [Prop. 2.7]. And since the (squares)
ἐστὶνἡΔΗκαὶἀσύμμετροςτῇΔΙμήκει.πάλιν,ἐπεὶμέσον on AB and BC are medial and commensurable (with
ἐστὶτὸὑπὸτῶνΑΒ,ΒΓ,καὶτὸδὶςἄραὑπὸτῶνΑΒ, one another), DE (is) thus also medial [Props. 10.15,
ΒΓμέσονἐστίν. καίἐστινἴσοντῷΔΘ·καὶτὸΔΘἄρα 10.23 corr.]. And it is applied to the rational (straightμέσονἐστίν.
καὶπαρὰῥητὴντὴνΔΙπαραβέβληταιπλάτος line) DI, producing DG as breadth. Thus, DG is rational,
ποιοῦντὴνΔΖ·ῥητὴἄραἐστὶνἡΔΖκαὶἀσύμμετροςτῇΔΙ and incommensurable in length with DI [Prop. 10.22].
μήκει. καὶἐπεὶαἱΑΒ,ΒΓδυνάμειμόνονσύμμετροίεἰσιν, Again, since the (rectangle contained) by AB and BC is
ἀσύμμετροςἄραἐστὶνἡΑΒτῇΒΓμήκει·ἀσύμμετρονἄρα medial, twice the (rectangle contained) by AB and BC
καὶτὸἀπὸτῆςΑΒτετράγωνοντῷὑπὸτῶνΑΒ,ΒΓ.ἀλλὰ is thus also medial [Prop. 10.23 corr.]. And it is equal
τῷμὲνἀπὸτῆςΑΒσύμμετράἐστιτὰἀπὸτῶνΑΒ,ΒΓ,τῷ to DH. Thus, DH is also medial. And it has been apδὲὑπὸτῶνΑΒ,ΒΓσύμμετρόνἐστιτὸδὶςὑπὸτῶνΑΒ,
plied to the rational (straight-line) DI, producing DF as
ΒΓ·ἀσύμμετρονἄραἐστὶτὸδὶςὑπὸτῶνΑΒ,ΒΓτοῖςἀπὸ breadth. DF is thus rational, and incommensurable in
τῶνΑΒ,ΒΓ.ἴσονδὲτοῖςμὲνἀπὸτῶνΑΒ,ΒΓτὸΔΕ, length with DI [Prop. 10.22]. And since AB and BC are
τῷδὲδὶςὑπὸτῶνΑΒ,ΒΓτὸΔΘ·ἀσύμμετρονἄρα[ἐστὶ] commensurable in square only, AB is thus incommensuτὸΔΕτῷΔΘ.ὡςδὲτὸΔΕπρὸςτὸΔΘ,οὕτωςἡΗΔ
rable in length with BC. Thus, the square on AB (is)
πρὸςτὴνΔΖ·ἀσύμμετροςἄραἐστὶνἡΗΔτῇΔΖ.καίεἰσιν also incommensurable with the (rectangle contained) by
ἀμφότεραιῥηταί·αἱἄραΗΔ,ΔΖῥηταίεἰσιδυνάμειμόνον AB and BC [Props. 10.21 lem., 10.11]. But, the (sum
σύμμετροι·ἡΖΗἄραἀποτομήἐστιν. ῥητὴδὲἡΔΙ·τὸδὲ of the squares) on AB and BC is commensurable with
ὑπὸῥητῆςκαὶἀλόγουπεριεχόμενονἄλογόνἐστιν,καὶἡ the (square) on AB [Prop. 10.15], and twice the (rectanδυναμένηαὐτὸἄλογόςἐστιν.
καὶδύναταιτὸΖΕἡΑΓ·ἡ gle contained) by AB and BC is commensurable with the
ΑΓἄραἄλογόςἐστιν·καλείσθωδὲμέσηςἀποτομὴδευτέρα. (rectangle contained) by AB and BC [Prop. 10.6]. Thus,
ὅπερἔδειδεῖξαι.
twice the (rectangle contained) by AB and BC is incommensurable
with the (sum of the squares) on AB and
BC [Prop. 10.13]. And DE is equal to the (sum of the
squares) on AB and BC, and DH to twice the (rectangle
contained) by AB and BC. Thus, DE [is] incommensurable
with DH. And as DE (is) to DH, so GD (is) to
DF [Prop. 6.1]. Thus, GD is incommensurable with DF
[Prop. 10.11]. And they are both rational (straight-lines).
Thus, GD and DF are rational (straight-lines which are)
commensurable in square only. Thus, FG is an apotome
[Prop. 10.73]. And DI (is) rational. And the (area) contained
by a rational and an irrational (straight-line) is
irrational [Prop. 10.20], and its square-root is irrational.
I
H
E
369

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
And AC is the square-root of FE. Thus, AC is an irrational
(straight-line) [Def. 10.4]. And let it be called the
second apotome of a medial (straight-line). † (Which is)
the very thing it was required to show.
† See footnote to Prop. 10.38.Ó¦. Proposition 76
᾿Εὰνἀπὸεὐθείαςεὐθεῖαἀφαιρεθῂδυνάμειἀσύμμετρος If a straight-line, which is incommensurable in square
οὖσατῇὅλῃ,μετὰδὲτῆςὅληςποιοῦσατὰμὲνἀπ᾿αὐτῶν with the whole, and with the whole makes the (squares)
ἅμαῥητόν,τὸδ᾿ὑπ᾿αὐτῶνμέσον,ἡλοιπὴἄλογόςἐστιν· on them (added) together rational, and the (rectangle
καλείσθωδὲἐλάσσων.
contained) by them medial, is subtracted from a(nother)
straight-line then the remainder is an irrational (straightline).
Let it be called a minor (straight-line).
Α
Γ
Β
Ἀπὸ γὰρ εὐθείας τῆς ΑΒ εὐθεῖα ἀφῃρήσθω ἡ ΒΓ For let the straight-line BC, which is incommensuδυνάμειἀσύμμετροςοὖσατῇὅλῃποιοῦσατὰπροκείμενα.
rable in square with the whole, and fulfils the (other)
λέγω,ὅτιἡλοιπὴἡΑΓἄλογόςἐστινἡκαλουμένηἐλάσσων. prescribed (conditions), have been subtracted from the
᾿ΕπεὶγὰρτὸμὲνσυγκείμενονἐκτῶνἀπὸτῶνΑΒ,ΒΓ straight-line AB [Prop. 10.33]. I say that the remainder
τετραγώνωνῥητόνἐστιν,τὸδὲδὶςὑπὸτῶνΑΒ,ΒΓμέσον, AC is that irrational (straight-line) called minor.
ἀσύμμετραἄραἐστὶτὰἀπὸτῶνΑΒ,ΒΓτῷδὶςὑπὸτῶν For since the sum of the squares on AB and BC is
ΑΒ,ΒΓ·καὶἀναστρέψαντιλοιπῷτῷἀπὸτῆςΑΓἀσύμμετρά rational, and twice the (rectangle contained) by AB and
ἐστιτὰἀπὸτῶνΑΒ,ΒΓ.ῥητὰδὲτὰἀπὸτῶνΑΒ,ΒΓ· BC (is) medial, the (sum of the squares) on AB and BC
ἄλογονἄρατὸἀπὸτῆςΑΓ·ἄλογοςἄραἡΑΓ·καλείσθωδὲ is thus incommensurable with twice the (rectangle con-
ἐλάσσων.ὅπερἔδειδεῖξαι.
tained) by AB and BC. And, via conversion, the (sum
of the squares) on AB and BC is incommensurable with
the remaining (square) on AC [Props. 2.7, 10.16]. And
the (sum of the squares) on AB and BC (is) rational.
The (square) on AC (is) thus irrational. Thus, AC (is)
an irrational (straight-line) [Def. 10.4]. Let it be called
a minor (straight-line). † (Which is) the very thing it was
required to show.
A
C
B
† See footnote to Prop. 10.39.ÓÞ. Proposition 77
᾿Εὰνἀπὸεὐθείαςεὐθεῖαἀφαιρεθῇδυνάμειἀσύμμετρος If a straight-line, which is incommensurable in square
οὖσατῇὅλῃ,μετὰδὲτῆςὅληςποιοῦσατὸμὲνσυγκείμενον with the whole, and with the whole makes the sum of the
ἐκτῶνἀπ᾿αὐτῶντετραγώνωνμέσον,τὸδὲδὶςὑπ᾿αὐτῶν squares on them medial, and twice the (rectangle con-
ῥητόν,ἡλοιπὴἄλογόςἐστιν· καλείσθωδὲἡμετὰῥητοῦ tained) by them rational, is subtracted from a(nother)
μέσοντὸὅλονποιοῦσα.
straight-line then the remainder is an irrational (straightline).
Let it be called that which makes with a rational
(area) a medial whole.
Α
Γ
Β
Ἀπὸ γὰρ εὐθείας τῆς ΑΒ εὐθεῖα ἀφῃρήσθω ἡ ΒΓ For let the straight-line BC, which is incommensuδυνάμειἀσύμμετοςοὖσατῇΑΒποιοῦσατὰπροκείμενα·
rable in square with AB, and fulfils the (other) prescribed
λέγω,ὅτιἡλοιπὴἡΑΓἄλογόςἐστινἡπροειρημένη. (conditions), have been subtracted from the straight-line
᾿ΕπεὶγὰρτὸμὲνσυγκείμενονἐκτῶνἀπὸτῶνΑΒ,ΒΓ AB [Prop. 10.34]. I say that the remainder AC is the
A
C
B
370

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
τετραγώνωνμέσονἐστίν,τὸδὲδὶςὑπὸτῶνΑΒ,ΒΓῥητόν, aforementioned irrational (straight-line).
ἀσύμμετραἄραἐστὶτὰἀπὸτῶνΑΒ,ΒΓτῷδὶςὑπὸτῶν For since the sum of the squares on AB and BC is
ΑΒ,ΒΓ·καὶλοιπὸνἄρατὸἀπὸτῆςΑΓἀσύμμετρόνἐστι medial, and twice the (rectangle contained) by AB and
τῷδὶςὑπὸτῶνΑΒ,ΒΓ.καίἐστιτὸδὶςὑπὸτῶνΑΒ,ΒΓ BC rational, the (sum of the squares) on AB and BC
ῥητόν·τὸἄραἀπὸτῆςΑΓἄλογόνἐστιν·ἄλογοςἄραἐστὶν is thus incommensurable with twice the (rectangle con-
ἡΑΓ·καλείσθωδὲἡμετὰῥητοῦμέσοντὸὅλονποιοῦσα. tained) by AB and BC. Thus, the remaining (square)
ὅπερἔδειδεῖξαι.
on AC is also incommensurable with twice the (rectangle
contained) by AB and BC [Props. 2.7, 10.16]. And
twice the (rectangle contained) by AB and BC is rational.
Thus, the (square) on AC is irrational. Thus, AC
is an irrational (straight-line) [Def. 10.4]. And let it be
called that which makes with a rational (area) a medial
whole. † (Which is) the very thing it was required to show.
† See footnote to Prop. 10.40.Ó. Proposition 78
᾿Εὰνἀπὸεὐθείαςεὐθεῖαἀφαιρεθῇδυνάμειἀσύμμετρος If a straight-line, which is incommensurable in square
οὖσατῇὅλῃ,μετὰδὲτῆςὅληςποιοῦσατότεσυγκείμενον with the whole, and with the whole makes the sum of the
ἐκτῶνἀπ᾿αὐτῶντετραγώνωνμέσοντότεδὶςὑπ᾿αὐτῶν squares on them medial, and twice the (rectangle conμέσονκαὶἔτιτὰἀπ᾿αὐτῶντετράγωναἀσύμμετρατῷδὶςὑπ᾿
tained) by them medial, and, moreover, the (sum of the)
αὐτῶν,ἡλοιπὴἄλογόςἐστιν·καλείσθωδὲἡμετὰμέσου squares on them incommensurable with twice the (rectμέσοντὸὅλονποιοῦσα.
angle contained) by them, is subtracted from a(nother)
straight-line then the remainder is an irrational (straightline).
Let it be called that which makes with a medial
(area) a medial whole.
∆ Ζ Η
D
F
G
Ι Θ Ε
I
H
E
Α
Γ
Β
Ἀπὸ γὰρ εὐθείας τῆς ΑΒ εὐθεῖα ἀφῃρήσθω ἡ ΒΓ For let the straight-line BC, which is incommensuδυνάμειἀσύμμετροςοὖσατῇΑΒποιοῦσατὰπροκείμενα·
rable in square AB, and fulfils the (other) prescribed
λέγω,ὅτιἡλοιπὴἡΑΓἄλογόςἐστινἡκαλουμένηἡμετὰ (conditions), have been subtracted from the (straightμέσουμέσοντὸὅλονποιοῦσα.
line) AB [Prop. 10.35]. I say that the remainder AC is
᾿ΕκκείσθωγὰρῥητὴἡΔΙ,καὶτοῖςμὲνἀπὸτῶνΑΒ, the irrational (straight-line) called that which makes with
ΒΓἴσονπαρὰτὴνΔΙπαραβεβλήσθωτὸΔΕπλάτοςποιοῦν a medial (area) a medial whole.
τὴνΔΗ,τῷδὲδὶςὑπὸτῶνΑΒ,ΒΓἴσονἀφῃρήσθωτὸ For let the rational (straight-line) DI be laid down.
ΔΘ[πλάτοςποιοῦντὴνΔΖ].λοιπὸνἄρατὸΖΕἴσονἐστὶ And let DE, equal to the (sum of the squares) on AB and
τῷἀπὸτῆςΑΓ·ὥστεἡΑΓδύναταιτὸΖΕ.καὶἐπεὶτὸ BC, have been applied to DI, producing DG as breadth.
συγκείμενονἐκτῶνἀπὸτῶνΑΒ,ΒΓτετραγώνωνμέσον And let DH, equal to twice the (rectangle contained) by
ἐστὶκαίἐστινἴσοντῷΔΕ,μέσονἄρα[ἐστὶ]τὸΔΕ.καὶπαρὰ AB and BC, have been subtracted (from DE) [produc-
ῥητὴντὴνΔΙπαράκειταιπλάτοςποιοῦντὴνΔΗ·ῥητὴἄρα ing DF as breadth]. Thus, the remainder FE is equal
ἐστὶνἡΔΗκαὶἀσύμμετροςτῇΔΙμήκει.πάλιν,ἐπεὶτὸδὶς to the (square) on AC [Prop. 2.7]. Hence, AC is the
ὑπὸτῶνΑΒ,ΒΓμέσονἐστὶκαίἐστινἴσοντῷΔΘ,τὸἄρα square-root of FE. And since the sum of the squares on
A
C
B
371

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ΔΘμέσονἐστίν.καὶπαρὰῥητὴντὴνΔΙπαράκειταιπλάτος AB and BC is medial, and is equal to DE, DE [is] thus
ποιοῦντὴνΔΖ·ῥητὴἄραἐστὶκαὶἡΔΖκαὶἀσύμμετρος medial. And it is applied to the rational (straight-line)
τῇΔΙμήκει.καὶἐπεὶἀσύμμετράἐστιτὰἀπὸτῶνΑΒ,ΒΓ DI, producing DG as breadth. Thus, DG is rational, and
τῷδὶςὑπὸτῶνΑΒ,ΒΓ,ἀσύμμετρονἄρακαὶτὸΔΕτῷ incommensurable in length with DI [Prop 10.22]. Again,
ΔΘ.ὡςδὲτὸΔΕπρὸςτὸΔΘ,οὕτωςἐστὶκαὶἡΔΗπρὸς since twice the (rectangle contained) by AB and BC is
τὴνΔΖ·ἀσύμμετροςἄραἡΔΗτῇΔΖ.καίεἰσινἀμφότεραι medial, and is equal to DH, DH is thus medial. And it is
ῥηταί·αἱΗΔ,ΔΖἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι. applied to the rational (straight-line) DI, producing DF
ἀποτομὴἄραἐστίνἡΖΗ·ῥητὴδὲἡΖΘ.τὸδὲὑπὸῥητῆς as breadth. Thus, DF is also rational, and incommenκαὶἀποτομῆςπεριεχόμενον[ὀρθογώνιον]ἄλογόνἐστιν,καὶ
surable in length with DI [Prop. 10.22]. And since the
ἡδυναμένηαὐτὸἄλογόςἐστιν·καὶδύναταιτὸΖΕἡΑΓ·ἡ (sum of the squares) on AB and BC is incommensurable
ΑΓἄραἄλογόςἐστιν·καλείσθωδὲἡμετὰμέσουμέσοντὸ with twice the (rectangle contained) by AB and BC, DE
ὅλονποιοῦσα.ὅπερἔδειδεῖξαι.
(is) also incommensurable with DH. And as DE (is) to
DH, so DG also is to DF [Prop. 6.1]. Thus, DG (is) incommensurable
(in length) with DF [Prop. 10.11]. And
they are both rational. Thus, GD and DF are rational
(straight-lines which are) commensurable in square
only. Thus, FG is an apotome [Prop. 10.73]. And FH
(is) rational. And the [rectangle] contained by a rational
(straight-line) and an apotome is irrational [Prop. 10.20],
and its square-root is irrational. And AC is the squareroot
of FE. Thus, AC is irrational. Let it be called
that which makes with a medial (area) a medial whole. †
(Which is) the very thing it was required to show.
† See footnote to Prop. 10.41.Ó. Proposition 79
Τῇ ἀποτομῇ μία [μόνον] προσαρμόζει εὐθεῖα ῥητὴ [Only] one rational straight-line, which is commensuδυνάμειμόνονσύμμετροςοὖσατῇὅλῃ.
rable in square only with the whole, can be attached to
an apotome. †
Α Β Γ ∆ A B
C D
῎ΕστωἀποτομὴἡΑΒ,προσαρμόζουσαδὲαὐτῇἡΒΓ·αἱ Let AB be an apotome, with BC (so) attached to it.
ΑΓ,ΓΒἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι·λέγω,ὅτι AC and CB are thus rational (straight-lines which are)
τῇΑΒἑτέραοὐπροσαρμόζειῥητὴδυνάμειμόνονσύμμετρος commensurable in square only [Prop. 10.73]. I say that
οὖσατῇὅλῇ.
another rational (straight-line), which is commensurable
Εἰ γὰρ δυνατόν, προσαρμοζέτω ἡ ΒΔ· καὶ αἱ ΑΔ, in square only with the whole, cannot be attached to AB.
ΔΒἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι. καὶἐπεί,ᾧ For, if possible, let BD be (so) attached (to AB).
ὑπερέχειτὰἀπὸτῶνΑΔ,ΔΒτοῦδὶςὑπὸτῶνΑΔ,ΔΒ, Thus, AD and DB are also rational (straight-lines which
τούτῳὑπερέχεικαὶτὰἀπὸτῶνΑΓ,ΓΒτοῦδὶςὑπὸτῶν are) commensurable in square only [Prop. 10.73]. And
ΑΓ,ΓΒ·τῷγὰραὐτῷτῷἀπὸτῆςΑΒἀμφότεραὑπερέχει· since by whatever (area) the (sum of the squares) on AD
ἐναλλὰξἄρα,ᾧὑπερέχειτὰἀπὸτῶνΑΔ,ΔΒτῶνἀπὸτῶν and DB exceeds twice the (rectangle contained) by AD
ΑΓ,ΓΒ,τούτῳὑπερέχει[καὶ]τὸδὶςὑπὸτῶνΑΔ,ΔΒτοῦ and DB, the (sum of the squares) on AC and CB also exδὶςὑπὸτῶνΑΓ,ΓΒ.τὰδὲἀπὸτῶνΑΔ,ΔΒτῶνἀπὸτῶν
ceeds twice the (rectangle contained) by AC and CB by
ΑΓ,ΓΒὑπερέχειῥητῷ·ῥητὰγὰρἀμφότερα.καὶτὸδὶςἄρα this (same area). For both exceed by the same (area)—
ὑπὸτῶνΑΔ,ΔΒτοῦδὶςὑπὸτῶνΑΓ,ΓΒὑπερέχειῥητῷ· (namely), the (square) on AB [Prop. 2.7]. Thus, alter-
ὅπερἐστὶνἀδύνατον·μέσαγὰρἀμφότερα,μέσονδὲμέσου nately, by whatever (area) the (sum of the squares) on
οὐχὑπερέχειῥητῷ.τῇἄραΑΒἑτέραοὐπροσαρμόζειῥητὴ AD and DB exceeds the (sum of the squares) on AC
δυνάμειμόνονσύμμετροςοὖσατῇὅλῃ.
and CB, twice the (rectangle contained) by AD and DB
Μίαἄραμόνητῇἀποτομῇπροσαρμόζειῥητὴδυνάμει [also] exceeds twice the (rectangle contained) by AC and
372

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
μόνονσύμμετροςοὖσατῇὅλῃ·ὅπερἔδειδεῖξαι.
CB by this (same area). And the (sum of the squares) on
AD and DB exceeds the (sum of the squares) on AC
and CB by a rational (area). For both (are) rational (areas).
Thus, twice the (rectangle contained) by AD and
DB also exceeds twice the (rectangle contained) by AC
and CB by a rational (area). The very thing is impossible.
For both are medial (areas) [Prop. 10.21], and a
medial (area) cannot exceed a(nother) medial (area) by
a rational (area) [Prop. 10.26]. Thus, another rational
(straight-line), which is commensurable in square only
with the whole, cannot be attached to AB.
Thus, only one rational (straight-line), which is commensurable
in square only with the whole, can be attached
to an apotome. (Which is) the very thing it was
required to show.
† This proposition is equivalent to Prop. 10.42, with minus signs instead of plus signs.
Ô. Proposition 80
Τῇμέσηςἀποτομῇπρώτῃμίαμόνονπροσαρμόζειεὐθεῖα Only one medial straight-line, which is commensuμέσηδυνάμειμόνονσύμμετροςοὖσατῇὅλῃ,μετὰδὲτῆς
rable in square only with the whole, and contains a ra-
ὅληςῥητὸνπεριέχουσα.
tional (area) with the whole, can be attached to a first
apotome of a medial (straight-line). †
Α Β Γ ∆ A B
C D
῎Εστω γὰρ μέσης ἀποτομὴπρώτη ἡ ΑΒ, καὶ τῇ ΑΒ For let AB be a first apotome of a medial (straightπροσαρμοζέτωἡΒΓ·
αἱΑΓ, ΓΒἄραμέσαιεἰσὶδυνάμει line), and let BC be (so) attached to AB. Thus, AC
μόνονσύμμετροιῥητὸνπεριέχουσαιτὸὑπὸτῶνΑΓ,ΓΒ· and CB are medial (straight-lines which are) commenλέγω,
ὅτι τῇ ΑΒ ἑτέρα οὐ προσαρμόζει μέση δυνάμει surable in square only, containing a rational (area)—
μόνον σύμμετροςοὖσατῇὅλῃ, μετὰδὲτῆςὅληςῥητὸν (namely, that contained) by AC and CB [Prop. 10.74].
περιέχουσα.
I say that a(nother) medial (straight-line), which is com-
Εἰγὰρδυνατόν,προσαρμοζέτωκαὶἡΔΒ·αἱἄραΑΔ, mensurable in square only with the whole, and contains
ΔΒμέσαιεἰσὶδυνάμειμόνονσύμμετροιῥητὸνπεριέχουσαι a rational (area) with the whole, cannot be attached to
τὸὑπὸτῶνΑΔ,ΔΒ.καὶἐπεί,ᾧὑπερέχειτὰἀπὸτῶνΑΔ, AB.
ΔΒτοῦδὶςὑπὸτῶνΑΔ,ΔΒ,τούτῳὑπερέχεικαὶτὰἀπὸ For, if possible, let DB also be (so) attached to
τῶνΑΓ,ΓΒτοῦδὶςὑπὸτῶνΑΓ,ΓΒ·τῷγὰραὐτῷ[πάλιν] AB. Thus, AD and DB are medial (straight-lines which
ὑπερέχουσιτῷἀπὸτῆςΑΒ·ἐναλλὰξἄρα,ᾧὑπερέχειτὰ are) commensurable in square only, containing a ratio-
ἀπὸτῶνΑΔ,ΔΒτῶνἀπὸτῶνΑΓ,ΓΒ,τούτῳὑπερέχει nal (area)—(namely, that) contained by AD and DB
καὶτὸδὶςὑπὸτῶνΑΔ,ΔΒτοῦδὶςὑπὸτῶνΑΓ,ΓΒ.τὸ [Prop. 10.74]. And since by whatever (area) the (sum of
δὲδὶςὑπὸτῶνΑΔ,ΔΒτοῦδὶςὑπὸτῶνΑΓ,ΓΒὑπερέχει the squares) on AD and DB exceeds twice the (rectangle
ῥητῷ·ῥητὰγὰρἀμφότερα. καὶτὰἀπὸτῶνΑΔ,ΔΒἄρα contained) by AD and DB, the (sum of the squares) on
τῶνἀπὸτῶνΑΓ,ΓΒ[τετραγώνων]ὑπερέχειῥητῷ·ὅπερ AC and CB also exceeds twice the (rectangle contained)
ἐστὶνἀδύνατον·μέσαγάρἐστινἀμφότερα,μέσονδὲμέσου by AC and CB by this (same area). For [again] both exοὐχὑπερέχειῥητῷ.
ceed by the same (area)—(namely), the (square) on AB
Τῇἄραμέσηςἀποτομῇπρώτῃμίαμόνονπροσαρμόζει [Prop. 2.7]. Thus, alternately, by whatever (area) the
εὐθεῖαμέσηδυνάμειμόνονσύμμετροςοὖσατῇὅλῃ,μετὰ (sum of the squares) on AD and DB exceeds the (sum
δὲτῆςὅληςῥητὸνπεριέχουσα·ὅπερἔδειδεῖξαι.
of the squares) on AC and CB, twice the (rectangle contained)
by AD and DB also exceeds twice the (rectangle
contained) by AC and CB by this (same area). And twice
the (rectangle contained) by AD and DB exceeds twice
373

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
† This proposition is equivalent to Prop. 10.43, with minus signs instead of plus signs.
the (rectangle contained) by AC and CB by a rational
(area). For both (are) rational (areas). Thus, the (sum
of the squares) on AD and DB also exceeds the (sum
of the) [squares] on AC and CB by a rational (area).
The very thing is impossible. For both are medial (areas)
[Props. 10.15, 10.23 corr.], and a medial (area) cannot
exceed a(nother) medial (area) by a rational (area)
[Prop. 10.26].
Thus, only one medial (straight-line), which is commensurable
in square only with the whole, and contains
a rational (area) with the whole, can be attached to a first
apotome of a medial (straight-line). (Which is) the very
thing it was required to show.
Ô. Proposition 81
Τῇ μέσης ἀποτομῇ δευτέρᾳ μία μόνον προσαρμόζει Only one medial straight-line, which is commensuεὐθεῖαμέσηδυνάμειμόνονσύμμετροςτῇὅλῃ,μετὰδὲτῆς
rable in square only with the whole, and contains a me-
ὅληςμέσονπεριέχουσα.
dial (area) with the whole, can be attached to a second
apotome of a medial (straight-line). †
Α Β Γ ∆
A B C D
Ε Θ Μ Ν
E H
M N
Ζ Λ Η Ι
F L
G I
῎ΕστωμέσηςἀποτομὴδευτέραἡΑΒκαὶτῇΑΒπρο- Let AB be a second apotome of a medial (straightσαρμόζουσαἡΒΓ·αἱἄραΑΓ,ΓΒμέσαιεἰσὶδυνάμειμόνον
line), with BC (so) attached to AB. Thus, AC and CB
σύμμετροιμέσονπεριέχουσαιτὸὑπὸτῶνΑΓ,ΓΒ·λέγω, are medial (straight-lines which are) commensurable in
ὅτι τῇ ΑΒ ἑτέρα οὐ προσαρμόσει εὐθεῖα μέση δυνάμει square only, containing a medial (area)—(namely, that
μόνονσύμμετροςοὖσατῇὅλῃ, μετὰδὲτῆςὅληςμέσον contained) by AC and CB [Prop. 10.75]. I say that
περιέχουσα.
a(nother) medial straight-line, which is commensurable
Εἰγὰρδυνατόν,προσαρμοζέτωἡΒΔ·καὶαἱΑΔ,ΔΒ in square only with the whole, and contains a medial
ἄραμέσαιεἰσὶδυνάμειμόνονσύμμετροιμέσονπεριέχουσαι (area) with the whole, cannot be attached to AB.
τὸὑπὸτῶνΑΔ,ΔΒ.καὶἐκκείσθωῥητὴἡΕΖ,καὶτοῖς For, if possible, let BD be (so) attached. Thus, AD
μὲνἀπὸτῶνΑΓ,ΓΒἴσονπαρὰτὴνΕΖπαραβεβλήσθωτὸ and DB are also medial (straight-lines which are) com-
ΕΗπλάτοςποιοῦντὴνΕΜ·τῷδὲδὶςὑπὸτῶνΑΓ,ΓΒἴσον mensurable in square only, containing a medial (area)—
ἀφῃρήσθωτὸΘΗπλάτοςποιοῦντὴνΘΜ·λοιπὸνἄρατὸΕΛ (namely, that contained) by AD and DB [Prop. 10.75].
ἴσονἐστὶτῷἀπὸτῆςΑΒ·ὥστεἡΑΒδύναταιτὸΕΛ.πάλιν And let the rational (straight-line) EF be laid down. And
δὴτοῖςἀπὸτῶνΑΔ,ΔΒἴσονπαρὰτὴνΕΖπαραβεβλήσθω let EG, equal to the (sum of the squares) on AC and
τὸΕΙπλάτοςποιοῦντὴνΕΝ·ἔστιδὲκαὶτὸΕΛἴσοντῷἀπὸ CB, have been applied to EF , producing EM as breadth.
τῆςΑΒτετραγώνῳ·λοιπὸνἄρατὸΘΙἴσονἐστὶτῷδὶςὑπὸ And let HG, equal to twice the (rectangle contained) by
τῶνΑΔ,ΔΒ.καὶἐπεὶμέσαιεἰσὶναἱΑΓ,ΓΒ,μέσαἄραἐστὶ AC and CB, have been subtracted (from EG), producκαὶτὰἀπὸτῶνΑΓ,ΓΒ.καίἐστινἴσατῷΕΗ·μέσονἄρακαὶ
ing HM as breadth. The remainder EL is thus equal
τὸΕΗ.καὶπαρὰῥητὴντὴνΕΖπαράκειταιπλάτοςποιοῦν to the (square) on AB [Prop. 2.7]. Hence, AB is the
374

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
τὴνΕΜ·ῥητὴἄραἐστὶνἡΕΜκαὶἀσύμμετροςτῇΕΖμήκει. square-root of EL. So, again, let EI, equal to the (sum
πάλιν,ἐπεὶμέσονἐστὶτὸὑπὸτῶνΑΓ,ΓΒ,καὶτὸδὶςὑπὸ of the squares) on AD and DB have been applied to EF ,
τῶνΑΓ,ΓΒμέσονἐστίν.καίἐστινἴσοντῷΘΗ·καὶτὸΘΗ producing EN as breadth. And EL is also equal to the
ἄραμέσονἐστίν.καὶπαρὰῥητὴντὴνΕΖπαράκειταιπλάτος square on AB. Thus, the remainder HI is equal to twice
ποιοῦντὴνΘΜ·ῥητὴἄραἐστὶκαὶἡΘΜκαὶἀσύμμετρος the (rectangle contained) by AD and DB [Prop. 2.7].
τῇΕΖμήκει.καὶἐπεὶαἱΑΓ,ΓΒδυνάμειμόνονσύμμετροί And since AC and CB are (both) medial (straight-lines),
εἰσιν,ἀσύμμετροςἄραἐστὶνἡΑΓτῇΓΒμήκει.ὡςδὲἡΑΓ the (sum of the squares) on AC and CB is also meπρὸςτὴνΓΒ,οὕτωςἐστὶτὸἀπὸτῆςΑΓπρὸςτὸὑπὸτῶν
dial. And it is equal to EG. Thus, EG is also medial
ΑΓ,ΓΒ·ἀσύμμετρονἄραἐστὶτὸἀπὸτῆςΑΓτῷὑπὸτῶν [Props. 10.15, 10.23 corr.]. And it is applied to the ratio-
ΑΓ,ΓΒ.ἀλλὰτῷμὲνἀπὸτῆςΑΓσύμμετράἐστιτὰἀπὸ nal (straight-line) EF , producing EM as breadth. Thus,
τῶνΑΓ,ΓΒ,τῷδὲὑπὸτῶνΑΓ,ΓΒσύμμετρόνἐστιτὸ EM is rational, and incommensurable in length with EF
δὶςὑπὸτῶνΑΓ,ΓΒ·ἀσύμμετραἄραἐστὶτὰἀπὸτῶνΑΓ, [Prop. 10.22]. Again, since the (rectangle contained) by
ΓΒτῷδὶςὑπὸτῶνΑΓ,ΓΒ.καίἐστιτοῖςμὲνἀπὸτῶνΑΓ, AC and CB is medial, twice the (rectangle contained)
ΓΒἴσοντὸΕΗ,τῷδὲδὶςὑπὸτῶνΑΓ,ΓΒἴσοντὸΗΘ· by AC and CB is also medial [Prop. 10.23 corr.]. And it
ἀσύμμετρονἄραἐστὶτὸΕΗτῷΘΗ.ὡςδὲτὸΕΗπρὸςτὸ is equal to HG. Thus, HG is also medial. And it is ap-
ΘΗ,οὕτωςἐστὶνἡΕΜπρὸςτὴνΘΜ·ἀσύμμετροςἄραἐστὶν plied to the rational (straight-line) EF , producing HM
ἡΕΜτῇΜΘμήκει.καίεἰσινἀμφότεραιῥηταί·αἱΕΜ,ΜΘ as breadth. Thus, HM is also rational, and incommen-
ἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι·ἀποτομὴἄραἐστὶν surable in length with EF [Prop. 10.22]. And since AC
ἡΕΘ,προσαρμόζουσαδὲαὐτῇἡΘΜ.ὁμοίωςδὴδείξομεν, and CB are commensurable in square only, AC is thus
ὅτικαὶἡΘΝαὐτῇπροσαρμόζει·τῇἄραἀποτομῇἄλληκαὶ incommensurable in length with CB. And as AC (is)
ἄλληπροσαρμόζειεὐθεῖαδυνάμειμόνονσύμμετροςοὖσα to CB, so the (square) on AC is to the (rectangle conτῇὅλῃ·ὅπερἐστὶνἀδύνατον.
tained) by AC and CB [Prop. 10.21 corr.]. Thus, the
Τῇἄραμέσηςἀποτομῇδευτέρᾳμίαμόνονπροσαρμόζει (square) on AC is incommensurable with the (rectanεὐθεῖαμέσηδυνάμειμόνονσύμμετροςοὖσατῇὅλῃ,μετὰ
gle contained) by AC and CB [Prop. 10.11]. But, the
δὲτῆςὅληςμέσονπεριέχουσα·ὅπερἔδειδεῖξαι. (sum of the squares) on AC and CB is commensurable
with the (square) on AC, and twice the (rectangle contained)
by AC and CB is commensurable with the (rectangle
contained) by AC and CB [Prop. 10.6]. Thus,
the (sum of the squares) on AC and CB is incommensurable
with twice the (rectangle contained) by AC and
CB [Prop. 10.13]. And EG is equal to the (sum of the
squares) on AC and CB. And GH is equal to twice the
(rectangle contained) by AC and CB. Thus, EG is incommensurable
with HG. And as EG (is) to HG, so EM
is to HM [Prop. 6.1]. Thus, EM is incommensurable
in length with MH [Prop. 10.11]. And they are both
rational (straight-lines). Thus, EM and MH are rational
(straight-lines which are) commensurable in square
only. Thus, EH is an apotome [Prop. 10.73], and HM
(is) attached to it. So, similarly, we can show that HN
(is) also (commensurable in square only with EN and is)
attached to (EH). Thus, different straight-lines, which
are commensurable in square only with the whole, are
attached to an apotome. The very thing is impossible
[Prop. 10.79].
Thus, only one medial straight-line, which is commensurable
in square only with the whole, and contains a medial
(area) with the whole, can be attached to a second
apotome of a medial (straight-line). (Which is) the very
thing it was required to show.
375

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
† This proposition is equivalent to Prop. 10.44, with minus signs instead of plus signs.
Ô. Proposition 82
Τῇ ἐλάσσονι μία μόνον προσαρμόζει εὐθεῖα δυνάμει Only one straight-line, which is incommensurable in
ἀσύμμετροςοὖσατῇὅλῃποιοῦσαμετὰτῆςὅληςτὸμὲν square with the whole, and (together) with the whole
ἐκτῶνἀπ᾿αὐτῶντετραγώνωνῥητόν,τὸδὲδὶςὑπ᾿αὐτῶν makes the (sum of the) squares on them rational, and
μέσον.
twice the (rectangle contained) by them medial, can be
attached to a minor (straight-line).
Α Β Γ ∆ A B
C D
῎ΕστωἡἐλάσσωνἡΑΒ,καὶτῇΑΒπροσαρμόζουσα Let AB be a minor (straight-line), and let BC be (so)
ἔστω ἡ ΒΓ· αἱ ἄρα ΑΓ, ΓΒ δυνάμει εἰσὶν ἀσύμμετροι attached to AB. Thus, AC and CB are (straight-lines
ποιοῦσαιτὸμὲνσυγκείμενονἐκτῶνἀπ᾿αὐτῶντετραγώνων which are) incommensurable in square, making the sum
ῥητόν,τὸδὲδὶςὑπ᾿αὐτῶνμέσον·λέγω,ὅτιτῇΑΒἑτέρα of the squares on them rational, and twice the (rectanεὐθεῖαοὐπροσαρμόσειτὰαὐτὰποιοῦσα.
gle contained) by them medial [Prop. 10.76]. I say that
Εἰγὰρδυνατόν,προσαρμοζέτωἡΒΔ·καὶαἱΑΔ,ΔΒ another another straight-line fulfilling the same (condi-
ἄραδυνάμειεἰσὶνἀσύμμετροιποιοῦσαιτὰπροειρημένα.καὶ tions) cannot be attached to AB.
ἐπεί,ᾧὑπερέχειτὰἀπὸτῶνΑΔ,ΔΒτῶνἀπὸτῶνΑΓ,ΓΒ, For, if possible, let BD be (so) attached (to AB).
τούτῳὑπερέχεικαὶτὸδὶςὑπὸτῶνΑΔ,ΔΒτοῦδὶςὑπὸ Thus, AD and DB are also (straight-lines which are)
τῶνΑΓ,ΓΒ,τὰδὲἀπὸτῶνΑΔ,ΔΒτετράγωνατῶνἀπὸ incommensurable in square, fulfilling the (other) aforeτῶνΑΓ,ΓΒτετραγώνωνὑπερέχειῥητῷ·ῥητὰγάρἐστιν
mentioned (conditions) [Prop. 10.76]. And since by
ἀμφότερα· καὶτὸδὶςὑπὸτῶνΑΔ,ΔΒἄρατοῦδὶςὑπὸ whatever (area) the (sum of the squares) on AD and DB
τῶνΑΓ,ΓΒὑπερέχειῥητῷ· ὅπερἐστὶνἀδύνατον· μέσα exceeds the (sum of the squares) on AC and CB, twice
γάρἐστινἀμφότερα.
the (rectangle contained) by AD and DB also exceeds
Τῇἄραἐλάσσονιμίαμόνονπροσαρμόζειεὐθεῖαδυνάμει twice the (rectangle contained) by AC and CB by this
ἀσύμμετροςοὖσατῇὅλῃκαὶποιοῦσατὰμὲνἀπ᾿αὐτῶν (same area) [Prop. 2.7]. And the (sum of the) squares
τετράγωναἅμαῥητόν,τὸδὲδὶςὑπ᾿αὐτῶνμέσον·ὅπερἔδει on AD and DB exceeds the (sum of the) squares on AC
δεῖξαι.
and CB by a rational (area). For both are rational (areas).
Thus, twice the (rectangle contained) by AD and
DB also exceeds twice the (rectangle contained) by AC
and CB by a rational (area). The very thing is impossible.
For both are medial (areas) [Prop. 10.26].
Thus, only one straight-line, which is incommensurable
in square with the whole, and (with the whole)
makes the squares on them (added) together rational,
and twice the (rectangle contained) by them medial, can
be attached to a minor (straight-line). (Which is) the very
thing it was required to show.
† This proposition is equivalent to Prop. 10.45, with minus signs instead of plus signs.
Ô. Proposition 83
Τῇμετὰῥητοῦμέσοντὸὅλονποιούσῃμίαμόνονπρο- Only one straight-line, which is incommensurable in
σαρμόζειεὐθεῖαδυνάμειἀσύμμετροςοὖσατῇὅλῃ,μετὰδὲ square with the whole, and (together) with the whole
τῆςὅληςποιοῦσατὸμὲνσυγκείμενονἐκτῶνἀπ᾿αὐτῶν makes the sum of the squares on them medial, and twice
τετραγώνωνμέσον,τὸδὲδὶςὑπ᾿αὐτῶνῥητόν.
the (rectangle contained) by them rational, can be attached
to that (straight-line) which with a rational (area)
makes a medial whole. †
Α Β Γ ∆ A B
C D
376

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
῎ΕστωἡμετὰῥητοῦμέσοντὸὅλονποιοῦσαἡΑΒ,καὶ Let AB be a (straight-line) which with a rational
τῇΑΒπροσαρμοζέτωἡΒΓ·αἱἄραΑΓ,ΓΒδυνάμειεἰσὶν (area) makes a medial whole, and let BC be (so) at-
ἀσύμμετροιποιοῦσαιτὰπροκείμενα·λέγω,ὅτιτῇΑΒἑτέρα tached to AB. Thus, AC and CB are (straight-lines
οὐπροσαρμόσειτὰαὐτὰποιοῦσα.
which are) incommensurable in square, fulfilling the
Εἰ γὰρ δυνατόν, προσαρμοζέτω ἡ ΒΔ· καὶ αἱ ΑΔ, (other) proscribed (conditions) [Prop. 10.77]. I say that
ΔΒ ἄρα εὐθεῖαι δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὰ another (straight-line) fulfilling the same (conditions)
προκείμενα. ἐπεὶοὐν, ᾧὑπερέχειτὰἀπὸτῶνΑΔ,ΔΒ cannot be attached to AB.
τῶνἀπὸτῶνΑΓ,ΓΒ,τούτῳὑπερέχεικαὶτὸδὶςὑπὸτῶν For, if possible, let BD be (so) attached (to AB).
ΑΔ,ΔΒτοῦδὶςὑπὸτῶνΑΓ,ΓΒἀκολούθωςτοῖςπρὸ Thus, AD and DB are also straight-lines (which are)
αὐτοῦ,τὸδὲδὶςὑπὸτῶνΑΔ,ΔΒτοῦδὶςὑπὸτῶνΑΓ,ΓΒ incommensurable in square, fulfilling the (other) pre-
ὑπερέχειῥητῷ·ῥητὰγάρἐστινἀμφότερα·καὶτὰἀπὸτῶν scribed (conditions) [Prop. 10.77]. Therefore, analo-
ΑΔ,ΔΒἄρατῶνἀπὸτῶνΑΓ,ΓΒὑπερέχειῥητῷ·ὅπερ gously to the (propositions) before this, since by what-
ἐστὶνἀδύνατον·μέσαγάρἐστινἀμφότερα.
ever (area) the (sum of the squares) on AD and DB ex-
Οὐκ ἄρα τῇ ΑΒ ἑτέρα προσαρμόσει εὐθεῖα δυνάμει ceeds the (sum of the squares) on AC and CB, twice the
ἀσύμμετροςοὖσατῇὅλῃ, μετὰδὲ τῆςὅληςποιοῦσατὰ (rectangle contained) by AD and DB also exceeds twice
προειρημένα·μίαἄραμόνονπροσαρμόσει·ὅπερἔδειδεῖξαι. the (rectangle contained) by AC and CB by this (same
area). And twice the (rectangle contained) by AD and
DB exceeds twice the (rectangle contained) by AC and
CB by a rational (area). For they are (both) rational (areas).
Thus, the (sum of the squares) on AD and DB also
exceeds the (sum of the squares) on AC and CB by a rational
(area). The very thing is impossible. For both are
medial (areas) [Prop. 10.26].
Thus, another straight-line cannot be attached to AB,
which is incommensurable in square with the whole, and
fulfills the (other) aforementioned (conditions) with the
whole. Thus, only one (such straight-line) can be (so)
attached. (Which is) the very thing it was required to
show.
† This proposition is equivalent to Prop. 10.46, with minus signs instead of plus signs.
Ô. Proposition 84
Τῇμετὰμέσουμέσοντὸὅλονποιούσῃμίαμόνηπρο- Only one straight-line, which is incommensurable in
σαρμόζειεὐθεῖαδυνάμειἀσύμμετροςοὖσατῇὅλῃ,μετὰδὲ square with the whole, and (together) with the whole
τῆς ὅληςποιοῦσατό τε συγκείμενον ἐκτῶν ἀπ᾿ αὐτῶν makes the sum of the squares on them medial, and twice
τετραγώνων μέσον τό τε δὶς ὑπ᾿ αὐτῶν μέσον καὶ ἔτι the (rectangle contained) by them medial, and, more-
ἀσύμμετροντῷσυγκειμένῳἐκτῶνἀπ᾿αὐτῶν.
over, incommensurable with the sum of the (squares) on
῎ΕστωἡμετὰμέσουμέσοντὸὅλονποιοῦσαἡΑΒ,προ- them, can be attached to that (straight-line) which with
σαρμόζουσαδὲαὐτῇἡΒΓ·αἱἄραΑΓ,ΓΒδυνάμειεἰσὶν a medial (area) makes a medial whole. †
ἀσύμμετροι ποιοῦσαιτὰ προειρημένα. λέγω, ὅτι τῇ ΑΒ Let AB be a (straight-line) which with a medial
ἑτέραοὐπροσαρμόσειποιοῦσαπροειρημένα.
(area) makes a medial whole, BC being (so) attached
to it. Thus, AC and CB are incommensurable in
square, fulfilling the (other) aforementioned (conditions)
[Prop. 10.78]. I say that a(nother) (straight-line) fulfilling
the aforementioned (conditions) cannot be attached
to AB.
377

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
Α Β Γ ∆
Ε Θ Μ Ν
A B C D
E H
M N
Ζ Λ Η Ι
F L
G I
Εἰγὰρδυνατόν, προσαρμοζέτωἡΒΔ,ὥστεκαὶτὰς For, if possible, let BD be (so) attached. Hence,
ΑΔ,ΔΒδυνάμειἀσυμμέτρουςεἶναιποιούσαςτάτεἀπὸ AD and DB are also (straight-lines which are) incomτῶνΑΔ,ΔΒτετράγωναἅμαμέσονκαὶτὸδὶςὑπὸτῶν
mensurable in square, making the squares on AD and
ΑΔ,ΔΒμέσονκαὶἔτιτὰἀπὸτῶνΑΔ,ΔΒἀσύμμετρατῷ DB (added) together medial, and twice the (rectangle
δὶςὑπὸτῶνΑΔ,ΔΒ·καὶἐκκείσθωῥητὴἡΕΖ,καὶτοῖς contained) by AD and DB medial, and, moreover, the
μὲνἀπὸτῶνΑΓ,ΓΒἴσονπαρὰτὴνΕΖπαραβεβλήσθωτὸ (sum of the squares) on AD and DB incommensurable
ΕΗπλάτοςποιοῦντὴνΕΜ,τῷδὲδὶςὑπὸτῶνΑΓ,ΓΒ with twice the (rectangle contained) by AD and DB
ἴσονπαρὰτὴνΕΖπαραβεβλήσθωτὸΘΗπλάτοςποιοῦν [Prop. 10.78]. And let the rational (straight-line) EF be
τὴνΘΜ·λοιπὸνἄρατὸἀπὸτῆςΑΒἴσονἐστὶτῷΕΛ·ἡ laid down. And let EG, equal to the (sum of the squares)
ἄραΑΒδύναταιτὸΕΛ.πάλιντοῖςἀπὸτῶνΑΔ,ΔΒἴσον on AC and CB, have been applied to EF , producing EM
παρὰτὴνΕΖπαραβεβλήσθωτὸΕΙπλάτοςποιοῦντὴνΕΝ. as breadth. And let HG, equal to twice the (rectangle
ἔστιδὲκαὶτὸἀπὸτῆςΑΒἴσοντῷΕΛ·λοιπὸνἄρατὸδὶς contained) by AC and CB, have been applied to EF ,
ὑπὸτῶνΑΔ,ΔΒἴσον[ἐστὶ]τῷΘΙ.καὶἐπεὶμέσονἐστὶ producing HM as breadth. Thus, the remaining (square)
τὸσυγκείμενονἐκτῶνἀπὸτῶνΑΓ,ΓΒκαίἐστινἴσον on AB is equal to EL [Prop. 2.7]. Thus, AB is the squareτῷΕΗ,μέσονἄραἐστὶκαὶτὸΕΗ.καὶπαρὰῥητὴντὴνΕΖ
root of EL. Again, let EI, equal to the (sum of the
παράκειταιπλάτοςποιοῦντὴνΕΜ·ῥητὴἄραἐστὶνἡΕΜκαὶ squares) on AD and DB, have been applied to EF , pro-
ἀσύμμετροςτῇΕΖμήκει.πάλιν,ἐπεὶμέσονἐστὶτὸδὶςὑπὸ ducing EN as breadth. And the (square) on AB is also
τῶνΑΓ,ΓΒκαίἐστινἴσοντῷΘΗ,μέσονἄρακαὶτὸΘΗ. equal to EL. Thus, the remaining twice the (rectangle
καὶπαρὰῥητὴντὴνΕΖπαράκειταιπλάτοςποιοῦντὴνΘΜ· contained) by AD and DB [is] equal to HI [Prop. 2.7].
ῥητὴἄραἐστὶνἡΘΜκαὶἀσύμμετροςτῇΕΖμήκει.καὶἐπεὶ And since the sum of the (squares) on AC and CB is me-
ἀσύμμετράἐστιτὰἀπὸτῶνΑΓ,ΓΒτῷδὶςὑπὸτῶνΑΓ, dial, and is equal to EG, EG is thus also medial. And
ΓΒ,ἀσύμμετρόνἐστικαὶτὸΕΗτῷΘΗ·ἀσύμμετροςἄρα it is applied to the rational (straight-line) EF , producing
ἐστὶκαὶἡΕΜτῇΜΘμήκει.καίεἰσινἀμφότεραιῥηταί·αἱ EM as breadth. EM is thus rational, and incommen-
ἄραΕΜ,ΜΘῥηταίεἰσιδυνάμειμόνονσύμμετροι·ἀποτομὴ surable in length with EF [Prop. 10.22]. Again, since
ἄραἐστὶνἡΕΘ,προσαρμόζουσαδὲαὐτῇἡΘΜ.ὁμοίωςδὴ twice the (rectangle contained) by AC and CB is meδείξομεν,ὅτιἡΕΘπάλινἀποτομήἐστιν,προσαρμόζουσα
dial, and is equal to HG, HG is thus also medial. And
δὲαὐτῇἡΘΝ.τῇἄραἀποτομῇἄλληκαὶἄλληπροσαρμόζει it is applied to the rational (straight-line) EF , produc-
ῥητὴδυνάμειμόνονσύμμετροςοὖσατῇὅλῃ·ὅπερἐδείχθη ing HM as breadth. HM is thus rational, and incom-
ἀδύνατον.οὐκἄρατῇΑΒἑτέραπροσαρμόσειεὐθεῖα. mensurable in length with EF [Prop. 10.22]. And since
Τῇ ἄρα ΑΒ μία μόνον προσαρμόζει εὐθεῖα δυνάμει the (sum of the squares) on AC and CB is incommen-
ἀσύμμετροςοὖσατῇὅλῃ, μετὰδὲ τῆςὅληςποιοῦσατά surable with twice the (rectangle contained) by AC and
τεἀπ᾿αὐτῶντετράγωναἅμαμέσονκαὶτὸδὶςὑπ᾿αὐτῶν CB, EG is also incommensurable with HG. Thus, EM
μέσονκαὶἔτιτὰἀπ᾿αὐτῶντετράγωναἀσύμμετρατῷδὶς is also incommensurable in length with MH [Props. 6.1,
ὑπ᾿αὐτῶν·ὅπερἔδειδεῖξαι.
10.11]. And they are both rational (straight-lines). Thus,
EM and MH are rational (straight-lines which are) commensurable
in square only. Thus, EH is an apotome
[Prop. 10.73], with HM attached to it. So, similarly,
we can show that EH is again an apotome, with HN
attached to it. Thus, different rational (straight-lines),
which are commensurable in square only with the whole,
are attached to an apotome. The very thing was shown
378

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
† This proposition is equivalent to Prop. 10.47, with minus signs instead of plus signs.
(to be) impossible [Prop. 10.79]. Thus, another straightline
cannot be (so) attached to AB.
Thus, only one straight-line, which is incommensurable
in square with the whole, and (together) with the
whole makes the squares on them (added) together medial,
and twice the (rectangle contained) by them medial,
and, moreover, the (sum of the) squares on them incommensurable
with the (rectangle contained) by them, can
be attached to AB. (Which is) the very thing it was required
to show.
ÎÇÖÓØÖØÓ.
ιαʹ.῾Υποκειμένηςῥητῆςκαὶἀποτομῆς,ἐὰνμὲνἡὅλητῆς 11. Given a rational (straight-line) and an apotome, if
Definitions III
προσαρμοζούσηςμεῖζονδύνηταιτῷἀπὸσυμμέτρουἑαυτῇ the square on the whole is greater than the (square on a
μήκει, καὶἡὅλησύμμετρος ᾖτῇἐκκειμένῃῥητῇμήκει, straight-line) attached (to the apotome) by the (square)
καλείσθωἀποτομὴπρώτη.
on (some straight-line) commensurable in length with
ιβʹ.᾿Εὰνδὲἡπροσαρμόζουσασύμμετροςᾖτῇἐκκειμένῃ (the whole), and the whole is commensurable in length
ῥητῇμήκει,καὶἡὅλητῆςπροσαρμοζούσηςμεῖζονδύνηται with the (previously) laid down rational (straight-line),
τῷἀπὸσυμμέτρουἑαυτῇ,καλείσθωἀποτομὴδευτέρα. then let the (apotome) be called a first apotome.
ιγʹ.᾿Εὰνδὲμηδετέρασύμμετροςᾖτῇἐκκειμένῃῥητῇ 12. And if the attached (straight-line) is commenμήκει,
ἡδὲὅλητῆςπροσαρμοζούσηςμεῖζονδύνηταιτῷ surable in length with the (previously) laid down ra-
ἀπὸσυμμέτρουἑαυτῇ,καλείσθωἀποτομὴτρίτη. tional (straight-line), and the square on the whole is
ιδʹ. Πάλιν, ἐὰν ἡ ὅλη τῆς προσαρμοζούσης μεῖζον greater than (the square on) the attached (straight-line)
δύνηταιτῷἀπὸἀσυμμέτρουἑαυτῇ[μήκει],ἐὰνμὲνἡὅλη by the (square) on (some straight-line) commensurable
σύμμετροςᾖτῇἐκκειμένῃῥητῇμήκει,καλείσθωἀποτομὴ (in length) with (the whole), then let the (apotome) be
τετάρτη.
ιεʹ.᾿Εὰνδὲἡπροσαρμόζουσα,πέμπτη.
ιϛʹ.᾿Εὰνδὲμηδετέρα,ἕκτη.
Ô. Proposition
Εὑρεῖντὴνπρώτηνἀποτομήν.
called a second apotome.
13. And if neither of (the whole or the attached
straight-line) is commensurable in length with the (previously)
laid down rational (straight-line), and the square
on the whole is greater than (the square on) the attached
(straight-line) by the (square) on (some straight-line)
commensurable (in length) with (the whole), then let the
(apotome) be called a third apotome.
14. Again, if the square on the whole is greater
than (the square on) the attached (straight-line) by the
(square) on (some straight-line) incommensurable [in
length] with (the whole), and the whole is commensurable
in length with the (previously) laid down rational
(straight-line), then let the (apotome) be called a fourth
apotome.
15. And if the attached (straight-line is commensurable),
a fifth (apotome).
16. And if neither (the whole nor the attached
straight-line is commensurable), a sixth (apotome).
To find a first apotome.
85
379

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
Α
Θ
Β Γ
Ε Ζ
∆
Η
᾿ΕκκείσθωῥητὴἡΑ,καὶτῇΑμήκεισύμμετροςἔστω Let the rational (straight-line) A be laid down. And
ἡ ΒΗ· ῥητὴ ἄρα ἐστὶ καὶ ἡ ΒΗ. καὶ ἐκκείσθωσαν δύο let BG be commensurable in length with A. BG is thus
τετράγωνοιἀριθμοὶοἱΔΕ,ΕΖ,ὧνἡὑπεροχὴὁΖΔμὴ also a rational (straight-line). And let two square num-
ἔστωτετράγωνος·οὐδ᾿ἄραὁΕΔπρὸςτὸνΔΖλόγονἔχει, bers DE and EF be laid down, and let their difference
ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν.καὶπε- FD be not square [Prop. 10.28 lem. I]. Thus, ED does
ποιήσθωὡςὁΕΔπρὸςτὸνΔΖ,οὕτωςτὸἀπὸτῆςΒΗ not have to DF the ratio which (some) square number
τετράγωνονπρὸςτὸἀπὸτὴςΗΓτετράγωνον·σύμμετρον (has) to (some) square number. And let it have been
ἄραἐστὶτὸἀπὸτῆςΒΗτῷἀπὸτῆςΗΓ.ῥητὸνδὲτὸἀπὸ contrived that as ED (is) to DF , so the square on BG
τῆςΒΗ·ῥητὸνἄρακαὶτὸἀπὸτῆςΗΓ·ῥητὴἄραἐστὶκαὶ (is) to the square on GC [Prop. 10.6. corr.]. Thus, the
ἡΗΓ.καὶἐπεὶὁΕΔπρὸςτὸνΔΖλόγονοὐκἔχει, ὃν (square) on BG is commensurable with the (square) on
τετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν,οὐδ᾿ἄρατὸ GC [Prop. 10.6]. And the (square) on BG (is) ratio-
ἀπὸτῆςΒΗπρὸςτὸἀπὸτῆςΗΓλόγονἔχει,ὃντετράγωνος nal. Thus, the (square) on GC (is) also rational. Thus,
ἀριθμὸςπρὸςτετράγωνονἀριθμόν·ἀσύμμετροςἄραἐστὶν GC is also rational. And since ED does not have to DF
ἡΒΗτῇΗΓμήκει. καίεἰσινἀμφότεραιῥηταί·αἱΒΗ,ΗΓ the ratio which (some) square number (has) to (some)
ἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι·ἡἄραΒΓἀποτομή square number, the (square) on BG thus does not have to
ἐστιν.λέγωδή,ὅτικαὶπρώτη.
the (square) on GC the ratio which (some) square num-
῟ΩιγὰρμεῖζόνἐστιτὸἀπὸτῆςΒΗτοῦἀπὸτῆςΗΓ, ber (has) to (some) square number either. Thus, BG is
ἔστω τὸ ἀπὸ τῆς Θ. καὶἐπεί ἐστιν ὡς ὁ ΕΔ πρὸς τὸν incommensurable in length with GC [Prop. 10.9]. And
ΖΔ,οὕτωςτὸἀπὸτῆςΒΗπρὸςτὸἀπὸτῆςΗΓ,καὶἀνα- they are both rational (straight-lines). Thus, BG and GC
στρέψαντιἄραἐστὶνὡςὁΔΕπρὸςτὸνΕΖ,οὕτωςτὸἀπὸ are rational (straight-lines which are) commensurable in
τῆςΗΒπρὸςτὸἀπὸτῆςΘ.ὁδὲΔΕπρὸςτὸνΕΖλόγον square only. Thus, BC is an apotome [Prop. 10.73]. So,
ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· I say that (it is) also a first (apotome).
ἑκάτεροςγὰρτετράγωνόςἐστιν·καὶτὸἀπὸτῆςΗΒἄρα Let the (square) on H be that (area) by which the
πρὸςτὸἀπὸτῆςΘλόγονἔχει, ὃντετράγωνοςἀριθμὸς (square) on BG is greater than the (square) on GC
πρὸςτετράγωνονἀριθμόν·σύμμετροςἄραἐστὶνἡΒΗτῇ [Prop. 10.13 lem.]. And since as ED is to FD, so the
Θμήκει. καὶδύναταιἡΒΗτῆςΗΓμεῖζοντῷἀπὸτῆςΘ· (square) on BG (is) to the (square) on GC, thus, via con-
ἡΒΗἄρατῆςΗΓμεῖζονδύναταιτῷἀπὸσυμμέτρουἑαυτῇ version, as DE is to EF , so the (square) on GB (is) to
μήκει.καίἐστινἡὅληἡΒΗσύμμετροςτῇἐκκειμένῃῥητῇ the (square) on H [Prop. 5.19 corr.]. And DE has to EF
μήκειτῇΑ.ἡΒΓἄραἀποτομήἐστιπρώτη.
the ratio which (some) square-number (has) to (some)
ΕὕρηταιἄραἡπρώτηἀποτομὴἡΒΓ·ὅπερἔδειεὑρεῖν. square-number. For each is a square (number). Thus, the
(square) on GB also has to the (square) on H the ratio
which (some) square number (has) to (some) square
number. Thus, BG is commensurable in length with H
[Prop. 10.9]. And the square on BG is greater than (the
square on) GC by the (square) on H. Thus, the square on
BG is greater than (the square on) GC by the (square)
on (some straight-line) commensurable in length with
(BG). And the whole, BG, is commensurable in length
with the (previously) laid down rational (straight-line) A.
Thus, BC is a first apotome [Def. 10.11].
Thus, the first apotome BC has been found. (Which
is) the very thing it was required to find.
A
H
B
E
C
F
D
G
† See footnote to Prop. 10.48.Ô¦. Proposition 86
Εὑρεῖντὴνδευτέρανἀποτομήν.
To find a second apotome.
380

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
᾿ΕκκείσθωῥητὴἡΑκαὶτῇΑσύμμετροςμήκειἡΗΓ. Let the rational (straight-line) A, and GC (which is)
ῥητὴ ἄραἐστὶν ἡΗΓ. καὶἐκκείσθωσαν δύοτετράγωνοι commensurable in length with A, be laid down. Thus,
ἀριθμοὶοἱΔΕ,ΕΖ,ὧνἡὑπεροχὴὁΔΖμὴἔστωτετράγωνος. GC is a rational (straight-line). And let the two square
καὶ πεποιήσθω ὡς ὁ ΖΔ πρὸς τὸν ΔΕ, οὕτως τὸ ἀπὸ numbers DE and EF be laid down, and let their differτῆς
ΓΗ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΗΒ τετράγωνον. ence DF be not square [Prop. 10.28 lem. I]. And let it
σύμμετρονἄραἐστὶτὸἀπὸτῆςΓΗτετράγωνοντῷἀπὸ have been contrived that as FD (is) to DE, so the square
τῆςΗΒτετραγώνῳ. ῥητὸνδὲτὸἀπὸτῆςΓΗ.ῥητὸνἄρα on CG (is) to the square on GB [Prop. 10.6 corr.]. Thus,
[ἐστὶ]καὶτὸἀπὸτῆςΗΒ·ῥητὴἄραἐστὶνἡΒΗ.καὶἐπεὶτὸ the square on CG is commensurable with the square on
ἀπὸτῆςΗΓτετράγωνονπρὸςτὸἀπὸτῆςΗΒλόγονοὐκ GB [Prop. 10.6]. And the (square) on CG (is) rational.
ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, Thus, the (square) on GB [is] also rational. Thus, BG is a
ἀσύμμετρόςἐστινἡΓΗτῇΗΒμήκει. καίεἰσινἀμφότεραι rational (straight-line). And since the square on GC does
ῥηταί·αἱΓΗ,ΗΒἄραρηταίεἰσιδυνάμειμόνονσύμμετροι· not have to the (square) on GB the ratio which (some)
ἡΒΓἄραἀποτομήἐστιν.λέγωδή,ὅτικαὶδευτέρα. square number (has) to (some) square number, CG is
incommensurable in length with GB [Prop. 10.9]. And
they are both rational (straight-lines). Thus, CG and GB
are rational (straight-lines which are) commensurable in
square only. Thus, BC is an apotome [Prop. 10.73]. So,
I say that it is also a second (apotome).
Α
Θ
Β Γ
Ε Ζ
∆
Η
῟ΩιγὰρμεῖζόνἐστιτὸἀπὸτῆςΒΗτοῦἀπὸτῆςΗΓ, For let the (square) on H be that (area) by which
ἔστω τὸἀπὸτῆςΘ.ἐπεὶοὖνἐστινὡς τὸἀπὸτῆςΒΗ the (square) on BG is greater than the (square) on GC
πρὸςτὸἀπὸτῆςΗΓ,οὕτωςὁΕΔἀριθμὸςπρὸςτὸνΔΖ [Prop. 10.13 lem.]. Therefore, since as the (square) on
ἀριθμόν,ἀναστρέψαντιἄραἐστὶνὡςτὸἀπὸτῆςΒΗπρὸς BG is to the (square) on GC, so the number ED (is)
τὸἀπὸτῆςΘ,οὕτωςὁΔΕπρὸςτὸνΕΖ.καίἐστινἑκάτερος to the number DF , thus, also, via conversion, as the
τῶνΔΕ,ΕΖτετράγωνος·τὸἄραἀπὸτῆςΒΗπρὸςτὸἀπὸ (square) on BG is to the (square) on H, so DE (is)
τῆςΘλόγονἔχει,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον to EF [Prop. 5.19 corr.]. And DE and EF are each
ἀριθμόν·σύμμετροςἄραἐστὶνἡΒΗτῇΘμήκει.καὶδύναται square (numbers). Thus, the (square) on BG has to the
ἡΒΗτῆςΗΓμεῖζοντῷἀπὸτῆςΘ·ἡΒΗἄρατῆςΗΓ (square) on H the ratio which (some) square number
μεῖζονδύναταιτῷἀπὸσυμμέτρουἑαυτῇμήκει.καίἐστινἡ (has) to (some) square number. Thus, BG is commenπροσαρμόζουσαἡΓΗτῇἐκκειμένῃῥητῇσύμμετροςτῇΑ.
surable in length with H [Prop. 10.9]. And the square on
ἡΒΓἄραἀποτομήἐστιδευτέτα.
BG is greater than (the square on) GC by the (square)
ΕὕρηταιἄραδευτέραἀποτομὴἡΒΓ·ὅπερἔδειδεῖξαι. on H. Thus, the square on BG is greater than (the square
on) GC by the (square) on (some straight-line) commensurable
in length with (BG). And the attachment CG
is commensurable (in length) with the (prevously) laid
down rational (straight-line) A. Thus, BC is a second
apotome [Def. 10.12]. †
Thus, the second apotome BC has been found.
(Which is) the very thing it was required to show.
A
H
B
E
C
F
D
G
† See footnote to Prop. 10.49.ÔÞ. Proposition 87
Εὑρεῖντὴντρίτηνἀποτομήν.
To find a third apotome.
381

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
Α
Ε
Β
∆
Ζ Θ
Γ
Η
A
E
B D C
F H G
Κ
᾿Εκκείσθω ῥητὴἡΑ, καὶἐκκείσθωσαντρεῖς ἀριθμοὶ Let the rational (straight-line) A be laid down. And
οἱ Ε, ΒΓ, ΓΔ λόγον μὴ ἔχοντες πρὸς ἀλλήλους, ὃν let the three numbers, E, BC, and CD, not having to one
τετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν,ὁδὲΓΒ another the ratio which (some) square number (has) to
πρὸςτὸνΒΔλόγονἐχέτω,ὃντετράγωνοςἀριθμὸςπρὸς (some) square number, be laid down. And let CB have
τετράγωνονἀριθμόν,καὶπεποιήσθωὡςμὲνὁΕπρὸςτὸν to BD the ratio which (some) square number (has) to
ΒΓ, οὕτωςτὸ ἀπὸτῆς Α τετράγωνον πρὸς τὸ ἀπὸτῆς (some) square number. And let it have been contrived
ΖΗτετράγωνον,ὡςδὲὁΒΓπρὸςτὸνΓΔ,οὕτωςτὸἀπὸ that as E (is) to BC, so the square on A (is) to the
τῆςΖΗτετράγωνονπρὸςτὸἀπὸτὴςΗΘ.ἐπεὶοὖνἐστιν square on FG, and as BC (is) to CD, so the square on
ὡςὁΕπρὸςτὸνΒΓ,οὕτωςτὸἀπὸτῆςΑτετράγωνον FG (is) to the (square) on GH [Prop. 10.6 corr.]. Thereπρὸς
τὸ ἀπὸ τῆς ΖΗ τετράγωνον, σύμμετρον ἄρα ἐστὶ fore, since as E is to BC, so the square on A (is) to the
τὸ ἀπὸ τῆς Α τετράγωνον τῷ ἀπὸ τῆς ΖΗ τετραγώνῳ. square on FG, the square on A is thus commensurable
ῥητὸνδὲτὸἀπὸτῆςΑτετράγωνον.ῥητὸνἄρακαὶτὸἀπὸ with the square on FG [Prop. 10.6]. And the square on
τῆςΖΗ·ῥητὴἄραἐστὶνἡΖΗ.καὶἐπεὶὁΕπρὸςτὸνΒΓ A (is) rational. Thus, the (square) on FG (is) also raλόγονοὐκἔχει,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον
tional. Thus, FG is a rational (straight-line). And since
ἀριθμόν,οὐδ᾿ἄρατὸἀπὸτῆςΑτετράγωνονπρὸςτὸἀπὸ E does not have to BC the ratio which (some) square
τῆςΖΗ[τετράγωνον]λόγονἔχει,ὅντετράγωνοςἀριθμὸς number (has) to (some) square number, the square on A
πρὸςτετράγωνονἀριθμόν·ἀσύμμετροςἄραἐστὶνἡΑτῇ thus does not have to the [square] on FG the ratio which
ΖΗμήκει.πάλιν,ἐπείἐστινὡςὁΒΓπρὸςτὸνΓΔ,οὕτωςτὸ (some) square number (has) to (some) square number
ἀπὸτῆςΖΗτετράγωνονπρὸςτὸἀπὸτῆςΗΘ,σύμμετρον either. Thus, A is incommensurable in length with FG
ἄραἐστὶτὸἀπὸτῆςΖΗτῷἀπὸτῆςΗΘ.ῥητὸνδὲτὸἀπὸ [Prop. 10.9]. Again, since as BC is to CD, so the square
τῆςΖΗ·ῥητὸνἄρακαὶτὸἀπὸτῆςΗΘ·ῥητὴἄραἐστὶνἡΗΘ. on FG is to the (square) on GH, the square on FG is thus
καὶἐπεὶὁΒΓπρὸςτὸνΓΔλόγονοὐκἔχει,ὃντετράγωνος commensurable with the (square) on GH [Prop. 10.6].
ἀριθμὸςπρὸςτετράγωνονἀριθμόν,οὐδ᾿ἄρατὸἀπὸτῆςΖΗ And the (square) on FG (is) rational. Thus, the (square)
πρὸςτὸἀπὸτῆςΗΘλόγονἔχει,ὃντετράγωνοςἀριθμὸς on GH (is) also rational. Thus, GH is a rational (straightπρὸςτετράγωνονἀριθμόν·ἀσύμμετροςἄραἐστὶνἡΖΗτῇ
line). And since BC does not have to CD the ratio which
ΗΘμήκει.καίεἰσινἀμφότεραιῥηταί·αἱΖΗ,ΗΘἄραῥηταί (some) square number (has) to (some) square number,
εἰσιδυνάμειμόνονσύμμετροι· ἀποτομὴἄραἐστὶνἡΖΘ. the (square) on FG thus does not have to the (square)
λέγωδή,ὅτικαὶτρίτη.
on GH the ratio which (some) square number (has) to
᾿ΕπεὶγάρἐστινὡςμὲνὁΕπρὸςτὸνΒΓ,οὕτωςτὸἀπὸ (some) square number either. Thus, FG is incommenτῆςΑτετράγωνονπρὸςτὸἀπὸτῆςΖΗ,ὡςδὲὁΒΓπρὸς
surable in length with GH [Prop. 10.9]. And both are
τὸνΓΔ,οὕτωςτὸἀπὸτῆςΖΗπρὸςτὸἀπὸτῆςΘΗ,δι᾿ἴσου rational (straight-lines). FG and GH are thus rational
ἄραἐστὶνὡςὁΕπρὸςτὸνΓΔ,οὕτωςτὸἀπὸτῆςΑπρὸς (straight-lines which are) commensurable in square only.
τὸἀπὸτῆςΘΗ.ὁδὲΕπρὸςτὸνΓΔλόγονοὐκἔχει,ὃν Thus, FH is an apotome [Prop. 10.73]. So, I say that (it
τετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν·οὐδ᾿ἄρατὸ is) also a third (apotome).
ἀπὸτῆςΑπρὸςτὸἀπὸτῆςΗΘλόγονἔχει,ὃντετράγωνος For since as E is to BC, so the square on A (is) to the
ἀριθμὸςπρὸςτετράγωνονἀριθμόν·ἀσύμμετροςἄραἡΑτῇ (square) on FG, and as BC (is) to CD, so the (square)
ΗΘμήκει. οὐδετέραἄρατῶνΖΗ,ΗΘσύμμετρόςἐστιτῇ on FG (is) to the (square) on HG, thus, via equality, as
ἐκκειμένῃῥητῇτῇΑμήκει. ᾧοὖνμεῖζόνἐστιτὸἀπὸτῆς E is to CD, so the (square) on A (is) to the (square) on
ΖΗτοῦἀπὸτῆςΗΘ,ἔστωτὸἀπὸτῆςΚ.ἐπεὶοὖνἐστιν HG [Prop. 5.22]. And E does not have to CD the ra-
ὡςὁΒΓπρὸςτὸνΓΔ,οὕτωςτὸἀπὸτῆςΖΗπρὸςτὸἀπὸ tio which (some) square number (has) to (some) square
τῆςΗΘ,ἀναστρέψαντιἄραἐστὶνὡςὁΒΓπρὸςτὸνΒΔ, number. Thus, the (square) on A does not have to the
οὕτωςτὸἀπὸτῆςΖΗτετράγωνονπρὸςτὸἀπὸτῆςΚ.ὁδὲ (square) on GH the ratio which (some) square number
ΒΓπρὸςτὸνΒΔλόγονἔχει,ὃντετράγωνοςἀριθμὸςπρὸς (has) to (some) square number either. A (is) thus incomτετράγωνονἀριθμόν.
καὶτὸἁπὸτῆςΖΗἄραπρὸςτὸἀπὸ mensurable in length with GH [Prop. 10.9]. Thus, neiτῆςΚλόγονἔχει,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον
ther of FG and GH is commensurable in length with the
K
382

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ἀριθμόν.σύμμετρόςἄραἐστὶνἡΖΗτῇΚμήκει,καὶδύναται (previously) laid down rational (straight-line) A. There-
ἡΖΗτῆςΗΘμεῖζοντῷἀπὸσυμμέτρουἑαυτῇ.καὶοὐδετέρα fore, let the (square) on K be that (area) by which the
τῶνΖΗ,ΗΘσύμμετρόςἐστιτῇἐκκειμένῃῥητῇτῇΑμήκει· (square) on FG is greater than the (square) on GH
ἡΖΘἄραἀποτομήἐστιτρίτη.
[Prop. 10.13 lem.]. Therefore, since as BC is to CD, so
ΕὕρηταιἄραἡτρίτηἀποτομὴἡΖΘ·ὅπερἔδειδεῖξαι. the (square) on FG (is) to the (square) on GH, thus, via
conversion, as BC is to BD, so the square on FG (is) to
the square on K [Prop. 5.19 corr.]. And BC has to BD
the ratio which (some) square number (has) to (some)
square number. Thus, the (square) on FG also has to
the (square) on K the ratio which (some) square number
(has) to (some) square number. FG is thus commensurable
in length with K [Prop. 10.9]. And the square
on FG is (thus) greater than (the square on) GH by
the (square) on (some straight-line) commensurable (in
length) with (FG). And neither of FG and GH is commensurable
in length with the (previously) laid down rational
(straight-line) A. Thus, FH is a third apotome
[Def. 10.13].
Thus, the third apotome FH has been found. (Which
is) very thing it was required to show.
† See footnote to Prop. 10.50.Ô. Proposition 88
Εὑρεῖντὴντετάρτηνἀποτομήν.
Α
Θ
Β
∆
Γ
Ζ
Ε
Η
To find a fourth apotome.
᾿ΕκκείσθωῥητὴἡΑκαὶτῇΑμήκεισύμμετροςἡΒΗ· Let the rational (straight-line) A, and BG (which is)
ῥητὴἄραἐστὶκαὶἡΒΗ.καὶἐκκείσθωσανδύοἀριθμοὶοἱ commensurable in length with A, be laid down. Thus,
ΔΖ,ΖΕ,ὥστετὸνΔΕὅλονπρὸςἑκάτεροντῶνΔΖ,ΕΖ BG is also a rational (straight-line). And let the two
λόγονμὴἔχειν,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον numbers DF and FE be laid down such that the whole,
ἀριθμόν. καὶπεποιήσθωὡςὁΔΕπρὸςτὸνΕΖ,οὕτωςτὸ DE, does not have to each of DF and EF the ratio
ἀπὸτῆςΒΗτετράγωνονπρὸςτὸἀπὸτῆςΗΓ·σύμμετρον which (some) square number (has) to (some) square
ἄραἐστὶτὸἀπὸτῆςΒΗτῷἀπὸτῆςΗΓ.ῥητὸνδὲτὸἀπὸ number. And let it have been contrived that as DE (is) to
τῆςΒΗ·ῥητὸνἄρακαὶτὸἀπὸτῆςΗΓ·ῥητὴἄραἐστὶνἡΗΓ. EF , so the square on BG (is) to the (square) on GC
καὶἐπεὶὁΔΕπρὸςτὸνΕΖλόγονοὐκἔχει,ὃντετράγωνος [Prop. 10.6 corr.]. The (square) on BG is thus com-
ἀριθμὸςπρὸςτετράγωνονἀριθμόν,οὐδ᾿ἄρατὸἀπὸτῆςΒΗ mensurable with the (square) on GC [Prop. 10.6]. And
πρὸςτὸἀπὸτῆςΗΓλόγονἔχει,ὃντετράγωνοςἀριθμὸς the (square) on BG (is) rational. Thus, the (square) on
πρὸςτετράγωνονἀριθμόν·ἀσύμμετροςἄραἐστὶνἡΒΗτῇ GC (is) also rational. Thus, GC (is) a rational (straight-
ΗΓμήκει.καίεἰσινἀμφότεραιῥηταί·αἱΒΗ,ΗΓἄραῥηταί line). And since DE does not have to EF the ratio which
εἰσιδυνάμειμόνονσύμμετροι· ἀποτομὴἄραἐστὶνἡΒΓ. (some) square number (has) to (some) square number,
[λέγωδή,ὅτικαὶτετάρτη.]
the (square) on BG thus does not have to the (square)
῟ΩιοὖνμεῖζόνἐστιτὸἀπὸτῆςΒΗτοῦἀπὸτῆςΗΓ, on GC the ratio which (some) square number (has) to
ἔστωτὸἀπὸτῆςΘ.ἐπεὶοὖνἐστιν ὡςὁΔΕπρὸςτὸν (some) square number either. Thus, BG is incommensu-
ΕΖ,οὕτωςτὸἀπὸτῆςΒΗπρὸςτὸἀπὸτῆςΗΓ,καὶἀνα- rable in length with GC [Prop. 10.9]. And they are both
στρέψαντι ἄρα ἐστὶν ὡς ὁ ΕΔ πρὸς τὸν ΔΖ, οὕτως τὸ rational (straight-lines). Thus, BG and GC are rational
ἀπὸτῆςΗΒπρὸςτὸἀπὸτῆςΘ.ὁδὲΕΔπρὸςτὸνΔΖ (straight-lines which are) commensurable in square only.
λόγονοὐκἔχει,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον Thus, BC is an apotome [Prop. 10.73]. [So, I say that (it
A
H
B
E
C
F
D
G
383

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ἀριθμόν·οὐδ᾿ἄρατὸἀπὸτῆςΗΒπρὸςτὸἀπὸτῆςΘλόγον is) also a fourth (apotome).]
ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· Now, let the (square) on H be that (area) by which
ἀσύμμετροςἄραἐστὶνἡΒΗτῇΘμήκει. καὶδύναταιἡ the (square) on BG is greater than the (square) on GC
ΒΗτῆςΗΓμεῖζοντῷἀπὸτῆςΘ·ἡἄραΒΗτῆςΗΓμεῖζον [Prop. 10.13 lem.]. Therefore, since as DE is to EF ,
δύναταιτῷἀπὸἀσυμμέτρουἑαυτῇ. καὶἐστινὅληἡΒΗ so the (square) on BG (is) to the (square) on GC, thus,
σύμμετροςτῇἐκκειμένῃῥητῇμήκειτῇΑ.ἡἄραΒΓἀπο- also, via conversion, as ED is to DF , so the (square) on
τομήἐστιτετάρτη.
Εὕρηταιἄραἡτετάρτηἀποτομή·ὅπερἔδειδεῖξαι.
GB (is) to the (square) on H [Prop. 5.19 corr.]. And ED
does not have to DF the ratio which (some) square number
(has) to (some) square number. Thus, the (square) on
GB does not have to the (square) on H the ratio which
(some) square number (has) to (some) square number
either. Thus, BG is incommensurable in length with H
[Prop. 10.9]. And the square on BG is greater than (the
square on) GC by the (square) on H. Thus, the square on
BG is greater than (the square) on GC by the (square) on
(some straight-line) incommensurable (in length) with
(BG). And the whole, BG, is commensurable in length
with the the (previously) laid down rational (straightline)
A. Thus, BC is a fourth apotome [Def. 10.14]. †
Thus, a fourth apotome has been found. (Which is)
the very thing it was required to show.
† See footnote to Prop. 10.51.Ô. Proposition 89
Εὐρεῖντὴνπέμπτηνἀποτομήν.
To find a fifth apotome.
Β Γ Η
B C G
Α
A
∆ Ζ Ε
D F E
Θ
H
᾿ΕκκείσθωῥητὴἡΑ,καὶτῇΑμήκεισύμμετροςἔστωἡ Let the rational (straight-line) A be laid down, and let
ΓΗ·ῥητὴἄρα[ἐστὶν]ἡΓΗ.καὶἐκκείσθωσανδύοἀριθμοὶοἱ CG be commensurable in length with A. Thus, CG [is]
ΔΖ,ΖΕ,ὥστετὸνΔΕπρὸςἑκάτεροντῶνΔΖ,ΖΕλόγον a rational (straight-line). And let the two numbers DF
πάλινμὴἔχειν,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον and FE be laid down such that DE again does not have
ἀριθμόν·καὶπεποιήσθωὡςὁΖΕπρὸςτὸνΕΔ,οὕτωςτὸ to each of DF and FE the ratio which (some) square
ἀπὸτῆςΓΗπρὸςτὸἀπὸτῆςΗΒ.ῥητὸνἄρακαὶτὸἀπὸτῆς number (has) to (some) square number. And let it have
ΗΒ·ῥητὴἄραἐστὶκαὶἡΒΗ.καὶἐπείἐστινὡςὁΔΕπρὸς been contrived that as FE (is) to ED, so the (square) on
τὸνΕΖ,οὕτωςτὸἀπὸτῆςΒΗπρὸςτὸἀπὸτῆςΗΓ,ὁδὲ CG (is) to the (square) on GB. Thus, the (square) on GB
ΔΕπρὸςτὸνΕΖλόγονοὐκἔχει,ὃντετράγωνοςἀριθμὸς (is) also rational [Prop. 10.6]. Thus, BG is also rational.
πρὸςτετράγωνονἀριθμόν,οὐδ᾿ἄρατὸἀπὸτῆςΒΗπρὸς And since as DE is to EF , so the (square) on BG (is) to
τὸἀπὸτῆςΗΓλόγονἔχει,ὃντετράγωνοςἀριθμὸςπρὸς the (square) on GC. And DE does not have to EF the raτετράγωνονἀριθμόν·ἀσύμμετροςἄραἐστὶνἡΒΗτῇΗΓ
tio which (some) square number (has) to (some) square
μήκει.καίεἰσινἀμφότεραιῥηταί·αἱΒΗ,ΗΓἄραῥηταίεἰσι number. The (square) on BG thus does not have to the
δυνάμειμόνονσύμμετροι·ἡΒΓἄραἀποτομήἐστιν. λέγω (square) on GC the ratio which (some) square number
δή,ὅτικαὶπέμπτη.
(has) to (some) square number either. Thus, BG is in-
῟ΩιγὰρμεῖζόνἐστιτὸἀπὸτῆςΒΗτοῦἀπὸτῆςΗΓ, commensurable in length with GC [Prop. 10.9]. And
ἔστωτὸἀπὸτῆςΘ.ἐπεὶοὖνἐστινὡςτὸἀπὸτὴςΒΗπρὸς they are both rational (straight-lines). BG and GC are
τὸἀπὸτῆςΗΓ,οὕτωςὁΔΕπρὸςτὸνΕΖ,ἀναστρέψαντι thus rational (straight-lines which are) commensurable
ἄραἐστὶνὡςὁΕΔπρὸςτὸνΔΖ,οὕτωςτὸἀπὸτῆςΒΗπρὸς in square only. Thus, BC is an apotome [Prop. 10.73].
τὸἀπὸτῆςΘ,ὁδὲΕΔπρὸςτὸνΔΖλόγονοὐκἔχει,ὃν So, I say that (it is) also a fifth (apotome).
384

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
τετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν·οὐδ᾿ἄρατὸ For, let the (square) on H be that (area) by which
ἀπὸτῆςΒΗπρὸςτὸἀπὸτῆςΘλόγονἔχει,ὃντετράγωνος the (square) on BG is greater than the (square) on GC
ἀριθμὸςπρὸςτετράγωνονἀριθμόν·ἀσύμμετροςἄραἐστὶνἡ [Prop. 10.13 lem.]. Therefore, since as the (square) on
ΒΗτῇΘμήκει. καὶδύναταιἡΒΗτῆςΗΓμεῖζοντῷἀπὸ BG (is) to the (square) on GC, so DE (is) to EF , thus,
τῆςΘ·ἡΗΒἄρατῆςΗΓμεῖζονδύναταιτῷἀπὸἀσυμμέτρου via conversion, as ED is to DF , so the (square) on BG
ἑαυτῇμήκει. καίἐστινἡπροσαρμόζουσαἡΓΗσύμμετρος (is) to the (square) on H [Prop. 5.19 corr.]. And ED does
τῇ ἐκκειμένῃῥητῇτῇ Α μήκει· ἡἄραΒΓἀποτομήἐστι not have to DF the ratio which (some) square number
πέμπτη.
(has) to (some) square number. Thus, the (square) on
ΕὕρηταιἄραἡπέμπτηἀποτομὴἡΒΓ·ὅπερἔδειδεῖξαι. BG does not have to the (square) on H the ratio which
(some) square number (has) to (some) square number
either. Thus, BG is incommensurable in length with H
[Prop. 10.9]. And the square on BG is greater than (the
square on) GC by the (square) on H. Thus, the square on
GB is greater than (the square on) GC by the (square)
on (some straight-line) incommensurable in length with
(GB). And the attachment CG is commensurable in
length with the (previously) laid down rational (straightline)
A. Thus, BC is a fifth apotome [Def. 10.15]. †
Thus, the fifth apotome BC has been found. (Which
is) the very thing it was required to show.
† See footnote to Prop. 10.52.. Proposition 90
Εὑρεῖντὴνἕκτηνἀποτομήν.
To find a sixth apotome.
᾿ΕκκείσθωῥητὴἡΑκαὶτρεῖςἀριθμοὶοἱΕ,ΒΓ,ΓΔ Let the rational (straight-line) A, and the three numλόγονμὴἔχοντεςπρὸςἀλλήλους,ὃντετράγωνοςἀριθμὸς
bers E, BC, and CD, not having to one another the raπρὸςτετράγωνονἀριθμόν·ἔτιδὲκαὶὁΓΒπρὸςτὸνΒΔ
tio which (some) square number (has) to (some) square
λόγονμὴἐχετώ,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον number, be laid down. Furthermore, let CB also not have
ἀριθμόν·καὶπεποιήσθωὡςμὲνὁΕπρὸςτὸνΒΓ,οὕτως to BD the ratio which (some) square number (has) to
τὸἀπὸτῆςΑπρὸςτὸἀπὸτῆςΖΗ,ὡςδὲὁΒΓπρὸςτὸν (some) square number. And let it have been contrived
ΓΔ,οὕτωςτὸἀπὸτῆςΖΗπρὸςτὸἀπὸτῆςΗΘ. that as E (is) to BC, so the (square) on A (is) to the
(square) on FG, and as BC (is) to CD, so the (square)
on FG (is) to the (square) on GH [Prop. 10.6 corr.].
Α
Ε
Κ
Β
Ζ Θ
∆
Γ
Η
᾿ΕπεὶοὖνἐστινὡςὁΕπρὸςτὸνΒΓ,οὕτωςτὸἀπὸτῆς Therefore, since as E is to BC, so the (square) on A
ΑπρὸςτὸἀπὸτῆςΖΗ,σύμμετρονἄρατὸἀπὸτῆςΑτῷ (is) to the (square) on FG, the (square) on A (is) thus
ἀπὸτῆςΖΗ.ῥητὸνδὲτὸἀπὸτῆςΑ·ῥητὸνἄρακαὶτὸἀπὸ commensurable with the (square) on FG [Prop. 10.6].
τῆςΖΗ·ῥητὴἄραἐστὶκαὶἡΖΗ.καὶἐπεὶὁΕπρὸςτὸνΒΓ And the (square) on A (is) rational. Thus, the (square)
λόγονοὐκἔχει,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον on FG (is) also rational. Thus, FG is also a rational
ἀριθμόν,οὐδ᾿ἄρατὸἀπὸτῆςΑπρὸςτὸἀπὸτῆςΖΗλόγον (straight-line). And since E does not have to BC the ra-
ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· tio which (some) square number (has) to (some) square
ἀσύμμετροςἄραἐστὶνἡΑτῆΖΗμήκει.πάλιν,ἐπείἐστινὡς number, the (square) on A thus does not have to the
ὁΒΓπρὸςτὸνΓΔ,οὕτωςτὸἀπὸτῆςΖΗπρὸςτὸἀπὸτῆς (square) on FG the ratio which (some) square number
ΗΘ,σύμμετρονἄρατὸἀπὸτῆςΖΗτῷἀπὸτῆςΗΘ.ῥητὸν (has) to (some) square number either. Thus, A is in-
A
E
K
B
F
H
D
C
G
385

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
δὲτὸἀπὸτῆςΖΗ·ῥητὸνἄρακαὶτὸἀπὸτῆςΗΘ·ῥητὴἄρα commensurable in length with FG [Prop. 10.9]. Again,
καὶἡΗΘ.καὶἐπεὶὁΒΓπρὸςτὸνΓΔλόγονοὐκἔχει,ὃν since as BC is to CD, so the (square) on FG (is) to the
τετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν,οὐδ᾿ἄρατὸ (square) on GH, the (square) on FG (is) thus commen-
ἀπὸτῆςΖΗπρὸςτὸἀπὸτῆςΗΘλόγονἔχει,ὃντετράγωνος surable with the (square) on GH [Prop. 10.6]. And the
ἀριθμὸςπρὸςτετράγωνονἀριθμόν·ἀσύμμετροςἄραἐστὶν (square) on FG (is) rational. Thus, the (square) on GH
ἡΖΗτῇΗΘμήκει. καίεἰσινἀμφότεραιῥηταί·αἱΖΗ,ΗΘ (is) also rational. Thus, GH (is) also rational. And since
ἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι·ἡἄραΖΘἀποτομή BC does not have to CD the ratio which (some) square
ἐστιν.λέγωδή,ὅτικαὶἕκτη.
number (has) to (some) square number, the (square) on
᾿ΕπεὶγάρἐστινὡςμὲνὁΕπρὸςτὸνΒΓ,οὕτωςτὸἀπὸ FG thus does not have to the (square) on GH the raτῆςΑπρὸςτὸἀπὸτῆςΖΗ,ὡςδὲὁΒΓπρὸςτὸνΓΔ,οὕτως
tio which (some) square (number) has to (some) square
τὸἀπὸτῆςΖΗπρὸςτὸἀπὸτῆςΗΘ,δι᾿ἴσουἄραἐστὶνὡςὁ (number) either. Thus, FG is incommensurable in length
ΕπρὸςτὸνΓΔ,οὕτωςτὸἀπὸτῆςΑπρὸςτὸἀπὸτῆςΗΘ.ὁ with GH [Prop. 10.9]. And both are rational (straightδὲΕπρὸςτὸνΓΔλόγονοὐκἔχει,ὃντετράγωνοςἀριθμὸς
lines). Thus, FG and GH are rational (straight-lines
πρὸςτετράγωνονἀριθμόν· οὐδ᾿ἄρατὸἀπὸτῆςΑπρὸς which are) commensurable in square only. Thus, FH is
τὸἀπὸτῆςΗΘλόγονἔχει,ὃντετράγωνοςἀριθμὸςπρὸς an apotome [Prop. 10.73]. So, I say that (it is) also a
τετράγωνονἀριθμόν· ἀσύμμετροςἄραἐστὶνἡΑτῇΗΘ sixth (apotome).
μήκει·οὐδετέραἄρατῶνΖΗ,ΗΘσύμμετρόςἐστιτῇΑῥητῇ For since as E is to BC, so the (square) on A (is)
μήκει.ᾧοὖνμεῖζόνἐστιτὸἀπὸτῆςΖΗτοῦἀπὸτῆςΗΘ, to the (square) on FG, and as BC (is) to CD, so the
ἔστωτὸἀπὸτῆςΚ.ἐπεὶοὖνἐστινὡςὁΒΓπρὸςτὸνΓΔ, (square) on FG (is) to the (square) on GH, thus, via
οὕτωςτὸἀπὸτῆςΖΗπρὸςτὸἀπὸτῆςΗΘ,ἀναστρέψαντι equality, as E is to CD, so the (square) on A (is) to
ἄραἐστὶνὡςὁΓΒπρὸςτὸνΒΔ,οὕτωςτὸἀπὸτῆςΖΗπρὸς the (square) on GH [Prop. 5.22]. And E does not have
τὸἀπὸτῆςΚ.ὁδὲΓΒπρὸςτὸνΒΔλόγονοὐκἔχει,ὃν to CD the ratio which (some) square number (has) to
τετράγωνοςἀριθμὸςπρὸςτετράγωνονἀριθμόν·οὐδ᾿ἄρατὸ (some) square number. Thus, the (square) on A does not
ἀπὸτῆςΖΗπρὸςτὸἀπὸτῆςΚλόγονἔχει,ὃντετράγωνος have to the (square) GH the ratio which (some) square
ἀριθμὸςπρὸςτετράγωνονἀριθμόν·ἀσύμμετροςἄραἐστὶνἡ number (has) to (some) square number either. A is thus
ΖΗτῇΚμήκει. καὶδύναταιἡΖΗτῆςΗΘμεῖζοντῷἀπὸ incommensurable in length with GH [Prop. 10.9]. Thus,
τῆςΚ·ἡΖΗἄρατῆςΗΘμεῖζονδύναταιτῷἀπὸἀσυμμέτρου neither of FG and GH is commensurable in length with
ἑαυτῇμήκει.καὶοὐδετέρατῶνΖΗ,ΗΘσύμμετρόςἐστιτῇ the rational (straight-line) A. Therefore, let the (square)
ἐκκειμένῃῥητῇμήκειτῇΑ.ἡἄραΖΘἀποτομήἐστινἕκτη. on K be that (area) by which the (square) on FG is
ΕὕρηταιἄραἡἕκτηἀποτομὴἡΖΘ·ὅπερἔδειδεῖξαι. greater than the (square) on GH [Prop. 10.13 lem.].
Therefore, since as BC is to CD, so the (square) on FG
(is) to the (square) on GH, thus, via conversion, as CB is
to BD, so the (square) on FG (is) to the (square) on K
[Prop. 5.19 corr.]. And CB does not have to BD the ratio
which (some) square number (has) to (some) square
number. Thus, the (square) on FG does not have to the
(square) on K the ratio which (some) square number
(has) to (some) square number either. FG is thus incommensurable
in length with K [Prop. 10.9]. And the
square on FG is greater than (the square on) GH by the
(square) on K. Thus, the square on FG is greater than
(the square on) GH by the (square) on (some straightline)
incommensurable in length with (FG). And neither
of FG and GH is commensurable in length with the (previously)
laid down rational (straight-line) A. Thus, FH
is a sixth apotome [Def. 10.16].
Thus, the sixth apotome FH has been found. (Which
is) the very thing it was required to show.
† See footnote to Prop. 10.53.
386

ËÌÇÁÉÁÏƺ ELEMENTS BOOK 10
. Proposition 91
᾿Εὰνχωρίονπεριέχηταιὑπὸῥητῆςκαὶἀποτομῆςπρώτης, If an area is contained by a rational (straight-line) and
ἡτὸχωρίονδυναμένηἀπορομήἐστιν.
a first apotome then the square-root of the area is an apo-
ΠεριεχέσθωγὰρχωρίοντὸΑΒὑπὸῥητῆςτῆςΑΓκαὶ tome.
ἀποτομῆςπρώτηςτῆςΑΔ·λέγω,ὅτιἡτὸΑΒχωρίονδυ- For let the area AB have been contained by the ratioναμένηἀποτομήἐστιν.
nal (straight-line) AC and the first apotome AD. I say
that the square-root of area AB is an apotome.
Α ∆ Ε Ζ Η Ν
A D E F G
N
Λ Ο
L
P
Φ
V
Γ Β Θ Ι Κ
Σ
Υ
Χ
Ξ
Ρ Μ
R
M
Τ
T
᾿ΕπεὶγὰρἀποτομήἐστιπρώτηἡΑΔ,ἔστωαὐτῇπρο- For since AD is a first apotome, let DG be its atσαρμόζουσαἡΔΗ·αἱΑΗ,ΗΔἄραῥηταίεἰσιδυνάμειμόνον
tachment. Thus, AG and DG are rational (straight-lines
σύμμετροι.καὶὅληἡΑΗσύμμετρόςἐστιτῇἐκκειμένῃῥητῇ which are) commensurable in square only [Prop. 10.73].
τῇΑΓ,καὶἡΑΗτῆςΗΔμεῖζονδύναταιτῷἀπὸσυμμέτρου And the whole, AG, is commensurable (in length) with
ἑαυτῇμήκει·ἐὰνἄρατῷτετάρτῳμέρειτοῦἀπὸτῆςΔΗ the (previously) laid down rational (straight-line) AC,
ἴσονπαρὰτὴνΑΗπαραβληθῇἐλλεῖπονεἴδειτετραγώνῳ, and the square on AG is greater than (the square on)
εἰςσύμμετρααὐτὴνδιαιρεῖ. τετμήσθωἡΔΗδίχακατὰτὸ GD by the (square) on (some straight-line) commensu-
Ε,καὶτῷἀπὸτῆςΕΗἴσονπαρὰτὴνΑΗπαραβεβλήσθω rable in length with (AG) [Def. 10.11]. Thus, if (an area)
ἐλλεῖπονεἴδειτετραγώνῳ,καὶἔστωτὸὑπὸτῶνΑΖ,ΖΗ· equal to the fourth part of the (square) on DG is applied
σύμμετροςἄραἐστὶνἡΑΖτῇΖΗ.καὶδιὰτῶνΕ,Ζ,Η to AG, falling short by a square figure, then it divides
σημείωντῇΑΓπαράλληλοιἤχθωσαναἱΕΘ,ΖΙ,ΗΚ. (AG) into (parts which are) commensurable (in length)
ΚαὶἐπεὶσύμμετρόςἐστινἡΑΖτῇΖΗμήκει,καὶἡΑΗ [Prop. 10.17]. Let DG have been cut in half at E. And
ἄραἑκατέρᾳτῶνΑΖ,ΖΗσύμμετρόςἐστιμήκει. ἀλλὰἡ let (an area) equal to the (square) on EG have been ap-
ΑΗσύμμετρόςἐστιτῇΑΓ·καὶἑκατέραἄρατῶνΑΖ,ΖΗ plied to AG, falling short by a square figure. And let it
σύμμετρόςἐστιτῇΑΓμήκει. καίἐστιῥητὴἡΑΓ·ῥητὴ be the (rectangle contained) by AF and FG. AF is thus
ἄρακαὶἑκατέρατῶνΑΖ,ΖΗ·ὥστεκαὶἑκάτεροντῶνΑΙ, commensurable (in length) with FG. And let EH, FI,
ΖΚῥητόνἐστιν. καὶἐπεὶσύμμετρόςἐστινἡΔΕτῇΕΗ and GK have been drawn through points E, F , and G
μήκει,καὶἡΔΗἄραἑκατέρᾳτῶνΔΕ,ΕΗσύμμετρόςἐστι (respectively), parallel to AC.
μήκει. ῥητὴδὲἡΔΗκαὶἀσύμμετροςτῇΑΓμήκει·ῥητὴ And since AF is commensurable in length with FG,
ἄρακαὶἑκατέρατῶνΔΕ,ΕΗκαὶἀσύμμετροςτῇΑΓμήκει· AG is thus also commensurable in length with each of
ἑκάτερονἄρατῶνΔΘ,ΕΚμέσονἐστίν.
AF and FG [Prop. 10.15]. But AG is commensurable
ΚείσθωδὴτῷμὲνΑΙἴσοντετράγωνοντὸΛΜ,τῷδὲ (in length) with AC. Thus, each of AF and FG is also
ΖΚἴσοντετράγωνονἀφῃρήσθωκοινὴνγωνίανἔχοναὐτῷ commensurable in length with AC [Prop. 10.12]. And
τὴνὑπὸΛΟΜτὸΝΞ·περὶτὴναὐτὴνἄραδιάμετρόνἐστιτὰ AC is a rational (straight-line). Thus, AF and FG (are)
ΛΜ,ΝΞτετράγωνα.ἔστωαὐτῶνδιάμετροςἡΟΡ,καὶκα- each also rational (straight-lines). Hence, AI and FK
ταγεγράφθωτὸσχῆμα.ἐπεὶοὖνἴσονἐστὶτὸὑπὸτῶνΑΖ, are also each rational (areas) [Prop. 10.19]. And since
ΖΗπεριεχόμενονὀρθογώνιοντῷἀπὸτῆςΕΗτετραγώνῳ, DE is commensurable in length with EG, DG is thus
ἔστινἄραὡςἡΑΖπρὸςτὴνΕΗ,οὕτωςἡΕΗπρὸςτὴνΖΗ. also commensurable in length with each of DE and EG
ἀλλ᾿ὡςμὲνἡΑΖπρὸςτὴνΕΗ,οὕτωςτὸΑΙπρὸςτὸΕΚ, [Prop. 10.15]. And DG (is) rational, and incommen-
ὡςδὲἡΕΗπρὸςτὴνΖΗ,οὕτωςἐστὶτὸΕΚπρὸςτὸΚΖ· surable in length with AC. DE and EG (are) thus
τῶνἄραΑΙ,ΚΖμέσονἀνάλογόνἐστιτὸΕΚ.ἔστιδὲκαὶ each rational, and incommensurable in length with AC
τῶνΛΜ,ΝΞμέσονἀνάλογοντὸΜΝ,ὡςἐντοῖςἔμπρο- [Prop. 10.13]. Thus, DH and EK are each medial (arσθενἐδείχθη,καίἐστιτὸ[μὲν]ΑΙτῷΛΜτετραγώνῳἴσον,
eas) [Prop. 10.21].
τὸδὲΚΖτῷΝΞ·καὶτὸΜΝἄρατῷΕΚἴσονἐστίν.ἀλλὰ So let the square LM, equal to AI, be laid down.
τὸμὲνΕΚτῷΔΘἐστινἴσον,τὸδὲΜΝτῷΛΞ·τὸἄρα And let the square NO, equal to FK, have been sub-
C
B
H
I
K
S
U
W
O
387

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ΔΚἴσονἐστὶτῷΥΦΧγνώμονικαὶτῷΝΞ.ἔστιδὲκαὶτὸ tracted (from LM), having with it the common angle
ΑΚἴσοντοῖςΛΜ,ΝΞτετραγώνοις·λοιπὸνἄρατὸΑΒἴσον LPM. Thus, the squares LM and NO are about the
ἐστὶτῷΣΤ.τὸδὲΣΤτὸἀπὸτῆςΛΝἐστιτετράγωνον·τὸ same diagonal [Prop. 6.26]. Let PR be their (com-
ἄραἀπὸτῆςΛΝτετράγωνονἴσονἐστὶτῷΑΒ·ἡΛΝἄρα mon) diagonal, and let the (rest of the) figure have been
δύναταιτὸΑΒ.λέγωδή,ὅτιἡΛΝἀποτομήἐστιν. drawn. Therefore, since the rectangle contained by AF
᾿ΕπεὶγὰρῥητόνἐστινἑκάτεροντῶνΑΙ,ΖΚ,καίἐστιν and FG is equal to the square EG, thus as AF is to
ἴσοντοῖςΛΜ,ΝΞ,καὶἑκάτερονἄρατῶνΛΜ,ΝΞῥητόν EG, so EG (is) to FG [Prop. 6.17]. But, as AF (is)
ἐστιν,τουτέστιτὸἀπὸἑκατέραςτῶνΛΟ,ΟΝ·καὶἑκατέρα to EG, so AI (is) to EK, and as EG (is) to FG, so
ἄρατῶνΛΟ,ΟΝῥητήἐστιν.πάλιν,ἐπεὶμέσονἐστὶτὸΔΘ EK is to KF [Prop. 6.1]. Thus, EK is the mean proκαίἐστινἴσοντῷΛΞ,μέσονἄραἐστὶκαὶτὸΛΞ.ἐπεὶοὖντὸ
portional to AI and KF [Prop. 5.11]. And MN is also
μὲνΛΞμέσονἐστίν,τὸδὲΝΞῥητόν,ἀσύμμετρονἄραἐστὶ the mean proportional to LM and NO, as shown before
τὸΛΞτῷΝΞ·ὡςδὲτὸΛΞπρὸςτὸΝΞ,οὕτωςἐστὶνἡΛΟ [Prop. 10.53 lem.]. And AI is equal to the square LM,
πρὸςτὴνΟΝ·ἀσύμμετροςἄραἐστὶνἡΛΟτῇΟΝμήκει. and KF to NO. Thus, MN is also equal to EK. But, EK
καίεἰσινἀμφότεραιῥηταί·αἱΛΟ,ΟΝἄραῥηταίεἰσιδυνάμει is equal to DH, and MN to LO [Prop. 1.43]. Thus, DK
μόνονσύμμετροι·ἀποτομὴἄραἐστὶνἡΛΝ.καὶδύναταιτὸ is equal to the gnomon UV W and NO. And AK is also
ΑΒχωρίον·ἡἄρατὸΑΒχωρίονδυναμένηἀποτομήἐστιν. equal to (the sum of) the squares LM and NO. Thus, the
᾿Εὰνἄραχωρίονπεριέχηταιὑπὸῥητῆςκαὶτὰἑξῆς. remainder AB is equal to ST . And ST is the square on
LN. Thus, the square on LN is equal to AB. Thus, LN is
the square-root of AB. So, I say that LN is an apotome.
For since AI and FK are each rational (areas), and
are equal to LM and NO (respectively), thus LM and
NO—that is to say, the (squares) on each of LP and PN
(respectively)—are also each rational (areas). Thus, LP
and PN are also each rational (straight-lines). Again,
since DH is a medial (area), and is equal to LO, LO
is thus also a medial (area). Therefore, since LO is
medial, and NO rational, LO is thus incommensurable
with NO. And as LO (is) to NO, so LP is to PN
[Prop. 6.1]. LP is thus incommensurable in length with
PN [Prop. 10.11]. And they are both rational (straightlines).
Thus, LP and PN are rational (straight-lines
which are) commensurable in square only. Thus, LN is
an apotome [Prop. 10.73]. And it is the square-root of
area AB. Thus, the square-root of area AB is an apotome.
Thus, if an area is contained by a rational (straightline),
and so on . . . .
. Proposition 92
᾿Εὰνχωρίονπεριέχηταιὑπὸῥητῆςκαὶἀποτομῆςδευτέρας, If an area is contained by a rational (straight-line) and
ἡτὸχωρίονδυναμένημέσηςἀποτομήἐστιπρώτη. a second apotome then the square-root of the area is a
first apotome of a medial (straight-line).
Α ∆ Ε Ζ Η Ν
A D E F G
N
Λ Ο
L
O
Φ
V
Γ Β Θ Ι Κ
Σ
Χ
Υ
Π
Ξ
C
B
H
I
K
S
W
U
Q
P
Ρ
Τ
Μ
R
T
M
388

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ΧωρίονγὰρτὸΑΒπεριεχέσθωὑπὸῥητῆςτῆςΑΓκαὶ For let the area AB have been contained by the ratio-
ἀποτομῆςδευτέραςτῆςΑΔ·λέγω, ὅτιἡτὸΑΒχωρίον nal (straight-line) AC and the second apotome AD. I say
δυναμένημέσηςἀποτομήἐστιπρώτη.
that the square-root of area AB is the first apotome of a
῎Εστω γὰρ τῇ ΑΔ προσαρμόζουσα ἡ ΔΗ· αἱ ἄρα medial (straight-line).
ΑΗ,ΗΔῥηταίεἰσιδυνάμειμόνονσύμμετροι,καὶἡπρο- For let DG be an attachment to AD. Thus, AG and
σαρμόζουσαἡΔΗσύμμετρόςἐστιτῇἐκκειμένῃῥητῇτῇ GD are rational (straight-lines which are) commensu-
ΑΓ,ἡδὲὅληἡΑΗτῆςπροσαρμοζούσηςτῆςΗΔμεῖζον rable in square only [Prop. 10.73], and the attachment
δύναταιτῷἀπὸσυμμέτρουἑαυτῇμήκει. ἐπεὶοὖνἡΑΗ DG is commensurable (in length) with the (previously)
τῆςΗΔμεῖζονδύναταιτῷἀπὸσυμμέτρουἑαυτῇ,ἐὰνἄρα laid down rational (straight-line) AC, and the square on
τῷτετάρτῳμέρειτοῦἀπὸτῆςΗΔἴσονπαρὰτὴνΑΗπα- the whole, AG, is greater than (the square on) the atραβληθῇ
ἐλλεῖπον εἴδει τετραγώνῳ, εἰς σύμμετρα αὐτὴν tachment, GD, by the (square) on (some straight-line)
διαιρεῖ. τετμήσθωοὖνἡΔΗδίχακατὰτὸΕ·καὶτῷἀπὸ commensurable in length with (AG) [Def. 10.12]. ThereτῆςΕΗἴσονπαρὰτὴνΑΗπαραβεβλήσθωἐλλεῖπονεἴδει
fore, since the square on AG is greater than (the square
τετραγώνῳ,καὶἔστωτὸὑπὸτῶνΑΖ,ΖΗ·σύμμετροςἄρα on) GD by the (square) on (some straight-line) commen-
ἐστὶνἡΑΖτῇΖΗμήκει. καὶἡΑΗἄραἑκατέρᾳτῶνΑΖ, surable (in length) with (AG), thus if (an area) equal
ΖΗσύμμετρόςἐστιμήκει. ῥητὴδὲἡΑΗκαὶἀσύμμετρος to the fourth part of the (square) on GD is applied
τῇΑΓμήκει·καὶἑκατέραἄρατῶνΑΖ,ΖΗῥητήἐστικαὶ to AG, falling short by a square figure, then it divides
ἀσύμμετροςτῇΑΓμήκει·ἑκάτερονἄρατῶνΑΙ,ΖΚμέσον (AG) into (parts which are) commensurable (in length)
ἐστίν.πάλιν,ἐπεὶσύμμετρόςἐστινἡΔΕτῇΕΗ,καὶἡΔΗ [Prop. 10.17]. Therefore, let DG have been cut in half at
ἄραἑκατέρᾳτῶνΔΕ,ΕΗσύμμετρόςἐστιν. ἀλλ᾿ἡΔΗ E. And let (an area) equal to the (square) on EG have
σύμμετρόςἐστιτῇΑΓμήκει[ῥητὴἄρακαὶἑκατέρατῶν been applied to AG, falling short by a square figure. And
ΔΕ,ΕΗκαὶσύμμετροςτῇΑΓμήκει]. ἑκάτερονἄρατῶν let it be the (rectangle contained) by AF and FG. Thus,
ΔΘ,ΕΚῥητόνἐστιν.
AF is commensurable in length with FG. AG is thus
ΣυνεστάτωοὖντῷμὲνΑΙἴσοντετράγωνοντὸΛΜ,τῷ also commensurable in length with each of AF and FG
δὲΖΚἴσονἀφῃρήσθωτὸΝΞπερὶτὴναὐτὴνγωνίανὂντῷ [Prop. 10.15]. And AG (is) a rational (straight-line), and
ΛΜτὴνὑπὸτῶνΛΟΜ·περὶτὴναὐτὴνἄραἐστὶδιάμετρον incommensurable in length with AC. AF and FG are
τὰΛΜ,ΝΞτετράγωνα. ἔστωαὐτῶνδιάμετροςἡΟΡ,καὶ thus also each rational (straight-lines), and incommenκαταγεγράφθωτὸσχῆμα.
ἐπεὶοὖντὰΑΙ,ΖΚμέσαἐστὶ surable in length with AC [Prop. 10.13]. Thus, AI and
καίἐστινἴσατοῖςἀπὸτῶνΛΟ,ΟΝ,καὶτὰἀπὸτῶνΛΟ, FK are each medial (areas) [Prop. 10.21]. Again, since
ΟΝ[ἄρα]μέσαἐστίν·καὶαἱΛΟ,ΟΝἄραμέσαιεἰσὶδυνάμει DE is commensurable (in length) with EG, thus DG is
μόνονσύμμετροι.καὶἐπεὶτὸὑπὸτῶνΑΖ,ΖΗἴσονἐστὶτῷ also commensurable (in length) with each of DE and EG
ἀπὸτῆςΕΗ,ἔστινἄραὡςἡΑΖπρὸςτὴνΕΗ,οὕτωςἡΕΗ [Prop. 10.15]. But, DG is commensurable in length with
πρὸςτὴνΖΗ·ἀλλ᾿ὡςμὲνἡΑΖπρὸςτὴνΕΗ,οὕτωςτὸΑΙ AC [thus, DE and EG are also each rational, and comπρὸςτὸΕΚ·ὡςδὲἡΕΗπρὸςτὴνΖΗ,οὕτως[ἐστὶ]τὸΕΚ
mensurable in length with AC]. Thus, DH and EK are
πρὸςτὸΖΚ·τῶνἄραΑΙ,ΖΚμέσονἀνάλογόνἐστιτὸΕΚ. each rational (areas) [Prop. 10.19].
ἔστιδὲκαὶτῶνΛΜ,ΝΞτετραγώνωνμέσονἀνάλογοντὸ Therefore, let the square LM, equal to AI, have
ΜΝ·καίἐστινἴσοντὸμὲνΑΙτῷΛΜ,τὸδὲΖΚτῷΝΞ·καὶ been constructed. And let NO, equal to FK, which is
τὸΜΝἄραἴσονἐστὶτῷΕΚ.ἀλλὰτῷμὲνΕΚἴσον[ἐστὶ] about the same angle LPM as LM, have been subtracted
τὸΔΘ,τῷδὲΜΝἴσοντὸΛΞ·ὅλονἄρατὸΔΚἴσονἐστὶ (from LM). Thus, the squares LM and NO are about
τῷΥΦΧγνώμονικαὶτῷΝΞ.ἐπεὶοὖνὅλοντὸΑΚἴσον the same diagonal [Prop. 6.26]. Let PR be their (com-
ἐστὶτοῖςΛΜ,ΝΞ,ὧντὸΔΚἴσονἐστὶτῷΥΦΧγνώμονι mon) diagonal, and let the (rest of the) figure have been
καὶτῷΝΞ,λοιπὸνἄρατὸΑΒἴσονἐστὶτῷΤΣ.τὸδὲΤΣ drawn. Therefore, since AI and FK are medial (areas),
ἐστιτὸἀπὸτῆςΛΝ·τὸἀπὸτῆςΛΝἄραἴσονἐστὶτῷΑΒ and are equal to the (squares) on LP and PN (respecχωρίῳ·ἡΛΝἄραδύναταιτὸΑΒχωρίον.
λέγω[δή],ὅτιἡ tively), [thus] the (squares) on LP and PN are also me-
ΛΝμέσηςἀποτομήἐστιπρώτη.
dial. Thus, LP and PN are also medial (straight-lines
᾿ΕπεὶγὰρῥητόνἐστιτὸΕΚκαίἐστινἴσοντῷΛΞ,ῥητὸν which are) commensurable in square only. † And since
ἄραἐστὶτὸΛΞ,τουτέστιτὸὑπὸτῶνΛΟ,ΟΝ.μέσονδὲ the (rectangle contained) by AF and FG is equal to
ἐδείχθητὸΝΞ·ἀσύμμετρονἄραἐστὶτὸΛΞτῷΝΞ·ὡς the (square) on EG, thus as AF is to EG, so EG (is)
δὲτὸΛΞπρὸςτὸΝΞ,οὕτωςἐστὶνἡΛΟπρὸςΟΝ·αἱ to FG [Prop. 10.17]. But, as AF (is) to EG, so AI
ΛΟ,ΟΝἄραἀσύμμετροίεἰσιμήκει.αἱἄραΛΟ,ΟΝμέσαι (is) to EK. And as EG (is) to FG, so EK [is] to FK
εἰσὶδυνάμειμόνονσύμμετροιῥητὸνπεριέχουσαι·ἡΛΝἄρα [Prop. 6.1]. Thus, EK is the mean proportional to AI
389

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
μέσηςἀποτομήἐστιπρώτη·καὶδύναταιτὸΑΒχωρίον. and FK [Prop. 5.11]. And MN is also the mean pro-
῾Η ἄρατὸ ΑΒχωρίον δυναμένημέσηςἀποτομήἐστι portional to the squares LM and NO [Prop. 10.53 lem.].
πρώτη·ὅπερἔδειδεῖξαι.
And AI is equal to LM, and FK to NO. Thus, MN is
also equal to EK. But, DH [is] equal to EK, and LO
equal to MN [Prop. 1.43]. Thus, the whole (of) DK is
equal to the gnomon UV W and NO. Therefore, since the
whole (of) AK is equal to LM and NO, of which DK is
equal to the gnomon UV W and NO, the remainder AB
is thus equal to TS. And TS is the (square) on LN. Thus,
the (square) on LN is equal to the area AB. LN is thus
the square-root of area AB. [So], I say that LN is the
first apotome of a medial (straight-line).
For since EK is a rational (area), and is equal to
LO, LO—that is to say, the (rectangle contained) by LP
and PN—is thus a rational (area). And NO was shown
(to be) a medial (area). Thus, LO is incommensurable
with NO. And as LO (is) to NO, so LP is to PN
[Prop. 6.1]. Thus, LP and PN are incommensurable
in length [Prop. 10.11]. LP and PN are thus medial
(straight-lines which are) commensurable in square only,
and which contain a rational (area). Thus, LN is the first
apotome of a medial (straight-line) [Prop. 10.74]. And it
is the square-root of area AB.
Thus, the square root of area AB is the first apotome
of a medial (straight-line). (Which is) the very thing it
was required to show.
† There is an error in the argument here. It should just say that LP and PN are commensurable in square, rather than in square only, since LP
and PN are only shown to be incommensurable in length later on.
. Proposition 93
᾿Εὰνχωρίονπεριέχηταιὑπὸῥητῆςκαὶἀποτομῆςτρίτης, If an area is contained by a rational (straight-line) and
ἡτὸχωρίονδυναμένημέσηςἀποτομήἐστιδευτέρα. a third apotome then the square-root of the area is a second
apotome of a medial (straight-line).
Α ∆ Ε Ζ Η Ν
A D E F G
N
Λ Ο
L
O
Φ
V
Γ Β Θ Ι Κ
Σ
Π Υ
Χ
Ξ
Ρ Μ
R
M
Τ
T
ΧωρίονγὰρτὸΑΒπεριεχέσθωὑπὸῥητῆςτῆςΑΓκαὶ For let the area AB have been contained by the ratio-
ἀποτομῆςτρίτηςτῆςΑΔ·λέγω,ὅτιἡτὸΑΒχωρίονδυ- nal (straight-line) AC and the third apotome AD. I say
ναμένημέσηςἀποτομήἐστιδευτέρα.
that the square-root of area AB is the second apotome of
῎ΕστωγὰρτῇΑΔπροσαρμόζουσαἡΔΗ·αἱΑΗ,ΗΔ a medial (straight-line).
ἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι,καὶοὐδετέρατῶν For let DG be an attachment to AD. Thus, AG and
ΑΗ,ΗΔσύμμετρόςἐστιμήκειτῇἐκκειμένῃῥητῇτῇΑΓ,ἡ GD are rational (straight-lines which are) commensuδὲὅληἡΑΗτὴςπροσαρμοζούσηςτῆςΔΗμεῖζονδύναται
rable in square only [Prop. 10.73], and neither of AG
τῷἀπὸσυμμέτρουἑαυτῇ. ἐπεὶοὖνἡΑΗτῆςΗΔμεῖζον and GD is commensurable in length with the (previ-
C
B
H
I
K
S
U
W
Q
P
390

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
δύναταιτῷἀπὸσυμμέτρουἑαυτῇ,ἐὰνἄρατῷτετάρτῳμέρει ously) laid down rational (straight-line) AC, and the
τοῦἀπὸτῆςΔΗἴσονπαρὰτὴνΑΗπαραβληθῇἐλλεῖπον square on the whole, AG, is greater than (the square on)
εἴδειτετραγώνῳ,εἰςσύμμετρααὐτὴνδιελεῖ.τετμήσθωοὖν the attachment, DG, by the (square) on (some straight-
ἡΔΗδίχακατὰτὸΕ,καὶτῷἀπὸτῆςΕΗἴσονπαρὰτὴν line) commensurable (in length) with (AG) [Def. 10.13].
ΑΗπαραβεβλήσθωἐλλεῖπονεἴδειτετραγώνῳ,καὶἔστωτὸ Therefore, since the square on AG is greater than (the
ὑπὸτῶνΑΖ,ΖΗ.καὶἤχθωσανδιὰτῶνΕ,Ζ,Ησημείων square on) GD by the (square) on (some straight-line)
τῇΑΓπαράλληλοιαἱΕΘ,ΖΙ,ΗΚ·σύμμετροιἄραεἰσὶναἱ commensurable (in length) with (AG), thus if (an area)
ΑΖ,ΖΗ·σύμμετρονἄρακαὶτὸΑΙτῷΖΚ.καὶἐπεὶαἱΑΖ, equal to the fourth part of the square on DG is applied
ΖΗσύμμετροίεἰσιμήκει,καὶἡΑΗἄραἑκατέρᾳτῶνΑΖ, to AG, falling short by a square figure, then it divides
ΖΗσύμμετρόςἐστιμήκει. ῥητὴδὲἡΑΗκαὶἀσύμμετρος (AG) into (parts which are) commensurable (in length)
τῇΑΓμήκει·ὥστεκαὶαἱΑΖ,ΖΗ.ἑκάτερονἄρατῶνΑΙ, [Prop. 10.17]. Therefore, let DG have been cut in half
ΖΚμέσονἐστίν.πάλιν,ἐπεὶσύμμετρόςἐστινἡΔΕτῇΕΗ at E. And let (an area) equal to the (square) on EG
μήκει,καὶἡΔΗἄραἑκατέρᾳτῶνΔΕ,ΕΗσύμμετρόςἐστι have been applied to AG, falling short by a square figμήκει.
ῥητὴδὲἡΗΔκαὶἀσύμμετροςτῇΑΓμήκει·ῥητὴ ure. And let it be the (rectangle contained) by AF and
ἄρακαὶἑκατέρατῶνΔΕ,ΕΗκαὶἀσύμμετροςτῇΑΓμήκει· FG. And let EH, FI, and GK have been drawn through
ἑκάτερονἄρατῶνΔΘ,ΕΚμέσονἐστίν. καὶἐπεὶαἱΑΗ, points E, F , and G (respectively), parallel to AC. Thus,
ΗΔδυνάμειμόνονσύμμετροίεἰσιν,ἀσύμμετροςἄραἐστὶ AF and FG are commensurable (in length). AI (is) thus
μήκειἡΑΗτῇΗΔ.ἀλλ᾿ἡμὲνΑΗτῇΑΖσύμμετρόςἐστι also commensurable with FK [Props. 6.1, 10.11]. And
μήκειἡδὲΔΗτῇΕΗ·ἀσύμμετροςἄραἐστὶνἡΑΖτῇΕΗ since AF and FG are commensurable in length, AG is
μήκει.ὡςδὲἡΑΖπρὸςτὴνΕΗ,οὕτωςἐστὶτὸΑΙπρὸςτὸ thus also commensurable in length with each of AF and
ΕΚ·ἀσύμμετρονἄραἐστὶτὸΑΙτῷΕΚ.
FG [Prop. 10.15]. And AG (is) rational, and incommen-
ΣυνεστάτωοὖντῷμὲνΑΙἴσοντετράγωνοντὸΛΜ,τῷ surable in length with AC. Hence, AF and FG (are)
δὲΖΚἴσονἀφῇρήσθωτὸΝΞπερὶτὴναὐτὴνγωνίανὂντῷ also (rational, and incommensurable in length with AC)
ΛΜ·περὶτὴναὐτὴνἄραδιάμετρόνἐστιτὰΛΜ,ΝΞ.ἔστω [Prop. 10.13]. Thus, AI and FK are each medial (arαὐτῶνδιάμετροςἡΟΡ,καὶκαταγεγράφθωτὸσχῆμα.ἐπεὶ
eas) [Prop. 10.21]. Again, since DE is commensurable
οὖντὸὑπὸτῶνΑΖ,ΖΗἴσονἐστὶτῷἀπὸτῆςΕΗ,ἔστινἄρα in length with EG, DG is also commensurable in length
ὡςἡΑΖπρὸςτὴνΕΗ,οὕτωςἡΕΗπρὸςτὴνΖΗ.ἀλλ᾿ὡς with each of DE and EG [Prop. 10.15]. And GD (is)
μὲνἡΑΖπρὸςτὴνΕΗ,οὕτωςἐστὶτὸΑΙπρὸςτὸΕΚ·ὡς rational, and incommensurable in length with AC. Thus,
δὲἡΕΗπρὸςτὴνΖΗ,οὕτωςἐστὶτὸΕΚπρὸςτὸΖΚ·καὶ DE and EG (are) each also rational, and incommensu-
ὡςἄρατὸΑΙπρὸςτὸΕΚ,οὕτωςτὸΕΚπρὸςτὸΖΚ·τῶν rable in length with AC [Prop. 10.13]. DH and EK are
ἄραΑΙ,ΖΚμέσονἀνάλογόνἐστιτὸΕΚ.ἔστιδὲκαὶτῶν thus each medial (areas) [Prop. 10.21]. And since AG
ΛΜ,ΝΞτετραγώνωνμέσονἀνάλογοντὸΜΝ·καίἐστιν and GD are commensurable in square only, AG is thus
ἴσοντὸμὲνΑΙτῷΛΜ,τὸδὲΖΚτῷΝΞ·καὶτὸΕΚἄρα incommensurable in length with GD. But, AG is com-
ἴσονἐστὶτῷΜΝ.ἀλλὰτὸμὲνΜΝἴσονἐστὶτῷΛΞ,τὸ mensurable in length with AF , and DG with EG. Thus,
δὲΕΚἴσον[ἐστὶ]τῷΔΘ·καὶὅλονἄρατὸΔΚἴσονἐστὶ AF is incommensurable in length with EG [Prop. 10.13].
τῷΥΦΧγνώμονικαὶτῷΝΞ.ἔστιδὲκαὶτὸΑΚἴσοντοῖς And as AF (is) to EG, so AI is to EK [Prop. 6.1]. Thus,
ΛΜ,ΝΞ·λοιπὸνἄρατὸΑΒἴσονἐστὶτῷΣΤ,τουτέστιτῷ AI is incommensurable with EK [Prop. 10.11].
ἀπὸτῆςΛΝτετραγώνῳ·ἡΛΝἄραδύναταιτὸΑΒχωρίον. Therefore, let the square LM, equal to AI, have
λέγω,ὅτιἡΛΝμέσηςἀποτομήἐστιδευτέρα.
been constructed. And let NO, equal to FK, which is
᾿ΕπεὶγὰρμέσαἐδείχθητὰΑΙ,ΖΚκαίἐστινἴσατοῖςἀπὸ about the same angle as LM, have been subtracted (from
τῶνΛΟ,ΟΝ,μέσονἄρακαὶἑκάτεροντῶνἀπὸτῶνΛΟ, LM). Thus, LM and NO are about the same diagonal
ΟΝ·μέσηἄραἑκατέρατῶνΛΟ,ΟΝ.καὶἐπεὶσύμμετρόν [Prop. 6.26]. Let PR be their (common) diagonal, and
ἐστιτὸΑΙτῷΖΚ,σύμμετρονἄρακαὶτὸἀπὸτῆςΛΟτῷ let the (rest of the) figure have been drawn. Therefore,
ἀπὸτῆςΟΝ.πάλιν,ἐπεὶἀσύμμετρονἐδείχθητὸΑΙτῷΕΚ, since the (rectangle contained) by AF and FG is equal
ἀσύμμετρονἄραἐστὶκαὶτὸΛΜτῷΜΝ,τουτέστιτὸἀπὸ to the (square) on EG, thus as AF is to EG, so EG (is)
τῆςΛΟτῷὑπὸτῶνΛΟ,ΟΝ·ὥστεκαὶἡΛΟἀσύμμετρός to FG [Prop. 6.17]. But, as AF (is) to EG, so AI is to
ἐστιμήκειτῇΟΝ·αἱΛΟ,ΟΝἄραμέσαιεἰσὶδυνάμειμόνον EK [Prop. 6.1]. And as EG (is) to FG, so EK is to FK
σύμμετροι.λέγωδή,ὅτικαὶμέσονπεριέχουσιν.
[Prop. 6.1]. And thus as AI (is) to EK, so EK (is) to
᾿ΕπεὶγὰρμέσονἐδείχθητὸΕΚκαίἐστινἴσοντῷὑπὸ FK [Prop. 5.11]. Thus, EK is the mean proportional to
τῶνΛΟ,ΟΝ,μέσονἄραἐστὶκαὶτὸὑπὸτῶνΛΟ,ΟΝ· AI and FK. And MN is also the mean proportional to
ὥστεαἱΛΟ,ΟΝμέσαιεἰσὶδυνάμειμόνονσύμμετροιμέσον the squares LM and NO [Prop. 10.53 lem.]. And AI is
391

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
περιέχουσαι. ἡΛΝἄραμέσηςἀποτομήἐστιδευτέρα·καὶ equal to LM, and FK to NO. Thus, EK is also equal to
δύναταιτὸΑΒχωρίον.
MN. But, MN is equal to LO, and EK [is] equal to DH
῾Η ἄρατὸ ΑΒχωρίον δυναμένημέσηςἀποτομήἐστι [Prop. 1.43]. And thus the whole of DK is equal to the
δευτέρα·ὅπερἔδειδεῖξαι.
gnomon UV W and NO. And AK (is) also equal to LM
and NO. Thus, the remainder AB is equal to ST —that is
to say, to the square on LN. Thus, LN is the square-root
of area AB. I say that LN is the second apotome of a
medial (straight-line).
For since AI and FK were shown (to be) medial (areas),
and are equal to the (squares) on LP and PN (respectively),
the (squares) on each of LP and PN (are)
thus also medial. Thus, LP and PN (are) each medial
(straight-lines). And since AI is commensurable with
FK [Props. 6.1, 10.11], the (square) on LP (is) thus
also commensurable with the (square) on PN. Again,
since AI was shown (to be) incommensurable with EK,
LM is thus also incommensurable with MN—that is to
say, the (square) on LP with the (rectangle contained)
by LP and PN. Hence, LP is also incommensurable in
length with PN [Props. 6.1, 10.11]. Thus, LP and PN
are medial (straight-lines which are) commensurable in
square only. So, I say that they also contain a medial
(area).
For since EK was shown (to be) a medial (area), and
is equal to the (rectangle contained) by LP and PN, the
(rectangle contained) by LP and PN is thus also medial.
Hence, LP and PN are medial (straight-lines which are)
commensurable in square only, and which contain a medial
(area). Thus, LN is the second apotome of a medial
(straight-line) [Prop. 10.75]. And it is the square-root of
area AB.
Thus, the square-root of area AB is the second apotome
of a medial (straight-line). (Which is) the very thing
it was required to show.
. Proposition 94
᾿Εὰνχωρίονπεριέχηταιὑπὸῥητῆςκαὶἀποτομῆςτετάρτης, If an area is contained by a rational (straight-line) and
ἡτὸχωρίονδυναμένηἐλάσσωνἐστίν.
a fourth apotome then the square-root of the area is a
minor (straight-line).
Α ∆ Ε Ζ Η Ν
A D E F G
N
Λ Ο
L
O
Φ
V
Γ Β Θ Ι Κ
Σ
Π Υ
Χ
Ξ
Ρ Μ
R
M
Τ
T
ΧωρίονγὰρτὸΑΒπεριεχέσθωὑπὸῥητῆςτῆςΑΓκαὶ For let the area AB have been contained by the ra-
ἀποτομῆςτετάρτηςτῆςΑΔ·λέγω, ὅτιἡτὸΑΒχωρίον tional (straight-line) AC and the fourth apotome AD. I
δυναμένηἐλάσσωνἐστίν.
say that the square-root of area AB is a minor (straight-
C
B
H
I
K
S
U
W
Q
P
392

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
῎Εστω γὰρ τῇ ΑΔ προσαρμόζουσα ἡ ΔΗ· αἱ ἄρα line). For let DG be an attachment to AD. Thus, AG
ΑΗ,ΗΔῥηταίεἰσιδυνάμειμόνονσύμμετροι, καὶἡΑΗ and DG are rational (straight-lines which are) commenσύμμετρόςἐστιτῇἐκκειμένῃῥητῇτῇΑΓμήκει,ἡδὲὅλη
surable in square only [Prop. 10.73], and AG is com-
ἡ ΑΗ τῆς προσαρμοζούσης τῆς ΔΗ μεῖζον δύναται τῷ mensurable in length with the (previously) laid down ra-
ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει. ἐπεὶ οὖν ἡ ΑΗ τῆς ΗΔ tional (straight-line) AC, and the square on the whole,
μεῖζονδύναταιτῷἀπὸἀσυμμέτρουἑαυτῇμήκει,ἐὰνἄρα AG, is greater than (the square on) the attachment, DG,
τῷτετάρτῳμέρειτοῦἀπὸτῆςΔΗἴσονπαρὰτὴνΑΗπα- by the square on (some straight-line) incommensurable
ραβληθῇἐλλεῖπονεἴδειτετραγώνῳ, εἰςἀσύμμετρααὐτὴν in length with (AG) [Def. 10.14]. Therefore, since the
διελεῖ. τετμήσθωοὖνἡΔΗδίχακατὰτὸΕ,καὶτῷἀπὸ square on AG is greater than (the square on) GD by
τῆςΕΗἴσονπαρὰτὴνΑΗπαραβεβλήσθωἐλλεῖπονεἴδει the (square) on (some straight-line) incommensurable in
τετραγώνῳ, καὶἔστω τὸ ὑπὸτῶν ΑΖ, ΖΗ· ἀσύμμετρος length with (AG), thus if (some area), equal to the fourth
ἄραἐστὶμήκειἡΑΖτῇΖΗ.ἤχθωσανοὖνδιὰτῶνΕ,Ζ,Η part of the (square) on DG, is applied to AG, falling short
παράλληλοιταῖςΑΓ,ΒΔαἱΕΘ,ΖΙ,ΗΚ.ἐπεὶοὖνῥητήἐστιν by a square figure, then it divides (AG) into (parts which
ἡΑΗκαὶσύμμετροςτῇΑΓμήκει,ῥητὸνἄραἐστὶνὅλοντὸ are) incommensurable (in length) [Prop. 10.18]. There-
ΑΚ.πάλιν,ἐπεὶἀσύμμετρόςἐστινἡΔΗτῇΑΓμήκει,καί fore, let DG have been cut in half at E, and let (some
εἰσινἀμφότεραιῥηταί,μέσονἄραἐστὶτὸΔΚ.πάλιν,ἐπεὶ area), equal to the (square) on EG, have been applied
ἀσύμμετρόςἐστὶνἡΑΖτῇΖΗμήκει,ἀσύμμετρονἄρακαὶ to AG, falling short by a square figure, and let it be the
τὸΑΙτῷΖΚ.
(rectangle contained) by AF and FG. Thus, AF is in-
ΣυνεστάτωοὖντῷμὲνΑΙἴσοντετράγωνοντὸΛΜ,τῷ commensurable in length with FG. Therefore, let EH,
δὲΖΚἴσονἀφῃρήσθωπερὶτὴναὐτὴνγωνίαντὴνὑπὸτῶν FI, and GK have been drawn through E, F , and G (re-
ΛΟΜτὸΝΞ.περὶτὴναὐτὴνἄραδιάμετόνἐστιτὰΛΜ,ΝΞ spectively), parallel to AC and BD. Therefore, since AG
τετράγωνα.ἔστωαὐτῶνδιάμετοςἡΟΡ,καὶκαταγεγράφθω is rational, and commensurable in length with AC, the
τὸσχῆμα. ἐπεὶοὖντὸὑπὸτῶνΑΖ,ΖΗἴσονἐστὶτῷἀπὸ whole (area) AK is thus rational [Prop. 10.19]. Again,
τῆςΕΗ,ἀνάλογονἄραἐστὶνὡςἡΑΖπρὸςτὴνΕΗ,οὕτως since DG is incommensurable in length with AC, and
ἡΕΗπρὸςτὴνΖΗ.ἀλλ᾿ὡςμὲνἡΑΖπρὸςτὴνΕΗ,οὕτως both are rational (straight-lines), DK is thus a medial
ἐστὶτὸΑΙπρὸςτὸΕΚ,ὡςδὲἡΕΗπρὸςτὴνΖΗ,οὕτως (area) [Prop. 10.21]. Again, since AF is incommensu-
ἐστὶτὸΕΚπρὸςτὸΖΚ·τῶνἄραΑΙ,ΖΚμέσονἀνάλογόν rable in length with FG, AI (is) thus also incommensu-
ἐστιτὸΕΚ.ἔστιδὲκαὶτῶνΛΜ,ΝΞτετραγώνωνμέσον rable with FK [Props. 6.1, 10.11].
ἀνάλογοντὸΜΝ,καίἐστινἴσοντὸμὲνΑΙτῷΛΜ,τὸδὲ Therefore, let the square LM, equal to AI, have been
ΖΚτῷΝΞ·καὶτὸΕΚἄραἴσονἐστὶτῷΜΝ.ἀλλὰτῷμὲν constructed. And let NO, equal to FK, (and) about the
ΕΚἴσονἐστὶτὸΔΘ,τῷδὲΜΝἴσονἐστὶτὸΛΞ·ὅλον same angle, LPM, have been subtracted (from LM).
ἄρατὸΔΚἴσονἐστὶτῷΥΦΧγνώμονικαὶτῷΝΞ.ἐπεὶ Thus, the squares LM and NO are about the same diagοὐνὅλοντὸΑΚἴσονἐστὶτοῖςΛΜ,ΝΞτετραγώνοις,ὧν
onal [Prop. 6.26]. Let PR be their (common) diagonal,
τὸΔΚἴσονἐστὶτῷΥΦΧγνώμονικαὶτῷΝΞτετραγώνῳ, and let the (rest of the) figure have been drawn. ThereλοιπὸνἄρατὸΑΒἴσονἐστὶτῷΣΤ,τουτέστιτῷἀπὸτῆς
fore, since the (rectangle contained) by AF and FG is
ΛΝτετραγώνῳ·ἡΛΝἄραδύναταιτὸΑΒχωρίον. λέγω, equal to the (square) on EG, thus, proportionally, as AF
ὅτιἡΛΝἄλογόςἐστινἡκαλουμένηἐλάσσων.
is to EG, so EG (is) to FG [Prop. 6.17]. But, as AF (is)
᾿ΕπεὶγὰρῥητόνἐστιτὸΑΚκαίἐστινἴσοντοῖςἀπὸτῶν to EG, so AI is to EK, and as EG (is) to FG, so EK is
ΛΟ,ΟΝτετράγωνοις,τὸἄρασυγκείμενονἐκτῶνἀπὸτῶν to FK [Prop. 6.1]. Thus, EK is the mean proportional to
ΛΟ,ΟΝῥητόνἐστιν. πάλιν,ἐπεὶτὸΔΚμέσονἐστίν,καί AI and FK [Prop. 5.11]. And MN is also the mean pro-
ἐστινἴσοντὸΔΚτῷδὶςὑπὸτῶνΛΟ,ΟΝ,τὸἄραδὶςὑπὸ portional to the squares LM and NO [Prop. 10.13 lem.],
τῶνΛΟ,ΟΝμέσονἐστίν.καὶἐπεὶἀσύμμετρονἐδείχθητὸ and AI is equal to LM, and FK to NO. EK is thus
ΑΙτῷΖΚ,ἀσύμμετρονἄρακαὶτὸἀπὸτῆςΛΟτετράγωνον also equal to MN. But, DH is equal to EK, and LO is
τῷἀπὸτῆςΟΝτετραγώνῳ. αἱΛΟ,ΟΝἄραδυνάμειεἰσὶν equal to MN [Prop. 1.43]. Thus, the whole of DK is
ἀσύμμετροιποιοῦσαιτὸμὲνσυγκείμενονἐκτῶνἀπ᾿αὐτῶν equal to the gnomon UV W and NO. Therefore, since
τετραγώνωνῥητόν,τὸδὲδὶςὑπ᾿αὐτῶνμέσον. ἡΛΝἄρα the whole of AK is equal to the (sum of the) squares LM
ἄλογόςἐστινἡκαλουμένηἐλάσσων· καὶδύναταιτὸΑΒ and NO, of which DK is equal to the gnomon UV W
χωρίον.
and the square NO, the remainder AB is thus equal to
῾ΗἄρατὸΑΒχωρίονδυναμένηἐλάσσωνἐστίν· ὅπερ ST —that is to say, to the square on LN. Thus, LN is the
ἔδειδεῖξαι.
square-root of area AB. I say that LN is the irrational
(straight-line which is) called minor.
393

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
For since AK is rational, and is equal to the (sum of
the) squares LP and PN, the sum of the (squares) on
LP and PN is thus rational. Again, since DK is medial,
and DK is equal to twice the (rectangle contained)
by LP and PN, thus twice the (rectangle contained) by
LP and PN is medial. And since AI was shown (to be)
incommensurable with FK, the square on LP (is) thus
also incommensurable with the square on PN. Thus, LP
and PN are (straight-lines which are) incommensurable
in square, making the sum of the squares on them rational,
and twice the (rectangle contained) by them medial.
LN is thus the irrational (straight-line) called minor
[Prop. 10.76]. And it is the square-root of area AB.
Thus, the square-root of area AB is a minor (straightline).
(Which is) the very thing it was required to show.
. Proposition 95
᾿Εὰνχωρίονπεριέχηταιὑπὸῥητῆςκαὶἀποτομῆςπέμπτης, If an area is contained by a rational (straight-line) and
ἡτὸχωρίονδυναμένη[ἡ]μετὰῥητοῦμέσοντὸὅλονποιοῦσά a fifth apotome then the square-root of the area is that
ἐστιν.
(straight-line) which with a rational (area) makes a medial
whole.
Α ∆ Ε Ζ Η Ν
A D E F G
N
Λ Ο
L
O
Φ
V
Γ Β Θ Ι Κ
Σ
Π Υ
Χ
Ξ
Ρ Μ
R
M
Τ
T
ΧωρίονγὰρτὸΑΒπεριεχέσθωὑπὸῥητῆςτῆςΑΓκαὶ For let the area AB have been contained by the ra-
ἀποτομῆςπέμπτηςτῆςΑΔ·λέγω,ὅτιἡτὸΑΒχωρίονδυ- tional (straight-line) AC and the fifth apotome AD. I
ναμένη[ἡ]μετὰῥητοῦμέσοντὸὅλονποιοῦσάἐστιν. say that the square-root of area AB is that (straight-line)
῎Εστω γὰρ τῇ ΑΔ προσαρμόζουσα ἡ ΔΗ· αἱ ἄρα which with a rational (area) makes a medial whole.
ΑΗ,ΗΔῥηταίεἰσιδυνάμειμόνονσύμμετροι,καὶἡπρο- For let DG be an attachment to AD. Thus, AG and
σαρμόζουσα ἡ ΗΔ σύμμετρός ἐστι μήκει τῇ ἐκκειμένῃ DG are rational (straight-lines which are) commensu-
ῥητῇ τῇ ΑΓ, ἡ δὲ ὅλη ἡ ΑΗ τῆς προσαρμοζούσης τῆς rable in square only [Prop. 10.73], and the attachment
ΔΗμεῖζονδύναταιτῷἀπὸἀσυμμέτρουἑαυτῇ. ἐὰνἄρα GD is commensurable in length the the (previously) laid
τῷτετάρτῳμέρειτοῦἀπὸτῆςΔΗἴσονπαρὰτὴνΑΗπα- down rational (straight-line) AC, and the square on the
ραβληθῇἐλλεῖπονεἴδειτετραγώνῳ, εἰςἀσύμμετρααὐτὴν whole, AG, is greater than (the square on) the attachδιελεῖ.
τετμήσθωοὖνἡΔΗδίχακατὰτὸΕσημεῖον,καὶ ment, DG, by the (square) on (some straight-line) incomτῷἀπὸτῆςΕΗἴσονπαρὰτὴνΑΗπαραβεβλήσθωἐλλεῖπον
mensurable (in length) with (AG) [Def. 10.15]. Thus, if
εἴδειτετραγώνῳκαὶἔστωτὸὑπὸτῶνΑΖ,ΖΗ·ἀσύμμετρος (some area), equal to the fourth part of the (square) on
ἄραἐστὶνἡΑΖτῇΖΗμήκει.καὶἐπεὶἀσύμμετρόςἐστὶνἡ DG, is applied to AG, falling short by a square figure,
ΑΗτῇΓΑμήκει,καίεἰσινἀμφότεραιῥηταί,μέσονἄραἐστὶ then it divides (AG) into (parts which are) incommensuτὸΑΚ.πάλιν,ἐπεὶῥητήἐστινἡΔΗκαὶσύμμετροςτῇΑΓ
rable (in length) [Prop. 10.18]. Therefore, let DG have
μήκει,ῥητόνἐστιτὸΔΚ.
been divided in half at point E, and let (some area), equal
ΣυνεστάτωοὖντῷμὲνΑΙἴσοντετράγωνοντὸΛΜ,τῷ to the (square) on EG, have been applied to AG, falling
δὲΖΚἴσοντετράγωνονἀφῃρήσθωτὸΝΞπερὶτὴναὐτὴν short by a square figure, and let it be the (rectangle conγωνίαντὴνὑπὸΛΟΜ·περὶτὴναὐτὴνἄραδιάμετρόνἐστι
tained) by AF and FG. Thus, AF is incommensurable
τὰΛΜ,ΝΞτετράγωνα. ἔστωαὐτῶνδιάμετροςἡΟΡ,καὶ in length with FG. And since AG is incommensurable
C
B
H
I
K
S
U
W
Q
P
394

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
καταγεγράφθωτὸσχῆμα. ὁμοίωςδὴδείξομεν,ὅτιἡΛΝ in length with CA, and both are rational (straight-lines),
δύναταιτὸΑΒχωρίον.λέγω,ὅτιἡΛΝἡμετὰῥητοῦμέσον AK is thus a medial (area) [Prop. 10.21]. Again, since
τὸὅλονποιοῦσάἐστιν.
DG is rational, and commensurable in length with AC,
᾿ΕπεὶγὰρμέσονἐδείχθητὸΑΚκαίἐστινἴσοντοῖςἀπὸ DK is a rational (area) [Prop. 10.19].
τῶνΛΟ,ΟΝ,τὸἄρασυγκείμενονἐκτῶνἀπὸτῶνΛΟ,ΟΝ Therefore, let the square LM, equal to AI, have been
μέσονἐστίν. πάλιν,ἐπεὶῥητόνἐστιτὸΔΚκαίἐστινἴσον constructed. And let the square NO, equal to FK, (and)
τῷδὶςὑπὸτῶνΛΟ,ΟΝ,καὶαὑτὸῥητόνἐστιν. καὶἐπεὶ about the same angle, LPM, have been subtracted (from
ἀσύμμετρόνἐστιτὸΑΙτῷΖΚ,ἀσύμμετρονἄραἐστὶκαὶτὸ NO). Thus, the squares LM and NO are about the same
ἀπὸτῆςΛΟτῷἀπὸτῆςΟΝ·αἱΛΟ,ΟΝἄραδυνάμειεἰσὶν diagonal [Prop. 6.26]. Let PR be their (common) diag-
ἀσύμμετροιποιοῦσαιτὸμὲνσυγκείμενονἐκτῶνἀπ᾿αὐτῶν onal, and let (the rest of) the figure have been drawn.
τετραγώνωνμέσον,τὸδὲδὶςὑπ᾿αὐτῶνῥητόν.ἡλοιπὴἄρα So, similarly (to the previous propositions), we can show
ἡΛΝἄλογόςἐστινἡκαλουμένημετὰῥητοῦμέσοντὸὅλον that LN is the square-root of area AB. I say that LN is
ποιοῦσα·καὶδύναταιτὸΑΒχωρίον.
that (straight-line) which with a rational (area) makes a
῾ΗτὸΑΒἄραχωρίονδυναμένημετὰῥητοῦμέσοντὸ medial whole.
ὅλονποιοῦσάἐστιν·ὅπερἔδειδεῖξαι.
For since AK was shown (to be) a medial (area), and
is equal to (the sum of) the squares on LP and PN,
the sum of the (squares) on LP and PN is thus medial.
Again, since DK is rational, and is equal to twice the
(rectangle contained) by LP and PN, (the latter) is also
rational. And since AI is incommensurable with FK, the
(square) on LP is thus also incommensurable with the
(square) on PN. Thus, LP and PN are (straight-lines
which are) incommensurable in square, making the sum
of the squares on them medial, and twice the (rectangle
contained) by them rational. Thus, the remainder LN is
the irrational (straight-line) called that which with a rational
(area) makes a medial whole [Prop. 10.77]. And it
is the square-root of area AB.
Thus, the square-root of area AB is that (straightline)
which with a rational (area) makes a medial whole.
(Which is) the very thing it was required to show.
¦. Proposition 96
᾿Εὰνχωρίονπεριέχηταιὑπὸῥητῆςκαὶἀποτομῆςἕκτης, If an area is contained by a rational (straight-line) and
ἡτὸχωρίονδυναμένημετὰμέσουμέσοντὸὅλονποιοῦσά a sixth apotome then the square-root of the area is that
ἐστιν.
(straight-line) which with a medial (area) makes a medial
whole.
Α ∆ Ε Ζ Η Ν
A D E F G
N
Λ Ο
L
O
Φ
V
Γ Β Θ Ι Κ
Σ
Π Υ
Χ
Ξ
Ρ Μ
R
M
Τ
T
ΧωρίονγὰρτὸΑΒπεριεχέσθωὑπὸῥητῆςτῆςΑΓκαὶ For let the area AB have been contained by the ra-
ἀποτομῆςἕκτηςτῆςΑΔ·λέγω,ὅτιἡτὸΑΒχωρίονδυ- tional (straight-line) AC and the sixth apotome AD. I
ναμένη[ἡ]μετὰμέσουμέσοντὸὅλονποιοῦσάἐστιν. say that the square-root of area AB is that (straight-line)
῎ΕστωγὰρτῇΑΔπροσαρμόζουσαἡΔΗ·αἱἄραΑΗ, which with a medial (area) makes a medial whole.
ΗΔ ῥηταί εἰσι δυνάμει μόνον σύμμετροι, καὶ οὐδετέρα For let DG be an attachment to AD. Thus, AG and
C
B
H
I
K
S
U
W
Q
P
395

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
αὐτῶνσύμμετρόςἐστιτῇἐκκειμένῃῥητῇτῇΑΓμήκει,ἡ GD are rational (straight-lines which are) commensuδὲὅληἡΑΗτῆςπροσαρμοζούσηςτῆςΔΗμεῖζονδύναται
rable in square only [Prop. 10.73], and neither of them is
τῷἀπὸἀσυμμέτρουἑαυτῇμήκει. ἐπεὶοὖνἡΑΗτῆςΗΔ commensurable in length with the (previously) laid down
μεῖζονδύναταιτῷἀπὸἁσυμμέτρουἐαυτῇμήκει,ἑὰνἄρα rational (straight-line) AC, and the square on the whole,
τῷτετάρτῳμέρειτοῦἀπὸτῆςΔΗἴσονπαρὰτὴνΑΗπα- AG, is greater than (the square on) the attachment, DG,
ραβληθῇἐλλεῖπονεἴδειτετραγώνῳ, εἰςἀσύμμετρααὐτὴν by the (square) on (some straight-line) incommensurable
διελεῖ.τετμήσθωοὖνἡΔΗδίχακατὰτὸΕ[σημεῖον],καὶ in length with (AG) [Def. 10.16]. Therefore, since the
τῷἀπὸτῆςΕΗἴσονπαρὰτὴνΑΗπαραβεβλήσθωἐλλεῖπον square on AG is greater than (the square on) GD by
εἴδειτετραγώνῳ,καὶἔστωτὸὑπὸτῶνΑΖ,ΖΗ·ἀσύμμετρος the (square) on (some straight-line) incommensurable in
ἄραἐστὶνἡΑΖτῇΖΗμήκει. ὡςδὲἡΑΖπρὸςτὴνΖΗ, length with (AG), thus if (some area), equal to the fourth
οὕτωςἐστὶτὸΑΙπρὸςτὸΖΚ·ἀσύμμετρονἄραἐστὶτὸ part of square on DG, is applied to AG, falling short by
ΑΙτῷΖΚ.καὶἐπεὶαἱΑΗ,ΑΓῥηταίεἰσιδυνάμειμόνον a square figure, then it divides (AG) into (parts which
σύμμετροι,μέσονἐστὶτὸΑΚ.πάλιν,ἐπεὶαἱΑΓ,ΔΗῥηταί are) incommensurable (in length) [Prop. 10.18]. Thereεἰσικαὶἀσύμμετροιμήκει,μέσονἐστὶκαὶτὸΔΚ.ἐπεὶοὖν
fore, let DG have been cut in half at [point] E. And let
αἱΑΗ,ΗΔδυνάμειμόνονσύμμετροίεἰσιν,ἀσύμμετροςἄρα (some area), equal to the (square) on EG, have been ap-
ἐστὶνἡΑΗτῇΗΔμήκει.ὡςδὲἡΑΗπρὸςτὴνΗΔ,οὕτως plied to AG, falling short by a square figure. And let it
ἐστὶτὸΑΚπρὸςτὸΚΔ·ἀσύμμετρονἄραἐστὶτὸΑΚτῷ be the (rectangle contained) by AF and FG. AF is thus
ΚΔ.
incommensurable in length with FG. And as AF (is) to
ΣυνεστάτωοὖντῷμὲνΑΙἴσοντετράγωνοντὸΛΜ, FG, so AI is to FK [Prop. 6.1]. Thus, AI is incommenτῷδὲΖΚἴσονἀφῃρήσθωπερὶτὴναὐτὴνγωνίαντὸΝΞ·
surable with FK [Prop. 10.11]. And since AG and AC
περὶτὴναὐτὴνἄραδιάμετρόνἐστιτὰΛΜ,ΝΞτετράγωνα. are rational (straight-lines which are) commensurable in
ἔστωαὐτῶνδιάμετροςἡΟΡ,καὶκαταγεγράφθωτὸσχῆμα. square only, AK is a medial (area) [Prop. 10.21]. Again,
ὁμοίωςδὴτοῖςἐπάνωδείξομεν,ὅτιἡΛΝδύναταιτὸΑΒ since AC and DG are rational (straight-lines which are)
χωρίον. λέγω,ὅτιἡΛΝ[ἡ]μετὰμέσουμέσοντὸὅλον incommensurable in length, DK is also a medial (area)
ποιοῦσάἐστιν.
[Prop. 10.21]. Therefore, since AG and GD are com-
᾿ΕπεὶγὰρμέσονἐδείχθητὸΑΚκαίἐστινἴσοντοῖςἀπὸ mensurable in square only, AG is thus incommensurable
τῶνΛΟ,ΟΝ,τὸἄρασυγκείμενονἐκτῶνἀπὸτῶνΛΟ,ΟΝ in length with GD. And as AG (is) to GD, so AK is to
μέσονἐστίν. πάλιν,ἐπεὶμέσονἐδείχθητὸΔΚκαίἐστιν KD [Prop. 6.1]. Thus, AK is incommensurable with KD
ἴσοντῷδὶςὑπὸτῶνΛΟ,ΟΝ,καὶτὸδὶςὑπὸτῶνΛΟ,ΟΝ [Prop. 10.11].
μέσονἐστίν. καὶἐπεὶἀσύμμετρονἐδείχθητὸΑΚτῷΔΚ, Therefore, let the square LM, equal to AI, have been
ἀσύμμετρα[ἄρα]ἐστὶκαὶτὰἀπὸτῶνΛΟ,ΟΝτετράγωνα constructed. And let NO, equal to FK, (and) about the
τῷδὶςὑπὸτῶνΛΟ,ΟΝ.καὶἐπεὶἀσύμμετρόνἐστιτὸΑΙτῷ same angle, have been subtracted (from LM). Thus,
ΖΚ,ἀσύμμετρονἄρακαὶτὸἀπὸτῆςΛΟτῷἀπὸτῆςΟΝ· the squares LM and NO are about the same diagonal
αἱΛΟ,ΟΝἄραδυνάμειεἰσὶνἀσύμμετροιποιοῦσαιτότε [Prop. 6.26]. Let PR be their (common) diagonal, and
συγκείμενονἐκτῶνἀπ᾿αὐτῶντετραγώνωνμέσονκαὶτὸδὶς let (the rest of) the figure have been drawn. So, similarly
ὑπ᾿αὐτῶνμέσονἔτιτετὰἀπ᾿αὐτῶντετράγωναἀσύμμετρα to the above, we can show that LN is the square-root of
τῷδὶςὑπ᾿αὐτῶν.ἡἄραΛΝἄλογόςἐστινἡκαλουμέμημετὰ area AB. I say that LN is that (straight-line) which with
μέσουμέσοντὸὅλονποιοῦσα·καὶδύναταιτὸΑΒχωρίον. a medial (area) makes a medial whole.
῾Ηἄρατὸχωρίονδυναμένημετὰμέσουμέσοντὸὅλον For since AK was shown (to be) a medial (area), and
ποιοῦσάἐστιν·ὅπερἔδειδεῖξαι.
is equal to the (sum of the) squares on LP and PN, the
sum of the (squares) on LP and PN is medial. Again,
since DK was shown (to be) a medial (area), and is
equal to twice the (rectangle contained) by LP and PN,
twice the (rectangle contained) by LP and PN is also
medial. And since AK was shown (to be) incommensurable
with DK, [thus] the (sum of the) squares on LP
and PN is also incommensurable with twice the (rectangle
contained) by LP and PN. And since AI is incommensurable
with FK, the (square) on LP (is) thus
also incommensurable with the (square) on PN. Thus,
LP and PN are (straight-lines which are) incommensu-
396

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
rable in square, making the sum of the squares on them
medial, and twice the (rectangle contained) by medial,
and, furthermore, the (sum of the) squares on them incommensurable
with twice the (rectangle contained) by
them. Thus, LN is the irrational (straight-line) called
that which with a medial (area) makes a medial whole
[Prop. 10.78]. And it is the square-root of area AB.
Thus, the square-root of area (AB) is that (straightline)
which with a medial (area) makes a medial whole.
(Which is) the very thing it was required to show.
Þ. Proposition 97
Τὸἀπὸἀποτομῆςπαρὰῥητὴνπαραβαλλόμενονπλάτος The (square) on an apotome, applied to a rational
ποιεῖἀποτομὴνπρώτην.
(straight-line), produces a first apotome as breadth.
Æ
A B G
C F N K M
À
à Å
Â
Ä
D E O H L
῎ΕστωἀποτομὴὴΑΒ,ῥητὴδὲἡΓΔ,καὶτῷἀπὸτῆς Let AB be an apotome, and CD a rational (straight-
ΑΒἴσονπαρὰτὴνΓΔπαραβεβλήσθωτὸΓΕπλάτοςποιοῦν line). And let CE, equal to the (square) on AB, have
τὴνΓΖ·λέγω,ὅτιἡΓΖἀποτομήἐστιπρώτη.
been applied to CD, producing CF as breadth. I say that
῎ΕστωγὰρτῇΑΒπροσαρμόζουσαἡΒΗ·αἱἄραΑΗ, CF is a first apotome.
ΗΒῥηταίεἰσιδυνάμειμόνονσύμμετροι. καὶτῷμὲνἀπὸ For let BG be an attachment to AB. Thus, AG and
τῆςΑΗἴσονπαρὰτὴνΓΔπαραβεβλήσθωτὸΓΘ,τῷδὲ GB are rational (straight-lines which are) commensu-
ἀπὸτῆςΒΗτὸΚΛ.ὅλονἄρατὸΓΛἴσονἐστὶτοῖςἀπὸ rable in square only [Prop. 10.73]. And let CH, equal
τῶνΑΗ,ΗΒ·ὧντὸΓΕἴσονἐστὶτῷἀπὸτῆςΑΒ·λοιπὸν to the (square) on AG, and KL, (equal) to the (square)
ἄρατὸΖΛἴσονἐστὶτῷδὶςὑπὸτῶνΑΗ,ΗΒ.τετμήσθω on BG, have been applied to CD. Thus, the whole of
ἡΖΜδίχακατὰτὸΝσημεῖον,καὶἤχθωδιὰτοῦΝτῇΓΔ CL is equal to the (sum of the squares) on AG and GB,
παράλληλοςἡΝΞ·ἑκάτερονἄρατῶνΖΞ,ΛΝἴσονἐστὶτῷ of which CE is equal to the (square) on AB. The re-
ὑπὸτῶνΑΗ,ΗΒ.καὶἐπεὶτὰἀπὸτῶνΑΗ,ΗΒῥητάἐστιν, mainder FL is thus equal to twice the (rectangle conκαίἐστιτοῖςἀπὸτῶνΑΗ,ΗΒἴσοντὸΔΜ,ῥητὸνἄρα
tained) by AG and GB [Prop. 2.7]. Let FM have been
ἐστὶτὸΔΜ.καὶπαρὰῥητὴντὴνΓΔπαραβέβληταιπλάτος cut in half at point N. And let NO have been drawn
ποιοῦντὴνΓΜ·ῥητὴἄραἐστὶνἡΓΜκαὶσύμμετροςτῇΓΔ through N, parallel to CD. Thus, FO and LN are each
μήκει.πάλιν,ἐπεὶμέσονἐστὶτὸδὶςὑπὸτῶνΑΗ,ΗΒ,καὶ equal to the (rectangle contained) by AG and GB. And
τῷδὶςὑπὸτῶνΑΗ,ΗΒἴσοντὸΖΛ,μέσονἄρατὸΖΛ.καὶ since the (sum of the squares) on AG and GB is rational,
παρὰῥητὴντὴνΓΔπαράκειταιπλάτοςποιοῦντὴνΖΜ·ῥητὴ and DM is equal to the (sum of the squares) on AG and
ἄραἐστὶνἡΖΜκαὶἀσύμμετροςτῇΓΔμήκει. καὶἐπεὶτὰ GB, DM is thus rational. And it has been applied to the
μὲνἀπὸτῶνΑΗ,ΗΒῥητάἐστιν,τὸδὲδὶςὑπὸτῶνΑΗ, rational (straight-line) CD, producing CM as breadth.
ΗΒμέσον,ἀσύμμετραἄραἐστὶτὰἀπὸτῶνΑΗ,ΗΒτῷδὶς Thus, CM is rational, and commensurable in length with
ὑπὸτῶνΑΗ,ΗΒ.καὶτοῖςμὲνἀπὸτῶνΑΗ,ΗΒἴσονἐστὶ CD [Prop. 10.20]. Again, since twice the (rectangle conτὸΓΛ,τῷδὲδὶςὑπὸτῶνΑΗ,ΗΒτὸΖΛ·ἀσύμμετρονἄρα
tained) by AG and GB is medial, and FL (is) equal to
ἐστὶτὸΔΜτῷΖΛ.ὡςδὲτὸΔΜπρὸςτὸΖΛ,οὕτωςἐστὶν twice the (rectangle contained) by AG and GB, FL (is)
ἡΓΜπρὸςτὴνΖΜ.ἀσύμμετροςἄραἐστὶνἡΓΜτῇΖΜ thus a medial (area). And it is applied to the rational
μήκει.καίεἰσινἀμφότεραιῥηταί·αἱἄραΓΜ,ΜΖῥηταίεἰσι (straight-line) CD, producing FM as breadth. FM is
397

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
δυνάμειμόνονσύμμετροι·ἡΓΖἄραἀποτομήἐστιν. λέγω thus rational, and incommensurable in length with CD
δή,ὅτικαὶπρώτη.
[Prop. 10.22]. And since the (sum of the squares) on AG
᾿ΕπεὶγὰρτῶνἀπὸτῶνΑΗ,ΗΒμέσονἀνάλογόνἐστι and GB is rational, and twice the (rectangle contained)
τὸὑπὸτῶνΑΗ,ΗΒ,καίἐστιτῷμὲνἀπὸτῆςΑΗἴσοντὸ by AG and GB medial, the (sum of the squares) on AG
ΓΘ,τῷδὲἀπὸτῆςΒΗἴσοντὸΚΛ,τῷδὲὐπὸτῶνΑΗ, and GB is thus incommensurable with twice the (rectan-
ΗΒτὸΝΛ,καὶτῶνΓΘ,ΚΛἄραμέσονἀνάλογόνἐστιτὸ gle contained) by AG and GB. And CL is equal to the
ΝΛ·ἔστινἄραὡςτὸΓΘπρὸςτὸΝΛ,οὕτωςτὸΝΛπρὸς (sum of the squares) on AG and GB, and FL to twice the
τὸΚΛ.ἀλλ᾿ὡςμὲντὸΓΘπρὸςτὸΝΛ,οὕτωςἐστὶνἡ (rectangle contained) by AG and GB. DM is thus incom-
ΓΚπρὸςτὴνΝΜ·ὡςδὲτὸΝΛπρὸςτὸΚΛ,οὕτωςἐστὶν mensurable with FL. And as DM (is) to FL, so CM is to
ἡΝΜπρὸςτὴνΚΜ·τὸἄραὑπὸτῶνΓΚ,ΚΜἴσονἐστὶ FM [Prop. 6.1]. CM is thus incommensurable in length
τῷἀπὸτῆςΝΜ,τουτέστιτῷτετάρτῳμέρειτοῦἀπὸτῆς with FM [Prop. 10.11]. And both are rational (straight-
ΖΜ.καὶεπεὶσύμμετρόνἐστιτὸἀπὸτῆςΑΗτῷἀπὸτῆς lines). Thus, CM and MF are rational (straight-lines
ΗΒ,σύμμετρόν[ἐστι]καὶτὸΓΘτῷΚΛ.ὡςδὲτὸΓΘπρὸς which are) commensurable in square only. CF is thus an
τὸΚΛ,οὕτωςἡΓΚπρὸςτὴνΚΜ·σύμμετροςἄραἐστὶνἡ apotome [Prop. 10.73]. So, I say that (it is) also a first
ΓΚτῇΚΜ.ἐπεὶοὖνδύοεὐθεῖαιἄνισοίεἰσιναἱΓΜ,ΜΖ, (apotome).
καὶτῷτετάρτῳμέρειτοῦἀπὸτῆςΖΜἴσονπαρὰτὴνΓΜ For since the (rectangle contained) by AG and GB is
παραβέβληταιἐλλεῖπονεἴδειτετραγώνῳτὸὑπὸτῶνΓΚ, the mean proportional to the (squares) on AG and GB
ΚΜ,καίἐστισύμμετροςἡΓΚτῇΚΜ,ἡἄραΓΜτῆςΜΖ [Prop. 10.21 lem.], and CH is equal to the (square) on
μεῖζονδύναταιτῷἀπὸσυμμέτρουἑαυτῇμήκει.καίἐστινἡ AG, and KL equal to the (square) on BG, and NL to
ΓΜσύμμετροςτῇἐκκειμένῃῥητῇτῇΓΔμήκει·ἡἄραΓΖ the (rectangle contained) by AG and GB, NL is thus
ἀποτομήἐστιπρώτη.
also the mean proportional to CH and KL. Thus, as
Τὸ ἄρα ἀπὸ ἀποτομῆς παρὰ ῥητὴν παραβαλλόμενον CH is to NL, so NL (is) to KL. But, as CH (is) to
πλάτοςποιεῖἀποτομὴνπρώτην·ὅπερἔδειδεῖξαι. NL, so CK is to NM, and as NL (is) to KL, so NM
is to KM [Prop. 6.1]. Thus, the (rectangle contained)
by CK and KM is equal to the (square) on NM—
that is to say, to the fourth part of the (square) on FM
[Prop. 6.17]. And since the (square) on AG is commensurable
with the (square) on GB, CH [is] also commensurable
with KL. And as CH (is) to KL, so CK (is) to
KM [Prop. 6.1]. CK is thus commensurable (in length)
with KM [Prop. 10.11]. Therefore, since CM and MF
are two unequal straight-lines, and the (rectangle contained)
by CK and KM, equal to the fourth part of the
(square) on FM, has been applied to CM, falling short
by a square figure, and CK is commensurable (in length)
with KM, the square on CM is thus greater than (the
square on) MF by the (square) on (some straight-line)
commensurable in length with (CM) [Prop. 10.17]. And
CM is commensurable in length with the (previously)
laid down rational (straight-line) CD. Thus, CF is a first
apotome [Def. 10.15].
Thus, the (square) on an apotome, applied to a rational
(straight-line), produces a first apotome as breadth.
(Which is) the very thing it was required to show.
. Proposition 98
Τὸ ἀπὸ μέσης ἀποτομῆς πρώτης παρὰ ῥητὴν παρα- The (square) on a first apotome of a medial (straightβαλλόμενονπλάτοςποιεῖἀποτομὴνδευτέραν.
line), applied to a rational (straight-line), produces a sec-
῎ΕστωμέσηςἀποτομὴπρώτηἡΑΒ,ῥητὴδὲἡΓΔ,καὶτῷ ond apotome as breadth.
ἀπὸτῆςΑΒἴσονπαρὰτὴνΓΔπαραβεβλήσθωτὸΓΕπλάτος Let AB be a first apotome of a medial (straight-line),
398

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ποιοῦντὴνΓΖ·λέγω,ὅτιἡΓΖἀποτομήἐστιδευτέρα. and CD a rational (straight-line). And let CE, equal to
῎ΕστωγὰρτῇΑΒπροσαρμόζουσαἡΒΗ·αἱἄραΑΗ, the (square) on AB, have been applied to CD, producing
ΗΒμέσαιεἰσὶδυνάμειμόνονσύμμετροιῥητὸνπεριέχουσαι. CF as breadth. I say that CF is a second apotome.
καὶτῷμὲνἀπὸτῆςΑΗἴσονπαρὰτὴνΓΔπαραβεβλήσθω For let BG be an attachment to AB. Thus, AG and
τὸΓΘπλάτοςποιοῦντὴνΓΚ,τῷδὲἀπὸτῆςΗΒἴσοντὸ GB are medial (straight-lines which are) commensurable
ΚΛπλάτοςποιοῦντὴνΚΜ·ὅλονἄρατὸΓΛἴσονἐστὶτοῖς in square only, containing a rational (area) [Prop. 10.74].
ἀπὸτῶνΑΗ,ΗΒ·μέσονἄρακαὶτὸΓΛ.καὶπαρὰῥητὴν And let CH, equal to the (square) on AG, have been apτὴνΓΔπαράκειταιπλάτοςποιοῦντὴνΓΜ·ῥητὴἄραἐστὶν
plied to CD, producing CK as breadth, and KL, equal
ἡΓΜκαὶἀσύμμετροςτῇΓΔμήκει. καὶἐπεὶτὸΓΛἴσον to the (square) on GB, producing KM as breadth. Thus,
ἐστὶτοῖςἀπὸτῶνΑΗ,ΗΒ,ὧντὸἀπὸτῆςΑΒἴσονἐστὶ the whole of CL is equal to the (sum of the squares)
τῷΓΕ,λοιπὸνἄρατὸδὶςὑπὸτῶνΑΗ,ΗΒἴσονἐστὶτῷ on AG and GB. Thus, CL (is) also a medial (area)
ΖΛ.ῥητὸνδέ[ἐστι]τὸδὶςὑπὸτῶνΑΗ,ΗΒ·ῥητὸνἄρατὸ [Props. 10.15, 10.23 corr.]. And it is applied to the ratio-
ΖΛ.καὶπαρὰῥητὴντὴνΖΕπαράκειταιπλάτοςποιοῦντὴν nal (straight-line) CD, producing CM as breadth. CM
ΖΜ·ῥητὴἄραἐστὶκαὶἡΖΜκαὶσύμμετροςτῇΓΔμήκει. is thus rational, and incommensurable in length with CD
ἐπεὶοὖντὰμὲνἀπὸτῶνΑΗ,ΗΒ,τουτέστιτὸΓΛ,μέσον [Prop. 10.22]. And since CL is equal to the (sum of the
ἐστίν,τὸδὲδὶςὑπὸτῶνΑΗ,ΗΒ,τουτέστιτὸΖΛ,ῥητόν squares) on AG and GB, of which the (square) on AB
ἀσύμμετρονἄραἐστὶτὸΓΛτῷΖΛ.ὡςδὲτὸΓΛπρὸςτὸ is equal to CE, the remainder, twice the (rectangle con-
ΖΛ,οὕτωςἐστὶνἡΓΜπρὸςτὴνΖΜ·ἀσύμμετροςἄραἡ tained) by AG and GB, is thus equal to FL [Prop. 2.7].
ΓΜτῇΖΜμήκει. καίεἰσινἀμφότεραιῥηταί·αἱἄραΓΜ, And twice the (rectangle contained) by AG and GB [is]
ΜΖῥηταίεἰσιδυνάμειμόνονσύμμετροι·ἡΓΖἄραἀποτομή rational. Thus, FL (is) rational. And it is applied to the
ἐστιν.λέγωδή,ὅτικαὶδευτέρα.
rational (straight-line) FE, producing FM as breadth.
FM is thus also rational, and commensurable in length
with CD [Prop. 10.20]. Therefore, since the (sum of the
squares) on AG and GB—that is to say, CL—is medial,
and twice the (rectangle contained) by AG and GB—
that is to say, FL—(is) rational, CL is thus incommensurable
with FL. And as CL (is) to FL, so CM is to
À
Æ Ã Å
 Ä
FM [Prop. 6.1]. Thus, CM (is) incommensurable in
length with FM [Prop. 10.11]. And they are both rational
(straight-lines). Thus, CM and MF are rational
(straight-lines which are) commensurable in square only.
CF is thus an apotome [Prop. 10.73]. So, I say that (it
is) also a second (apotome).
D E O H L
ΤετμήσθωγὰρἡΖΜδίχακατὰτὸΝ,καὶἤχθωδιὰτοῦ For let FM have been cut in half at N. And let
ΝτῇΓΔπαράλληλοςἡΝΞ·ἑκάτερονἄρατῶνΖΞ,ΝΛἴσον NO have been drawn through (point) N, parallel to
ἐστὶτῷὑπὸτῶνΑΗ,ΗΒ.καὶἐπεὶτῶνἀπὸτῶνΑΗ,ΗΒ CD. Thus, FO and NL are each equal to the (rectanτετραγώνωνμέσονἀνάλογόνἐστιτὸὑπὸτῶνΑΗ,ΗΒ,καί
gle contained) by AG and GB. And since the (rectan-
ἐστινἴσοντὸμὲνἀπὸτῆςΑΗτῷΓΘ,τὸδὲὑπὸτῶνΑΗ, gle contained) by AG and GB is the mean proportional
ΗΒτῷΝΛ,τὸδὲἀπὸτῆςΒΗτῷΚΛ,καὶτῶνΓΘ,ΚΛ to the squares on AG and GB [Prop. 10.21 lem.], and
ἄραμέσονἀνάλογόνἐστιτὸΝΛ·ἔστινἄραὡςτὸΓΘπρὸς the (square) on AG is equal to CH, and the (rectangle
τὸΝΛ,οὕτωςτὸΝΛπρὸςτὸΚΛ.ἀλλ᾿ὡςμὲντὸΓΘπρὸς contained) by AG and GB to NL, and the (square) on
C
A
F
B
N
G
K
M
399

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
τὸΝΛ,οὕτωςἐστὶνἡΓΚπρὸςτὴνΝΜ,ὡςδὲτὸΝΛπρὸς BG to KL, NL is thus also the mean proportional to
τὸΚΛ,οὕτωςἐστὶνἡΝΜπρὸςτὴνΜΚ·ὡςἄραἡΓΚ CH and KL. Thus, as CH is to NL, so NL (is) to KL
πρὸςτὴνΝΜ,οὕτωςἐστὶνἡΝΜπρὸςτὴνΚΜ·τὸἄρα [Prop. 5.11]. But, as CH (is) to NL, so CK is to NM,
ὑπὸτῶνΓΚ,ΚΜἴσονἐστὶτῷἀπὸτῆςΝΜ,τουτέστιτῷ and as NL (is) to KL, so NM is to MK [Prop. 6.1].
τετάρτῳμέρειτοῦἀπὸτῆςΖΜ[καὶἐπεὶσύμμετρόνἐστιτὸ Thus, as CK (is) to NM, so NM is to KM [Prop. 5.11].
ἀπὸτῆςΑΗτῷἀπὸτῆςΒΗ,σύμμετρόνἐστικαὶτὸΓΘτῷ The (rectangle contained) by CK and KM is thus equal
ΚΛ,τουτέστινἡΓΚτῇΚΜ].ἐπεὶοὖνδύοεὐθεῖαιἄνισοί to the (square) on NM [Prop. 6.17]—that is to say, to
εἰσιναἱΓΜ,ΜΖ,καὶτῷτετάτρῳμέρειτοῦἀπὸτῆςΜΖ the fourth part of the (square) on FM [and since the
ἴσονπαρὰτὴνμείζονατὴνΓΜπαραβέβληταιἐλλεῖπονεἴδει (square) on AG is commensurable with the (square) on
τετραγώνῳτὸὑπὸτῶνΓΚ,ΚΜκαὶεἰςσύμμετρααὐτὴν BG, CH is also commensurable with KL—that is to say,
διαιρεῖ,ἡἄραΓΜτῆςΜΖμεῖζονδύναταιτῷἀπὸσυμμέτρου CK with KM]. Therefore, since CM and MF are two
ἑαυτῇμήκει.καίἐστινἡπροσαρμόζουσαἡΖΜσύμμετρος unequal straight-lines, and the (rectangle contained) by
μήκειτῇἐκκειμένῃῥητῇτῇΓΔ·ἡἄραΓΖἀποτομήἐστι CK and KM, equal to the fourth part of the (square)
δευτέρα.
on MF , has been applied to the greater CM, falling
Τὸἄραἀπὸμέσηςἀποτομῆςπρώτηςπαρὰῥητὴνπα- short by a square figure, and divides it into commensuραβαλλόμενονπλάτοςποιεῖἀποτομὴνδευτέραν·ὅπερἔδει
rable (parts), the square on CM is thus greater than (the
δεῖξαι.
square on) MF by the (square) on (some straight-line)
commensurable in length with (CM) [Prop. 10.17]. The
attachment FM is also commensurable in length with the
(previously) laid down rational (straight-line) CD. CF is
thus a second apotome [Def. 10.16].
Thus, the (square) on a first apotome of a medial
(straight-line), applied to a rational (straight-line), produces
a second apotome as breadth. (Which is) the very
thing it was required to show.
. Proposition 99
Τὸ ἀπὸ μέσης ἀποτομῆς δευτέρας παρὰ ῥητὴν παρα- The (square) on a second apotome of a medial
βαλλόμενονπλάτοςποιεῖἀποτομὴντρίτην.
(straight-line), applied to a rational (straight-line), produces
a third apotome as breadth.
Æ
A B G
C F N K M
À
à Å
Â
Ä
D E O H L
῎ΕστωμέσηςἀποτομὴδευτέραἡΑΒ,ῥητὴδὲἡΓΔ,καὶ Let AB be the second apotome of a medial (straightτῷἀπὸτῆςΑΒἴσονπαρὰτὴνΓΔπαραβεβλήσθωτὸΓΕ
line), and CD a rational (straight-line). And let CE,
πλάτοςποιοῦντὴνΓΖ·λέγω,ὅτιἡΓΖἀποτομήἐστιτρίτη. equal to the (square) on AB, have been applied to CD,
῎ΕστωγὰρτῇΑΒπροσαρμόζουσαἡΒΗ·αἱἄραΑΗ,ΗΒ producing CF as breadth. I say that CF is a third apoμέσαιεἰσὶδυνάμειμόνονσύμμετροιμέσονπεριέχουσαι.καὶ
tome.
τῷμὲνἀπὸτῆςΑΗἴσονπαρὰτὴνΓΔπαραβεβλήσθωτὸΓΘ For let BG be an attachment to AB. Thus, AG and
πλάτοςποιοῦντὴνΓΚ,τῷδὲἀπὸτῆςΒΗἴσονπαρὰτὴνΚΘ GB are medial (straight-lines which are) commensurable
παραβεβλήσθωτὸΚΛπλάτοςποιοῦντὴνΚΜ·ὅλονἄρατὸ in square only, containing a medial (area) [Prop. 10.75].
ΓΛἴσονἐστὶτοῖςἀπὸτῶνΑΗ,ΗΒ[καίἐστιμέσατὰἀπὸ And let CH, equal to the (square) on AG, have been
τῶνΑΗ,ΗΒ]·μέσονἄρακαὶτὸΓΛ.καὶπαρὰῥητὴντὴν applied to CD, producing CK as breadth. And let KL,
400

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ΓΔπαραβέβληταιπλάτοςποιοῦντὴνΓΜ·ῥητὴἄραἐστὶν equal to the (square) on BG, have been applied to KH,
ἡΓΜκαὶἀσύμμετροςτῇΓΔμήκει. καὶἐπεὶὅλοντὸΓΛ producing KM as breadth. Thus, the whole of CL is
ἴσονἐστὶτοῖςἀπὸτῶνΑΗ,ΗΒ,ὧντὸΓΕἴσονἐστὶτῷ equal to the (sum of the squares) on AG and GB [and
ἀπὸτῆςΑΒ,λοιπὸνἄρατὸΛΖἴσονἐστὶτῷδὶςὑπὸτῶν the (sum of the squares) on AG and GB is medial]. CL
ΑΗ,ΗΒ.τετμήσθωοὖνἡΖΜδίχακατὰτὸΝσημεῖον, (is) thus also medial [Props. 10.15, 10.23 corr.]. And it
καὶτῇΓΔπαράλληλοςἤχθωἡΝΞ·ἑκάτερονἄρατῶνΖΞ, has been applied to the rational (straight-line) CD, pro-
ΝΛἴσονἐστὶτῷὑπὸτῶνΑΗ,ΗΒ.μέσονδὲτὸὑπὸτῶν ducing CM as breadth. Thus, CM is rational, and incom-
ΑΗ,ΗΒ·μέσονἄραἐστὶκαὶτὸΖΛ.καὶπαρὰῥητὴντὴν mensurable in length with CD [Prop. 10.22]. And since
ΕΖπαράκειταιπλάτοςποιοῦντὴνΖΜ·ῥητὴἄρακαὶἡΖΜ the whole of CL is equal to the (sum of the squares) on
καὶἀσύμμετροςτῇΓΔμήκει. καὶἐπεὶαἱΑΗ,ΗΒδυνάμει AG and GB, of which CE is equal to the (square) on
μόνονεἰσὶσύμμετροι,ἀσύμμετροςἄρα[ἐστὶ]μήκειἡΑΗ AB, the remainder LF is thus equal to twice the (rectτῇΗΒ·ἀσύμμετρονἄραἐστὶκαὶτὸἀπὸτῆςΑΗτῷὑπὸ
angle contained) by AG and GB [Prop. 2.7]. Therefore,
τῶνΑΗ,ΗΒ.ἀλλὰτῷμὲνἀπὸτῆςΑΗσύμμετράἐστιτὰ let FM have been cut in half at point N. And let NO
ἀπὸτῶνΑΗ,ΗΒ,τῷδὲὑπὸτῶνΑΗ,ΗΒτὸδὶςὑπὸτῶν have been drawn parallel to CD. Thus, FO and NL are
ΑΗ,ΗΒ·ἀσύμμετραἄραἐστὶτὰἀπὸτῶνΑΗ,ΗΒτῷδὶς each equal to the (rectangle contained) by AG and GB.
ὑπὸτῶνΑΗ,ΗΒ.ἀλλὰτοῖςμὲνἀπὸτῶνΑΗ,ΗΒἴσον And the (rectangle contained) by AG and GB (is) me-
ἐστὶτὸΓΛ,τῷδὲδὶςὑπὸτῶνΑΗ,ΗΒἴσονἐστὶτὸΖΛ· dial. Thus, FL is also medial. And it is applied to the
ἀσύμμετρονἄραἐστὶτὸΓΛτῷΖΛ.ὡςδὲτὸΓΛπρὸςτὸ rational (straight-line) EF , producing FM as breadth.
ΖΛ,οὕτωςἐστὶνἡΓΜπρὸςτὴνΖΜ·ἀσύμμετροςἄραἐστὶν FM is thus rational, and incommensurable in length with
ἡΓΜτῇΖΜμήκει.καίεἰσινἀμφότεραιῥηταί·αἱἄραΓΜ, CD [Prop. 10.22]. And since AG and GB are commen-
ΜΖῥηταίεἰσιδυνάμειμόνονσύμμετροι·ἀποτομὴἄραἐστὶν surable in square only, AG [is] thus incommensurable in
ἡΓΖ.λέγωδή,ὅτικαὶτρίτη.
length with GB. Thus, the (square) on AG is also incom-
᾿ΕπεὶγὰρσύμμετρόνἐστιτὸἀπὸτῆςΑΗτῷἀπὸτῆς mensurable with the (rectangle contained) by AG and
ΗΒ,σύμμετρονἄρακαὶτὸΓΘτῷΚΛ·ὥστεκαὶἡΓΚτῇ GB [Props. 6.1, 10.11]. But, the (sum of the squares)
ΚΜ.καὶἐπεὶτῶνἀπὸτῶνΑΗ,ΗΒμέσονἀνάλογόνἐστι on AG and GB is commensurable with the (square) on
τὸὑπὸτῶνΑΗ,ΗΒ,καίἐστιτῷμὲνἀπὸτῆςΑΗἴσον AG, and twice the (rectangle contained) by AG and GB
τὸΓΘ,τῷδὲἀπὸτῆςΗΒἴσοντὸΚΛ,τῷδὲὑπὸτῶν with the (rectangle contained) by AG and GB. The
ΑΗ,ΗΒἴσοντὸΝΛ,καὶτῶνΓΘ,ΚΛἄραμέσονἀνάλογόν (sum of the squares) on AG and GB is thus incommen-
ἐστιτὸΝΛ·ἔστινἄραὡςτὸΓΘπρὸςτὸΝΛ,οὕτωςτὸ surable with twice the (rectangle contained) by AG and
ΝΛπρὸςτὸΚΛ.ἀλλ᾿ὡςμὲντὸΓΘπρὸςτὸΝΛ,οὕτως GB [Prop. 10.13]. But, CL is equal to the (sum of the
ἐστὶνἡΓΚπρὸςτὴνΝΜ,ὡςδὲτὸΝΛπρὸςτὸΚΛ,οὕτως squares) on AG and GB, and FL is equal to twice the
ἐστὶνἡΝΜπρὸςτὴνΚΜ·ὡςἄραἡΓΚπρὸςτὴνΜΝ, (rectangle contained) by AG and GB. Thus, CL is inοὕτωςἐστὶνἡΜΝπρὸςτὴνΚΜ·τὸἄραὑπὸτῶνΓΚ,ΚΜ
commensurable with FL. And as CL (is) to FL, so CM
ἴσονἐστὶτῷ[ἀπὸτῆςΜΝ,τουτέστιτῷ]τετάρτῳμέρειτοῦ is to FM [Prop. 6.1]. CM is thus incommensurable in
ἀπὸτῆςΖΜ.ἐπεὶοὖνδύοεὐθεῖαιἄνισοίεἰσιναἱΓΜ,ΜΖ, length with FM [Prop. 10.11]. And they are both raκαὶτῷτετάρτῳμέρειτοῦἀπὸτῆςΖΜἴσονπαρὰτὴνΓΜ
tional (straight-lines). Thus, CM and MF are rational
παραβέβληταιἐλλεῖπονεἴδειτετραγώνῳκαὶεἰςσύμμετρα (straight-lines which are) commensurable in square only.
αὐτὴνδιαιρεῖ,ἡΓΜἄρατῆςΜΖμεῖζονδύναταιτῷἀπὸ CF is thus an apotome [Prop. 10.73]. So, I say that (it
συμμέτρουἑαυτῇ. καὶοὐδετέρατῶνΓΜ,ΜΖσύμμετρός is) also a third (apotome).
ἐστιμήκειτῇἐκκειμένῃῥητῇτῇΓΔ·ἡἄραΓΖἀποτομήἐστι For since the (square) on AG is commensurable with
τρίτη.
the (square) on GB, CH (is) thus also commensu-
Τὸἄραἀπὸμέσηςἀποτομῆςδευτέραςπαρὰῥητὴνπαρα- rable with KL. Hence, CK (is) also (commensurable
βαλλόμενονπλάτοςποιεῖἀποτομὴντρίτην·ὅπερἔδειδεῖξαι. in length) with KM [Props. 6.1, 10.11]. And since the
(rectangle contained) by AG and GB is the mean proportional
to the (squares) on AG and GB [Prop. 10.21 lem.],
and CH is equal to the (square) on AG, and KL equal
to the (square) on GB, and NL equal to the (rectangle
contained) by AG and GB, NL is thus also the mean
proportional to CH and KL. Thus, as CH is to NL, so
NL (is) to KL. But, as CH (is) to NL, so CK is to NM,
and as NL (is) to KL, so NM (is) to KM [Prop. 6.1].
401

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
Thus, as CK (is) to MN, so MN is to KM [Prop. 5.11].
Thus, the (rectangle contained) by CK and KM is equal
to the [(square) on MN—that is to say, to the] fourth
part of the (square) on FM [Prop. 6.17]. Therefore,
since CM and MF are two unequal straight-lines, and
(some area), equal to the fourth part of the (square) on
FM, has been applied to CM, falling short by a square
figure, and divides it into commensurable (parts), the
square on CM is thus greater than (the square on) MF
by the (square) on (some straight-line) commensurable
(in length) with (CM) [Prop. 10.17]. And neither of
CM and MF is commensurable in length with the (previously)
laid down rational (straight-line) CD. CF is
thus a third apotome [Def. 10.13].
Thus, the (square) on a second apotome of a medial
(straight-line), applied to a rational (straight-line), produces
a third apotome as breadth. (Which is) the very
thing it was required to show.
Ö. Proposition 100
Τὸἀπὸἐλάσσονοςπαρὰῥητὴνπαραβαλλόμενονπλάτος The (square) on a minor (straight-line), applied to
ποιεῖἀποτομὴντετάρτην.
a rational (straight-line), produces a fourth apotome as
breadth.
Æ
A B G
C F N K M
À
à Å
Â
Ä
D E O H L
῎ΕστωἐλάσσωνἡΑΒ,ῥητὴδὲἡΓΔ,καὶτῷἀπὸτῆς Let AB be a minor (straight-line), and CD a rational
ΑΒἴσονπαρὰῥητὴντὴνΓΔπαραβεβλήσθωτὸΓΕπλάτος (straight-line). And let CE, equal to the (square) on AB,
ποιοῦντὴνΓΖ·λέγω,ὅτιἡΓΖἀποτομήἐστιτετάρτη. have been applied to the rational (straight-line) CD, pro-
῎ΕστωγὰρτῇΑΒπροσαρμόζουσαἡΒΗ·αἱἄραΑΗ, ducing CF as breadth. I say that CF is a fourth apotome.
ΗΒδυνάμειεἰσὶνἀσύμμετροιποιοῦσαιτὸμὲνσυγκείμενον For let BG be an attachment to AB. Thus, AG and
ἐκτῶνἀπὸτῶνΑΗ,ΗΒτετραγώνωνῥητόν,τὸδὲδὶςὑπὸ GB are incommensurable in square, making the sum of
τῶνΑΗ,ΗΒμέσον. καὶτῷμὲνἀπὸτῆςΑΗἴσονπαρὰ the squares on AG and GB rational, and twice the (rectτὴνΓΔπαραβεβλήσθωτὸΓΘπλάτοςποιοῦντὴνΓΚ,τῷ
angle contained) by AG and GB medial [Prop. 10.76].
δὲἀπὸτῆςΒΗἴσοντὸΚΛπλάτοςποιοῦντὴνΚΜ·ὅλον And let CH, equal to the (square) on AG, have been ap-
ἄρατὸΓΛἴσονἐστὶτοῖςἀπὸτῶνΑΗ,ΗΒ.καίἐστιτὸ plied to CD, producing CK as breadth, and KL, equal
συγκείμενονἐκτῶνἀπὸτῶνΑΗ,ΗΒῥητόν· ῥητὸνἄρα to the (square) on BG, producing KM as breadth. Thus,
ἐστὶκαὶτὸΓΛ.καὶπαρὰῥητὴντὴνΓΔπαράκειταιπλάτος the whole of CL is equal to the (sum of the squares) on
ποιοῦντὴνΓΜ·ῥητὴἄρακαὶἡΓΜκαὶσύμμετροςτῇΓΔ AG and GB. And the sum of the (squares) on AG and
μήκει.καὶἐπεὶὅλοντὸΓΛἴσονἐστὶτοῖςἀπὸτῶνΑΗ,ΗΒ, GB is rational. CL is thus also rational. And it is ap-
ὧντὸΓΕἴσονἐστὶτῷἀπὸτῆςΑΒ,λοιπὸνἄρατὸΖΛἴσον plied to the rational (straight-line) CD, producing CM
ἐστὶτῷδὶςὑπὸτῶνΑΗ,ΗΒ.τετμήσθωοὖνἡΖΜδίχα as breadth. Thus, CM (is) also rational, and commenκατὰτὸΝσημεῖον,καὶἤχθωδὶατοῦΝὁποτέρᾳτῶνΓΔ,
surable in length with CD [Prop. 10.20]. And since the
402

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ΜΛπαράλληλοςἡΝΞ·ἑκάτερονἄρατῶνΖΞ,ΝΛἴσονἐστὶ whole of CL is equal to the (sum of the squares) on AG
τῷὑπὸτῶνΑΗ,ΗΒ.καὶἐπεὶτὸδὶςὑπὸτῶνΑΗ,ΗΒμέσον and GB, of which CE is equal to the (square) on AB,
ἐστὶκαίἐστινἴσοντῷΖΛ,καὶτὸΖΛἄραμέσονἐστίν.καὶ the remainder FL is thus equal to twice the (rectangle
παρὰῥητὴντὴνΖΕπαράκειταιπλάτοςποιοῦντὴνΖΜ·ῥητὴ contained) by AG and GB [Prop. 2.7]. Therefore, let
ἄραἐστὶνἡΖΜκαὶἀσύμμετροςτῇΓΔμήκει. καὶἐπεὶτὸ FM have been cut in half at point N. And let NO have
μὲνσυγκείμενονἐκτῶνἀπὸτῶνΑΗ,ΗΒῥητόνἐστιν,τὸ been drawn through N, parallel to either of CD or ML.
δὲδὶςὑπὸτῶνΑΗ,ΗΒμέσον,ἀσύμμετρα[ἄρα]ἐστὶτὰἀπὸ Thus, FO and NL are each equal to the (rectangle conτῶνΑΗ,ΗΒτῷδὶςὑπὸτῶνΑΗ,ΗΒ.ἴσονδέ[ἐστι]τὸΓΛ
tained) by AG and GB. And since twice the (rectangle
τοῖςἀπὸτῶνΑΗ,ΗΒ,τῷδὲδὶςὑπὸτῶνΑΗ,ΗΒἴσοντὸ contained) by AG and GB is medial, and is equal to FL,
ΖΛ·ἀσύμμετρονἄρα[ἐστὶ]τὸΓΛτῷΖΛ.ὡςδὲτὸΓΛπρὸς FL is thus also medial. And it is applied to the ratioτὸΖΛ,οὕτωςἐστὶνἡΓΜπρὸςτὴνΜΖ·ἀσύμμετροςἄρα
nal (straight-line) FE, producing FM as breadth. Thus,
ἐστὶνἡΓΜτῇΜΖμήκει.καίεἰσινἀμφότεραιῥηταί·αἱἄρα FM is rational, and incommensurable in length with CD
ΓΜ,ΜΖῥηταίεἰσιδυνάμειμόνονσύμμετροι·ἀποτομὴἄρα [Prop. 10.22]. And since the sum of the (squares) on AG
ἐστὶνἡΓΖ.λέγω[δή],ὅτικαὶτετάρτη.
and GB is rational, and twice the (rectangle contained)
᾿ΕπεὶγὰραἱΑΗ,ΗΒδυνάμειεἰσὶνἀσύμμετροι,ἀσύμμετ- by AG and GB medial, the (sum of the squares) on AG
ρονἄρακαὶτὸἀπὸτῆςΑΗτῷἀπὸτῆςΗΒ.καίἐστιτῷ and GB is [thus] incommensurable with twice the (rectμὲνἀπὸτῆςΑΗἴσοντὸΓΘ,τῷδὲἀπὸτῆςΗΒἴσοντὸ
angle contained) by AG and GB. And CL (is) equal to
ΚΛ·ἀσύμμετρονἄραἐστὶτὸΓΘτῷΚΛ.ὡςδὲτὸΓΘπρὸς the (sum of the squares) on AG and GB, and FL equal
τὸΚΛ,οὕτωςἐστὶνἡΓΚπρὸςτὴνΚΜ·ἀσύμμετροςἄρα to twice the (rectangle contained) by AG and GB. CL
ἐστὶνἡΓΚτῇΚΜμήκει. καὶἐπεὶτῶνἀπὸτῶνΑΗ,ΗΒ [is] thus incommensurable with FL. And as CL (is) to
μέσονἀνάλογόνἐστιτὸὑπὸτῶνΑΗ,ΗΒ,καίἐστινἴσον FL, so CM is to MF [Prop. 6.1]. CM is thus incommenτὸμὲνἀπὸτῆςΑΗτῷΓΘ,τὸδὲἀπὸτῆςΗΒτῷΚΛ,
surable in length with MF [Prop. 10.11]. And both are
τὸδὲὑπὸτῶνΑΗ,ΗΒτῷΝΛ,τῶνἄραΓΘ,ΚΛμέσον rational (straight-lines). Thus, CM and MF are rational
ἀνάλογόνἐστιτὸΝΛ·ἔστινἄραὡςτὸΓΘπρὸςτὸΝΛ, (straight-lines which are) commensurable in square only.
οὕτωςτὸΝΛπρὸςτὸΚΛ.ἀλλ᾿ὡςμὲντὸΓΘπρὸςτὸ CF is thus an apotome [Prop. 10.73]. [So], I say that (it
ΝΛ,οὕτωςἐστίνἡΓΚπρὸςτὴνΝΜ,ὡςδὲτὸΝΛπρὸς is) also a fourth (apotome).
τὸΚΛ,οὕτωςἐστὶνἡΝΜπρὸςτὴνΚΜ·ὡςἄραἡΓΚ For since AG and GB are incommensurable in square,
πρὸςτὴνΜΝ,οὕτωςἐστὶνἡΜΝπρὸςτὴνΚΜ·τὸἄρα the (square) on AG (is) thus also incommensurable with
ὑπὸτῶνΓΚ,ΚΜἴσονἐστὶτῷἀπὸτῆςΜΝ,τουτέστιτῷ the (square) on GB. And CH is equal to the (square) on
τετάρτῳμέρειτοῦἀπὸτῆςΖΜ.ἐπεὶοὖνδύοεὐθεῖαιἄνισοί AG, and KL equal to the (square) on GB. Thus, CH is
εἰσιναἱΓΜ,ΜΖ,καὶτῷτετράρτῳμέρειτοῦἀπὸτῆςΜΖ incommensurable with KL. And as CH (is) to KL, so
ἴσονπαρὰτὴνΓΜπαραβέβληταιἐλλεῖπονεἴδειτετραγώνῳ CK is to KM [Prop. 6.1]. CK is thus incommensurable
τὸὑπὸτῶνΓΚ,ΚΜκαὶεἰςἀσύμμετρααὐτὴνδιαιρεῖ,ἡἄρα in length with KM [Prop. 10.11]. And since the (rectan-
ΓΜτῆςΜΖμεῖζονδύναταιτῷἀπὸἀσυμμέτρουἑαυτῇ.καί gle contained) by AG and GB is the mean proportional
ἐστινὅληἡΓΜσύμμετροςμήκειτῇἐκκειμένῃῥητῇτῇΓΔ· to the (squares) on AG and GB [Prop. 10.21 lem.], and
ἡἄραΓΖἀποτομήἐστιτετάρτη.
Τὸἄραἀπὸἐλάσσονοςκαὶτὰἑξῆς.
the (square) on AG is equal to CH, and the (square)
on GB to KL, and the (rectangle contained) by AG and
GB to NL, NL is thus the mean proportional to CH and
KL. Thus, as CH is to NL, so NL (is) to KL. But,
as CH (is) to NL, so CK is to NM, and as NL (is) to
KL, so NM is to KM [Prop. 6.1]. Thus, as CK (is) to
MN, so MN is to KM [Prop. 5.11]. The (rectangle contained)
by CK and KM is thus equal to the (square) on
MN—that is to say, to the fourth part of the (square) on
FM [Prop. 6.17]. Therefore, since CM and MF are two
unequal straight-lines, and the (rectangle contained) by
CK and KM, equal to the fourth part of the (square) on
MF , has been applied to CM, falling short by a square
figure, and divides it into incommensurable (parts), the
square on CM is thus greater than (the square on) MF
by the (square) on (some straight-line) incommensurable
403

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
(in length) with (CM) [Prop. 10.18]. And the whole of
CM is commensurable in length with the (previously)
laid down rational (straight-line) CD. Thus, CF is a
fourth apotome [Def. 10.14].
Thus, the (square) on a minor, and so on . . .
Ö. Proposition 101
Τὸἀπὸτῆςμετὰῥητοῦμέσοντὸὅλονποιούσηςπαρὰ The (square) on that (straight-line) which with a ra-
ῥητὴνπαραβαλλόμενονπλάτοςποιεῖἀποτομὴνπέμπτην. tional (area) makes a medial whole, applied to a rational
(straight-line), produces a fifth apotome as breadth.
Æ
A B G
C F N K M
À
à Å
Â
Ä
D E O H L
῎ΕστωἡμετὰῥητοῦμέσοντὸὅλονποιοῦσαἡΑΒ,ῥητὴ Let AB be that (straight-line) which with a ratioδὲ
ἡΓΔ, καὶτῷ ἀπὸτῆς ΑΒ ἴσον παρὰτὴν ΓΔ παρα- nal (area) makes a medial whole, and CD a rational
βεβλήσθωτὸΓΕπλάτοςποιοῦντὴνΓΖ·λέγω,ὅτιἡΓΖ (straight-line). And let CE, equal to the (square) on AB,
ἀποτομήἐστιπέμπτη.
have been applied to CD, producing CF as breadth. I
῎ΕστωγὰρτῇΑΒπροσαρμόζουσαἡΒΗ·αἱἄραΑΗ, say that CF is a fifth apotome.
ΗΒ εὐθεῖαι δυνάμει εἰσὶν ἀσύμμετροι ποιοῦσαι τὸ μὲν Let BG be an attachment to AB. Thus, the straightσυγκείμενονἐκτῶνἀπ᾿αὐτῶντετραγώνωνμέσον,τὸδὲ
lines AG and GB are incommensurable in square, makδὶςὑπ᾿αὐτῶνῥητόν,καὶτῷμὲνἀπὸτῆςΑΗἴσονπαρὰ
ing the sum of the squares on them medial, and twice
τὴνΓΔπαραβεβλήσθωτὸΓΘ,τῷδὲἀπὸτῆςΗΒἵσον the (rectangle contained) by them rational [Prop. 10.77].
τὸΚΛ·ὅλονἄρατὸΓΛἴσονἐστὶτοῖςἀπὸτῶνΑΗ,ΗΒ. And let CH, equal to the (square) on AG, have been apτὸδὲσυγκείμενονἐκτῶνἀπὸτῶνΑΗ,ΗΒἅμαμέσον
plied to CD, and KL, equal to the (square) on GB. The
ἐστίν· μέσον ἄρα ἐστὶ τὸ ΓΛ. καὶ παρὰ ῥητὴν τὴν ΓΔ whole of CL is thus equal to the (sum of the squares) on
παράκειταιπλάτοςποιοῦντὴνΓΜ·ῥητὴἄραἐστὶνἡΓΜκαὶ AG and GB. And the sum of the (squares) on AG and
ἀσύμμετροςτῇΓΔ.καὶἐπεὶὅλοντὸΓΛἴσονἐστὶτοῖςἀπὸ GB together is medial. Thus, CL is medial. And it has
τῶνΑΗ,ΗΒ,ὧντὸΓΕἴσονἐστὶτῷἀπὸτῆςΑΒ,λοιπὸν been applied to the rational (straight-line) CD, produc-
ἄρατὸΖΛἴσονἐστὶτῷδὶςὑπὸτῶνΑΗ,ΗΒ.τετμήσθω ing CM as breadth. CM is thus rational, and incommenοὖνἡΖΜδίχακατὰτὸΝ,καὶἤχθωδιὰτοῦΝὁποτέρᾳ
surable (in length) with CD [Prop. 10.22]. And since
τῶνΓΔ,ΜΛπαράλληλοςἡΝΞ·ἑκάτερονἄρατῶνΖΞ,ΝΛ the whole of CL is equal to the (sum of the squares) on
ἴσονἐστὶτῷὑπὸτῶνΑΗ,ΗΒ,καὶἐπεὶτὸδὶςὑπὸτῶνΑΗ, AG and GB, of which CE is equal to the (square) on
ΗΒῥητόνἐστικαί[ἐστιν]ἴσοντῷΖΛ,ῥητὸνἄραἐστὶτὸ AB, the remainder FL is thus equal to twice the (rect-
ΖΛ.καὶπαρὰῥητὴντὴνΕΖπαράκειταιπλάτοςποιοῦντὴν angle contained) by AG and GB [Prop. 2.7]. Therefore,
ΖΜ·ῥητὴἄραἐστὶνἡΖΜκαὶσύμμετροςτῇΓΔμήκει.καὶ let FM have been cut in half at N. And let NO have
ἐπεὶτὸμὲνΓΛμέσονἐστίν,τὸδὲΖΛῥητόν,ἀσύμμετρον been drawn through N, parallel to either of CD or ML.
ἄραἐστὶτὸΓΛτῷΖΛ.ὡςδὲτὸΓΛπρὸςτὸΖΛ,οὕτως Thus, FO and NL are each equal to the (rectangle con-
ἡΓΜπρὸςτὴνΜΖ·ἀσύμμετροςἄραἐστὶνἡΓΜτῇΜΖ tained) by AG and GB. And since twice the (rectangle
μήκει.καίεἰσινἀμφότεραιῥηταί·αἱἄραΓΜ,ΜΖῥηταίεἰσι contained) by AG and GB is rational, and [is] equal to
δυνάμειμόνονσύμμετροι·ἀποτομὴἄραἐστὶνἡΓΖ.λὲγω FL, FL is thus rational. And it is applied to the ratioδή,ὅτικαὶπέμπτη.
nal (straight-line) EF , producing FM as breadth. Thus,
῾Ομοίωςγὰρδείξομεν,ὅτιτὸὑπὸτῶνΓΚΜἴσονἐστὶ FM is rational, and commensurable in length with CD
τῷἀπὸτῆςΝΜ,τουτέστιτῷτετάρτῳμέρειτοῦἀπὸτῆς [Prop. 10.20]. And since CL is medial, and FL rational,
404

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ΖΜ.καὶἐπεὶἀσύμμετρόνἐστιτὸἀπὸτῆςΑΗτῷἀπὸτῆς CL is thus incommensurable with FL. And as CL (is) to
ΗΒ,ἴσονδὲτὸμὲνἀπὸτῆςΑΗτῷΓΘ,τὸδὲἀπὸτῆς FL, so CM (is) to MF [Prop. 6.1]. CM is thus incom-
ΗΒτῷΚΛ,ἀσύμμετρονἄρατὸΓΘτῷΚΛ.ὡςδὲτὸΓΘ mensurable in length with MF [Prop. 10.11]. And both
πρὸςτὸΚΛ,οὕτωςἡΓΚπρὸςτὴνΚΜ·ἀσύμμετροςἄρα are rational. Thus, CM and MF are rational (straight-
ἡΓΚτῇΚΜμήκει. ἐπεὶοὖνδύοεὐθεῖαιἄνισοίεἰσιναἱ lines which are) commensurable in square only. CF is
ΓΜ,ΜΖ,καὶτῷτετάρτῳμέρειτοῦἀπὸτῆςΖΜἴσονπαρὰ thus an apotome [Prop. 10.73]. So, I say that (it is) also
τὴν ΓΜ παραβέβληταιἐλλεῖπον εἴδει τετραγώνῳ καὶ εἰς a fifth (apotome).
ἀσύμμετρααὐτὴνδιαιρεῖ,ἡἄραΓΜτῆςΜΖμεῖζονδύναται For, similarly (to the previous propositions), we can
τῷἀπὸἀσύμμέτρουἑαυτῇ. καίἐστινἡπροσαρμόζουσαἡ show that the (rectangle contained) by CKM is equal to
ΖΜσύμμετροςτῇἐκκειμένῃῥητῇτῇΓΔ·ἡἄραΓΖἀποτομή the (square) on NM—that is to say, to the fourth part
ἐστιπέμπτη·ὅπερἔδειδεῖξαι.
of the (square) on FM. And since the (square) on AG
is incommensurable with the (square) on GB, and the
(square) on AG (is) equal to CH, and the (square) on
GB to KL, CH (is) thus incommensurable with KL.
And as CH (is) to KL, so CK (is) to KM [Prop. 6.1].
Thus, CK (is) incommensurable in length with KM
[Prop. 10.11]. Therefore, since CM and MF are two unequal
straight-lines, and (some area), equal to the fourth
part of the (square) on FM, has been applied to CM,
falling short by a square figure, and divides it into incommensurable
(parts), the square on CM is thus greater
than (the square on) MF by the (square) on (some
straight-line) incommensurable (in length) with (CM)
[Prop. 10.18]. And the attachment FM is commensurable
with the (previously) laid down rational (straightline)
CD. Thus, CF is a fifth apotome [Def. 10.15].
(Which is) the very thing it was required to show.
Ö. Proposition 102
Τὸἀπὸτῆςμετὰμέσουμέσοντὸὅλονποιούσηςπαρὰ The (square) on that (straight-line) which with a me-
ῥητὴνπαραβαλλόμενονπλάτοςποιεῖἀποτομὴνἕκτην. dial (area) makes a medial whole, applied to a rational
(straight-line), produces a sixth apotome as breadth.
Æ
A B G
C F N K M
À
à Å
Â
Ä
D E O H L
῎ΕστωἡμετὰμέσουμέσοντὸὅλονποιοῦσαἡΑΒ,ῥητὴ Let AB be that (straight-line) which with a meδὲ
ἡΓΔ, καὶτῷ ἀπὸτῆς ΑΒ ἴσον παρὰτὴν ΓΔ παρα- dial (area) makes a medial whole, and CD a rational
βεβλήσθωτὸΓΕπλάτοςποιοῦντὴνΓΖ·λέγω,ὅτιἡΓΖ (straight-line). And let CE, equal to the (square) on AB,
ἀποτομήἐστινἕκτη.
have been applied to CD, producing CF as breadth. I
῎ΕστωγὰρτῇΑΒπροσαρμόζουσαἡΒΗ·αἱἄραΑΗ, say that CF is a sixth apotome.
ΗΒδυνάμειεἰσὶνἀσύμμετροιποιοῦσαιτότεσυγκείμενον For let BG be an attachment to AB. Thus, AG and
ἐκτῶνἀπ᾿αὐτῶντετραγώνωνμέσονκαὶτὸδὶςὑπὸτῶν GB are incommensurable in square, making the sum of
ΑΗ,ΗΒμέσονκαὶἀσύμμετροντὰἀπὸτῶνΑΗ,ΗΒτῷ the squares on them medial, and twice the (rectangle
405

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
δὶςὑπὸτῶνΑΗ,ΗΒ.παραβεβλήσθωοὖνπαρὰτὴνΓΔτῷ contained) by AG and GB medial, and the (sum of the
μὲνἀπὸτῆςΑΗἴσοντὸΓΘπλάτοςποιοῦντὴνΓΚ,τῷ squares) on AG and GB incommensurable with twice
δὲἀπὸτῆςΒΗτὸΚΛ·ὅλονἄρατὸΓΛἴσονἐστὶτοῖςἀπὸ the (rectangle contained) by AG and GB [Prop. 10.78].
τῶνΑΗ,ΗΒ·μέσονἄρα[ἐστὶ]καὶτὸΓΛ.καὶπαρὰῥητὴν Therefore, let CH, equal to the (square) on AG, have
τὴνΓΔπαράκειταιπλάτοςποιοῦντὴνΓΜ·ῥητὴἄραἐστὶν been applied to CD, producing CK as breadth, and KL,
ἡΓΜκαὶἀσύμμετροςτῇΓΔμήκει. ἐπεὶοὖντὸΓΛἴσον equal to the (square) on BG. Thus, the whole of CL
ἐστὶτοῖςἀπὸτῶνΑΗ,ΗΒ,ὧντὸΓΕἴσοντῷἀπὸτῆςΑΒ, is equal to the (sum of the squares) on AG and GB.
λοιπὸνἄρατὸΖΛἴσονἐστὶτῷδὶςὑπὸτῶνΑΗ,ΗΒ.καί CL [is] thus also medial. And it is applied to the ratio-
ἐστιτὸδὶςὑπὸτῶνΑΗ,ΗΒμέσον·καὶτὸΖΛἄραμέσον nal (straight-line) CD, producing CM as breadth. Thus,
ἐστίν.καὶπαρὰῥητὴντὴνΖΕπαράκειταιπλάτοςποιοῦντὴν CM is rational, and incommensurable in length with
ΖΜ·ῥητὴἄραἐστὶνἡΖΜκαὶἀσύμμετροςτῇΓΔμήκει.καὶ CD [Prop. 10.22]. Therefore, since CL is equal to the
ἐπεὶτὰἀπὸτῶνΑΗ,ΗΒἀσύμμετράἐστιτῷδὶςὑπὸτῶν (sum of the squares) on AG and GB, of which CE (is)
ΑΗ,ΗΒ,καίἐστιτοῖςμὲνἀπὸτῶνΑΗ,ΗΒἴσοντὸΓΛ, equal to the (square) on AB, the remainder FL is thus
τῷδὲδὶςὑπὸτῶνΑΗ,ΗΒἴσοντὸΖΛ,ἀσύμμετροςἄρα equal to twice the (rectangle contained) by AG and GB
[ἐστὶ]τὸΓΛτῷΖΛ.ὡςδὲτὸΓΛπρὸςτὸΖΛ,οὕτωςἐστὶν [Prop. 2.7]. And twice the (rectangle contained) by AG
ἡΓΜπρὸςτὴνΜΖ·ἀσύμμετροςἄραἐστὶνἡΓΜτῇΜΖ and GB (is) medial. Thus, FL is also medial. And it is
μήκει.καίεἰσινἀμφότεραιῥηταί.αἱΓΜ,ΜΖἄραῥηταίεἰσι applied to the rational (straight-line) FE, producing FM
δυνάμειμόνονσύμμετροι·ἀποτομὴἄραἐστὶνἡΓΖ.λέγω as breadth. FM is thus rational, and incommensurable
δή,ὅτικαὶἕκτη.
in length with CD [Prop. 10.22]. And since the (sum
᾿ΕπεὶγὰρτὸΖΛἴσονἐστὶτῷδὶςὑπὸτῶνΑΗ,ΗΒ, of the squares) on AG and GB is incommensurable with
τετμήσθωδίχαἡΖΜκατὰτὸΝ,καὶἤχθωδιὰτοῦΝτῇ twice the (rectangle contained) by AG and GB, and CL
ΓΔπαράλληλοςἡΝΞ·ἑκάτερονἄρατῶνΖΞ,ΝΛἴσονἐστὶ equal to the (sum of the squares) on AG and GB, and
τῷὑπὸτῶνΑΗ,ΗΒ.καὶἐπεὶαἱΑΗ,ΗΒδυνάμειεἰσὶν FL equal to twice the (rectangle contained) by AG and
ἀσύμμετροι,ἀσύμμετρονἄραἐστὶτὸἀπὸτῆςΑΗτῷἀπὸ GB, CL [is] thus incommensurable with FL. And as CL
τῆςΗΒ.ἀλλὰτῷμὲνἀπὸτῆςΑΗἴσονἐστὶτὸΓΘ,τῷδὲ (is) to FL, so CM is to MF [Prop. 6.1]. Thus, CM is
ἀπὸτῆςΗΒἴσονἐστὶτὸΚΛ·ἀσύμμετρονἄραἐστὶτὸΓΘ incommensurable in length with MF [Prop. 10.11]. And
τῷΚΛ.ὡςδὲτὸΓΘπρὸςτὸΚΛ,οὕτωςἐστὶνἡΓΚπρὸς they are both rational. Thus, CM and MF are rational
τὴνΚΜ·ἀσύμμετροςἄραἐστὶνἡΓΚτῇΚΜ.καὶἐπεὶτῶν (straight-lines which are) commensurable in square only.
ἀπὸτῶνΑΗ,ΗΒμέσονἀνάλογόνἐστιτὸὑπὸτῶνΑΗ, CF is thus an apotome [Prop. 10.73]. So, I say that (it
ΗΒ,καὶἐστιτῷμὲνἀπὸτῆςΑΗἴσοντὸΓΘ,τῷδὲἀπὸ is) also a sixth (apotome).
τῆςΗΒἴσοντὸΚΛ,τῷδὲὑπὸτῶνΑΗ,ΗΒἴσοντὸΝΛ, For since FL is equal to twice the (rectangle conκαὶτῶνἄραΓΘ,ΚΛμέσονἀνάλογόνἐστιτὸΝΛ·ἔστινἄρα
tained) by AG and GB, let FM have been cut in half
ὡςτὸΓΘπρὸςτὸΝΛ,οὕτωςτὸΝΛπρὸςτὸΚΛ.καὶδιὰ at N, and let NO have been drawn through N, parallel
τὰαὐτὰἡΓΜτῆςΜΖμεῖζονδύναταιτῷἀπὸἀσυμμέτρου to CD. Thus, FO and NL are each equal to the (rect-
ἑαυτῇ. καὶοὐδετέρααὐτῶνσύμμετρόςἐστιτῇἐκκειμένῃ angle contained) by AG and GB. And since AG and GB
ῥητῇτῇΓΔ·ἡΓΖἄραἀποτομήἐστινἕκτη·ὅπερἔδειδεῖξαι. are incommensurable in square, the (square) on AG is
thus incommensurable with the (square) on GB. But,
CH is equal to the (square) on AG, and KL is equal
to the (square) on GB. Thus, CH is incommensurable
with KL. And as CH (is) to KL, so CK is to KM
[Prop. 6.1]. Thus, CK is incommensurable (in length)
with KM [Prop. 10.11]. And since the (rectangle contained)
by AG and GB is the mean proportional to the
(squares) on AG and GB [Prop. 10.21 lem.], and CH
is equal to the (square) on AG, and KL equal to the
(square) on GB, and NL equal to the (rectangle contained)
by AG and GB, NL is thus also the mean proportional
to CH and KL. Thus, as CH is to NL, so NL
(is) to KL. And for the same (reasons as the preceding
propositions), the square on CM is greater than (the
square on) MF by the (square) on (some straight-line)
406

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
incommensurable (in length) with (CM) [Prop. 10.18].
And neither of them is commensurable with the (previously)
laid down rational (straight-line) CD. Thus, CF
is a sixth apotome [Def. 10.16]. (Which is) the very thing
it was required to show.
Ö. Proposition 103
῾Ητῇἀποτομῇμήκεισύμμετροςἀποτομήἐστικαὶτῇ A (straight-line) commensurable in length with an
τάξειἡαὐτή.
apotome is an apotome, and (is) the same in order.
Α
Β
Ε
A
B
E
Γ
∆
Ζ
C
D
F
῎ΕστωἀποτομὴἡΑΒ,καὶτῇΑΒμήκεισύμμετροςἔστω Let AB be an apotome, and let CD be commensu-
ἡΓΔ·λέγω,ὅτικαὶἡΓΔἀποτομήἐστικαὶτῇτάξειἡαὐτὴ rable in length with AB. I say that CD is also an apoτῇΑΒ.
tome, and (is) the same in order as AB.
᾿Επεὶ γὰρ ἀποτομή ἐστιν ἡ ΑΒ, ἔστω αὐτῇ προ- For since AB is an apotome, let BE be an attachment
σαρμόζουσαἡΒΕ·αἱΑΕ,ΕΒἄραῥηταίεἰσιδυνάμειμόνον to it. Thus, AE and EB are rational (straight-lines which
σύμμετροι. καὶτῷτῆςΑΒπρὸςτὴνΓΔλόγῳὁαὐτὸς are) commensurable in square only [Prop. 10.73]. And
γεγονέτωὁτῆςΒΕπρὸςτὴνΔΖ·καὶὡςἓνἄραπρὸςἕν, let it have been contrived that the (ratio) of BE to DF
πάντα[ἐστὶ]πρὸςπάντα·ἔστινἄρακαὶὡςὅληἡΑΕπρὸς is the same as the ratio of AB to CD [Prop. 6.12]. Thus,
ὅληντὴνΓΖ,οὕτωςἡΑΒπρὸςτὴνΓΔ.σύμμετροςδὲἡΑΒ also, as one is to one, (so) all [are] to all [Prop. 5.12].
τῇΓΔμήκει·σύμμετροςἄρακαὶἡΑΕμὲντῇΓΖ,ἡδὲΒΕ And thus as the whole AE is to the whole CF , so AB
τῇΔΖ.καὶαἱΑΕ,ΕΒῥηταίεἰσιδυνάμειμόνονσύμμετροι· (is) to CD. And AB (is) commensurable in length with
καὶαἱΓΖ,ΖΔἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι[ἀπο- CD. AE (is) thus also commensurable (in length) with
τομὴἄραἐστὶνἡΓΔ.λέγωδή,ὅτικαὶτῇτάξειἡαὐτὴτῇ CF , and BE with DF [Prop. 10.11]. And AE and
ΑΒ].
BE are rational (straight-lines which are) commensu-
᾿ΕπεὶοὖνἐστινὡςἡΑΕπρὸςτὴνΓΖ,οὕτωςἡΒΕπρὸς rable in square only. Thus, CF and FD are also rational
τὴνΔΖ,ἐναλλὰξἄραἐστὶνὡςἡΑΕπρὸςτὴνΕΒ,οὕτως (straight-lines which are) commensurable in square only
ἡΓΖπρὸςτὴνΖΔ.ἤτοιδὴἡΑΕτῆςΕΒμεῖζονδύναται [Prop. 10.13]. [CD is thus an apotome. So, I say that (it
τῷἀπὸσυμμέτρουἑαυτῇἢτῷἀπὸἀσυμμέτρου.εἰμὲνοὖν is) also the same in order as AB.]
ἡΑΕτῆςΕΒμεῖζονδύναταιτῷἀπὸσυμμέτρουἑαυτῇ,καὶ Therefore, since as AE is to CF , so BE (is) to
ἡΓΖτῆςΖΔμεῖζονδυνήσεταιτῷἀπὸσυμμέτρουἑαυτῇ. DF , thus, alternately, as AE is to EB, so CF (is) to
καὶεἰμὲνσύμμετρόςἐστινἡΑΕτῇἐκκειμένῃῥητῇμήκει, FD [Prop. 5.16]. So, the square on AE is greater
καὶἡΓΖ,εἰδὲἡΒΕ,καὶἡΔΖ,εἰδὲοὐδετέρατῶνΑΕ, than (the square on) EB either by the (square) on
ΕΒ,καὶοὐδετέρατῶνΓΖ,ΖΔ.εἰδὲἡΑΕ[τῆςΕΒ]μεῖζον (some straight-line) commensurable, or by the (square)
δύναταιτῷἀπὸἀσυμμέτρουἑαυτῇ,καὶἡΓΖτῆςΖΔμεῖζον on (some straight-line) incommensurable, (in length)
δυνήσεταιτῷἀπὸἀσυμμέτρουἑαυτῇ.καὶεἰμὲνσύμμετρός with (AE). Therefore, if the (square) on AE is greater
ἐστινἡΑΕτῇἐκκειμένῃῥητῇμήκει,καὶἡΓΖ,εἰδὲἡΒΕ, than (the square on) EB by the (square) on (some
καὶἡΔΖ,εἰδὲοὐδετέρατῶνΑΕ,ΕΒ,οὐδετέρατῶνΓΖ, straight-line) commensurable (in length) with (AE) then
ΖΔ.
the square on CF will also be greater than (the square
ἈποτομὴἄραἐστὶνἡΓΔκαὶτῇτάξειἡαὐτὴτῇΑΒ· on) FD by the (square) on (some straight-line) commen-
ὅπερἔδειδεῖξαι.
surable (in length) with (CF ) [Prop. 10.14]. And if AE
is commensurable in length with a (previously) laid down
rational (straight-line) then so (is) CF [Prop. 10.12],
and if BE (is commensurable), so (is) DF , and if neither
of AE or EB (are commensurable), neither (are)
either of CF or FD [Prop. 10.13]. And if the (square)
on AE is greater [than (the square on) EB] by the
(square) on (some straight-line) incommensurable (in
407

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
length) with (AE) then the (square) on CF will also
be greater than (the square on) FD by the (square) on
(some straight-line) incommensurable (in length) with
(CF ) [Prop. 10.14]. And if AE is commensurable in
length with a (previously) laid down rational (straightline),
so (is) CF [Prop. 10.12], and if BE (is commensurable),
so (is) DF , and if neither of AE or EB
(are commensurable), neither (are) either of CF or FD
[Prop. 10.13].
Thus, CD is an apotome, and (is) the same in order
as AB [Defs. 10.11—10.16]. (Which is) the very thing it
was required to show.
Ö. Proposition 104
῾Ητῇμέσηςἀποτομῇσύμμετροςμέσηςἀποτομήἐστι A (straight-line) commensurable (in length) with an
καὶτῇτάξειἡαὐτή.
apotome of a medial (straight-line) is an apotome of a
medial (straight-line), and (is) the same in order.
Α
Β
Ε
A
B
E
Γ
∆
Ζ
C
D
F
῎ΕστωμέσηςἀποτομὴἡΑΒ,καὶτῇΑΒμήκεισύμμετρος Let AB be an apotome of a medial (straight-line), and
ἔστωἡΓΔ·λέγω,ὅτικαὶἡΓΔμέσηςἀποτομήἐστικαὶτῇ let CD be commensurable in length with AB. I say that
τάξειἡαὐτὴτῇΑΒ.
CD is also an apotome of a medial (straight-line), and
᾿ΕπεὶγὰρμέσηςἀποτομήἐστινἡΑΒ,ἔστωαὐτῇπρο- (is) the same in order as AB.
σαρμόζουσαἡΕΒ.αἱΑΕ,ΕΒἄραμέσαιεἰσὶδυνάμειμόνον For since AB is an apotome of a medial (straightσύμμετροι.
καὶγεγονέτωὡςἡΑΒπρὸςτὴνΓΔ,οὕτωςἡ line), let EB be an attachment to it. Thus, AE and
ΒΕπρὸςτὴνΔΖ·σύμμετροςἄρα[ἐστὶ]καὶἡΑΕτῇΓΖ, EB are medial (straight-lines which are) commensurable
ἡδὲΒΕτῇΔΖ.αἱδὲΑΕ,ΕΒμέσαιεἰσὶδυνάμειμόνον in square only [Props. 10.74, 10.75]. And let it have
σύμμετροι· καὶαἱΓΖ,ΖΔἄραμέσαιεἰσὶδυνάμειμόνον been contrived that as AB is to CD, so BE (is) to DF
σύμμετροι·μέσηςἄραἀποτομήἐστινἡΓΔ.λέγωδή,ὅτι [Prop. 6.12]. Thus, AE [is] also commensurable (in
καὶτῇτάξειἐστὶνἡαὐτὴτῇΑΒ. length) with CF , and BE with DF [Props. 5.12, 10.11].
᾿Επεὶ[γάρ]ἐστινὡςἡΑΕπρὸςτὴνΕΒ,οὕτωςἡΓΖ And AE and EB are medial (straight-lines which are)
πρὸςτὴνΖΔ[ἀλλ᾿ὡςμὲνἡΑΕπρὸςτὴνΕΒ,οὕτωςτὸ commensurable in square only. CF and FD are thus
ἀπὸτῆςΑΕπρὸςτὸὑπὸτῶνΑΕ,ΕΒ,ὡςδὲἡΓΖπρὸςτὴν also medial (straight-lines which are) commensurable in
ΖΔ,οὕτωςτὸἀπὸτῆςΓΖπρὸςτὸὑπὸτῶνΓΖ,ΖΔ],ἔστιν square only [Props. 10.23, 10.13]. Thus, CD is an apo-
ἄρακαὶὡςτὸἀπὸτῆςΑΕπρὸςτὸὑπὸτῶνΑΕ,ΕΒ,οὕτως tome of a medial (straight-line) [Props. 10.74, 10.75].
τὸἀπὸτῆςΓΖπρὸςτὸὑπὸτῶνΓΖ,ΖΔ[καὶἐναλλὰξὡς So, I say that it is also the same in order as AB.
τὸἀπὸτῆςΑΕπρὸςτὸἀπὸτῆςΓΖ,οὕτωςτὸὑπὸτῶνΑΕ, [For] since as AE is to EB, so CF (is) to FD
ΕΒπρὸςτὸὑπὸτῶνΓΖ,ΖΔ].σύμμετρονδὲτὸἀπὸτῆςΑΕ [Props. 5.12, 5.16] [but as AE (is) to EB, so the (square)
τῷἀπὸτῆςΓΖ·σύμμετρονἄραἐστὶκαὶτὸὑπὸτῶνΑΕ, on AE (is) to the (rectangle contained) by AE and EB,
ΕΒτῷὑπὸτῶνΓΖ,ΖΔ.εἴτεοὖνῥητόνἐστιτὸὑπὸτῶν and as CF (is) to FD, so the (square) on CF (is) to
ΑΕ,ΕΒ,ῥητὸνἔσταικαὶτὸὑπὸτῶνΓΖ,ΖΔ,εἴτεμέσον the (rectangle contained) by CF and FD], thus as the
[ἐστὶ]τὸὑπὸτῶνΑΕ,ΕΒ,μέσον[ἐστὶ]καὶτὸὑπὸτῶνΓΖ, (square) on AE is to the (rectangle contained) by AE
ΖΔ.
and EB, so the (square) on CF also (is) to the (rectan-
ΜέσηςἄραἀποτομήἐστινἡΓΔκαὶτῇτάξειἡαὐτὴτῇ gle contained) by CF and FD [Prop. 10.21 lem.] [and,
ΑΒ·ὅπερἔδειδεῖξαι.
alternately, as the (square) on AE (is) to the (square)
on CF , so the (rectangle contained) by AE and EB (is)
to the (rectangle contained) by CF and FD]. And the
(square) on AE (is) commensurable with the (square)
408

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
Ö. Proposition
῾Ητῇἐλάσσονισύμμετροςἐλάσσωνἐστίν.
Α Β Ε
on CF . Thus, the (rectangle contained) by AE and EB
is also commensurable with the (rectangle contained) by
CF and FD [Props. 5.16, 10.11]. Therefore, either the
(rectangle contained) by AE and EB is rational, and the
(rectangle contained) by CF and FD will also be rational
[Def. 10.4], or the (rectangle contained) by AE and
EB [is] medial, and the (rectangle contained) by CF and
FD [is] also medial [Prop. 10.23 corr.].
Therefore, CD is the apotome of a medial (straightline),
and is the same in order as AB [Props. 10.74,
10.75]. (Which is) the very thing it was required to show.
105
A (straight-line) commensurable (in length) with a
minor (straight-line) is a minor (straight-line).
A
B
E
Γ
∆
Ζ
C
D
F
῎ΕστωγὰρἐλάσσωνἡΑΒκαὶτῇΑΒσύμμετροςἡΓΔ· For let AB be a minor (straight-line), and (let) CD
λέγω,ὅτικαὶἡΓΔἐλάσσωνἐστίν.
(be) commensurable (in length) with AB. I say that CD
Γεγονέτωγὰρτὰαὐτά·καὶἐπεὶαἱΑΕ,ΕΒδυνάμειεἰσὶν is also a minor (straight-line).
ἀσύμμετροι,καὶαἱΓΖ,ΖΔἄραδυνάμειεἰσὶνἀσύμμετροι. For let the same things have been contrived (as in
ἐπεὶοὖνἐστινὡςἡΑΕπρὸςτὴνΕΒ,οὕτωςἡΓΖπρὸς the former proposition). And since AE and EB are
τὴνΖΔ,ἔστινἄρακαὶὡςτὸἀπὸτῆςΑΕπρὸςτὸἀπὸτῆς (straight-lines which are) incommensurable in square
ΕΒ,οὕτωςτὸἀπὸτῆςΓΖπρὸςτὸἀπὸτῆςΖΔ.συνθέντι [Prop. 10.76], CF and FD are thus also (straight-lines
ἄραἐστὶνὡςτὰἀπὸτῶνΑΕ,ΕΒπρὸςτὸἀπὸτῆςΕΒ, which are) incommensurable in square [Prop. 10.13].
οὕτως τὰ ἀπὸ τῶν ΓΖ, ΖΔ πρὸς τὸ ἀπὸ τῆς ΖΔ [καὶ Therefore, since as AE is to EB, so CF (is) to FD
ἐναλλάξ]·σύμμετρονδέἐστιτὸἀπὸτῆςΒΕτῷἀπὸτῆςΔΖ· [Props. 5.12, 5.16], thus also as the (square) on AE is
σύμμετρονἄρακαὶτὸσυγκείμενονἐκτῶνἀπὸτῶνΑΕ,ΕΒ to the (square) on EB, so the (square) on CF (is) to the
τετραγώνωντῷσυγκειμένῳἐκτῶνἀπὸτῶνΓΖ,ΖΔτε- (square) on FD [Prop. 6.22]. Thus, via composition, as
τραγώνων. ῥητὸνδέἐστιτὸσυγκείμενονἐκτῶνἀπὸτῶν the (sum of the squares) on AE and EB is to the (square)
ΑΕ,ΕΒτετραγώνων·ῥητὸνἄραἐστὶκαὶτὸσυγκείμενον on EB, so the (sum of the squares) on CF and FD (is) to
ἐκτῶνἀπὸτῶνΓΖ,ΖΔτετραγώνων.πάλιν,ἐπείἐστινὡς the (square) on FD [Prop. 5.18], [also alternately]. And
τὸἀπὸτῆςΑΕπρὸςτὸὑπὸτῶνΑΕ,ΕΒ,οὕτωςτὸἀπὸ the (square) on BE is commensurable with the (square)
τῆςΓΖπρὸςτὸὑπὸτῶνΓΖ,ΖΔ,σύμμετρονδὲτὸἀπὸτῆς on DF [Prop. 10.104]. The sum of the squares on AE
ΑΕτετράγωνοντῷἀπὸτῆςΓΖτετραγώνῳ,σύμμετρονἄρα and EB (is) thus also commensurable with the sum of the
ἐστὶκαὶτὸὑπὸτῶνΑΕ,ΕΒτῷὑπὸτῶνΓΖ,ΖΔ.μέσον squares on CF and FD [Prop. 5.16, 10.11]. And the sum
δὲτὸὑπὸτῶνΑΕ,ΕΒ·μέσονἄρακαὶτὸὑπὸτῶνΓΖ,ΖΔ· of the (squares) on AE and EB is rational [Prop. 10.76].
αἱΓΖ,ΖΔἄραδυνάμειεἰσὶνἀσύμμετροιποιοῦσαιτὸμὲν Thus, the sum of the (squares) on CF and FD is also
συγκείμενονἐκτῶνἀπ᾿αὐτῶντετραγώνωνῥητόν,τὸδ᾿ rational [Def. 10.4]. Again, since as the (square) on
ὑπ᾿αὐτῶνμέσον.
AE is to the (rectangle contained) by AE and EB, so
᾿ΕλάσσωνἄραἐστὶνἡΓΔ·ὅπερἔδειδεῖξαι.
the (square) on CF (is) to the (rectangle contained) by
CF and FD [Prop. 10.21 lem.], and the square on AE
(is) commensurable with the square on CF , the (rectangle
contained) by AE and EB is thus also commensurable
with the (rectangle contained) by CF and FD.
And the (rectangle contained) by AE and EB (is) medial
[Prop. 10.76]. Thus, the (rectangle contained) by
CF and FD (is) also medial [Prop. 10.23 corr.]. CF and
409

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
FD are thus (straight-lines which are) incommensurable
in square, making the sum of the squares on them rational,
and the (rectangle contained) by them medial.
Thus, CD is a minor (straight-line) [Prop. 10.76].
(Which is) the very thing it was required to show.
Ö¦. Proposition 106
῾Ητῇμετὰῥητοῦμέσοντὸὅλονποιούσῃσύμμετρος A (straight-line) commensurable (in length) with a
μετὰῥητοῦμέσοντὸὅλονποιοῦσάἐστιν.
(straight-line) which with a rational (area) makes a medial
whole is a (straight-line) which with a rational (area)
makes a medial whole.
Α
Β
Ε
A
B
E
Γ
∆
Ζ
C
D
F
῎ΕστωμετὰῥητοῦμέσοντὸὅλονποιοῦσαἡΑΒκαὶτῇ Let AB be a (straight-line) which with a rational
ΑΒσύμμετροςἡΓΔ·λέγω,ὅτικαὶἡΓΔμετὰῥητοῦμέσον (area) makes a medial whole, and (let) CD (be) comτὸὅλονποιοῦσάἐστιν.
mensurable (in length) with AB. I say that CD is also a
῎ΕστωγὰρτῇΑΒπροσαρμόζουσαἡΒΕ·αἱΑΕ,ΕΒἄρα (straight-line) which with a rational (area) makes a meδυνάμειεἰσὶνἀσύμμετροιποιοῦσαιτὸμὲνσυγκείμενονἐκ
dial (whole).
τῶνἀπὸτῶνΑΕ,ΕΒτετραγώνωνμέσον,τὸδ᾿ὑπ᾿αὐτῶν For let BE be an attachment to AB. Thus, AE and
ῥητόν.καὶτὰαὐτὰκατεσκευάσθω.ὁμοίωςδὴδείξομεντοῖς EB are (straight-lines which are) incommensurable in
πρότερον,ὅτιαἱΓΖ,ΖΔἐντῷαὐτῷλόγῳεἰσὶταῖςΑΕ, square, making the sum of the squares on AE and EB
ΕΒ,καὶσύμμετρόνἐστιτὸσυγκείμενονἐκτῶνἀπὸτῶν medial, and the (rectangle contained) by them rational
ΑΕ,ΕΒτετραγώνωντῷσυγκειμένῳἐκτῶνἀπὸτῶνΓΖ, [Prop. 10.77]. And let the same construction have been
ΖΔτετραγώνων,τὸδὲὑπὸτῶνΑΕ,ΕΒτῷὑπὸτῶνΓΖ, made (as in the previous propositions). So, similarly to
ΖΔ·ὥστεκαὶαἱΓΖ,ΖΔδυνάμειεἰσὶνἀσύμμετροιποιοῦσαι the previous (propositions), we can show that CF and
τὸμὲνσυγκείμενονἐκτῶνἀπὸτῶνΓΖ,ΖΔτετραγώνων FD are in the same ratio as AE and EB, and the sum of
μέσον,τὸδ᾿ὑπ᾿αὐτῶνῥητόν.
the squares on AE and EB is commensurable with the
῾ΗΓΔἄραμετὰῥητοῦμέσοντὸὅλονποιοῦσάἐστιν· sum of the squares on CF and FD, and the (rectangle
ὅπερἔδειδεῖξαι.
contained) by AE and EB with the (rectangle contained)
by CF and FD. Hence, CF and FD are also (straightlines
which are) incommensurable in square, making the
sum of the squares on CF and FD medial, and the (rectangle
contained) by them rational.
CD is thus a (straight-line) which with a rational
(area) makes a medial whole [Prop. 10.77]. (Which is)
the very thing it was required to show.
ÖÞ. Proposition 107
῾Ητῇμετὰμέσουμέσοντὸὅλονποιούσῃσύμμετρος A (straight-line) commensurable (in length) with a
καὶαὐτὴμετὰμέσουμέσοντὸὅλονποιοῦσάἐστιν. (straight-line) which with a medial (area) makes a medial
whole is itself also a (straight-line) which with a medial
(area) makes a medial whole.
Α
Β
Ε
A
B
E
Γ
∆
Ζ
C
D
F
῎ΕστωμετὰμέσουμέσοντὸὅλονποιοῦσαἡΑΒ,καὶτῇ Let AB be a (straight-line) which with a medial (area)
410

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ΑΒἔστωσύμμετροςἡΓΔ·λέγω,ὅτικαὶἡΓΔμετὰμέσου makes a medial whole, and let CD be commensurable (in
μέσοντὸὅλονποιοῦσάἐστιν.
length) with AB. I say that CD is also a (straight-line)
῎ΕστωγὰρτῇΑΒπροσαρμόζουσαἡΒΕ,καὶτὰαὐτὰ which with a medial (area) makes a medial whole.
κατεσκευάσθω·αἱΑΕ,ΕΒἄραδυνάμειεἱσὶνἀσύμμετροι For let BE be an attachment to AB. And let the same
ποιοῦσαιτότεσυγκείμενονἐκτῶνἀπ᾿αὐτῶντετραγώνων construction have been made (as in the previous propoμέσον
καὶ τὸ ὑπ᾿ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τὸ sitions). Thus, AE and EB are (straight-lines which
συγκέιμενονἐκτῶνἀπ᾿αὐτῶντετραγώνωντῷὑπ᾿αὐτῶν. are) incommensurable in square, making the sum of the
καίεἰσιν,ὡςἐδείχθη,αἱΑΕ,ΕΒσύμμετροιταῖςΓΖ,ΖΔ, squares on them medial, and the (rectangle contained)
καὶτὸσυγκείμενονἐκτῶνἀπὸτῶνΑΕ,ΕΒτετραγώνων by them medial, and, further, the sum of the squares on
τῷσυγκειμένῳἐκτῶνἀπὸτῶνΓΖ,ΖΔ,τὸδὲὑπὸτῶν them incommensurable with the (rectangle contained) by
ΑΕ,ΕΒτῷὑπὸτῶνΓΖ,ΖΔ·καὶαἱΓΖ,ΖΔἄραδυνάμει them [Prop. 10.78]. And, as was shown (previously), AE
εἰσὶνἀσύμμετροιποιοῦσαιτότεσυγκείμενονἐκτῶνἀπ᾿ and EB are commensurable (in length) with CF and FD
αὐτῶν τετραγώνων μέσον καὶ τὸ ὑπ᾿ ἀὐτῶν μέσον καὶ (respectively), and the sum of the squares on AE and
ἔτι ἀσύμμετρον τὸ συγκείμενον ἐκ τῶν ἀπ᾿ αὐτῶν [τε- EB with the sum of the squares on CF and FD, and
τραγώνων]τῷὑπ᾿αὐτῶν.
the (rectangle contained) by AE and EB with the (rect-
῾ΗΓΔἄραμετὰμέσουμέσοντὸὅλονποιοῦσάἐστιν· angle contained) by CF and FD. Thus, CF and FD
ὅπερἔδειδεῖξαι.
are also (straight-lines which are) incommensurable in
square, making the sum of the squares on them medial,
and the (rectangle contained) by them medial, and, further,
the sum of the [squares] on them incommensurable
with the (rectangle contained) by them.
Thus, CD is a (straight-line) which with a medial
(area) makes a medial whole [Prop. 10.78]. (Which is)
the very thing it was required to show.
Ö. Proposition 108
Ἀπὸ ῥητοῦμέσουἀφαιρουμένουἡτὸλοιπὸνχωρίον A medial (area) being subtracted from a rational
δυναμένημίαδύοἀλόγωνγίνεταιἤτοιἀποτομὴἢἐλάσσων. (area), one of two irrational (straight-lines) arise (as) the
square-root of the remaining area—either an apotome, or
a minor (straight-line).
Α Ε Β
A E B
Λ
Η
L
G
Θ
Κ
Ζ
H
K
F
Γ ∆
C D
ἈπὸγὰρῥητοῦτοῦΒΓμέσονἀφῃρήσθωτὸΒΔ·λέγω, For let the medial (area) BD have been subtracted
ὅτιἡτὸλοιπὸνδυναμένητὸΕΓμίαδύοἀλόγωνγίνεται from the rational (area) BC. I say that one of two ir-
ἤτοιἀποτομὴἢἐλάσσων.
rational (straight-lines) arise (as) the square-root of the
᾿ΕκκείσθωγὰρῥητὴἡΖΗ,καὶτῷμὲνΒΓἴσονπαρὰτὴν remaining (area), EC—either an apotome, or a minor
ΖΗπαραβεβλήσθωὀρθογώνιονπαραλληλόγραμμοντὸΗΘ, (straight-line).
τῷδὲΔΒἴσονἀφῃρήσθωτὸΗΚ·λοιπὸνἄρατὸΕΓἴσον For let the rational (straight-line) FG have been laid
ἐστὶτῷΛΘ.ἐπεὶοὖνῥητὸνμένἐστιτὸΒΓ,μέσονδὲτὸ out, and let the right-angled parallelogram GH, equal to
ΒΔ,ἴσονδὲτὸμὲνΒΓτῷΗΘ,τὸδὲΒΔτῷΗΚ,ῥητὸνμὲν BC, have been applied to FG, and let GK, equal to DB,
ἄραἐστὶτὸΗΘ,μέσονδὲτὸΗΚ.καὶπαρὰῥητὴντὴνΖΗ have been subtracted (from GH). Thus, the remainder
παράκειται·ῥητὴμὲνἄραἡΖΘκαὶσύμμετροςτῇΖΗμήκει, EC is equal to LH. Therefore, since BC is a rational
ῥητὴδὲἡΖΚκαὶἀσύμμετροςτῇΖΗμήκει·ἀσύμμετροςἄρα (area), and BD a medial (area), and BC (is) equal to
411

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ἐστὶνἡΖΘτῇΖΚμήκει.αἱΖΘ,ΖΚἄραῥηταίεἰσιδυνάμει GH, and BD to GK, GH is thus a rational (area), and
μόνονσύμμετροι·ἀποτομὴἄραἐστὶνἡΚΘ,προσαρμόζουσα GK a medial (area). And they are applied to the rational
δὲαὐτῇἡΚΖ.ἤτοιδὴἡΘΖτῆςΖΚμεῖζονδύναταιτῷἀπὸ (straight-line) FG. Thus, FH (is) rational, and commenσυμμέτρουἢοὔ.
surable in length with FG [Prop. 10.20], and FK (is)
Δυνάσθωπρότεροντῷἀπὸσυμμέτρου.καίἐστινὅληἡ also rational, and incommensurable in length with FG
ΘΖσύμμετροςτῇἐκκειμένῃῥητῇμήκειτῇΖΗ·ἀποτομὴἄρα [Prop. 10.22]. Thus, FH is incommensurable in length
πρώτηἐστὶνἡΚΘ.τὸδ᾿ὑπὸῥητῆςκαὶἀποτομῆςπρώτης with FK [Prop. 10.13]. FH and FK are thus rational
περιεχόμενονἡδυναμένηἀποτομήἐστιν. ἡἄρατὸΛΘ, (straight-lines which are) commensurable in square only.
τουτέστιτὸΕΓ,δυναμένηἀποτομήἐστιν.
Thus, KH is an apotome [Prop. 10.73], and KF an at-
ΕἰδὲἡΘΖτῆςΖΚμεῖζονδύναταιτῷἀπὸἀσυμμέτρου tachment to it. So, the square on HF is greater than
ἑαυτῇ,καίἐστινὅληἡΖΘσύμμετροςτῇἐκκειμένῃῥητῇ (the square on) FK by the (square) on (some straightμήκειτῇΖΗ,ἀποτομὴτετάρτηἐστὶνἡΚΘ.τὸδ᾿ὑπὸῥητῆς
line which is) either commensurable, or not (commensuκαὶἀποτομῆςτετάρτηςπεριεχόμενονἡδυναμένηἐλάσσων
rable), (in length with HF ).
ἐστίν·ὅπερἔδειδεῖξαι.
First, let the square (on it) be (greater) by the
(square) on (some straight-line which is) commensurable
(in length with HF ). And the whole of HF is commensurable
in length with the (previously) laid down
rational (straight-line) FG. Thus, KH is a first apotome
[Def. 10.1]. And the square-root of an (area) contained
by a rational (straight-line) and a first apotome is an apotome
[Prop. 10.91]. Thus, the square-root of LH—that
is to say, (of) EC—is an apotome.
And if the square on HF is greater than (the square
on) FK by the (square) on (some straight-line which is)
incommensurable (in length) with (HF ), and (since) the
whole of FH is commensurable in length with the (previously)
laid down rational (straight-line) FG, KH is a
fourth apotome [Prop. 10.14]. And the square-root of an
(area) contained by a rational (straight-line) and a fourth
apotome is a minor (straight-line) [Prop. 10.94]. (Which
is) the very thing it was required to show.
Ö. Proposition 109
Ἀπὸ μέσου ῥητοῦ ἀφαιρουμένου ἄλλαι δύο ἄλογοι A rational (area) being subtracted from a medial
γίνονταιἤτοιμέσηςἀποτομὴπρώτηἢμετὰῥητοῦμέσον (area), two other irrational (straight-lines) arise (as the
τὸὅλονποιοῦσα.
square-root of the remaining area)—either a first apo-
ἈπὸγὰρμέσουτοῦΒΓῥητὸνἀφῃρήσθωτὸΒΔ.λέγω, tome of a medial (straight-line), or that (straight-line)
ὅτιἡτὸλοιπὸντὸΕΓδυναμένημίαδύοἀλόγωνγίνεται which with a rational (area) makes a medial whole.
ἤτοιμέσηςἀποτομὴπρώτηἢμετὰῥητοῦμέσοντὸὅλον For let the rational (area) BD have been subtracted
ποιοῦσα.
from the medial (area) BC. I say that one of two ir-
᾿ΕκκείσθωγὰρῥητὴἡΖΗ,καὶπαραβεβλήσθωὁμοίωςτὰ rational (straight-lines) arise (as) the square-root of the
χωρία.ἔστιδὴἀκολούθωςῥητὴμὲνἡΖΘκαὶἀσύμμετρος remaining (area), EC—either a first apotome of a medial
τῇΖΗμήκει,ῥητὴδὲἡΚΖκαὶσύμμετροςτῇΖΗμήκει·αἱ (straight-line), or that (straight-line) which with a ratio-
ΖΘ,ΖΚἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι·ἀποτομὴ nal (area) makes a medial whole.
ἄραἐστὶνἡΚΘ,προσαρμόζουσαδὲταύτῃἡΖΚ.ἤτοιδὴἡ For let the rational (straight-line) FG be laid down,
ΘΖτῆςΖΚμεῖζονδύναταιτῷἀπὸσυμμέτρουἑαυτῇἢτῷ and let similar areas (to the preceding proposition) have
ἀπὸἀσυμμέτρου.
been applied (to it). So, accordingly, FH is rational, and
incommensurable in length with FG, and KF (is) also
rational, and commensurable in length with FG. Thus,
FH and FK are rational (straight-lines which are) com-
412

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
Β
Ε
Ζ
Κ
Θ
mensurable in square only [Prop. 10.13]. KH is thus
an apotome [Prop. 10.73], and FK an attachment to it.
So, the square on HF is greater than (the square on) FK
either by the (square) on (some straight-line) commensurable
(in length) with (HF ), or by the (square) on (some
straight-line) incommensurable (in length with HF ).
B E F K H
Α
∆
Γ
A
D
C
Η Λ
G L
Εἰ μὲν οὖν ἡ ΘΖ τῆς ΖΚ μεῖζον δύναται τῷ ἀπὸ Therefore, if the square on HF is greater than (the
συμμέτρου ἑαυτῇ, καί ἐστιν ἡ προσαρμόζουσα ἡ ΖΚ square on) FK by the (square) on (some straight-line)
σύμμετροςτῇἐκκειμένῃῥητῇμήκειτῇΖΗ,ἀποτομὴδευτέρα commensurable (in length) with (HF ), and (since) the
ἐστὶνἡΚΘ.ῥητὴδὲἡΖΗ·ὥστεἡτὸΛΘ,τουτέστιτὸΕΓ, attachment FK is commensurable in length with the
δυναμένημέσηςἀποτομὴπρώτηἐστίν.
(previously) laid down rational (straight-line) FG, KH
ΕἰδὲἡΘΖτῆςΖΚμεῖζονδύναταιτῷἀπὸἀσυμμέτρου, is a second apotome [Def. 10.12]. And FG (is) rational.
καίἐστινἡπροσαρμόζουσαἡΖΚσύμμετροςτῇἐκκειμένῃ Hence, the square-root of LH—that is to say, (of) EC—is
ῥητῇμήκειτῇΖΗ,ἀποτομὴπέμπτηἐστὶνἡΚΘ·ὥστεἡ a first apotome of a medial (straight-line) [Prop. 10.92].
τὸΕΓδυναμένημετὰῥητοῦμέσοντὸὅλονποιοῦσάἐστιν· And if the square on HF is greater than (the square
ὅπερἔδειδεῖξαι.
on) FK by the (square) on (some straight-line) incommensurable
(in length with HF ), and (since) the attachment
FK is commensurable in length with the (previously)
laid down rational (straight-line) FG, KH is a
fifth apotome [Def. 10.15]. Hence, the square-root of EC
is that (straight-line) which with a rational (area) makes
a medial whole [Prop. 10.95]. (Which is) the very thing
it was required to show.
Ö. Proposition 110
Ἀπὸμέσουμέσουἀφαιρουμένουἀσυμμέτρουτῷὅλῳαἱ A medial (area), incommensurable with the whole,
λοιπαὶδύοἄλογοιγίνονταιἤτοιμέσηςἀποτομὴδευτέραἢ being subtracted from a medial (area), the two remaining
μετὰμέσουμέσοντὸὅλονποιοῦσα.
irrational (straight-lines) arise (as) the (square-root of
Ἀφῃρήσθωγὰρὡςἐπὶτῶνπροκειμένωνκαταγραφῶν the area)—either a second apotome of a medial (straight-
ἀπὸμέσουτοῦΒΓμέσοντὸΒΔἀσύμμετροντῷὅλῳ·λέγω, line), or that (straight-line) which with a medial (area)
ὅτιἡτὸΕΓδυναμένημίαἐστὶδύοἀλόγωνἤτοιμέσηςἀπο- makes a medial whole.
τομὴδευτέραἢμετὰμέσουμέσοντὸὅλονποιοῦσα. For, as in the previous figures, let the medial (area)
BD, incommensurable with the whole, have been subtracted
from the medial (area) BC. I say that the squareroot
of EC is one of two irrational (straight-lines)—
either a second apotome of a medial (straight-line), or
that (straight-line) which with a medial (area) makes a
medial whole.
413

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
Β
Ε
Ζ
Κ
Θ
B
E
F
K
H
Α
∆
Γ
A
D
C
Η Λ
G L
᾿Επεὶ γὰρ μέσον ἐστὶν ἑκάτερον τῶν ΒΓ, ΒΔ, καὶ For since BC and BD are each medial (areas), and
ἀσύμμετροντὸΒΓτῷΒΔ,ἔσταιἀκολούθωςῥητὴἑκατέρα BC (is) incommensurable with BD, accordingly, FH and
τῶν ΖΘ, ΖΚ καὶ ἀσύμμετρος τῇ ΖΗ μήκει. καὶ ἐπεὶ FK will each be rational (straight-lines), and incommen-
ἀσύμμετρόνἐστιτὸΒΓτῷΒΔ,τουτέστιτὸΗΘτῷΗΚ, surable in length with FG [Prop. 10.22]. And since BC
ἀσύμμετροςκαὶἡΘΖτῇΖΚ·αἱΖΘ,ΖΚἄραῥηταίεἰσι is incommensurable with BD—that is to say, GH with
δυνάμειμόνονσύμμετροι·ἀποτομὴἄραἐστὶνἡΚΘ[προ- GK—HF (is) also incommensurable (in length) with
σαρμόζουσαδὲἡΖΚ.ἤτοιδὴἡΖΘτῆςΖΚμεῖζονδύναται FK [Props. 6.1, 10.11]. Thus, FH and FK are ratioτῷἀπὸσυμμέτρουἢτῷἀπὸἀσυμμέτρουἑαυτῇ].
nal (straight-lines which are) commensurable in square
Εἰ μὲν δὴ ἡ ΖΘ τῆς ΖΚ μεῖζον δύναται τῷ ἀπὸ only. KH is thus as apotome [Prop. 10.73], [and FK
συμμέτρουἑαυτῇ,καὶοὐθετέρατῶνΖΘ,ΖΚσύμμετρός an attachment (to it). So, the square on FH is greater
ἐστιτῇἐκκειμέμνῃῥητῇμήκειτῇΖΗ,ἀποτομὴτρίτηἐστὶν than (the square on) FK either by the (square) on
ἡΚΘ.ῥητὴδὲἡΚΛ,τὸδ᾿ὑπὸῥητῆςκαὶἀποτομῆςτρίτης (some straight-line) commensurable, or by the (square)
περιεχόμενον ὀρθογώνιον ἄλογόν ἐστιν, καὶ ἡ δυναμένη on (some straight-line) incommensurable, (in length)
αὐτὸ ἄλογόςἐστιν, καλεῖταιδὲ μέσης ἀποτομὴδευτέρα· with (FH).]
ὥστεἡτὸΛΘ,τουτέστιτὸΕΓ,δυναμένημέσηςἀποτομή So, if the square on FH is greater than (the square
ἐστιδευτερά.
on) FK by the (square) on (some straight-line) com-
ΕἰδὲἡΖΘτῆςΖΚμεῖζονδύναταιτῷἀπὸἀσυμμέτρου mensurable (in length) with (FH), and (since) neither of
ἑαυτῇ[μήκει],καὶοὐθετέρατῶνΘΖ,ΖΚσύμμετρόςἐστι FH and FK is commensurable in length with the (preτῇΖΗμήκει,ἀποτομὴἕκτηἐστὶνἡΚΘ.τὸδ᾿ὑπὸῥητῆς
viously) laid down rational (straight-line) FG, KH is a
καὶἀποτομῆςἕκτηςἡδυναμένηἐστὶμετὰμέσουμέσοντὸ third apotome [Def. 10.3]. And KL (is) rational. And
ὅλονποιοῦσα. ἡτὸΛΘἄρα,τουτέστιτὸΕΓ,δυναμένη the rectangle contained by a rational (straight-line) and
μετὰμέσουμέσοντὸὅλονποιοῦσάἐστιν·ὅπερἔδειδεῖξαι. a third apotome is irrational, and the square-root of it is
that irrational (straight-line) called a second apotome of
a medial (straight-line) [Prop. 10.93]. Hence, the squareroot
of LH—that is to say, (of) EC—is a second apotome
of a medial (straight-line).
And if the square on FH is greater than (the square
on) FK by the (square) on (some straight-line) incommensurable
[in length] with (FH), and (since) neither of
HF and FK is commensurable in length with FG, KH
is a sixth apotome [Def. 10.16]. And the square-root of
the (rectangle contained) by a rational (straight-line) and
a sixth apotome is that (straight-line) which with a medial
(area) makes a medial whole [Prop. 10.96]. Thus,
the square-root of LH—that is to say, (of) EC—is that
(straight-line) which with a medial (area) makes a medial
whole. (Which is) the very thing it was required to
414

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
Ö. Proposition
῾Ηἀποτομὴοὐκἔστινἡαὐτὴτῇἐκδύοὀνομάτων.
Α Β
show.
111
An apotome is not the same as a binomial.
A
B
∆
Η
Ε
Ζ
D
G
E
F
Γ
῎ΕστωἀποτομὴἡΑΒ·λέγω,ὅτιἡΑΒοὐκἔστινἡαὐτὴ Let AB be an apotome. I say that AB is not the same
τῇἐκδύοὀνομάτων.
as a binomial.
Εἰγὰρδυνατόν,ἔστω·καὶἐκκείσθωῥητὴἡΔΓ,καὶτῷ For, if possible, let it be (the same). And let a rational
ἀπὸτῆςΑΒἴσονπαρὰτὴνΓΔπαραβεβλήσθωὀρθογώνιον (straight-line) DC be laid down. And let the rectangle
τὸΓΕπλάτοςποιοῦντὴνΔΕ.ἐπεὶοὖνἀποτομήἐστινἡΑΒ, CE, equal to the (square) on AB, have been applied to
ἀποτομὴπρώτηἐστὶνἡΔΕ.ἔστωαὐτῇπροσαρμόζουσαἡ CD, producing DE as breadth. Therefore, since AB is an
ΕΖ·αἱΔΖ,ΖΕἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι,καὶ apotome, DE is a first apotome [Prop. 10.97]. Let EF
ἡΔΖτῆςΖΕμεῖζονδύναταιτῷἀπὸσυμμέτρουἑαυτῇ,καὶἡ be an attachment to it. Thus, DF and FE are rational
ΔΖσύμμετρόςἐστιτῇἐκκειμένῃῥητῇμήκειτῇΔΓ.πάλιν, (straight-lines which are) commensurable in square only,
ἐπεὶἐκδύοὀνομάτωνἐστὶνἡΑΒ,ἐκδύοἄραὀνομάτων and the square on DF is greater than (the square on) FE
πρώτηἐστὶνἡΔΕ.διῃρήσθωεἰςτὰὀνόματακατὰτὸΗ, by the (square) on (some straight-line) commensurable
καὶἔστωμεῖζονὄνοματὸΔΗ·αἱΔΗ,ΗΕἄραῥηταίεἰσι (in length) with (DF ), and DF is commensurable in
δυνάμειμόνονσύμμετροι,καὶἡΔΗτῆςΗΕμεῖζονδύναται length with the (previously) laid down rational (straightτῷἀπὸσυμμέτρουἑαυτῇ,καὶτὸμεῖζονἡΔΗσύμμετρός
line) DC [Def. 10.10]. Again, since AB is a binomial,
ἐστιτῇἐκκειμένῃῥητῇμήκειτῇΔΓ.καὶἡΔΖἄρατῇΔΗ DE is thus a first binomial [Prop. 10.60]. Let (DE) have
σύμμετρόςἐστιμήκει·καὶλοιπὴἄραἡΗΖσύμμετρόςἐστι been divided into its (component) terms at G, and let
τῇΔΖμήκει.[ἐπεὶοῦνσύμμετρόςἐστινἡΔΖτῇΗΖ,ῥητὴ DG be the greater term. Thus, DG and GE are rational
δέἐστινἡΔΖ,ῥητὴἄραἐστὶκαὶἡΗΖ.ἐπεὶοὖνσύμμετρός (straight-lines which are) commensurable in square only,
ἐστινἡΔΖτῇΗΖμήκει]ἀσύμμετροςδὲἡΔΖτῇΕΖμήκει. and the square on DG is greater than (the square on)
ἀσύμμετροςἄραἐστὶκαὶἡΖΗτῇΕΖμήκει.αἱΗΖ,ΖΕἄρα GE by the (square) on (some straight-line) commensu-
ῥηταί[εἰσι]δυνάμειμόνονσύμμετροι·ἀποτομὴἄραἐστὶνἡ rable (in length) with (DG), and the greater (term) DG
ΕΗ.ἀλλὰκαὶῥητή·ὅπερἐστὶνἀδύνατον.
is commensurable in length with the (previously) laid
῾Ηἄραἀποτομὴοὐκἔστινἡαὐτὴτῇἐκδύοὀνομάτων· down rational (straight-line) DC [Def. 10.5]. Thus, DF
ὅπερἔδειδεῖξαι. is also commensurable in length with DG [Prop. 10.12].
The remainder GF is thus commensurable in length with
DF [Prop. 10.15]. [Therefore, since DF is commensurable
with GF , and DF is rational, GF is thus also rational.
Therefore, since DF is commensurable in length
with GF ,] DF (is) incommensurable in length with EF .
Thus, FG is also incommensurable in length with EF
[Prop. 10.13]. GF and FE [are] thus rational (straightlines
which are) commensurable in square only. Thus,
C
415

ËÌÇÁÉÁÏƺ ELEMENTS
ÈÖ×Ѻ℄
῾Ηἀποτομὴκαὶαἱμετ᾿αὐτὴνἄλογοιοὔτετῇμέσῃοὔτε The apotome and the irrational (straight-lines) after it
BOOK 10
EG is an apotome [Prop. 10.73]. But, (it is) also rational.
The very thing is impossible.
Thus, an apotome is not the same as a binomial.
(Which is) the very thing it was required to show.
[Corollary]
ἀλλήλαιςεἰσὶναἱαὐταί.
are neither the same as a medial (straight-line) nor (the
Τὸ μὲν γὰρ ἀπὸ μέσης παρὰ ῥητὴν παραβαλλόμενον same) as one another.
πλάτοςποιεῖῥητὴνκαὶἀσύμμετροντῇ,παρ᾿ἣνπαράκειται, For the (square) on a medial (straight-line), applied
μήκει, τὸ δὲ ἀπὸἀποτομῆςπαρὰῥητὴν παραβαλλόμενον to a rational (straight-line), produces as breadth a ratioπλάτοςποιεῖἀποτομὴνπρώτην,τὸδὲἀπὸμέσηςἀποτομῆς
nal (straight-line which is) incommensurable in length
πρώτηςπαρὰῥητὴνπαραβαλλόμενονπλάτοςποιεῖἀποτομὴν with the (straight-line) to which (the area) is applied
δευτέραν,τὸδὲἀπὸμέσηςἀποτομῆςδευτέραςπαρὰῥητὴν [Prop. 10.22]. And the (square) on an apotome, apπαραβαλλόμενονπλάτοςποιεῖἀποτομὴντρίτην,τὸδὲἀπὸ
plied to a rational (straight-line), produces as breadth a
ἐλάσσονοςπαρὰῥητὴνπαραβαλλόμενονπλάτοςποιεῖἀπο- first apotome [Prop. 10.97]. And the (square) on a first
τομὴντετάρτην,τὸδὲἀπὸτῆςμετὰῥητοῦμέσοντὸὅλον apotome of a medial (straight-line), applied to a ratioποιούσηςπαρὰῥητὴνπαραβαλλόμενονπλάτοςποιεῖἀπο-
nal (straight-line), produces as breadth a second apotome
τομὴνπέμπτην,τὸδὲἀπὸτῆςμετὰμέσουμέσοντὸὅλον [Prop. 10.98]. And the (square) on a second apotome of
ποιούσηςπαρὰῥητὴνπαραβαλλόμενονπλάτοςποιεῖἀπο- a medial (straight-line), applied to a rational (straightτομὴν
ἕκτην. ἐπεὶ οὖν τὰ εἰρημένα πλάτη διαφέρει τοῦ line), produces as breadth a third apotome [Prop. 10.99].
τεπρώτουκαὶἀλλήλων,τοῦμὲνπρώτου,ὅτιῥητήἐστιν, And (square) on a minor (straight-line), applied to a ra-
ἀλλήλωνδὲ,ἐπεὶτῇτάξειοὐκεἰσὶναἱαὐταί,δῆλον,ὡςκαὶ tional (straight-line), produces as breadth a fourth apoαὐταὶαἱἄλογοιδιαφέρουσινἀλλήλων.
καὶἐπεὶδέδεικται tome [Prop. 10.100]. And (square) on that (straight-line)
ἡἀποτομὴοὐκοὖσαἡαὐτὴτῇἐκδύοὀνομάτων,ποιοῦσι which with a rational (area) produces a medial whole,
δὲπλάτηπαρὰῥητὴνπαραβαλλόμεναιαἱμετὰτὴνἀποτομὴν applied to a rational (straight-line), produces as breadth
ἀποτομὰςἀκολούθωςἑκάστητῇτάξειτῇκαθ᾿αὑτήν,αἱδὲ a fifth apotome [Prop. 10.101]. And (square) on that
μετὰτὴνἐκδύοὀνομάτωντὰςἐκδύοὀνομάτωνκαὶαὐταὶ (straight-line) which with a medial (area) produces a
τῇτάξειἀκολούθως,ἕτεραιἄραεἰσὶναἱμετὰτὴνἀποτομὴν medial whole, applied to a rational (straight-line), proκαὶἕτεραιαἱμετὰτὴνἐκδύοὀνομάτων,ὡςεἶναιτῇτάξει
duces as breadth a sixth apotome [Prop. 10.102]. Thereπάσαςἀλόγουςιγ,
fore, since the aforementioned breadths differ from the
first (breadth), and from one another—from the first, because
it is rational, and from one another since they are
not the same in order—clearly, the irrational (straightlines)
themselves also differ from one another. And since
it has been shown that an apotome is not the same
as a binomial [Prop. 10.111], and (that) the (irrational
straight-lines) after the apotome, being applied to a rational
(straight-line), produce as breadth, each according
to its own (order), apotomes, and (that) the (irrational
straight-lines) after the binomial themselves also (produce
as breadth), according (to their) order, binomials,
the (irrational straight-lines) after the apotome are thus
different, and the (irrational straight-lines) after the binomial
(are also) different, so that there are, in order, 13
irrational (straight-lines) in all:
416

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
Μέσην,
᾿Εκδύοὀνομάτων,
᾿Εκδύομέσωνπρώτην,
᾿Εκδύομέσωνδευτέραν,
Μείζονα,
῾Ρητὸνκαὶμέσονδυναμένην,
Δύομέσαδυναμένην,
Ἀποτομήν,
Μἑσηςἀποτομὴνπρώτην,
Μἑσηςἀποτομὴνδευτέραν,
᾿Ελάσσονα,
Μετὰῥητοῦμέσοντὸὅλονποιοῦσαν,
Medial,
Binomial,
First bimedial,
Second bimedial,
Major,
Square-root of a rational plus a medial (area),
Square-root of (the sum of) two medial (areas),
Apotome,
First apotome of a medial,
Second apotome of a medial,
Minor,
That which with a rational (area) produces a medial
whole,
Μετὰμέσουμέσοντὸὅλονποιοῦσαν.
That which with a medial (area) produces a medial
whole.
Ö. Proposition 112 †
Τὸ ἀπὸ ῥητῆς παρὰ τὴν ἐκ δύο ὀνομάτων παρα- The (square) on a rational (straight-line), applied to
βαλλόμενονπλάτοςποιεῖἀποτομήν,ἧςτὰὀνόματασύμμετρά a binomial (straight-line), produces as breadth an apo-
ἐστιτοῖςτῆςἐκδύοὀνομάτωνὀνόμασικαὶἔτιἐντῷαὐτῷ tome whose terms are commensurable (in length) with
λόγῳ,καὶἔτιἡγινομένηἀποτομὴτὴναὐτὴνἕξειτάξιντῇ the terms of the binomial, and, furthermore, in the same
ἐκδύοὀνομάτων.
ratio. Moreover, the created apotome will have the same
order as the binomial.
Α
Β ∆
Κ
Ε
Γ
Η
Ζ
Θ
A
B
D
C
G
K E F H
῎ΕστωῥητὴμὲνἡΑ,ἐκδύοὀνομάτωνδὲἡΒΓ,ἧς Let A be a rational (straight-line), and BC a binomial
μεῖζονὄνομαἔστωἡΔΓ,καὶτῷἀπὸτῆςΑἴσονἔστω (straight-line), of which let DC be the greater term. And
τὸὑπὸτῶνΒΓ,ΕΖ·λέγω,ὅτιἡΕΖἀποτομήἐστιν,ἧςτὰ let the (rectangle contained) by BC and EF be equal to
ὀνόματασύμμετράἐστιτοῖςΓΔ,ΔΒ,καὶἐντῷαὐτῷλόγῳ, the (square) on A. I say that EF is an apotome whose
καὶἔτιἡΕΖτὴναὐτὴνἕξειτάξιντῇΒΓ.
terms are commensurable (in length) with CD and DB,
῎ΕστωγὰρπάλιντῷἀπὸτῆςΑἴσοντὸὑπὸτῶνΒΔ, and in the same ratio, and, moreover, that EF will have
Η.ἐπεὶοὖντὸὑπὸτῶνΒΓ,ΕΖἴσονἐστὶτῷὑπὸτῶνΒΔ, the same order as BC.
Η,ἔστινἄραὡςἡΓΒπρὸςτὴνΒΔ,οὕτωςἡΗπρὸςτὴν For, again, let the (rectangle contained) by BD and G
ΕΖ.μείζωνδὲἡΓΒτῆςΒΔ·μείζωνἄραἐστὶκαὶἡΗτῆς be equal to the (square) on A. Therefore, since the (rect-
ΕΖ.ἔστωτῇΗἴσηἡΕΘ·ἔστινἄραὡςἡΓΒπρὸςτὴν angle contained) by BC and EF is equal to the (rectan-
ΒΔ,οὕτωςἡΘΕπρὸςτὴνΕΖ·διελόντιἄραἐστὶνὡςἡΓΔ gle contained) by BD and G, thus as CB is to BD, so G
πρὸςτὴνΒΔ,οὕτωςἡΘΖπρὸςτὴνΖΕ.γεγονέτωὡςἡ (is) to EF [Prop. 6.16]. And CB (is) greater than BD.
ΘΖπρὸςτὴνΖΕ,οὕτωςἡΖΕπρὸςτὴνΚΕ·καὶὅληἄραἡ Thus, G is also greater than EF [Props. 5.16, 5.14]. Let
ΘΚπρὸςὅληντὴνΚΖἐστιν,ὡςἡΖΚπρὸςΚΕ·ὡςγὰρἓν EH be equal to G. Thus, as CB is to BD, so HE (is) to
τῶνἡγουμένωνπρὸςἓντῶνἑπομένων,οὕτωςἅπαντατὰ EF . Thus, via separation, as CD is to BD, so HF (is)
ἡγούμεναπρὸςἅπαντατὰἑπόμενα.ὡςδὲἡΖΚπρὸςΚΕ, to FE [Prop. 5.17]. Let it have been contrived that as
οὕτωςἐστὶνἡΓΔπρὸςτὴνΔΒ·καὶὡςἄραἡΘΚπρὸςΚΖ, HF (is) to FE, so FK (is) to KE. And, thus, the whole
οὕτωςἡΓΔπρὸςτὴνΔΒ.σύμμετρονδὲτὸἀπὸτῆςΓΔτῷ HK is to the whole KF , as FK (is) to KE. For as one
ἀπὸτῆςΔΒ·σύμμετρονἄραἐστὶκαὶτὸἀπὸτῆςΘΚτῷ of the leading (proportional magnitudes is) to one of the
417

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
ἀπὸτῆςΚΖ.καίἐστινὡςτὸἀπὸτῆςΘΚπρὸςτὸἀπὸτῆς following, so all of the leading (magnitudes) are to all of
ΚΖ,οὕτωςἡΘΚπρὸςτὴνΚΕ,ἐπεὶαἱτρεῖςαἱΘΚ,ΚΖ, the following [Prop. 5.12]. And as FK (is) to KE, so CD
ΚΕἀνάλογόνεἰσιν. σύμμετροςἄραἡΘΚτῇΚΕμήκει. is to DB [Prop. 5.11]. And, thus, as HK (is) to KF , so
ὥστεκαὶἡΘΕτῇΕΚσύμμετρόςἐστιμήκει. καὶἐπεὶτὸ CD is to DB [Prop. 5.11]. And the (square) on CD (is)
ἀπὸτῆςΑἴσονἐστὶτῷὑπὸτῶνΕΘ,ΒΔ,ῥητὸνδέἐστι commensurable with the (square) on DB [Prop. 10.36].
τὸἀπὸτῆςΑ,ῥητὸνἄραἐστὶκαὶτὸὑπὸτῶνΕΘ,ΒΔ.καὶ The (square) on HK is thus also commensurable with
παρὰῥητὴντὴνΒΔπαράκειται·ῥητὴἄραἐστὶνἡΕΘκαὶ the (square) on KF [Props. 6.22, 10.11]. And as the
σύμμετροςτῇΒΔμήκει·ὥστεκαὶἡσύμμετροςαὐτῇἡΕΚ (square) on HK is to the (square) on KF , so HK (is) to
ῥητήἐστικαὶσύμμετροςτῇΒΔμήκει.ἐπεὶοὖνἐστινὡςἡ KE, since the three (straight-lines) HK, KF , and KE
ΓΔπρὸςΔΒ,οὕτωςἡΖΚπρὸςΚΕ,αἱδὲΓΔ,ΔΒδυνάμει are proportional [Def. 5.9]. HK is thus commensurable
μόνονεἰσὶσύμμετροι,καὶαἱΖΚ,ΚΕδυνάμειμόνονεἰσὶ in length with KE [Prop. 10.11]. Hence, HE is also comσύμμετροι.ῥητὴδέἐστινἡΚΕ·ῥητὴἄραἐστὶκαὶἡΖΚ.αἱ
mensurable in length with EK [Prop. 10.15]. And since
ΖΚ,ΚΕἄραῥηταὶδυνάμειμόνονεἰσὶσύμμετροι·ἀποτομὴ the (square) on A is equal to the (rectangle contained) by
ἄραἐστὶνἡΕΖ.
EH and BD, and the (square) on A is rational, the (rect-
῎ΗτοιδὲἡΓΔτῆςΔΒμεῖζονδύναταιτῷἀπὸσυμμέτρου angle contained) by EH and BD is thus also rational.
ἑαυτῇἢτῷἀπὸἀσυμμέτρου.
And it is applied to the rational (straight-line) BD. Thus,
Εἰ μὲν οὖν ἡ ΓΔ τῆς ΔΒ μεῖζον δύναται τῷ ἀπὸ EH is rational, and commensurable in length with BD
συμμέτρου[ἑαυτῇ],καὶἡΖΚτῆςΚΕμεῖζονδυνήσεταιτῷ [Prop. 10.20]. And, hence, the (straight-line) commensu-
ἀπὸσυμμέτρουἑαυτῇ.καὶεἰμὲνσύμμετρόςἐστινἡΓΔτῇ rable (in length) with it, EK, is also rational [Def. 10.3],
ἐκκειμένῃῥητῇμήκει,καὶἡΖΚ·εἰδὲἡΒΔ,καὶἡΚΕ·εἰ and commensurable in length with BD [Prop. 10.12].
δὲοὐδετέρατῶνΓΔ,ΔΒ,καὶοὐδετέρατῶνΖΚ,ΚΕ. Therefore, since as CD is to DB, so FK (is) to KE, and
ΕἰδὲἡΓΔτῆςΔΒμεῖζονδύναταιτῷἀπὸἀσυμμέτρου CD and DB are (straight-lines which are) commensu-
ἑαυτῇ, καὶ ἡ ΖΚ τῆς ΚΕ μεῖζον δυνήσεται τῷ ἀπὸ rable in square only, FK and KE are also commensu-
ἀσυμμέτρουἑαυτῇ. καὶεἰμὲνἡΓΔσύμμετρόςἐστιτῇ rable in square only [Prop. 10.11]. And KE is rational.
ἐκκειμένῃῥητῇμήκει,καὶἡΖΚ·εἰδὲἡΒΔ,καὶἡΚΕ·εἰ Thus, FK is also rational. FK and KE are thus rational
δὲοὐδετέρατῶνΓΔ,ΔΒ,καὶοὐδετέρατῶνΖΚ,ΚΕ·ὥστε (straight-lines which are) commensurable in square only.
ἀποτομήἐστινἡΖΕ,ἧςτὰὀνόματατὰΖΚ,ΚΕσύμμετρά Thus, EF is an apotome [Prop. 10.73].
ἐστιτοῖςτῆςἐκδύοὀνομάτωνὀνόμασιτοῖςΓΔ,ΔΒκαὶἐν And the square on CD is greater than (the square on)
τῷαὐτῷλόγῳ,καὶτὴναὐτῆντάξινἔχειτῇΒΓ·ὅπερἔδει DB either by the (square) on (some straight-line) comδεῖξαι.
mensurable, or by the (square) on (some straight-line)
incommensurable, (in length) with (CD).
Therefore, if the square on CD is greater than (the
square on) DB by the (square) on (some straight-line)
commensurable (in length) with [CD] then the square
on FK will also be greater than (the square on) KE by
the (square) on (some straight-line) commensurable (in
length) with (FK) [Prop. 10.14]. And if CD is commensurable
in length with a (previously) laid down rational
(straight-line), (so) also (is) FK [Props. 10.11,
10.12]. And if BD (is commensurable), (so) also (is)
KE [Prop. 10.12]. And if neither of CD or DB (is commensurable),
neither also (are) either of FK or KE.
And if the square on CD is greater than (the square
on) DB by the (square) on (some straight-line) incommensurable
(in length) with (CD) then the square on
FK will also be greater than (the square on) KE by
the (square) on (some straight-line) incommensurable
(in length) with (FK) [Prop. 10.14]. And if CD is commensurable
in length with a (previously) laid down rational
(straight-line), (so) also (is) FK [Props. 10.11,
10.12]. And if BD (is commensurable), (so) also (is) KE
418

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
[Prop. 10.12]. And if neither of CD or DB (is commensurable),
neither also (are) either of FK or KE. Hence,
FE is an apotome whose terms, FK and KE, are commensurable
(in length) with the terms, CD and DB, of
the binomial, and in the same ratio. And (FE) has the
same order as BC [Defs. 10.5—10.10]. (Which is) the
very thing it was required to show.
† Heiberg considers this proposition, and the succeeding ones, to be relatively early interpolations into the original text.
Ö. Proposition 113
Τὸἀπὸῥητῆςπαρὰἀποτομὴνπαραβαλλόμενονπλάτος The (square) on a rational (straight-line), applied to
ποιεῖτὴνἐκδύοὀνομάτων,ἧςτὰὀνόματασύμμετράἐστι an apotome, produces as breadth a binomial whose terms
τοῖςτῆςἀποτομῆςὀνόμασικαὶἐντῷαὐτῷλόγῳ,ἔτιδὲἡ are commensurable with the terms of the apotome, and
γινομένηἐκδύοὀνομάτωντὴναὐτὴντάξινἔχειτῇἀποτομῇ. in the same ratio. Moreover, the created binomial has the
same order as the apotome.
Α
A
Β ∆
Κ
Γ
Ε
Η
Ζ
Θ
῎ΕστωῥητὴμὲνἡΑ,ἀποτομὴδὲἡΒΔ,καὶτῷἀπὸτῆς Let A be a rational (straight-line), and BD an apo-
ΑἴσονἔστωτὸὑπὸτῶνΒΔ,ΚΘ,ὥστετὸἀπὸτῆςΑῥητῆς tome. And let the (rectangle contained) by BD and KH
παρὰτὴνΒΔἀποτομὴνπαραβαλλόμενονπλάτοςποιεῖτὴν be equal to the (square) on A, such that the square on the
ΚΘ·λέγω,ὅτιἐκδύοὀνομάτωνἐστὶνἡΚΘ,ἧςτὰὀνόματα rational (straight-line) A, applied to the apotome BD,
σύμμετράἐστιτοῖςτῆςΒΔὀνόμασικαὶἐντῷαὐτῷλόγῳ, produces KH as breadth. I say that KH is a binomial
καὶἔτιἡΚΘτὴναὐτὴνἔχειτάξιντῇΒΔ.
whose terms are commensurable with the terms of BD,
῎ΕστωγὰρτῇΒΔπροσαρμόζουσαἡΔΓ·αἱΒΓ,ΓΔ and in the same ratio, and, moreover, that KH has the
ἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι. καὶτῷἀπὸτῆς same order as BD.
ΑἴσονἔστωκαὶτὸὑπὸτῶνΒΓ,Η.ῥητὸνδὲτὸἀπὸτῆς For let DC be an attachment to BD. Thus, BC and
Α·ῥητὸνἄρακαὶτὸὑπὸτῶνΒΓ,Η.καὶπαρὰῥητὴντὴν CD are rational (straight-lines which are) commensu-
ΒΓπαραβέβληται·ῥητὴἄραἐστὶνἡΗκαὶσύμμετροςτῇΒΓ rable in square only [Prop. 10.73]. And let the (rectangle
μήκει. ἐπεὶοὖντὸὑπὸτῶνΒΓ,Ηἴσονἐστὶτῷὑπὸτῶν contained) by BC and G also be equal to the (square)
ΒΔ,ΚΘ,ἀνάλογονἄραἐστὶνὡςἡΓΒπρὸςΒΔ,οὕτωςἡ on A. And the (square) on A (is) rational. The (rect-
ΚΘπρὸςΗ.μείζωνδὲἡΒΓτῆςΒΔ·μείζωνἄρακαὶἡΚΘ angle contained) by BC and G (is) thus also rational.
τῆςΗ.κείσθωτῇΗἴσηἡΚΕ·σύμμετροςἄραἐστὶνἡΚΕ And it has been applied to the rational (straight-line)
τῇΒΓμήκει. καὶἐπείἐστινὡςἡΓΒπρὸςΒΔ,οὕτωςἡ BC. Thus, G is rational, and commensurable in length
ΘΚπρὸςΚΕ,ἀναστρέψαντιἄραἐστὶνὡςἡΒΓπρὸςτὴν with BC [Prop. 10.20]. Therefore, since the (rectangle
ΓΔ,οὕτωςἡΚΘπρὸςΘΕ.γεγονέτωὡςἡΚΘπρὸςΘΕ, contained) by BC and G is equal to the (rectangle conοὕτωςἡΘΖπρὸςΖΕ·καὶλοιπὴἄραἡΚΖπρὸςΖΘἐστιν,
tained) by BD and KH, thus, proportionally, as CB is to
ὡςἡΚΘπρὸςΘΕ,τουτέστιν[ὡς]ἡΒΓπρὸςΓΔ.αἱδὲ BD, so KH (is) to G [Prop. 6.16]. And BC (is) greater
ΒΓ,ΓΔδυνάμειμόνον[εἰσὶ]σύμμετροι·καὶαἱΚΖ,ΖΘἄρα than BD. Thus, KH (is) also greater than G [Prop. 5.16,
δυνάμειμόνονεἰσὶσύμμετροι·καὶἐπείἐστινὡςἡΚΘπρὸς 5.14]. Let KE be made equal to G. KE is thus com-
ΘΕ,ἡΚΖπρὸςΖΘ,ἀλλ᾿ὡςἡΚΘπρὸςΘΕ,ἡΘΖπρὸς mensurable in length with BC. And since as CB is to
ΖΕ,καὶὡςἄραἡΚΖπρὸςΖΘ,ἡΘΖπρὸςΖΕ·ὥστεκαὶ BD, so HK (is) to KE, thus, via conversion, as BC (is)
ὡςἡπρώτηπρὸςτὴντρίτην,τὸἀπὸτῆςπρώτηςπρὸςτὸ to CD, so KH (is) to HE [Prop. 5.19 corr.]. Let it have
ἀπὸτῆςδευτέρας·καὶὡςἄραἡΚΖπρὸςΖΕ,οὕτωςτὸἀπὸ been contrived that as KH (is) to HE, so HF (is) to
τῆςΚΖπρὸςτὸἀπὸτῆςΖΘ.σύμμετρονδέἐστιτὸἀπὸτῆς FE. And thus the remainder KF is to FH, as KH (is)
ΚΖτῷἀπὸτῆςΖΘ·αἱγὰρΚΖ,ΖΘδυνάμειεἰσὶσύμμετροι· to HE—that is to say, [as] BC (is) to CD [Prop. 5.19].
σύμμετροςἄραἐστὶκαὶἡΚΖτῇΖΕμήκει·ὥστεἡΚΖκαὶ And BC and CD [are] commensurable in square only.
B
K
D
C
E
G
F
H
419

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
τῇΚΕσύμμετρός[ἐστι]μήκει. ῥητὴδέἐστινἡΚΕκαὶ KF and FH are thus also commensurable in square only
σύμμετροςτῇΒΓμήκει.ῥητὴἄρακαὶἡΚΖκαὶσύμμετρος [Prop. 10.11]. And since as KH is to HE, (so) KF (is)
τῇΒΓμήκει.καὶἐπείἐστινὡςἡΒΓπρὸςΓΔ,οὕτωςἡΚΖ to FH, but as KH (is) to HE, (so) HF (is) to FE, thus,
πρὸςΖΘ,ἐναλλὰξὡςἡΒΓπρὸςΚΖ,οὕτωςἡΔΓπρὸςΖΘ. also as KF (is) to FH, (so) HF (is) to FE [Prop. 5.11].
σύμμετροςδὲἡΒΓτῇΚΖ·σύμμετροςἄρακαὶἡΖΘτῇΓΔ And hence as the first (is) to the third, so the (square) on
μήκει. αἱΒΓ,ΓΔδὲῥηταίεἰσιδυνάμειμόνονσύμμετροι· the first (is) to the (square) on the second [Def. 5.9]. And
καὶαἱΚΖ,ΖΘἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι·ἐκ thus as KF (is) to FE, so the (square) on KF (is) to the
δύοὀνομάτωνἐστὶνἄραἡΚΘ.
(square) on FH. And the (square) on KF is commen-
Εἰ μὲν οὖν ἡ ΒΓ τῆς ΓΔ μεῖζον δύναται τῷ ἀπὸ surable with the (square) on FH. For KF and FH are
συμμέτρουἑαυτῇ,καὶἡΚΖτῆςΖΘμεῖζονδυνήσεταιτῷ commensurable in square. Thus, KF is also commensu-
ἀπὸσυμμέτρουἑαυτῇ.καὶεἰμὲνσύμμετρόςἐστινἡΒΓτῇ rable in length with FE [Prop. 10.11]. Hence, KF [is]
ἐκκειμένῃῥητῇμήκει,καὶἡΚΖ,εἰδὲἡΓΔσύμμετρόςἐστι also commensurable in length with KE [Prop. 10.15].
τῇἐκκειμένῃῥητῇμήκει,καὶἡΖΘ,εἰδὲοὐδετέρατῶνΒΓ, And KE is rational, and commensurable in length with
ΓΔ,οὐδετέρατῶνΚΖ,ΖΘ.
BC. Thus, KF (is) also rational, and commensurable in
ΕἰδὲἡΒΓτῆςΓΔμεῖζονδύναταιτῷἀπὸἀσυμμέτρου length with BC [Prop. 10.12]. And since as BC is to
ἑαυτῇ,καὶἡΚΖτῆςΖΘμεῖζονδυνήσεταιτῷἀπὸἀσυμμέτρ- CD, (so) KF (is) to FH, alternately, as BC (is) to KF ,
ουἑαυτῇ. καὶεἰμὲνσύμμετρόςἐστινἡΒΓτῇἐκκειμένῃ so DC (is) to FH [Prop. 5.16]. And BC (is) commen-
ῥητῇμήκει,καὶἡΚΖ,εἰδὲἡΓΔ,καὶἡΖΘ,εἰδὲοὐδετέρα surable (in length) with KF . Thus, FH (is) also comτῶνΒΓ,ΓΔ,οὐδετέρατῶνΚΖ,ΖΘ.
mensurable in length with CD [Prop. 10.11]. And BC
᾿ΕκδύοἄραὀνομάτωνἐστὶνἡΚΘ,ἧςτὰὀνόματατὰ and CD are rational (straight-lines which are) commen-
ΚΖ,ΖΘσύμμετρά[ἐστι]τοῖςτῆςἀποτομῆςὀνόμασιτοῖς surable in square only. KF and FH are thus also ratio-
ΒΓ,ΓΔκαὶἐντῷαὐτῷλόγῳ,καὶἔτιἡΚΘτῇΒΓτὴν nal (straight-lines which are) commensurable in square
αὐτὴνἕξειτάξιν·ὅπερἔδειδεῖξαι.
only [Def. 10.3, Prop. 10.13]. Thus, KH is a binomial
[Prop. 10.36].
Therefore, if the square on BC is greater than (the
square on) CD by the (square) on (some straight-line)
commensurable (in length) with (BC), then the square
on KF will also be greater than (the square on) FH by
the (square) on (some straight-line) commensurable (in
length) with (KF ) [Prop. 10.14]. And if BC is commensurable
in length with a (previously) laid down rational
(straight-line), (so) also (is) KF [Prop. 10.12].
And if CD is commensurable in length with a (previously)
laid down rational (straight-line), (so) also (is)
FH [Prop. 10.12]. And if neither of BC or CD (are
commensurable), neither also (are) either of KF or FH
[Prop. 10.13].
And if the square on BC is greater than (the square
on) CD by the (square) on (some straight-line) incommensurable
(in length) with (BC) then the square on
KF will also be greater than (the square on) FH by
the (square) on (some straight-line) incommensurable
(in length) with (KF ) [Prop. 10.14]. And if BC is commensurable
in length with a (previously) laid down rational
(straight-line), (so) also (is) KF [Prop. 10.12]. And
if CD is commensurable, (so) also (is) FH [Prop. 10.12].
And if neither of BC or CD (are commensurable), neither
also (are) either of KF or FH [Prop. 10.13].
KH is thus a binomial whose terms, KF and FH,
[are] commensurable (in length) with the terms, BC and
CD, of the apotome, and in the same ratio. Moreover,
420

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
KH will have the same order as BC [Defs. 10.5—10.10].
(Which is) the very thing it was required to show.
Ö. Proposition 114
᾿Εὰνχωρίονπεριέχηταιὑπὸἀποτομῆςκαὶτῆςἐκδύο If an area is contained by an apotome, and a binomial
ὀνομάτων,ἧςτὰὀνόματασύμμετράτέἐστιτοῖςτῆςἀπο- whose terms are commensurable with, and in the same
τομῆςὀνόμασικαὶἐντῷαὐτῷλόγῳ,ἡτὸχωρίονδυναμένη ratio as, the terms of the apotome then the square-root of
ῥητήἐστιν.
the area is a rational (straight-line).
Α Β Ζ
A
B F
Γ Ε ∆
C
E
D
Η
G
Θ
Κ Λ Μ
H
K L M
Περιεχέσθω γὰρ χωρίον τὸ ὑπὸ τῶν ΑΒ, ΓΔ ὑπὸ For let an area, the (rectangle contained) by AB and
ἀποτομῆςτῆςΑΒκαὶτῆςἐκδύοὀνομάτωντῆςΓΔ,ἧς CD, have been contained by the apotome AB, and the
μεῖζονὄνομαἔστωτὸΓΕ,καὶἔστωτὰὀνόματατῆςἐκ binomial CD, of which let the greater term be CE. And
δύοὀνομάτωντὰΓΕ,ΕΔσύμμετράτετοῖςτῆςἀποτομῆς let the terms of the binomial, CE and ED, be commen-
ὀνόμασιτοῖςΑΖ,ΖΒκαὶἐντῷαὐτῷλόγῳ,καὶἔστωἡτὸ surable with the terms of the apotome, AF and FB (re-
ὑπὸτῶνΑΒ,ΓΔδυναμένηἡΗ·λέγω,ὅτιῥητήἐστινἡΗ. spectively), and in the same ratio. And let the square-root
᾿ΕκκείσθωγὰρῥητὴἡΘ,καὶτῷἀπὸτῆςΘἴσονπαρὰ of the (rectangle contained) by AB and CD be G. I say
τὴνΓΔπαραβεβλήσθωπλάτοςποιοῦντὴνΚΛ·ἀποτομὴἄρα that G is a rational (straight-line).
ἐστὶνἡΚΛ,ἧςτὰὀνόματαἔστωτὰΚΜ,ΜΛσύμμετρατοῖς For let the rational (straight-line) H be laid down.
τῆςἐκδύοὀνομάτωνὀνόμασιτοῖςΓΕ,ΕΔκαὶἐντῷαὐτῷ And let (some rectangle), equal to the (square) on H,
λόγῳ.ἀλλὰκαὶαἱΓΕ,ΕΔσύμμετροίτέεἰσιταῖςΑΖ,ΖΒ have been applied to CD, producing KL as breadth.
καὶἐντῷαὐτῷλόγῳ·ἔστινἄραὡςἡΑΖπρὸςτὴνΖΒ, Thus, KL is an apotome, of which let the terms, KM
οὕτωςἡΚΜπρὸςΜΛ.ἐναλλὰξἄραἐστὶνὡςἡΑΖπρὸς and ML, be commensurable with the terms of the binoτὴνΚΜ,οὕτωςἡΒΖπρὸςτὴνΛΜ·καὶλοιπὴἄραἡΑΒ
mial, CE and ED (respectively), and in the same ratio
πρὸςλοιπὴντὴνΚΛἐστινὡςἡΑΖπρὸςΚΜ.σύμμετρος [Prop. 10.112]. But, CE and ED are also commensuδὲἡΑΖτῇΚΜ·σύμμετροςἄραἐστὶκαὶἡΑΒτῇΚΛ.καί
rable with AF and FB (respectively), and in the same ra-
ἐστινὡςἡΑΒπρὸςΚΛ,οὕτωςτὸὑπὸτῶνΓΔ,ΑΒπρὸς tio. Thus, as AF is to FB, so KM (is) to ML. Thus, alterτὸὑπὸτῶνΓΔ,ΚΛ·σύμμετρονἄραἐστὶκαὶτὸὑπὸτῶν
nately, as AF is to KM, so BF (is) to LM [Prop. 5.16].
ΓΔ,ΑΒτῷὑπὸτῶνΓΔ,ΚΛ.ἴσονδὲτὸὑπὸτῶνΓΔ,ΚΛ Thus, the remainder AB is also to the remainder KL as
τῷἀπὸτῆςΘ·σύμμετρονἄραἐστὶτὸὑπὸτῶνΓΔ,ΑΒτῷ AF (is) to KM [Prop. 5.19]. And AF (is) commensu-
ἀπὸτῆςΘ.τῷδὲὑπὸτῶνΓΔ,ΑΒἴσονἐστὶτὸἀπὸτῆςΗ· rable with KM [Prop. 10.12]. AB is thus also commenσύμμετρονἄραἐστὶτὸἀπὸτῆςΗτῷἀπὸτῆςΘ.ῥητὸνδὲ
surable with KL [Prop. 10.11]. And as AB is to KL,
τὸἀπὸτῆςΘ·ῥητὸνἄραἐστὶκαὶτὸἀπὸτῆςΗ·ῥητὴἄρα so the (rectangle contained) by CD and AB (is) to the
ἐστὶνἡΗ.καὶδύναταιτὸὑπὸτῶνΓΔ,ΑΒ.
(rectangle contained) by CD and KL [Prop. 6.1]. Thus,
᾿Εὰνἄραχωρίονπεριέχηταιὑπὸἀποτομῆςκαὶτῆςἐκ the (rectangle contained) by CD and AB is also comδύοὀνομάτων,ἧςτὰὀνόματασύμμετράἐστιτοῖςτῆςἀπο-
mensurable with the (rectangle contained) by CD and
τομῆςὀνόμασικαὶἐντῷαὐτῷλόγῳ,ἡτὸχωρίονδυναμένη KL [Prop. 10.11]. And the (rectangle contained) by CD
ῥητήἐστιν.
and KL (is) equal to the (square) on H. Thus, the (rectangle
contained) by CD and AB is commensurable with
the (square) on H. And the (square) on G is equal to the
(rectangle contained) by CD and AB. The (square) on G
421

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 10
is thus commensurable with the (square) on H. And the
(square) on H (is) rational. Thus, the (square) on G is
also rational. G is thus rational. And it is the square-root
of the (rectangle contained) by CD and AB.
Thus, if an area is contained by an apotome, and a
binomial whose terms are commensurable with, and in
the same ratio as, the terms of the apotome, then the
square-root of the area is a rational (straight-line).
ÈÖ×Ñ.
Καὶγέγονενἡμῖνκαὶδιὰτούτουφανερόν,ὅτιδυνατόν And it has also been made clear to us, through this,
Corollary
ἐστιῥητὸνχωρίονὑπὸἀλόγωνεὐθειῶνπεριέχεσθαι.ὅπερ that it is possible for a rational area to be contained by
ἔδειδεῖξαι.
irrational straight-lines. (Which is) the very thing it was
required to show.
Ö. Proposition 115
Ἀπὸμέσηςἄπειροιἄλογοιγίνονται,καὶοὐδεμίαοὐδεμιᾷ An infinite (series) of irrational (straight-lines) can be
τῶνπρότερονἡαὐτή.
created from a medial (straight-line), and none of them
is the same as any of the preceding (straight-lines).
Α
Β
Γ
∆
῎ΕστωμέσηἡΑ·λέγω,ὅτιἀπὸτῆςΑἄπειροιἄλογοι Let A be a medial (straight-line). I say that an infiγίνονται,καὶοὐδεμίαοὐδεμιᾷτῶνπρότερονἡαὐτή.
nite (series) of irrational (straight-lines) can be created
᾿ΕκκείσθωῥητὴἡΒ,καὶτῷὑπὸτῶνΒ,Αἴσονἔστωτὸ from A, and that none of them is the same as any of the
ἀπὸτῆςΓ·ἄλογοςἄραἐστὶνἡΓ·τὸγὰρὑπὸἀλόγουκαὶ preceding (straight-lines).
ῥητῆςἄλογόνἐστιν. καὶοὐδεμιᾷτῶνπρότερονἡαὐτή·τὸ Let the rational (straight-line) B be laid down. And
γὰρἀπ᾿οὐδεμιᾶςτῶνπρότερονπαρὰῥητὴνπαραβαλλόμενον let the (square) on C be equal to the (rectangle conπλάτοςποιεῖμέσην.πάλινδὴτῷὑπὸτῶνΒ,Γἴσονἔστωτὸ
tained) by B and A. Thus, C is irrational [Def. 10.4].
ἀπὸτῆςΔ·ἄλογονἄραἐστὶτὸἀπὸτῆςΔ.ἄλογοςἄραἐστὶν For an (area contained) by an irrational and a rational
ἡΔ·καὶοὐδεμιᾷτῶνπρότερονἡαὐτή·τὸγὰρἀπ᾿οὐδεμιᾶς (straight-line) is irrational [Prop. 10.20]. And (C is) not
τῶνπρότερονπαρὰῥητὴνπαραβαλλόμενονπλάτοςποιεῖτὴν the same as any of the preceding (straight-lines). For
Γ.ὁμοίωςδὴτῆςτοιαύτηςτάξεωςἐπ᾿ἄπειρονπροβαινούσης the (square) on none of the preceding (straight-lines),
φανερόν, ὅτιἀπὸτῆςμέσηςἄπειροιἄλογοιγίνονται,καὶ applied to a rational (straight-line), produces a medial
οὐδεμίαοὐδεμιᾷτῶνπρότερονἡαὐτή·ὅπερἔδειδεῖξαι. (straight-line) as breadth. So, again, let the (square) on
D be equal to the (rectangle contained) by B and C.
Thus, the (square) on D is irrational [Prop. 10.20]. D
is thus irrational [Def. 10.4]. And (D is) not the same as
any of the preceding (straight-lines). For the (square) on
none of the preceding (straight-lines), applied to a rational
(straight-line), produces C as breadth. So, similarly,
this arrangement being advanced to infinity, it is clear
that an infinite (series) of irrational (straight-lines) can
be created from a medial (straight-line), and that none of
them is the same as any of the preceding (straight-lines).
(Which is) the very thing it was required to show.
A
B
C
D
422

ELEMENTS BOOK 11
Elementary Stereometry
423

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
ÎÇÖÓ.
αʹ.Στερεόνἐστιτὸμῆκοςκαὶπλάτοςκαὶβάθοςἔχον. 1. A solid is a (figure) having length and breadth and
Definitions
βʹ.Στερεοῦδὲπέραςἐπιφάνεια.
depth.
γʹ.Εὐθεῖαπρὸςἐπίπεδονὀρθήἐστιν,ὅτανπρὸςπάσας 2. The extremity of a solid (is) a surface.
τὰςἁπτομέναςαὐτῆςεὐθείαςκαὶοὔσαςἐντῷ[ὑποκειμένῳ] 3. A straight-line is at right-angles to a plane when it
ἐπιπέδῳὀρθὰςποιῇγωνίας.
makes right-angles with all of the straight-lines joined to
δʹ.᾿Επίπεδον πρὸς ἐπίπεδον ὀρθόν ἐστιν, ὅταν αἱ τῇ it which are also in the plane.
κοινῇτομῇτῶνἐπιπέδωνπρὸςὀρθὰςἀγόμεναιεὐθεῖαιἐν 4. A plane is at right-angles to a(nother) plane when
ἑνὶτῶνἐπιπέδωντῷλοιπῷἐπιπέδῳπρὸςὀρθὰςὦσιν. (all of) the straight-lines drawn in one of the planes, at
εʹ.Εὐθείαςπρὸςἐπίπεδονκλίσιςἐστίν, ὅτανἀπὸτοῦ right-angles to the common section of the planes, are at
μετεώρου πέρατος τῆς εὐθείας ἐπὶ τὸ ἐπίπεδον κάθετος right-angles to the remaining plane.
ἀχθῇ,καὶἀπὸτοῦγενομένουσημείουἐπὶτὸἐντῷἐπιπέδῳ 5. The inclination of a straight-line to a plane is the
πέραςτῆςεὐθείαςεὐθεῖαἐπιζευχθῇ,ἡπεριεχομένηγωνία angle contained by the drawn and standing (straight-
ὑπὸτῆςἀχθείσηςκαὶτῆςἐφεστώσης.
lines), when a perpendicular is lead to the plane from
ϛʹ.᾿Επιπέδουπρὸςἐπίπεδονκλίσιςἐστὶνἡπεριεχομένη the end of the (standing) straight-line raised (out of the
ὀξεῖαγωνίαὑπὸτῶνπρὸςὀρθὰςτῇκοινῇτομῇἀγομένων plane), and a straight-line is (then) joined from the point
πρὸςτῷαὐτῷσημείῳἐνἑκατέρῳτῶνἐπιπέδων. (so) generated to the end of the (standing) straight-line
ζʹ.᾿Επίπεδονπρὸςἐπίπεδονὁμοίωςκεκλίσθαιλέγεται (lying) in the plane.
καὶ ἕτερον πρὸς ἕτερον, ὅταν αἱ εἰρημέναι τῶν κλίσεων 6. The inclination of a plane to a(nother) plane is the
γωνίαιἴσαιἀλλήλαιςὦσιν.
acute angle contained by the (straight-lines), (one) in
ηʹ.Παράλληλαἐπίπεδάἐστιτὰἀσύμπτωτα.
each of the planes, drawn at right-angles to the common
θʹ.῞Ομοιαστερεὰσχήματάἐστιτὰὑπὸὁμοίωνἐπιπέδων segment (of the planes), at the same point.
περιεχόμεναἴσωντὸπλῆθος.
7. A plane is said to have been similarly inclined to a
ιʹ.῎Ισαδὲκαὶὅμοιαστερεὰσχήματάἐστιτὰὑπὸὁμοίων plane, as another to another, when the aforementioned
ἐπιπέδωνπεριεχόμεναἴσωντῷπλήθεικαὶτῷμεγέθει. angles of inclination are equal to one another.
ιαʹ.Στερεὰγωνίαἐστὶνἡὑπὸπλειόνωνἢδύογραμμῶν 8. Parallel planes are those which do not meet (one
ἁπτομένωνἀλλήλωνκαὶμὴἐντῇαὐτῇἐπιφανείᾳοὐσῶν another).
πρὸςπάσαιςταῖςγραμμαῖςκλίσις. ἄλλως· στερεὰγωνία 9. Similar solid figures are those contained by equal
ἐστὶνἡὑπὸπλειόνωνἢδύογωνιῶνἐπιπέδωνπεριεχομένη numbers of similar planes (which are similarly arranged).
μὴ οὐσῶν ἐν τῷ αὐτῷ ἐπιπέδῳ πρὸς ἑνὶ σημείῳ συνι- 10. But equal and similar solid figures are those conσταμένων.
tained by similar planes equal in number and in magniιβʹ.Πυραμίςἐστισχῆμαστερεὸνἐπιπέδοιςπεριχόμενον
tude (which are similarly arranged).
ἀπὸἑνὸςἐπιπέδουπρὸςἑνὶσημείῳσυνεστώς.
11. A solid angle is the inclination (constituted) by
ιγʹ.Πρίσμαἐστὶσχῆμαστερεὸνἐπιπέδοιςπεριεχόμενον, more than two lines joining one another (at the same
ὧνδύοτὰἀπεναντίονἴσατεκαὶὅμοιάἐστικαὶπαράλληλα, point), and not being in the same surface, to all of the
τὰδὲλοιπὰπαραλληλόγραμμα.
lines. Otherwise, a solid angle is that contained by more
ιδʹ.Σφαῖράἐστιν,ὅτανἡμικυκλίουμενούσηςτῆςδιαμέτ- than two plane angles, not being in the same plane, and
ρουπεριενεχθὲντὸἡμικύκλιονεἰςτὸαὐτὸπάλινἀποκατα- constructed at one point.
σταθῇ,ὅθενἤρξατοφέρεσθαι,τὸπεριληφθὲνσχῆμα. 12. A pyramid is a solid figure, contained by planes,
ιεʹ.῎Αξωνδὲτῆςσφαίραςἐστὶνἡμένουσαεὐθεῖα,περὶ (which is) constructed from one plane to one point.
ἣντὸἡμικύκλιονστρέφεται.
13. A prism is a solid figure, contained by planes, of
ιϛʹ.Κέντρονδὲτῆςσφαίραςἐστὶτὸαὐτό,ὃκαὶτοῦ which the two opposite (planes) are equal, similar, and
ἡμικυκλίου.
parallel, and the remaining (planes are) parallelograms.
ιζʹ.Διάμετροςδὲτῆςσφαίραςἐστὶνεὐθεῖάτιςδιὰτοῦ 14. A sphere is the figure enclosed when, the diamκέντρουἠγμένηκαὶπερατουμένηἐφ᾿ἑκάτερατὰμέρηὑπὸ
eter of a semicircle remaining (fixed), the semicircle is
τῆςἐπιφανείαςτῆςσφαίρας.
carried around, and again established at the same (posiιηʹ.Κῶνόςἐστιν,ὅτανὀρθογωνίουτριγώνουμενούσης
tion) from which it began to be moved.
μιᾶςπλευρᾶςτῶνπερὶτὴνὀρθὴνγωνίανπεριενεχθὲντὸ 15. And the axis of the sphere is the fixed straight-line
τρίγωνον εἰςτὸαὐτὸπάλινἀποκατασταθῇ,ὅθενἤρξατο about which the semicircle is turned.
424

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
φέρεσθαι, τὸ περιληφθὲν σχῆμα. κἂν μὲν ἡ μένουσα 16. And the center of the sphere is the same as that of
εὐθεῖαἴσηᾖτῇλοιπῇ[τῇ]περὶτὴνὀρθὴνπεριφερομένῃ, the semicircle.
ὀρθογώνιοςἔσταιὁκῶνος,ἐὰνδὲἐλάττων,ἀμβλυγώνιος, 17. And the diameter of the sphere is any straight-
ἐὰνδὲμείζων,ὀξυγώνιος.
line which is drawn through the center and terminated in
ιθʹ.῎Αξωνδὲτοῦκώνουἐστὶνἡμένουσαεὐθεῖα,περὶ both directions by the surface of the sphere.
ἣντὸτρίγωνονστρέφεται.
18. A cone is the figure enclosed when, one of the
κʹ.Βάσιςδὲὁκύκλοςὁὑπὸτῆςπεριφερομένηςεὐθείας sides of a right-angled triangle about the right-angle reγραφόμενος.
maining (fixed), the triangle is carried around, and again
καʹ.Κύλινδρόςἐστιν,ὅτανὀρθογωνίουπαραλληλογράμ- established at the same (position) from which it began to
μουμενούσηςμιᾶςπλευρᾶςτῶνπερὶτὴνὀρθὴνγωνίανπε- be moved. And if the fixed straight-line is equal to the reριενεχθὲντὸπαραλληλόγραμμονεἰςτὸαὐτὸπάλινἀποκα-
maining (straight-line) about the right-angle, (which is)
τασταθῇ,ὅθενἤρξατοφέρεσθαι,τὸπεριληφθὲνσχῆμα. carried around, then the cone will be right-angled, and if
κβʹ.῎Αξωνδὲτοῦκυλίνδρουἐστὶνἡμένουσαεὐθεῖα, less, obtuse-angled, and if greater, acute-angled.
περὶἣντὸπαραλληλόγραμμονστρέφεται.
19. And the axis of the cone is the fixed straight-line
κγʹ.Βάσειςδὲοἱκύκλοιοἱὑπὸτῶνἀπεναντίονπερια- about which the triangle is turned.
γομένωνδύοπλευρῶνγραφόμενοι.
20. And the base (of the cone is) the circle described
κδʹ.῞Ομοιοικῶνοικαὶκύλινδροίεἰσιν,ὧνοἵτεἄξονες by the (remaining) straight-line (about the right-angle
καὶαἱδιάμετροιτῶνβάσεωνἀνάλογόνεἰσιν.
which is) carried around (the axis).
κεʹ.Κύβοςἐστὶσχῆμαστερεὸνὑπὸἓξτετραγώνωνἴσων 21. A cylinder is the figure enclosed when, one of
περιεχόμενον.
the sides of a right-angled parallelogram about the rightκϛʹ.᾿Οκτάεδρόνἐστὶσχῆμαστερεὸνὑπὸὀκτὼτριγώνων
angle remaining (fixed), the parallelogram is carried
ἴσωνκαὶἰσοπλεύρωνπεριεχόμενον.
around, and again established at the same (position)
κζʹ. Εἰκοσάεδρόν ἐστι σχῆμα στερεὸν ὑπὸ εἴκοσι from which it began to be moved.
τριγώνωνἴσωνκαὶἰσοπλεύρωνπεριεχόμενον.
22. And the axis of the cylinder is the stationary
κηʹ.Δωδεκάεδρόνἐστισχῆμαστερεὸνὑπὸδώδεκαπεν- straight-line about which the parallelogram is turned.
ταγώνωνἴσωνκαὶἰσοπλεύρωνκαὶἰσογωνίωνπεριεχόμενον. 23. And the bases (of the cylinder are) the circles
described by the two opposite sides (which are) carried
around.
24. Similar cones and cylinders are those for which
the axes and the diameters of the bases are proportional.
25. A cube is a solid figure contained by six equal
squares.
26. An octahedron is a solid figure contained by eight
equal and equilateral triangles.
27. An icosahedron is a solid figure contained by
twenty equal and equilateral triangles.
28. A dodecahedron is a solid figure contained by
twelve equal, equilateral, and equiangular pentagons.
. Proposition 1 †
Εὐθείαςγραμμῆςμέροςμέντιοὐκἔστινἐντῷὑπο- Some part of a straight-line cannot be in a reference
κειμένῳἐπιπέδῳ,μέροςδέτιἐνμετεωροτέρῳ.
plane, and some part in a more elevated (plane).
Εἰγὰρδυνατόν,εὐθείαςγραμμῆςτῆςΑΒΓμέροςμέν For, if possible, let some part, AB, of the straight-line
τιτὸΑΒἔστωἐντῷὑποκειμένῳἐπιπέδῳ,μέροςδέτιτὸ ABC be in a reference plane, and some part, BC, in a
ΒΓἐνμετεωροτέρῳ.
more elevated (plane).
῎Εσται δή τις τῇ ΑΒ συνεχὴς εὐθεῖα ἐπ᾿ εὐθείας ἐν In the reference plane, there will be some straight-line
τῷὑποκειμένῳἐπιπέδῳ.ἔστωἡΒΔ·δύοἄραεὐθειῶντῶν continuous with, and straight-on to, AB. ‡ Let it be BD.
ΑΒΓ,ΑΒΔκοινὸντμῆμάἐστινἡΑΒ·ὅπερἐστὶνἀδύνατον, Thus, AB is a common segment of the two (different)
ἐπειδήπερἐὰνκέντρῳτῷΒκαὶδιαστήματιτῷΑΒκύκλον straight-lines ABC and ABD. The very thing is imposγράψωμεν,αἱδιάμετροιἀνίσουςἀπολήψονταιτοῦκύκλου
sible, inasmuch as if we draw a circle with center B and
425

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
περιφερείας.
Γ
radius AB then the diameters (ABD and ABC) will cut
off unequal circumferences of the circle.
C
∆
D
Β
B
Α
A
Εὐθείαςἄραγραμμῆςμέροςμέντιοὐκἔστινἐντῷὑπο- Thus, some part of a straight-line cannot be in a referκειμένῳἐπιπέδῳ,τὸδὲἐνμετεωροτέρῳ·ὅπερἔδειδεῖξαι.
ence plane, and (some part) in a more elevated (plane).
(Which is) the very thing it was required to show.
† The proofs of the first three propositions in this book are not at all rigorous. Hence, these three propositions should properly be regarded as
additional axioms.
‡ This assumption essentially presupposes the validity of the proposition under discussion.
. Proposition 2
᾿Εὰνδύοεὐθεῖαιτέμνωσινἀλλήλας,ἐνἑνίεἰσινἐπιπέδῳ, If two straight-lines cut one another then they are in
καὶπᾶντρίγωνονἐνἑνίἐστινἐπιπέδῳ.
one plane, and every triangle (formed using segments of
both lines) is in one plane.
Α
∆
A
D
Ε
E
Ζ
Η
F
G
Γ Θ Κ Β
ΔύογὰρεὐθεῖαιαἱΑΒ,ΓΔτεμνέτωσανἀλλήλαςκατὰ For let the two straight-lines AB and CD have cut
τὸΕσημεῖον. λέγω,ὅτιαἱΑΒ,ΓΔἐνἑνίεἰσινἐπιπέδῳ, one another at point E. I say that AB and CD are in one
καὶπᾶντρίγωνονἐνἑνίἐστινἐπιπέδῳ.
plane, and that every triangle (formed using segments of
ΕἰλήφθωγὰρἐπὶτῶνΕΓ,ΕΒτυχόντασημεῖατὰΖ,Η, both lines) is in one plane.
καὶἐπεζεύχθωσαναἱΓΒ,ΖΗ,καὶδιήχθωσαναἱΖΘ,ΗΚ· For let the random points F and G have been taken
λέγωπρῶτον,ὅτιτὸΕΓΒτρίγωνονἐνἑνίἐστινἐπιπέδῳ.εἰ on EC and EB (respectively). And let CB and FG
γάρἐστιτοῦΕΓΒτριγώνουμέροςἤτοιτὸΖΘΓἢτὸΗΒΚ have been joined, and let FH and GK have been drawn
ἐντῷὑποκειμένῳ[ἐπιπέδῳ],τὸδὲλοιπὸνἐνἄλλῳ,ἔσταικαὶ across. I say, first of all, that triangle ECB is in one (refμιᾶςτῶνΕΓ,ΕΒεὐθειῶνμέροςμέντιἐντῷὑποκειμένῳ
erence) plane. For if part of triangle ECB, either FHC
C
H
K
B
426

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
ἐπιπέδῳ,τὸδὲἐναλλῳ.εἰδὲτοῦΕΓΒτριγώνουτὸΖΓΒΗ or GBK, is in the reference [plane], and the remainder
μέροςᾖἐντῷὑποκειμένῳἐπιπέδῳ,τὸδὲλοιπὸνἐνἄλλῳ, in a different (plane) then a part of one the straight-lines
ἔσταικαὶἀμφοτέρωντῶνΕΓ,ΕΒεὐθειῶνμέροςμέντι EC and EB will also be in the reference plane, and (a
ἐντῷὑποκειμένῳἐπιπέδῳ,τὸδὲἐνἄλλω· ὅπερἄτοπον part) in a different (plane). And if the part FCBG of tri-
ἐδείχθη.τὸἄραΕΓΒτρίγωνονἐνἑνίἐστινἐπιπέδῳ.ἐνᾧ angle ECB is in the reference plane, and the remainder
δέἐστιτὸΕΓΒτρίγωνον,ἐντούτῳκαὶἑκατέρατῶνΕΓ, in a different (plane) then parts of both of the straight-
ΕΒ,ἐνᾧδὲἑκατέρατῶνΕΓ,ΕΒ,ἐντούτῳκαὶαἱΑΒ, lines EC and EB will also be in the reference plane,
ΓΔ.αἱΑΒ,ΓΔἄραεὐθεῖαιἐνἑνίεἰσινἐπιπέδῳ,καὶπᾶν and (parts) in a different (plane). The very thing was
τρίγωνονἐνἑνίἐστινἐπιπέδῳ·ὅπερἔδειδεῖξαι.
shown to be absurb [Prop. 11.1]. Thus, triangle ECB
is in one plane. And in whichever (plane) triangle ECB
is (found), in that (plane) EC and EB (will) each also
(be found). And in whichever (plane) EC and EB (are)
each (found), in that (plane) AB and CD (will) also (be
found) [Prop. 11.1]. Thus, the straight-lines AB and CD
are in one plane, and every triangle (formed using segments
of both lines) is in one plane. (Which is) the very
thing it was required to show.
. Proposition 3
᾿Εὰνδύοἐπίπεδατεμνῇἄλληλα,ἡκοινὴαὐτῶντομὴ If two planes cut one another then their common secεὐθεῖάἐστιν.
tion is a straight-line.
Β
B
Ζ
Ε
F
E
∆
Α
D
A
Γ
ΔύογὰρἐπίπεδατὰΑΒ,ΒΓτεμνέτωἄλληλα,κοινὴδὲ For let the two planes AB and BC cut one another,
αὐτῶντομὴἔστωἡΔΒγραμμή·λέγω,ὅτιἡΔΒγραμμὴ and let their common section be the line DB. I say that
εὐθεῖάἐστιν.
the line DB is straight.
Εἰγὰρμή,ἐπεζεύχθωἀπὸτοῦΔἐπὶτὸΒἐνμὲντῷΑΒ For, if not, let the straight-line DEB have been joined
ἐπιπέδῳεὐθεῖαἡΔΕΒ,ἐνδὲτῷΒΓἐπιπέδῳεὐθεῖαἡΔΖΒ. from D to B in the plane AB, and the straight-line DFB
ἔσταιδὴδύοεὐθειῶντῶνΔΕΒ,ΔΖΒτὰαὐτὰπέρατα,καὶ in the plane BC. So two straight-lines, DEB and DFB,
περιέξουσιδηλαδὴχωρίον·ὅπερἄτοπον.οὔκἄρααἰΔΕΒ, will have the same ends, and they will clearly enclose an
ΔΖΒεὐθεῖαίεἰσιν. ὁμοίωςδὴδείξομεν,ὅτιοὐδὲἄλλητις area. The very thing (is) absurd. Thus, DEB and DFB
ἀπὸτοῦΔἐπὶτὸΒἐπιζευγνυμένηεὐθεῖαἔσταιπλὴντῆς are not straight-lines. So, similarly, we can show than no
ΔΒκοινῆςτομῆςτῶνΑΒ,ΒΓἐπιπέδων.
other straight-line can be joined from D to B except DB,
᾿Εὰνἄραδύοἐπίπεδατέμνῃἄλληλα,ἡκοινὴαὐτῶντομὴ the common section of the planes AB and BC.
εὐθεῖάἐστιν·ὅπερἔδειδεῖξαι.
Thus, if two planes cut one another then their common
section is a straight-line. (Which is) the very thing it
was required to show.
C
427

ËÌÇÁÉÁÏƺ ELEMENTS BOOK 11
. Proposition 4
᾿Εὰν εὐθεῖα δύο εὐθείαις τεμνούσαις ἀλλήλας πρὸς If a straight-line is set up at right-angles to two
ὀρθὰς ἐπὶτῆς κοινῆς τομῆςἐπισταθῇ, καὶ τῷ δι᾿ αὐτῶν straight-lines cutting one another, at the common point
ἐπιπέδῳπρὸςὀρθὰςἔσται.
of section, then it will also be at right-angles to the plane
(passing) through them (both).
Ζ
F
Α
Γ
A
C
Η
∆
Ε
Θ
Β
G
D
E
H
B
Εὐθεῖα γάρ τις ἡ ΕΖ δύο εὐθείαιςταῖς ΑΒ, ΓΔ τε- For let some straight-line EF have (been) set up at
μνούσαιςἀλλήλαςκατὰτὸΕσημεῖονἀπὸτοῦΕπρὸςὀρθὰς right-angles to two straight-lines, AB and CD, cutting
ἐφεστάτω·λέγω,ὅτιἡΕΖκαὶτῷδιὰτῶνΑΒ,ΓΔἐπιπέδῳ one another at point E, at E. I say that EF is also at
πρὸςὀρθάςἐστιν.
right-angles to the plane (passing) through AB and CD.
ἈπειλήφθωσανγὰραἱΑΕ,ΕΒ,ΓΕ,ΕΔἴσαιἀλλήλαις, For let AE, EB, CE and ED have been cut off from
καὶ διήχθω τις διὰ τοῦ Ε, ὡς ἔτυχεν, ἡ ΗΕΘ, καὶ (the two straight-lines so as to be) equal to one another.
ἐπεζεύχθωσαν αἱΑΔ, ΓΒ, καὶἔτιἀπὸτυχόντοςτοῦΖ And let GEH have been drawn, at random, through E
ἐπεζεύχθωσαναἱΖΑ,ΖΗ,ΖΔ,ΖΓ,ΖΘ,ΖΒ.
(in the plane passing through AB and CD). And let AD
ΚαὶἐπεὶδύοαἱΑΕ,ΕΔδυσὶταῖςΓΕ,ΕΒἴσαιεἰσὶκαὶ and CB have been joined. And, furthermore, let FA,
γωνίαςἴσαςπεριέχουσιν,βάσιςἄραἡΑΔβάσειτῇΓΒἴση FG, FD, FC, FH, and FB have been joined from the
ἐστίν,καὶτὸΑΕΔτρίγωνοντῷΓΕΒτριγώνῳἴσονἔσται· random (point) F (on EF ).
ὥστεκαὶγωνίαἡὑπὸΔΑΕγωνίᾳτῇὑπὸΕΒΓἴση[ἐστίν]. For since the two (straight-lines) AE and ED are
ἔστιδὲκαὶἡὑπὸΑΕΗγωνίατῇὑπὸΒΕΘἴση. δύοδὴ equal to the two (straight-lines) CE and EB, and they
τρίγωνάἐστιτὰΑΗΕ,ΒΕΘτὰςδύογωνίαςδυσὶγωνίαις enclose equal angles [Prop. 1.15], the base AD is thus
ἴσαςἔχονταἑκατέρανἑκατέρᾳκαὶμίανπλευρὰνμιᾷπλευρᾷ equal to the base CB, and triangle AED will be equal
ἴσηντὴνπρὸςταῖςἴσαιςγωνίαιςτὴνΑΕτῇΕΒ·καὶτὰς to triangle CEB [Prop. 1.4]. Hence, the angle DAE
λοιπὰςἄραπλευρὰςταῖςλοιπαῖςπλευραῖςἴσαςἕξουσιν.ἴση [is] equal to the angle EBC. And the angle AEG (is)
ἄραἡμὲνΗΕτῇΕΘ,ἡδὲΑΗτῇΒΘ.καὶἐπεὶἴσηἐστὶνἡ also equal to the angle BEH [Prop. 1.15]. So AGE
ΑΕτῇΕΒ,κοινὴδὲκαὶπρὸςὀρθὰςἡΖΕ,βάσιςἄραἡΖΑ and BEH are two triangles having two angles equal to
βάσειτῇΖΒἐστινἴση.διὰτὰαὐτὰδὴκαὶἡΖΓτῇΖΔἐστιν two angles, respectively, and one side equal to one side—
ἴση.καὶἐπεὶἴσηἐστὶνἡΑΔτῇΓΒ,ἔστιδὲκαὶἡΖΑτῇΖΒ (namely), those by the equal angles, AE and EB. Thus,
ἴση,δύοδὴαἱΖΑ,ΑΔδυσὶταῖςΖΒ,ΒΓἴσαιεἰσὶνἑκατέρα they will also have the remaining sides equal to the re-
ἑκατέρᾳ·καὶβάσιςἡΖΔβάσειτῇΖΓἐδείχθηἴση·καὶγωνία maining sides [Prop. 1.26]. Thus, GE (is) equal to EH,
ἄραἡὑπὸΖΑΔγωνίᾳτῇὑπὸΖΒΓἴσηἐστίν.καὶἐπεὶπάλιν and AG to BH. And since AE is equal to EB, and FE
ἐδείχθηἡΑΗτῇΒΘἴση,ἀλλὰμὴνκαὶἡΖΑτῇΖΒἴση,δύο is common and at right-angles, the base FA is thus equal
δὴαἱΖΑ,ΑΗδυσὶταῖςΖΒ,ΒΘἴσαιεἰσίν.καὶγωνίαἡὑπὸ to the base FB [Prop. 1.4]. So, for the same (reasons),
ΖΑΗἐδείχθηἴσητῇὑπὸΖΒΘ·βάσιςἄραἡΖΗβάσειτῇΖΘ FC is also equal to FD. And since AD is equal to CB,
ἐστινἴση.καὶἐπεὶπάλινἴσηἐδείχθηἡΗΕτῇΕΘ,κοινὴδὲ and FA is also equal to FB, the two (straight-lines) FA
ἡΕΖ,δύοδὴαἱΗΕ,ΕΖδυσὶταῖςΘΕ,ΕΖἴσαιεἰσίν·καὶ and AD are equal to the two (straight-lines) FB and BC,
βάσιςἡΖΗβάσειτῇΖΘἴση·γωνίαἄραἡὑπὸΗΕΖγωνίᾳ respectively. And the base FD was shown (to be) equal
τῇὑπὸΘΕΖἴσηἐστίν. ὀρθὴἄραἑκατέρατῶνὑπὸΗΕΖ, to the base FC. Thus, the angle FAD is also equal to
ΘΕΖγωνιῶν. ἡΖΕἄραπρὸςτὴνΗΘτυχόντωςδιὰτοῦ the angle FBC [Prop. 1.8]. And, again, since AG was
Εἀχθεῖσανὀρθήἐστιν. ὁμοίωςδὴδείξομεν,ὅτιἡΖΕκαὶ shown (to be) equal to BH, but FA (is) also equal to
428

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
πρὸςπάσαςτὰςἁπτομέναςαὐτῆςεὐθείαςκαὶοὔσαςἐντῷ FB, the two (straight-lines) FA and AG are equal to the
ὑποκειμένῳἐπιπέδῳὀρθὰςποιήσειγωνίας.εὐθεῖαδὲπρὸς two (straight-lines) FB and BH (respectively). And the
ἐπίπεδονὀρθήἐστιν,ὅτανπρὸςπάσαςτὰςἁπτομέναςαὐτῆς angle FAG was shown (to be) equal to the angle FBH.
εὐθείαςκαὶοὔσαςἐντῷαὐτῷἐπιπέδῳὀρθὰςποιῇγωνίας· Thus, the base FG is equal to the base FH [Prop. 1.4].
ἡΖΕἄρατῷὑποκειμένῳἐπιπέδῳπρὸςὀρθάςἐστιν.τὸδὲ And, again, since GE was shown (to be) equal to EH,
ὑποκείμενονἐπίπεδόνἐστιτὸδιὰτῶνΑΒ,ΓΔεὐθειῶν. ἡ and EF (is) common, the two (straight-lines) GE and
ΖΕἄραπρὸςὀρθάςἐστιτῷδιὰτῶνΑΒ,ΓΔἐπιπέδῳ. EF are equal to the two (straight-lines) HE and EF
᾿Εὰνἄραεὐθεῖαδύοεὐθείαιςτεμνούσαιςἀλλήλαςπρὸς (respectively). And the base FG (is) equal to the base
ὀρθὰς ἐπὶτῆς κοινῆς τομῆςἐπισταθῇ, καὶ τῷ δι᾿ αὐτῶν FH. Thus, the angle GEF is equal to the angle HEF
ἐπιπέδῳπρὸςὀρθὰςἔσται·ὅπερἔδειδεῖξαι.
[Prop. 1.8]. Each of the angles GEF and HEF (are)
thus right-angles [Def. 1.10]. Thus, FE is at right-angles
to GH, which was drawn at random through E (in the
reference plane passing though AB and AC). So, similarly,
we can show that FE will make right-angles with
all straight-lines joined to it which are in the reference
plane. And a straight-line is at right-angles to a plane
when it makes right-angles with all straight-lines joined
to it which are in the plane [Def. 11.3]. Thus, FE is at
right-angles to the reference plane. And the reference
plane is that (passing) through the straight-lines AB and
CD. Thus, FE is at right-angles to the plane (passing)
through AB and CD.
Thus, if a straight-line is set up at right-angles to two
straight-lines cutting one another, at the common point
of section, then it will also be at right-angles to the plane
(passing) through them (both). (Which is) the very thing
it was required to show.
. Proposition 5
᾿Εὰνεὐθεῖατρισὶνεὐθείαιςἁπτομέναιςἀλλήλωνπρὸς If a straight-line is set up at right-angles to three
ὀρθὰςἐπὶτῆςκοινῆςτομῆςἐπισταθῇ,αἱτρεῖςεὐθεῖαιἐνἑνί straight-lines cutting one another, at the common point
εἰσινἐπιπέδῳ.
of section, then the three straight-lines are in one plane.
Α
Γ
A
C
Ζ
F
Β
∆
B
D
Ε
E
ΕὐθεῖαγάρτιςἡΑΒτρισὶνεὐθείαιςταῖςΒΓ,ΒΔ,ΒΕ For let some straight-line AB have been set up at
πρὸςὀρθὰςἐπὶτῆςκατὰτὸΒἁφῆςἐφεστάτω·λέγω,ὅτιαἱ right-angles to three straight-lines BC, BD, and BE, at
ΒΓ,ΒΔ,ΒΕἐνἑνίεἰσινἐπιπέδῳ.
the (common) point of section B. I say that BC, BD,
Μὴ γάρ, ἀλλ᾿ εἰ δυνατόν, ἔστωσαν αἱ μὲν ΒΔ, ΒΕ and BE are in one plane.
ἐντῷὑποκειμένῳἐπιπέδῳ,ἡδὲΒΓἐνμετεωροτέρῳ,καὶ For (if) not, and if possible, let BD and BE be in
ἐκβεβλήσθωτὸδὶατῶνΑΒ,ΒΓἐπίπεδον·κοινὴνδὴτομὴν the reference plane, and BC in a more elevated (plane).
429

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
ποιήσειἐντῷὑποκειμένῳἐπιπέδῳεὐθεῖαν. ποιείτωτὴν And let the plane through AB and BC have been pro-
ΒΖ.ἐν ἑνὶἄραεἰσὶνἐπιπέδῳτῷδιηγμένῳδιὰτῶν ΑΒ, duced. So it will make a straight-line as a common sec-
ΒΓαἱτρεῖςεὐθεῖαιαἱΑΒ,ΒΓ,ΒΖ.καὶἐπεὶἡΑΒὀρθή tion with the reference plane [Def. 11.3]. Let it make
ἐστι πρὸς ἑκατέραν τῶν ΒΔ, ΒΕ, καὶ τῷ διὰ τῶν ΒΔ, BF . Thus, the three straight-lines AB, BC, and BF
ΒΕἄραἐπιπέδῳὀρθήἐστινἡΑΒ.τὸδὲδιὰτῶνΒΔ,ΒΕ are in one plane—(namely), that drawn through AB and
ἐπίπεδοντὸὑποκείμενόνἐστιν·ἡΑΒἄραὀρθήἐστιπρὸςτὸ BC. And since AB is at right-angles to each of BD and
ὑποκείμενονἐπίπεδον.ὥστεκαὶπρὸςπάσαςτὰςἁπτομένας BE, AB is thus also at right-angles to the plane (passing)
αὐτῆςεὐθείαςκαὶοὔσαςἐντῷὑποκειμένῳἐπιπέδῳὀρθὰς through BD and BE [Prop. 11.4]. And the plane (passποιήσειγωνίαςἡΑΒ.ἅπτεταιδὲαὐτῆςἡΒΖοὖσαἐντῷ
ing) through BD and BE is the reference plane. Thus,
ὑποκειμένῳἐπιπέδῳ· ἡἄραὑπὸΑΒΖγωνίαὀρθήἐστιν. AB is at right-angles to the reference plane. Hence, AB
ὑπόκειταιδὲκαὶἡὑπὸΑΒΓὀρθή·ἴσηἄραἡὑπὸΑΒΖγωνία will also make right-angles with all straight-lines joined
τῇὑπὸΑΒΓ.καίεἰσινἐνἑνὶἐπιπέδῳ·ὅπερἐστὶνἀδύνατον. to it which are also in the reference plane [Def. 11.3].
οὐκἄραἡΒΓεὐθεῖαἐνμετεωροτέρῳἐστὶνἐπιπέδῳ·αἱτρεῖς And BF , which is in the reference plane, is joined to it.
ἄραεὐθεῖαιαἱΒΓ,ΒΔ,ΒΕἐνἑνίεἰσινἐπιπέδῳ. Thus, the angle ABF is a right-angle. And ABC was also
᾿Εὰνἄραεὐθεῖατρισὶνεὐθείαιςἁπτομέναιςἀλλήλωνἐπὶ assumed to be a right-angle. Thus, angle ABF (is) equal
τῆςἁφῆςπρὸςὀρθὰςἐπισταθῇ,αἱτρεῖςεὐθεῖαιἐνἑνίεἰσιν to ABC. And they are in one plane. The very thing is
ἐπιπέδῳ·ὅπερἔδειδεῖξαι.
impossible. Thus, BC is not in a more elevated plane.
Thus, the three straight-lines BC, BD, and BE are in
one plane.
Thus, if a straight-line is set up at right-angles to three
straight-lines cutting one another, at the (common) point
of section, then the three straight-lines are in one plane.
(Which is) the very thing it was required to show.
¦. Proposition 6
᾿Εὰνδύοεὐθεῖαιτῷαὐτῷἐπιπέδῳπρὸςὀρθὰςὦσιν, If two straight-lines are at right-angles to the same
παράλληλοιἔσονταιαἱεὐθεῖαι. plane then the straight-lines will be parallel. †
C
A
D
E
B
ΔύογὰρεὐθεῖαιαἱΑΒ,ΓΔτῷὑποκειμένῳἐπιπέδῳ For let the two straight-lines AB and CD be at rightπρὸςὀρθὰςἔστωσαν·λέγω,ὅτιπαράλληλόςἐστινἡΑΒτῇ
angles to a reference plane. I say that AB is parallel to
ΓΔ.
CD.
ΣυμβαλλέτωσανγὰρτῷὑποκειμένῳἐπιπέδῳκατὰτὰΒ, For let them meet the reference plane at points B and
Δσημεῖα,καὶἐπεζεύχθωἡΒΔεὐθεῖα,καὶἤχθωτῇΒΔ D (respectively). And let the straight-line BD have been
πρὸςὀρθὰςἐντῷὑποκειμένῳἐπιπέδῳἡΔΕ,καὶκείσθω joined. And let DE have been drawn at right-angles to
τῇΑΒἴσηἡΔΕ,καὶἐπεζεύχθωσαναἱΒΕ,ΑΕ,ΑΔ. BD in the reference plane. And let DE be made equal to
ΚαὶἐπεὶἡΑΒὀρθήἐστιπρὸςτὸὑποκείμενονἐπίπεδον, AB. And let BE, AE, and AD have been joined.
καὶ πρὸς πάσας [ἄρα] τὰς ἁπτομένας αὐτῆς εὐθείας καὶ And since AB is at right-angles to the reference plane,
οὔσαςἐντῷὑποκειμένῳἐπιπέδῳὀρθὰςποιήσειγωνίας. it will [thus] also make right-angles with all straight-lines
ἅπτεταιδὲτῆςΑΒἑκατέρατῶνΒΔ,ΒΕοὖσαἐντῷὑπο- joined to it which are in the reference plane [Def. 11.3].
430

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
κειμένῳἐπιπέδῳ·ὀρθὴἄραἐστὶνἑκατέρατῶνὑπὸΑΒΔ, And BD and BE, which are in the reference plane, are
ΑΒΕγωνιῶν. διὰτὰαὐτὰδὴκαὶἑκατέρατῶνὑπὸΓΔΒ, each joined to AB. Thus, each of the angles ABD and
ΓΔΕὀρθήἐστιν. καὶἐπεὶἴσηἐστὶνἡΑΒτῇΔΕ,κοινὴ ABE are right-angles. So, for the same (reasons), each
δὲἡΒΔ,δύοδὴαἱΑΒ,ΒΔδυσὶταῖςΕΔ,ΔΒἴσαιεἰσίν· of the angles CDB and CDE are also right-angles. And
καὶγωνίαςὀρθὰςπεριέχουσιν·βάσιςἄραἡΑΔβάσειτῇ since AB is equal to DE, and BD (is) common, the
ΒΕἐστινἴση. καὶἐπεὶἴσηἐστὶνἡΑΒτῇΔΕ,ἀλλὰκαὶ two (straight-lines) AB and BD are equal to the two
ἡΑΔτῇΒΕ,δύοδὴαἱΑΒ,ΒΕδυσὶταῖςΕΔ,ΔΑἴσαι (straight-lines) ED and DB (respectively). And they
εἰσίν·καὶβάσιςαὐτῶνκοινὴἡΑΕ·γωνίαἄραἡὑπὸΑΒΕ contain right-angles. Thus, the base AD is equal to the
γωνιᾴτῇὑπὸΕΔΑἐστινἴση. ὀρθὴδὲἡὑπὸΑΒΕ·ὀρθὴ base BE [Prop. 1.4]. And since AB is equal to DE, and
ἄρακαὶἡὑπὸΕΔΑ·ἡΕΔἄραπρὸςτὴνΔΑὀρθήἐστιν. AD (is) also (equal) to BE, the two (straight-lines) AB
ἔστιδὲκαὶπρὸςἑκατέραντῶνΒΔ,ΔΓὀρθή. ἡΕΔἄρα and BE are thus equal to the two (straight-lines) ED
τρισὶνεὐθείαιςταῖςΒΔ,ΔΑ,ΔΓπρὸςὀρθὰςἐπὶτῆςἁφῆς and DA (respectively). And their base AE (is) common.
ἐφέστηκεν·αἱτρεῖςἄραεὐθεῖαιαἱΒΔ,ΔΑ,ΔΓἐνἑνίεἰσιν Thus, angle ABE is equal to angle EDA [Prop. 1.8]. And
ἐπιπέδῳ.ἐνᾧδὲαἱΔΒ,ΔΑ,ἐντούτῳκαὶἡΑΒ·πᾶνγὰρ ABE (is) a right-angle. Thus, EDA (is) also a rightτρίγωνονἐνἑνίἐστινἐπιπέδῳ·αἱἄραΑΒ,ΒΔ,ΔΓεὐθεῖαι
angle. ED is thus at right-angles to DA. And it is also at
ἐνἑνίεἰσινἐπιπέδῳ.καίἐστινὀρθὴἑκατέρατῶνὑπὸΑΒΔ, right-angles to each of BD and DC. Thus, ED is stand-
ΒΔΓγωνιῶν·παράλληλοςἄραἐστὶνἡΑΒτῇΓΔ. ing at right-angles to the three straight-lines BD, DA,
᾿Εὰνἄραδύοεὐθεῖαιτῷαὐτῷἐπιπέδῳπρὸςὀρθὰςὦσιν, and DC at the (common) point of section. Thus, the
παράλληλοιἔσονταιαἱεὐθεῖαι·ὅπερἔδειδεῖξαι.
three straight-lines BD, DA, and DC are in one plane
[Prop. 11.5]. And in which(ever) plane DB and DA (are
found), in that (plane) AB (will) also (be found). For
every triangle is in one plane [Prop. 11.2]. And each of
the angles ABD and BDC is a right-angle. Thus, AB is
parallel to CD [Prop. 1.28].
Thus, if two straight-lines are at right-angles to
the same plane then the straight-lines will be parallel.
(Which is) the very thing it was required to show.
† In other words, the two straight-lines lie in the same plane, and never meet when produced in either direction.
Þ. Proposition 7
᾿Εὰνὦσιδύοεὐθεῖαιπαράλληλοι,ληφθῇδὲἐφ᾿ἑκατέρας If there are two parallel straight-lines, and random
αὐτῶν τυχόντα σημεῖα, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη points are taken on each of them, then the straight-line
εὐθεῖαἐντῷαὐτῷἐπιπέδῳἐστὶταῖςπαραλλήλοις. joining the two points is in the same plane as the parallel
(straight-lines).
À
A E B
G
C
F D
῎Εστωσαν δύο εὐθεῖαι παράλληλοι αἱ ΑΒ, ΓΔ, καὶ Let AB and CD be two parallel straight-lines, and let
εἰλήφθω ἐφ᾿ ἑκατέρας αὐτῶν τυνχόντα σημεῖα τὰ Ε, Ζ· the random points E and F have been taken on each of
λέγω,ὅτιἡἐπὶτὰΕ,Ζσημεῖαἐπιζευγνυμένηεὐθεῖαἐντῷ them (respectively). I say that the straight-line joining
αὐτῷἐπιπέδῳἐστὶταῖςπαραλλήλοις.
points E and F is in the same (reference) plane as the
Μὴγάρ,ἀλλ᾿εἰδυνατόν,ἔστωἐνμετεωροτέρῳὡςἡ parallel (straight-lines).
ΕΗΖ,καὶδιήχθωδιὰτῆςΕΗΖἐπίπεδον·τομὴνδὴποιήσει For (if) not, and if possible, let it be in a more elevated
431

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
ἐντῷὑποκειμένῳἐπιπέδῳεὐθεῖαν.ποιείτωὡςτὴνΕΖ·δύο (plane), such as EGF . And let a plane have been drawn
ἄραεὐθεῖαιαἱΕΗΖ,ΕΖχωρίονπεριέξουσιν· ὅπερἐστὶν through EGF . So it will make a straight cutting in the
ἀδύνατον. οὐκἄραἡἀπὸτοῦΕἐπὶτὸΖἐπιζευγνυμένη reference plane [Prop. 11.3]. Let it make EF . Thus, two
εὐθεῖαιἐνμετεωροτέρῳἐστὶνἐπιπέδῳ·ἐντῷδιὰτῶνΑΒ, straight-lines (with the same end-points), EGF and EF ,
ΓΒἄραπαραλλήλωνἐστὶνἐπιπέδῳἡἀπὸτοῦΕἐπὶτὸΖ will enclose an area. The very thing is impossible. Thus,
ἐπιζευγνυμένηεὐθεῖα.
the straight-line joining E to F is not in a more elevated
᾿Εὰν ἄρα ὦσι δύο εὐθεῖαι παράλληλοι, ληφθῇ δὲ ἐφ᾿ plane. The straight-line joining E to F is thus in the plane
ἑκατέραςαὐτῶντυχόντασημεῖα,ἡἐπὶτὰσημεῖαἐπιζευ- through the parallel (straight-lines) AB and CD.
γνυμένηεὐθεῖαἐντῷαὐτῷἐπιπέδῳἐστὶταῖςπαραλλήλοις· Thus, if there are two parallel straight-lines, and ran-
ὅπερἔδειδεῖξαι.
dom points are taken on each of them, then the straightline
joining the two points is in the same plane as the
parallel (straight-lines). (Which is) the very thing it was
required to show.
. Proposition 8
᾿Εὰν ὦσι δύο εὐθεῖαιπαράλληλοι, ἡ δὲ ἑτέρα αὐτῶν If two straight-lines are parallel, and one of them is at
ἐπιπέδῳτινὶπρὸςὀρθὰςᾖ,καὶἡλοιπὴτῷαὐτῷἐπιπέδῳ right-angles to some plane, then the remaining (one) will
πρὸςὀρθὰςἔσται.
also be at right-angles to the same plane.
A
C
῎ΕστωσανδύοεὐθεῖαιπαράλληλοιαἱΑΒ,ΓΔ,ἡδὲἑτέρα Let AB and CD be two parallel straight-lines, and let
αὐτῶν ἡ ΑΒ τῷ ὑποκειμένῳ ἐπιπέδῳ πρὸς ὀρθὰς ἔστω· one of them, AB, be at right-angles to a reference plane.
λέγω,ὅτικαὶἡλοιπὴἡΓΔτῷαὐτῷἐπιπέδῳπρὸςὀρθὰς I say that the remaining (one), CD, will also be at right-
ἔσται.
angles to the same plane.
ΣυμβαλλέτωσανγὰραἱΑΒ,ΓΔτῷὑποκειμένῳἐπιπέδῳ For let AB and CD meet the reference plane at points
κατὰτὰΒ,Δσημεῖα,καὶἐπεζέυχθωἡΒΔ·αἱΑΒ,ΓΔ,ΒΔ B and D (respectively). And let BD have been joined.
ἄραἐνἑνίεἰσινἐπιπέδῳ. ἤχθωτῇΒΑπρὸςὀρθὰςἐντῷ AB, CD, and BD are thus in one plane [Prop. 11.7].
ὑποκειμένῳἐπιπέδῳἡΔΕ,καὶκείσθωτῇΑΒἴσηἡΔΕ, Let DE have been drawn at right-angles to BD in the
καὶἐπεζεύχθωσαναἱΒΕ,ΑΕ,ΑΔ.
reference plane, and let DE be made equal to AB, and
ΚαὶἐπεὶἡΑΒὁρθήἐστιπρὸςτὸὑποκείμενονἐπίπεδον, let BE, AE, and AD have been joined.
καὶπρὸςπάσαςἄρατὰςἁπτομέναςαὐτῆςεὐθείαςκαὶοὔσας And since AB is at right-angles to the reference
ἐντῷὑποκειμένῳἐπιπέδῳπρὸςὀρθάςἐστινἡΑΒ·ὀρθὴἄρα plane, AB is thus also at right-angles to all of the
[ἐστὶν]ἑκατέρατῶνὑπὸΑΒΔ,ΑΒΕγωνιῶν. καὶἐπεὶεἰς straight-lines joined to it which are in the reference plane
παραλλήλουςτὰςΑΒ,ΓΔεὐθεῖαἐμπέπτωκενἡΒΔ,αἱἄρα [Def. 11.3]. Thus, the angles ABD and ABE [are] each
ὑπὸΑΒΔ,ΓΔΒγωνίαιδυσὶνὀρθαῖςἴσαιεἰσίν. ὀρθὴδὲἡ right-angles. And since the straight-line BD has met the
ὑπὸΑΒΔ·ὀρθὴἄρακαὶἡὑπὸΓΔΒ·ἡΓΔἄραπρὸςτὴνΒΔ parallel (straight-lines) AB and CD, the (sum of the)
ὀρθήἐστιν.καὶἐπεὶἴσηἐστὶνἡΑΒτῇΔΕ,κοινὴδὲἡΒΔ, angles ABD and CDB is thus equal to two right-angles
B
E
D
432

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
δύοδὴαἱΑΒ,ΒΔδυσὶταῖςΕΔ,ΔΒἴσαιεἰσίν·καὶγωνία [Prop. 1.29]. And ABD (is) a right-angle. Thus, CDB
ἡὑπὸΑΒΔγωνίᾳτῇὑπὸΕΔΒἴση·ὀρθὴγὰρἑκατέρα· (is) also a right-angle. CD is thus at right-angles to BD.
βάσιςἄραἡΑΔβάσειτῇΒΕἴση. καὶἐπεὶἴσηἐστὶνἡ And since AB is equal to DE, and BD (is) common,
μὲνΑΒτῇΔΕ,ἡδὲΒΕτῇΑΔ,δύοδὴαἱΑΒ,ΒΕδυσὶ the two (straight-lines) AB and BD are equal to the two
ταῖςΕΔ,ΔΑἴσαιεἰσὶνἑκατέραἑκατέρᾳ.καὶβάσιςαὐτῶν (straight-lines) ED and DB (respectively). And angle
κοινὴἡΑΕ·γωνίαἄραἡὑπὸΑΒΕγωνίᾳτῇὑπὸΕΔΑ ABD (is) equal to angle EDB. For each (is) a right-
ἐστινἴση.ὀρθὴδὲἡὑπὸΑΒΕ·ὀρθὴἄρακαὶἡὑπὸΕΔΑ· angle. Thus, the base AD (is) equal to the base BE
ἡΕΔἄραπρὸςτὴνΑΔὀρθήἐστιν. ἔστιδὲκαὶπρὸςτὴν [Prop. 1.4]. And since AB is equal to DE, and BE to
ΔΒὀρθή·ἡΕΔἄρακαὶτῲδιὰτῶνΒΔ,ΔΑἐπιπέδῳὀρθή AD, the two (sides) AB, BE are equal to the two (sides)
ἐστιν.καὶπρὸςπάσαςἄρατὰςἁπτομέναςαὐτῆςεὐθείαςκαὶ ED, DA, respectively. And their base AE is common.
οὔσαςἐντῷδιὰτῶνΒΔΑἐπιπέδῳὀρθὰςποιήσειγωνίαςἡ Thus, angle ABE is equal to angle EDA [Prop. 1.8].
ΕΔ.ἐνδὲτῷδιὰτῶνΒΔΑἐπιπέδῳἐστὶνἡΔΓ,ἐπειδήπερ And ABE (is) a right-angle. EDA (is) thus also a right-
ἐντῷδιὰτῶνΒΔΑἐπιπέδῳἐστὶναἱΑΒ,ΒΔ,ἐνᾧδὲ angle. Thus, ED is at right-angles to AD. And it is also
αἱΑΒ,ΒΔ,ἐντούτῳἐστὶκαὶἡΔΓ.ἡΕΔἄρατῇΔΓ at right-angles to DB. Thus, ED is also at right-angles
πρὸςὀρθάςἐστιν·ὥστεκαὶἡΓΔτῇΔΕπρὸςὀρθάςἐστιν. to the plane through BD and DA [Prop. 11.4]. And
ἔστιδὲκαὶἡΓΔτῇΒΔπρὸςὀρθάς.ἡΓΔἄραδύοεὐθείαις ED will thus make right-angles with all of the straightτεμνούσαιςἀλλήλαςταῖςΔΕ,ΔΒἀπὸτῆςκατὰτὸΔτομῆς
lines joined to it which are also in the plane through
πρὸςὀρθὰςἐφέστηκεν·ὥστεἡΓΔκαὶτῷδιὰτῶνΔΕ,ΔΒ BDA. And DC is in the plane through BDA, inas-
ἐπιπέδῳπρὸςὀρθάςἐστιν.τὸδὲδιὰτῶνΔΕ,ΔΒἐπίπεδον much as AB and BD are in the plane through BDA
τὸὑποκείμενόνἐστιν· ἡΓΔἄρατῷὑποκειμένῳἐπιπέδῳ [Prop. 11.2], and in which(ever plane) AB and BD (are
πρὸςὀρθάςἐστιν.
found), DC is also (found). Thus, ED is at right-angles
᾿Εὰνἄραὦσιδύοεὐθεῖαιπαράλληλοι,ἡδὲμίααὐτῶν to DC. Hence, CD is also at right-angles to DE. And
ἐπιπέδῳτινὶπρὸςὀρθὰςᾖ,καὶἡλοιπὴτῷαὐτῷἐπιπέδῳ CD is also at right-angles to BD. Thus, CD is standing
πρὸςὀρθὰςἔσται·ὅπερἔδειδεῖξαι.
at right-angles to two straight-lines, DE and DB, which
meet one another, at the (point) of section, D. Hence,
CD is also at right-angles to the plane through DE and
DB [Prop. 11.4]. And the plane through DE and DB is
the reference (plane). CD is thus at right-angles to the
reference plane.
Thus, if two straight-lines are parallel, and one of
them is at right-angles to some plane, then the remaining
(one) will also be at right-angles to the same plane.
(Which is) the very thing it was required to show.
. Proposition 9
Αἱτῇαὐτῇεὐθείᾳπαράλληλοικαὶμὴοὖσαιαὐτῇἐντῷ
Â
(Straight-lines) parallel to the same straight-line, and
αὐτῷἐπιπέδῳκαὶἀλλήλαιςεἰσὶπαράλληλοι.
which are not in the same plane as it, are also parallel to
one another.
B
H
A
F G
E
à D
C
K
῎Εστω γὰρἑκατέρατῶνΑΒ,ΓΔτῇΕΖπαράλληλος For let AB and CD each be parallel to EF , not being
μὴοὖσαιαὐτῇἐντῷαὐτῷἐπιπέδῳ·λέγω,ὅτιπαράλληλός in the same plane as it. I say that AB is parallel to CD.
À
433

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
ἐστινἡΑΒτῇΓΔ.
For let some point G have been taken at random on
ΕἰλήφθωγὰρἐπὶτῆςΕΖτυχὸνσημεῖοντὸΗ,καὶἀπ᾿ EF . And from it let GH have been drawn at right-angles
αὐτοῦτῇΕΖἐνμὲντῷδιὰτῶνΕΖ,ΑΒἐπιπέδῳπρὸςὀρθὰς to EF in the plane through EF and AB. And let GK
ἤχθωἡΗΘ,ἐνδὲτῷδιὰτῶνΖΕ,ΓΔτῇΕΖπάλινπρὸς have been drawn, again at right-angles to EF , in the
ὀρθὰςἤχθωἡΗΚ.
plane through FE and CD.
ΚαὶἐπεὶἡΕΖπρὸςἑκατέραντῶνΗΘ,ΗΚὀρθήἐστιν, And since EF is at right-angles to each of GH and
ἡΕΖἄρακαὶτῷδιὰτῶνΗΘ,ΗΚἐπιπέδῳπρὸςὀρθάς GK, EF is thus also at right-angles to the plane through
ἐστιν. καίἐστινἡΕΖτῇΑΒπαράλληλος·καὶἡΑΒἄρα GH and GK [Prop. 11.4]. And EF is parallel to AB.
τῷδιὰτῶνΘΗΚἐπιπέδῳπρὸςὀρθάςἐστιν. διὰτὰαὐτὰ Thus, AB is also at right-angles to the plane through
δὴκαὶἡΓΔτῷδιὰτῶνΘΗΚἐπιπέδῳπρὸςὀρθάςἐστιν· HGK [Prop. 11.8]. So, for the same (reasons), CD is
ἑκατέραἄρατῶνΑΒ,ΓΔτῷδιὰτῶνΘΗΚἐπιπέδῳπρὸς also at right-angles to the plane through HGK. Thus,
ὀρθάςἐστιν. ἐὰνδὲδύοεὐθεῖαιτῷαὐτῷἐπιπέδῳπρὸς AB and CD are each at right-angles to the plane through
ὀρθὰςὦσιν,παράλληλοίεἰσιναἱεὐθεῖαι·παράλληλοςἄρα HGK. And if two straight-lines are at right–angles
ἐστὶνἡΑΒτῇΓΔ·ὅπερἔδειδεῖξαι.
to the same plane then the straight-lines are parallel
[Prop. 11.6]. Thus, AB is parallel to CD. (Which is)
the very thing it was required to show.
. Proposition 10
᾿Εὰνδύοεὐθεῖαιἁπτόμεναιἀλλήλωνπαρὰδύοεὐθείας If two straight-lines joined to one another are (respec-
ἁπτομένας ἀλλήλων ὦσι μὴ ἐν τῷ αὐτῷ ἐπιπέδῳ, ἵσας tively) parallel to two straight-lines joined to one another,
γωνίαςπεριέξουσιν.
(but are) not in the same plane, then they will contain
equal angles.
B
A
E F
D
ΔύογὰρεὐθεῖαιαἱΑΒ,ΒΓἁπτόμεναιἀλλήλωνπαρὰ For let the two straight-lines joined to one another,
δύοεὐθείαςτὰςΔΕ,ΕΖἁπτομέναςἀλλήλωνἔστωσανμὴ AB and BC, be (respectively) parallel to the two
ἐντῷαὐτῷἐπιπέδῳ·λέγω,ὅτιἴσηἐστὶνἡὑπὸΑΒΓγωνία straight-lines joined to one another, DE and EF , (but)
τῇὑπὸΔΕΖ.
not in the same plane. I say that angle ABC is equal to
ἈπειλήφθωσανγὰραἱΒΑ,ΒΓ,ΕΔ,ΕΖἴσαιἀλλήλαις, (angle) DEF .
καὶἐπεζεύχθωσαναἱΑΔ,ΓΖ,ΒΕ,ΑΓ,ΔΖ.
For let BA, BC, ED, and EF have been cut off (so
ΚαὶἐπεὶἡΒΑτῇΕΔἴσηἐστὶκαὶπαράλληλος,καὶἡ as to be, respectively) equal to one another. And let AD,
ΑΔἄρατῇΒΕἴσηἐστὶκαὶπαράλληλος. διὰτὰαὐτὰδὴ CF , BE, AC, and DF have been joined.
καὶἡΓΖτῇΒΕἴσηἐστὶκαὶπαράλληλος·ἑκατέραἄρατῶν And since BA is equal and parallel to ED, AD is thus
ΑΔ,ΓΖτῇΒΕἴσηἐστὶκαὶπαράλληλος. αἱδὲτῇαὐτῇ also equal and parallel to BE [Prop. 1.33]. So, for the
εὐθείᾳπαράλληλοικαὶμὴοὖσαιαὐτῇἐντῷαὐτῷἐπιπέδῳ same reasons, CF is also equal and parallel to BE. Thus,
καὶἀλλήλαιςεἰσὶπαράλληλοι·παράλληλοςἄραἐστὶνἡΑΔ AD and CF are each equal and parallel to BE. And
τῇΓΖκαὶἴση. καὶἐπιζευγνύουσιναὐτὰςαἱΑΓ,ΔΖ·καὶ straight-lines parallel to the same straight-line, and which
ἡΑΓἄρατῇΔΖἴσηἐστὶκαὶπαράλληλος.καὶἐπεὶδύοαἱ are not in the same plane as it, are also parallel to one an-
ΑΒ,ΒΓδυσὶταῖςΔΕ,ΕΖἴσαιεἰσίν,καὶβάσιςἡΑΓβάσει other [Prop. 11.9]. Thus, AD is parallel and equal to CF .
τῇΔΖἴση,γωνίαἄραἡὑπὸΑΒΓγωνίᾳτῇὑπὸΔΕΖἐστιν And AC and DF join them. Thus, AC is also equal and
C
434

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
ἴση.
parallel to DF [Prop. 1.33]. And since the two (straight-
᾿Εὰν ἄρα δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων παρὰ δύο lines) AB and BC are equal to the two (straight-lines)
εὐθείαςἁπτομέναςἀλλήλωνὦσιμὴἐντῷαὐτῷἐπιπέδῳ, DE and EF (respectvely), and the base AC (is) equal to
ἵσαςγωνίαςπεριέξουσιν·ὅπερἔδειδεῖξαι.
the base DF , the angle ABC is thus equal to the (angle)
DEF [Prop. 1.8].
Thus, if two straight-lines joined to one another are
(respectively) parallel to two straight-lines joined to one
another, (but are) not in the same plane, then they will
contain equal angles. (Which is) the very thing it was
required to show.
. Proposition 11
Ἀπὸ τοῦ δοθέντος σημείου μετεώρου ἐπὶ τὸ δοθὲν To draw a perpendicular straight-line from a given
ἐπίπεδονκάθετονεὐθεῖανγραμμὴνἀγαγεῖν.
raised point to a given plane.
A
À
Â
῎ΕστωτὸμὲνδοθὲνσημεῖονμετέωροντὸΑ,τὸδὲδοθὲν Let A be the given raised point, and the given plane
ἐπίπεδοντὸὑποκείμενον·δεῖδὴἀπὸτοῦΑσημείουἐπὶτὸ the reference (plane). So, it is required to draw a perpen-
ὑποκείμενονἐπίπεδονκάθετονεὐθεῖανγραμμὴνἀγαγεῖν. dicular straight-line from point A to the reference plane.
Διήχθωγάρτιςἐντῷὑποκειμένῳἐπιπέδῳεὐθεῖα,ὡς Let some random straight-line BC have been drawn
ἔτυχεν,ἡΒΓ,καὶἤχθωἀπὸτοῦΑσημείουἐπὶτὴνΒΓ across in the reference plane, and let the (straight-line)
κάθετοςἡΑΔ.εἰμὲνοὖνἡΑΔκάθετόςἐστικαὶἐπὶτὸ AD have been drawn from point A perpendicular to BC
ὑποκείμενονἐπίπεδον,γεγονὸςἂνεἴητὸἐπιταχθέν. εἰδὲ [Prop. 1.12]. If, therefore, AD is also perpendicular to
οὔ, ἤχθωἀπὸτοῦΔσημείουτῇΒΓἐντῷὑποκειμένῳ the reference plane then that which was prescribed will
ἐπιπέδῳπρὸςὀρθὰςἡΔΕ,καὶἤχθωἀπὸτοῦΑἐπὶτὴνΔΕ have occurred. And, if not, let DE have been drawn in
κάθετοςἡΑΖ,καὶδιὰτοῦΖσημείουτῇΒΓπαράλληλος the reference plane from point D at right-angles to BC
ἤχθωἡΗΘ.
[Prop. 1.11], and let the (straight-line) AF have been
ΚαὶἐπεὶἡΒΓἑκατέρᾳτῶνΔΑ,ΔΕπρὸςὀρθάςἐστιν, drawn from A perpendicular to DE [Prop. 1.12], and let
ἡΒΓἄρακαὶτῷδιὰτῶνΕΔΑἐπιπέδῳπρὸςὀρθάςἐστιν. GH have been drawn through point F , parallel to BC
καίἐστιναὐτῇπαράλληλοςἡΗΘ·ἐὰνδὲὦσιδύοεὐθεῖαι [Prop. 1.31].
παράλληλοι,ἡδὲμίααὐτῶνἐπιπέδῳτινὶπρὸςὀρθὰςᾖ,καὶἡ And since BC is at right-angles to each of DA and
λοιπὴτῷαὐτῷἐπιπέδῳπρὸςὀρθὰςἔσται·καὶἡΗΘἄρατῷ DE, BC is thus also at right-angles to the plane through
διὰτῶνΕΔ,ΔΑἐπιπέδῳπρὸςὀρθάςἐστιν.καὶπρὸςπάσας EDA [Prop. 11.4]. And GH is parallel to it. And if two
ἄρατὰςἁπτομέναςαὐτῆςεὐθείαςκαὶοὔσαςἐντῷδιὰτῶν straight-lines are parallel, and one of them is at right-
ΕΔ,ΔΑἐπιπέδῳὀρθήἐστινἡΗΘ.ἅπτεταιδὲαὐτῆςἡΑΖ angles to some plane, then the remaining (straight-line)
οὖσαἐντῷδιὰτῶνΕΔ,ΔΑἐπιπέδῳ·ἡΗΘἄραὀρθήἐστι will also be at right-angles to the same plane [Prop. 11.8].
πρὸςτὴνΖΑ·ὥστεκαὶἡΖΑὀρθήἐστιπρὸςτὴνΘΗ.ἔστι Thus, GH is also at right-angles to the plane through
G
E
B
F
D
H
C
435

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
δὲἡΑΖκαὶπρὸςτὴνΔΕὀρθή·ἡΑΖἄραπρὸςἑκατέραν ED and DA. And GH is thus at right-angles to all of
τῶν ΗΘ,ΔΕὀρθήἐστιν. ἐὰνδὲεὐθεῖαδυσὶνεὐθείαις the straight-lines joined to it which are also in the plane
τεμνούσαιςἀλλήλαςἐπὶτῆςτομῆςπρὸςὀρθὰςἐπισταθῇ, through ED and AD [Def. 11.3]. And AF , which is in the
καὶτῷδι᾿αὐτῶνἐπιπέδῳπρὸςὀρθὰςἔσται·ἡΖΑἄρατῷ plane through ED and DA, is joined to it. Thus, GH is at
διὰτῶν ΕΔ, ΗΘἐπιπέδῳπρὸςὀρθάςἐστιν. τὸδὲ διὰ right-angles to FA. Hence, FA is also at right-angles to
τῶνΕΔ,ΗΘἐπίπεδόνἐστιτὸὑποκείμενον·ἡΑΖἄρατῷ HG. And AF is also at right-angles to DE. Thus, AF is
ὑποκειμένῳἐπιπέδῳπρὸςὀρθάςἐστιν.
at right-angles to each of GH and DE. And if a straight-
ἈπὸτοῦἄραδοθέντοςσημείουμετεώρουτοῦΑἐπὶτὸ line is set up at right-angles to two straight-lines cutting
ὑποκείμενονἐπίπεδονκάθετοςεὐθεῖαγραμμὴἦκταιἡΑΖ· one another, at the point of section, then it will also be
ὅπερἔδειποιῆσαι. at right-angles to the plane through them [Prop. 11.4].
Thus, FA is at right-angles to the plane through ED and
GH. And the plane through ED and GH is the reference
(plane). Thus, AF is at right-angles to the reference
plane.
Thus, the straight-line AF has been drawn from the
given raised point A perpendicular to the reference plane.
(Which is) the very thing it was required to do.
.
Proposition 12
Τῷ δοθέντι ἐπιπέδῳ ἀπὸ τοῦ πρὸς αὐτῷ δοθέντος To set up a straight-line at right-angles to a given
σημείουπρὸςὀρθὰςεὐθεῖανγραμμὴνἀναστῆσαι.
plane from a given point in it.
B
D
A
C
῎Εστωτὸμὲνδοθὲνἐπίπεδοντὸὑποκείμενον, τὸδὲ Let the given plane be the reference (plane), and A a
πρὸςαὐτῷσημεῖοντὸΑ·δεῖδὴἀπὸτοῦΑσημείουτῷὑπο- point in it. So, it is required to set up a straight-line at
κειμένῳἐπιπέδῳπρὸςὀρθὰςεὐθεῖανγραμμὴνἀναστῆσαι. right-angles to the reference plane at point A.
ΝενοήσθωτισημεῖονμετέωροντὸΒ,καὶἀπὸτοῦΒἐπὶ Let some raised point B have been assumed, and let
τὸὑποκείμενονἐπίπεδονκάθετοςἤχθωἡΒΓ,καὶδιὰτοῦ the perpendicular (straight-line) BC have been drawn
ΑσημείουτῇΒΓπαράλληλοςἤχθωἡΑΔ. from B to the reference plane [Prop. 11.11]. And
᾿ΕπεὶοὖνδύοεὐθεῖαιπαράλληλοίεἰσιναἱΑΔ,ΓΒ,ἡδὲ let AD have been drawn from point A parallel to BC
μίααὐτῶνἡΒΓτῷὑποκειμένῳἐπιπέδῳπρὸςὀρθάςἐστιν, [Prop. 1.31].
καὶἡλοιπὴἄραἡΑΔτῷὑποκειμένῳἐπιπέδῳπρὸςὀρθὰς Therefore, since AD and CB are two parallel straight-
ἐστιν.
lines, and one of them, BC, is at right-angles to the refer-
Τῷἄραδοθέντιἐπιπέδῳἀπὸτοῦπρὸςαὐτῷσημείου ence plane, the remaining (one) AD is thus also at rightτοῦΑπρὸςὀρθὰςἀνέσταταιἡΑΔ·ὅπερἔδειποιῆσαι.
angles to the reference plane [Prop. 11.8].
436

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
Thus, AD has been set up at right-angles to the given
plane, from the point in it, A. (Which is) the very thing it
was required to do.
. Proposition 13
Ἀπὸτοῦαὐτοῦσημείουτῷαὐτῷἐπιπέδῳδύοεὐθεῖαι
Two (different) straight-lines cannot be set up at the
πρὸςὀρθὰςοὐκἀναστήσονταιἐπὶτὰαὐτὰμέρη. same point at right-angles to the same plane, on the same
side.
B
C
D
A
E
Εἰ γὰρ δυνατόν, ἀπὸ τοῦ αὐτοῦ σημείου τοῦ Α τῷ For, if possible, let the two straight-lines AB and AC
ὑποκειμένῳἐπιπέδῳδύοεὐθεῖαιαἱΑΒ,ΒΓπρὸςὀρθὰς have been set up at the same point A at right-angles
ἀνεστάτωσαν ἐπὶ τὰ αὐτὰ μέρη, καὶ διήχθω τὸ διὰ τῶν to the reference plane, on the same side. And let the
ΒΑ,ΑΓἐπὶπεδον·τομὴνδὴποιήσειδιὰτοῦΑἐντῷὑπο- plane through BA and AC have been drawn. So it will
κειμένῳἐπιπέδῳεὐθεῖαν. ποιείτωτὴνΔΑΕ·αἱἄραΑΒ, make a straight cutting (passing) through (point) A in
ΑΓ,ΔΑΕεὐθεῖαιἐνἑνιεἰσινἐπιπέδῳ. καὶἐπεὶἡΓΑτῷ the reference plane [Prop. 11.3]. Let it have made DAE.
ὑποκειμένῳἐπιπέδῳπρὸςὀρθάςἐστιν,καὶπρὸςπάσαςἄρα Thus, AB, AC, and DAE are straight-lines in one plane.
τὰςἁπτομέναςαὐτῆςεὐθείαςκαὶοὔσαςἐντῷὑποκειμένῳ And since CA is at right-angles to the reference plane, it
ἐπιπέδῳὀρθὰςποιήσειγωνίας. ἅπτεταιδὲαὐτῆςἡΔΑΕ will thus also make right-angles with all of the straightοὖσαἐντῷὑποκειμένῳἐπιπέδῳ·
ἡἄραὑπὸΓΑΕγωνία lines joined to it which are also in the reference plane
ὀρθήἐστιν.διὰτὰαὐτὰδὴκαὶἡὑπὸΒΑΕὀρθήἐστιν·ἴση [Def. 11.3]. And DAE, which is in the reference plane, is
ἄραἡὑπὸΓΑΕτῇὑπὸΒΑΕκαίεἰσινἐνἑνὶἐπιπέδῳ·ὅπερ joined to it. Thus, angle CAE is a right-angle. So, for the
ἐστὶνἀδύνατον.
same (reasons), BAE is also a right-angle. Thus, CAE
Οὐκἄραἀπὸτοῦαὐτοῦσημείουτῷαὐτῷἐπιπέδῳδύο (is) equal to BAE. And they are in one plane. The very
εὐθεῖαιπρὸςὀρθὰςἀνασταθήσονταιἐπὶτὰαὐτὰμέρη·ὅπερ thing is impossible.
ἔδειδεῖξαι.
Thus, two (different) straight-lines cannot be set up
at the same point at right-angles to the same plane, on
the same side. (Which is) the very thing it was required
to show.
. Proposition 14
Πρὸςἃἐπίπεδαἡαὐτὴεὐθεῖαὀρθήἐστιν,παράλληλα Planes to which the same straight-line is at right-
ἔσταιτὰἐπίπεδα.
angles will be parallel planes.
Εὐθεῖα γάρ τις ἡ ΑΒ πρὸς ἑκάτερον τῶν ΓΔ, ΕΖ For let some straight-line AB be at right-angles to
ἐπιπέδωνπρὸςὀρθὰςἔστω·λέγω,ὅτιπαράλληλάἐστιτὰ each of the planes CD and EF . I say that the planes
ἐπίπεδα.
are parallel.
437

ËÌÇÁÉÁÏƺ ELEMENTS
À
Ã
Â
Εἰ γὰρ μή, ἐκβαλλόμενα συμπεσοῦνται. συμπιπτέτ- For, if not, being produced, they will meet. Let them
ωσαν·ποιήσουσιδὴκοινὴντομὴνεὐθεῖαν.ποιείτωσαντὴν have met. So they will make a straight-line as a common
D
E
A
B
C
F
BOOK 11
ΗΘ, καὶεἰλήφθω ἐπὶτῆς ΗΘτυχὸν σημεῖοντὸ Κ, καὶ section [Prop. 11.3]. Let them have made GH. And let
ἐπεζεύχθωσαναἱΑΚ,ΒΚ.
some random point K have been taken on GH. And let
ΚαὶἐπεὶἡΑΒὀρθήἐστιπρὸςτὸΕΖἐπίπεδον,καὶπρὸς AK and BK have been joined.
τὴνΒΚἄραεὐθεῖανοὖσανἐντῷΕΖἐκβληθέντιἐπιπέδῳ And since AB is at right-angles to the plane EF , AB
ὀρθήἐστινἡΑΒ·ἡἄραὑπὸΑΒΚγωνίαὀρθήἐστιν. διὰ is thus also at right-angles to BK, which is a straight-line
τὰαὐτὰδὴκαὶἡὑπὸΒΑΚὀρθήἐστιν. τριγώνουδὴτοῦ in the produced plane EF [Def. 11.3]. Thus, angle ABK
ΑΒΚαἱδύογωνίαιαἱὑπὸΑΒΚ,ΒΑΚδυσὶνὀρθαῖςεἰσιν is a right-angle. So, for the same (reasons), BAK is also
ἴσαι· ὅπερἐστὶνἀδύνατον. οὐκἄρατὰΓΔ,ΕΖἐπίπεδα a right-angle. So the (sum of the) two angles ABK and
ἐκβαλλόμενασυμπεσοῦνται· παράλληλαἄραἐστὶτὰΓΔ, BAK in the triangle ABK is equal to two right-angles.
ΕΖἐπίπεδα.
The very thing is impossible [Prop. 1.17]. Thus, planes
Πρὸςἃἐπίπεδαἄραἡαὐτὴεὐθεῖαὀρθήἐστιν,παράλληλά CD and EF , being produced, will not meet. Planes CD
ἐστιτὰἐπίπεδα·ὅπερἔδειδεῖξαι. and EF are thus parallel [Def. 11.8].
Thus, planes to which the same straight-line is at
right-angles are parallel planes. (Which is) the very thing
it was required to show.
. Proposition 15
᾿Εὰνδύοεὐθεῖαιἁπτόμεναιἀλλήλωνπαρὰδύοεὐθείας If two straight-lines joined to one another are parallel
ἁπτομένας ἀλλήλων ὦσι μὴ ἐν τῷ αὐτῷ ἐπιπέδῳ οὖσαι, (respectively) to two straight-lines joined to one another,
παράλληλάἐστιτὰδι᾿αὐτῶνἐπίπεδα.
which are not in the same plane, then the planes through
ΔύογὰρεὐθεῖαιἁπτόμεναιἀλλήλωναἱΑΒ,ΒΓπαρὰ them are parallel (to one another).
δύοεὐθείαςἁπτομέναςἀλλήλωντὰςΔΕ,ΕΖἔστωσανμὴ For let the two straight-lines joined to one another,
ἐντῷαὐτῷἐπιπέδῳοὖσαι·λέγω,ὅτιἐκβαλλόμενατὰδιὰ AB and BC, be parallel to the two straight-lines joined to
τῶνΑΒ,ΒΓ,ΔΕ,ΕΖἐπίπεδαοὐσυμπεσεῖταιἀλλήλοις. one another, DE and EF (respectively), not being in the
῎ΗχθωγὰρἀπὸτοῦΒσημείουἐπὶτὸδιὰτῶνΔΕ,ΕΖ same plane. I say that the planes through AB, BC and
ἐπίπεδονκάθετοςἡΒΗκαὶσυμβαλλέτωτῷἐπιπέδῳκατὰ DE, EF will not meet one another (when) produced.
τὸΗσημεῖον,καὶδιὰτοῦΗτῇμὲνΕΔπαράλληλοςἤχθω For let BG have been drawn from point B perpendic-
ἡΗΘ,τῇδὲΕΖἡΗΚ.
ular to the plane through DE and EF [Prop. 11.11], and
let it meet the plane at point G. And let GH have been
drawn through G parallel to ED, and GK (parallel) to
EF [Prop. 1.31].
H
K
G
438

ËÌÇÁÉÁÏƺ ELEMENTS
A
B
BOOK 11
C
Â
À Ã
ΚαὶἐπεὶἡΒΗ ὀρθή ἐστι πρὸςτὸ διὰτῶν ΔΕ, ΕΖ And since BG is at right-angles to the plane through
ἐπίπεδον,καὶπρὸςπάσαςἄρατὰςἁπτομέναςαὐτῆςεὐθείας DE and EF , it will thus also make right-angles with all
καὶοὔσαςἐντῷδιὰτῶνΔΕ,ΕΖἐπιπέδῳὀρθὰςποιήσει of the straight-lines joined to it, which are also in the
γωνίας. ἅπτεταιδὲαὐτῆςἑκατέρατῶνΗΘ,ΗΚοὖσαἐν plane through DE and EF [Def. 11.3]. And each of
τῷδιὰτῶνΔΕ,ΕΖἐπιπέδῳ·ὀρθὴἄραἐστὶνἑκατέρατῶν GH and GK, which are in the plane through DE and
ὑπὸΒΗΘ,ΒΗΚγωνιῶν.καὶἐπεὶπαράλληλόςἐστινἡΒΑ EF , are joined to it. Thus, each of the angles BGH and
τῇΗΘ,αἱἄραὑπὸΗΒΑ,ΒΗΘγωνίαιδυσὶνὀρθαῖςἴσαι BGK are right-angles. And since BA is parallel to GH
εἰσίν.ὀρθὴδὲἡὑπὸΒΗΘ·ὀρθὴἄρακαὶἡὑπὸΗΒΑ·ἡΗΒ [Prop. 11.9], the (sum of the) angles GBA and BGH is
ἄρατῇΒΑπρὸςὀρθάςἐστιν. διὰτὰαὐτὰδὴἡΗΒκαὶτῇ equal to two right-angles [Prop. 1.29]. And BGH (is)
ΒΓἐστιπρὸςὀρθάς.ἐπεὶοὖνεὐθεῖαἡΗΒδυσὶνεὐθείαις a right-angle. GBA (is) thus also a right-angle. Thus,
ταῖςΒΑ,ΒΓτεμνούσαιςἀλλήλαςπρὸςὀρθὰςἐφέστηκεν,ἡ GB is at right-angles to BA. So, for the same (reasons),
ΗΒἄρακαὶτῷδιὰτῶνΒΑ,ΒΓἐπιπέδῳπρὸςὀρθάςἐστιν. GB is also at right-angles to BC. Therefore, since the
[διὰτὰαὐτὰδὴἡΒΗκαὶτῷδιὰτῶνΗΘ,ΗΚἐπιπέδῳ straight-line GB has been set up at right-angles to two
πρὸςὀρθάςἐστιν.τὸδὲδιὰτῶνΗΘ,ΗΚἐπίπεδόνἐστιτὸ straight-lines, BA and BC, cutting one another, GB is
διὰτῶνΔΕ,ΕΖ·ἡΒΗἄρατῷδιὰτῶνΔΕ,ΕΖἐπιπέδῳ thus at right-angles to the plane through BA and BC
ἐστὶπρὸςὀρθάς.ἐδείχθηδὲἡΗΒκαὶτῷδιὰτῶνΑΒ,ΒΓ [Prop. 11.4]. [So, for the same (reasons), BG is also
ἐπιπέδῳπρὸςὀρθάς].πρὸςἃδὲἐπίπεδαἡαὐτὴεὐθεῖαὀρθή at right-angles to the plane through GH and GK. And
ἐστιν,παράλληλάἐστιτὰἐπίπεδα·παράλληλονἄραἐστὶτὸ the plane through GH and GK is the (plane) through
διὰτῶνΑΒ,ΒΓἐπίπεδοντῷδιὰτῶνΔΕ,ΕΖ. DE and EF . And it was also shown that GB is at right-
᾿Εὰν ἄρα δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων παρὰ δύο angles to the plane through AB and BC.] And planes
εὐθείαςἁπτομέναςἀλλήλωνὦσιμὴἐντῷαὐτῷἐπιπέδῳ, to which the same straight-line is at right-angles are parπαράλληλάἐστιτὰδι᾿αὐτῶνἐπίπεδα·ὅπερἔδειδεῖξαι.
allel planes [Prop. 11.14]. Thus, the plane through AB
and BC is parallel to the (plane) through DE and EF .
Thus, if two straight-lines joined to one another are
parallel (respectively) to two straight-lines joined to one
another, which are not in the same plane, then the planes
through them are parallel (to one another). (Which is)
the very thing it was required to show.
¦. Proposition 16
᾿Εὰνδύοἐπίπεδαπαράλληλαὑπὸἐπιπέδουτινὸςτέμνηται, If two parallel planes are cut by some plane then their
αἱκοιναὶαὐτῶντομαὶπαράλληλοίεἰσιν.
common sections are parallel.
ΔύογὰρἐπίπεδαπαράλληλατὰΑΒ,ΓΔὑπὸἐπιπέδου For let the two parallel planes AB and CD have been
τοῦΕΖΗΘτεμνέσθω,κοιναὶδὲαὐτῶντομαὶἔστωσαναἱ cut by the plane EFGH. And let EF and GH be their
ΕΖ,ΗΘ·λέγω,ὅτιπαράλληλόςἐστινἡΕΖτῇΗΘ. common sections. I say that EF is parallel to GH.
D
H
E
G
F
K
439

ËÌÇÁÉÁÏƺ ELEMENTS
À
Â
Ã
A
E
C
G
B
F
H
D
BOOK 11
Εἰγὰρμή,ἐκβαλλόμεναιαἱΕΖ,ΗΘἤτοιἐπὶτὰΖ,Θ For, if not, being produced, EF and GH will meet eiμέρηἢἐπὶτὰΕ,Ησυμπεσοῦνται.
ἐκβεβλήσθωσανὡςἐπὶ ther in the direction of F , H, or of E, G. Let them be
τὰΖ,ΘμέρηκαὶσυμπιπτέτωσανπρότερονκατὰτὸΚ.καὶ produced, as in the direction of F , H, and let them, first
ἐπεὶἡΕΖΚἐντῷΑΒἐστινἐπιπέδῳ,καὶπάνταἄρατὰἐπὶ of all, have met at K. And since EFK is in the plane
τῆςΕΖΚσημεῖαἐντῷΑΒἐστινἐπιπέδῳ.ἓνδὲτῶνἐπὶτῆς AB, all of the points on EFK are thus also in the plane
ΕΖΚεὐθείαςσημείωνἐστὶτὸΚ·τὸΚἄραἐντῷΑΒἐστιν AB [Prop. 11.1]. And K is one of the points on EFK.
ἐπιπέδῳ.διὰτὰαὐτὰδὴτὸΚκαὶἐντῷΓΔἐστινἐπιπέδῳ· Thus, K is in the plane AB. So, for the same (reasons),
τὰΑΒ,ΓΔἄραἐπίπεδαἐκβαλλόμενασυμπεσοῦνται. οὐ K is also in the plane CD. Thus, the planes AB and CD,
συμπίπτουσι δὲ διὰ τὸ παράλληλα ὑποκεῖσθαι· οὐκ ἄρα being produced, will meet. But they do not meet, on acαἱΕΖ,ΗΘεὐθεῖαιἐκβαλλόμεναιἐπὶτὰΖ,Θμέρησυμ-
count of being (initially) assumed (to be mutually) paralπεσοῦνται.
ὁμοίως δὴ δείξομεν, ὅτι αἱΕΖ, ΗΘ εὐθεῖαι lel. Thus, the straight-lines EF and GH, being produced
οὐδέ ἐπὶ τὰ Ε, Η μέρη ἐκβαλλόμεναισυμπεσοῦνται. αἱ in the direction of F , H, will not meet. So, similarly, we
δὲἐπὶμηδέτερατὰμέρησυμπίπτουσαιπαράλληλοίεἰσιν. can show that the straight-lines EF and GH, being proπαράλληλοςἄραἐστὶνἡΕΖτῇΗΘ.
duced in the direction of E, G, will not meet either. And
᾿Εὰν ἄρα δύο ἐπίπεδαπαράλληλαὑπὸἐπιπέδουτινὸς (straight-lines in one plane which), being produced, do
τέμνηται,αἱκοιναὶαὐτῶντομαὶπαράλληλοίεἰσιν·ὅπερἔδει not meet in either direction are parallel [Def. 1.23]. EF
δεῖξαι.
is thus parallel to GH.
Thus, if two parallel planes are cut by some plane then
their common sections are parallel. (Which is) the very
thing it was required to show.
Þ. Proposition 17
᾿Εὰνδύοεὐθεῖαιὑπὸπαραλλήλωνἐπιπεδωντέμνωνται, If two straight-lines are cut by parallel planes then
εἰςτοὺςαὐτοὺςλόγουςτμηθήσονται.
they will be cut in the same ratios.
ΔύογὰρεὐθεῖαιαἱΑΒ,ΓΔὑπὸπαραλλήλωνἐπιπέδων For let the two straight-lines AB and CD be cut by the
τῶνΗΘ,ΚΛ,ΜΝτεμνέσθωσανκατὰτὰΑ,Ε,Β,Γ,Ζ, parallel planes GH, KL, and MN at the points A, E, B,
Δσημεῖα·λέγω,ὅτιἐστὶνὡςἡΑΕεὐθεῖαπρὸςτὴνΕΒ, and C, F , D (respectively). I say that as the straight-line
οὕτωςἡΓΖπρὸςτὴνΖΔ.
AE is to EB, so CF (is) to FD.
᾿ΕπεζεύχθωσανγὰραἱΑΓ,ΒΔ,ΑΔ,καὶσυμβαλλέτωἡ For let AC, BD, and AD have been joined, and let
ΑΔτῷΚΛἐπιπέδῳκατὰτὸΞσημεῖον,καὶἐπεζεύχθωσαν AD meet the plane KL at point O, and let EO and OF
αἱΕΞ,ΞΖ.
have been joined.
Καὶ ἐπεὶ δύο ἐπίπεδα παράλληλα τὰ ΚΛ, ΜΝ ὑπὸ And since two parallel planes KL and MN are cut
ἐπιπέδου τοῦ ΕΒΔΞ τέμνεται, αἱ κοιναὶ αὐτῶν τομαὶ αἱ by the plane EBDO, their common sections EO and BD
ΕΞ,ΒΔπαράλληλοίεἰσιν.διὰτὰαὐτὰδὴἐπεὶδύοἐπίπεδα are parallel [Prop. 11.16]. So, for the same (reasons),
παράλληλατὰΗΘ,ΚΛὑπὸἐπιπέδουτοῦΑΞΖΓτέμνεται, since two parallel planes GH and KL are cut by the
αἱκοιναὶαὐτῶντομαὶαἱΑΓ,ΞΖπαράλληλοίεἰσιν.καὶἐπεὶ plane AOFC, their common sections AC and OF are
τριγώνουτοῦΑΒΔπαρὰμίαντῶνπλευρῶντὴνΒΔεὐθεῖα parallel [Prop. 11.16]. And since the straight-line EO
ἦκταιἡΕΞ,ἀνάλογονἄραἐστὶνὡςἡΑΕπρὸςΕΒ,οὕτως has been drawn parallel to one of the sides BD of trian-
K
440

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
ἡΑΞπρὸςΞΔ.πάλινἐπεὶτριγώνουτοῦΑΔΓπαρὰμίαν gle ABD, thus, proportionally, as AE is to EB, so AO
τῶνπλευρῶντὴνΑΓεὐθεῖαἦκταιἡΞΖ,ἀνάλογόνἐστιν (is) to OD [Prop. 6.2]. Again, since the straight-line OF
ὡςἡΑΞπρὸςΞΔ,οὕτωςἡΓΖπρὸςΖΔ.ἐδείχθηδὲκαὶ has been drawn parallel to one of the sides AC of trian-
ὡςἡΑΞπρὸςΞΔ,οὕτωςἡΑΕπρὸςΕΒ·καὶὡςἄραἡΑΕ gle ADC, proportionally, as AO is to OD, so CF (is) to
πρὸςΕΒ,οὕτωςἡΓΖπρὸςΖΔ.
Â
FD [Prop. 6.2]. And it was also shown that as AO (is)
to OD, so AE (is) to EB. And thus as AE (is) to EB, so
CF (is) to FD [Prop. 5.11].
À
H
A
Ä
C
Ã
G
L
O
E
Æ
F
K
N
Å
B
D
M
᾿Εὰνἄραδύοεὐθεῖαιὑπὸπαραλλήλωνἐπιπέδωντέμνων- Thus, if two straight-lines are cut by parallel planes
ται,εἰςτοὺςαὐτοὺςλόγουςτμηθήσονται·ὅπερἔδειδειξαι. then they will be cut in the same ratios. (Which is) the
very thing it was required to show.
. Proposition 18
᾿Εὰνεὐθεῖαἐπιπέδῳτινὶπρὸςὀρθὰςᾖ,καὶπάντατὰδι᾿ If a straight-line is at right-angles to some plane then
αὐτῆςἐπίπεδατῷαὐτῷἐπιπέδῳπρὸςὀρθὰςἔσται. all of the planes (passing) through it will also be at right-
ΕὐθεῖαγάρτιςἡΑΒτῷὑποκειμένῳἐπιπέδῳπρὸςὀρθὰς angles to the same plane.
ἔστω·λέγω,ὅτικαὶπάντατὰδιὰτῆςΑΒἐπίπεδατῷὑπο- For let some straight-line AB be at right-angles to
κειμένῳἐπιπέδῳπρὸςὀρθάςἐστιν. a reference plane. I say that all of the planes (pass-
᾿ΕκβεβλήσθωγὰρδιὰτῆςΑΒἐπίπεδοντὸΔΕ,καὶἔστω ing) through AB are also at right-angles to the reference
κοινὴτομὴτοῦΔΕἐπιπέδουκαὶτοῦὑποκειμένουἡΓΕ,καὶ plane.
εἰλήφθωἐπὶτῆςΓΕτυχὸνσημεῖοντὸΖ,καὶἀπὸτοῦΖτῇ For let the plane DE have been produced through
ΓΕπρὸςὀρθὰςἤχθωἐντῷΔΕἐπιπέδῳἡΖΗ. AB. And let CE be the common section of the plane
Καὶ ἐπεὶ ἡ ΑΒ πρὸς τὸ ὑποκείμενον ἐπίπεδον ὀρθή DE and the reference (plane). And let some random
ἐστιν,καὶπρὸςπάσαςἄρατὰςἁπτομέναςαὐτῆςεὐθείαςκαὶ point F have been taken on CE. And let FG have been
οὔσαςἐντῷὑποκειμένῳἐπιπέδῳὀρθήἐστινἡΑΒ·ὥστε drawn from F , at right-angles to CE, in the plane DE
καὶπρὸςτὴνΓΕὀρθήἐστιν·ἡἄραὑπὸΑΒΖγωνίαὀρθή [Prop. 1.11].
ἐστιν.ἔστιδὲκαὶἡὑπὸΗΖΒὀρθὴ·παράλληλοςἄραἐστὶν And since AB is at right-angles to the reference plane,
ἡΑΒτῇΖΗ.ἡδὲΑΒτῷὑποκειμένῳἐπιπέδῳπρὸςὀρθάς AB is thus also at right-angles to all of the straight-
ἐστιν· καὶἡΖΗἄρατῷὑποκειμένῳἐπιπέδῳπρὸςὀρθάς lines joined to it which are also in the reference plane
ἐστιν. καὶἐπίπεδονπρὸςἐπίπεδονὀρθόνἐστιν,ὅταναἱτῇ [Def. 11.3]. Hence, it is also at right-angles to CE. Thus,
κοινῇτομῇτῶνἐπιπέδωνπρὸςὀρθὰςἀγόμεναιεὐθεῖαιἐν angle ABF is a right-angle. And GFB is also a right-
ἑνὶτῶνἐπιπέδωντῷλοιπῷἐπιπέδῳπρὸςὀρθὰςὦσιν. καὶ angle. Thus, AB is parallel to FG [Prop. 1.28]. And AB
τῇκοινῇτομῇτῶνἐπιπέδωντῇΓΕἐνἑνὶτῶνἐπιπέδων is at right-angles to the reference plane. Thus, FG is also
441

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
τῷΔΕπρὸςὀρθὰςἀχθεῖσαἡΖΗἐδείχθητῷὑποκειμένῳ at right-angles to the reference plane [Prop. 11.8]. And
ἐπιπέδῳπρὸςὀρθάς·τὸἄραΔΕἐπίπεδονὀρθόνἐστιπρὸς a plane is at right-angles to a(nother) plane when the
τὸὑποκείμενον. ὁμοίωςδὴδειχθήσεταικαὶπάντατὰδιὰ straight-lines drawn at right-angles to the common secτῆς
ΑΒ ἐπίπεδα ὀρθὰ τυγχανοντα πρὸς τὸ ὑποκείμενον tion of the planes, (and lying) in one of the planes, are
ἐπίπεδον.
at right-angles to the remaining plane [Def. 11.4]. And
FG, (which was) drawn at right-angles to the common
section of the planes, CE, in one of the planes, DE, was
shown to be at right-angles to the reference plane. Thus,
À
plane DE is at right-angles to the reference (plane). So,
similarly, it can be shown that all of the planes (passing)
at random through AB (are) at right-angles to the reference
plane.
D G A
C
F
B
E
᾿Εὰνἄραεὐθεῖαἐπιπέδῳτινὶπρὸςὀρθὰςᾖ,καὶπάντατὰ Thus, if a straight-line is at right-angles to some plane
δι᾿αὐτῆςἐπίπεδατῷαὐτῷἐπιπέδῳπρὸςὀρθὰςἔσται·ὅπερ then all of the planes (passing) through it will also be at
ἔδειδεῖξαι.
right-angles to the same plane. (Which is) the very thing
it was required to show.
. Proposition 19
᾿Εὰνδύοἐπίπεδατέμνονταἄλληλαἐπιπέδῳτινὶπρὸς If two planes cutting one another are at right-angles
ὀρθὰςᾖ,καὶἡκοινὴαὐτῶντομὴτῷαὐτῷἐπιπέδῳπρὸς to some plane then their common section will also be at
ὀρθὰςἔσται.
right-angles to the same plane.
B
E F
D
A C
ΔύογὰρἐπίπεδατὰΑΒ,ΒΓτῷὑποκειμένῳἐπιπέδῳ For let the two planes AB and BC be at right-angles
πρὸςὀρθὰςἔστω,κοινὴδὲαὐτῶντομὴἔστωἡΒΔ·λέγω, to a reference plane, and let their common section be
ὅτιἡΒΔτῷὑποκειμένῳἐπιπέδῳπρὸςὀρθάςἐστιν. BD. I say that BD is at right-angles to the reference
442

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
Μὴγάρ,καὶἤχθωσανἀπὸτοῦΔσημείουἐνμὲντῷ plane.
ΑΒἐπιπέδῳτῇΑΔεὐθείᾳπρὸςὀρθὰςἡΔΕ,ἐνδὲτῷΒΓ For (if) not, let DE also have been drawn from point
ἐπιπέδῳτῇΓΔπρὸςὀρθὰςἡΔΖ.
D, in the plane AB, at right-angles to the straight-line
ΚαὶἐπεὶτὸΑΒἐπίπεδονὀρθόνἐστιπρὸςτὸὑποκείμενον, AD, and DF , in the plane BC, at right-angles to CD.
καὶ τῇ κοινῇ αὐτῶν τομῇ τῇ ΑΔ πρὸς ὀρθὰς ἐν τῷ And since the plane AB is at right-angles to the refer-
ΑΒἐπιπέδῳἦκταιἡΔΕ,ἡΔΕἄραὀρθήἐστιπρὸςτὸ ence (plane), and DE has been drawn at right-angles to
ὑποκείμενονἐπίπεδον. ὁμοίωςδὴδείξομεν,ὅτικαὶἡΔΖ their common section AD, in the plane AB, DE is thus at
ὀρθήἐστιπρὸςτὸὑποκείμενονἐπίπεδον. ἀπὸτοῦαὐτοῦ right-angles to the reference plane [Def. 11.4]. So, simi-
ἄρα σημείου τοῦ Δ τῷ ὑποκειμένῳ ἐπιπέδῳ δύο εὐθεῖα larly, we can show that DF is also at right-angles to the
πρὸςὀρθὰςἀνεσταμέναιεἰσὶνἐπὶτὰαὐτὰμέρη·ὅπερἐστὶν reference plane. Thus, two (different) straight-lines are
ἀδύνατον. οὐκἄρα τῷ ὑποκειμένῳἐπιπέδῳ ἀπὸτοῦ Δ set up, at the same point D, at right-angles to the referσημείουἀνασταθήσεταιπρὸςὀρθὰςπλὴντῆςΔΒκοινῆς
ence plane, on the same side. The very thing is impossible
τομῆςτῶνΑΒ,ΒΓἐπιπέδων.
[Prop. 11.13]. Thus, no (other straight-line) except the
᾿Εὰνἄραδύοἐπίπεδατέμνονταἄλληλαἐπιπέδῳτινὶπρὸς common section DB of the planes AB and BC can be set
ὀρθὰςᾖ,καὶἡκοινὴαὐτῶντομὴτῷαὐτῷἐπιπέδῳπρὸς up at point D, at right-angles to the reference plane.
ὀρθὰςἔσται·ὅπερἔδειδεῖξαι.
Thus, if two planes cutting one another are at rightangles
to some plane then their common section will also
be at right-angles to the same plane. (Which is) the very
thing it was required to show.
. Proposition 20
᾿Εὰνστερεὰγωνίαὑπὸτριῶνγωνιῶνἐπιπέδωνπεριέχη-
If a solid angle is contained by three plane angles then
ται,δύοὁποιαιοῦντῆςλοιπῆςμείζονέςεἰσιπάντῃμεταλαμ- (the sum of) any two (angles) is greater than the remainβανόμεναι.
ing (one), (the angles) being taken up in any (possible
way).
D
A
B
E C
Στερεὰ γὰρ γωνία ἡ πρὸς τῷ Α ὑπὸ τριῶν γωνιῶν For let the solid angle A have been contained by the
ἐπιπέδωντῶνὑπὸΒΑΓ,ΓΑΔ,ΔΑΒπεριεχέσθω· λέγω, three plane angles BAC, CAD, and DAB. I say that (the
ὅτιτῶνὑπὸΒΑΓ,ΓΑΔ,ΔΑΒγωνιῶνδύοὁποιαιοῦντῆς sum of) any two of the angles BAC, CAD, and DAB
λοιπῆςμείζονέςεἰσιπάντῃμεταλαμβανόμεναι.
is greater than the remaining (one), (the angles) being
ΕἰμὲνοὖναἱὑπὸΒΑΓ,ΓΑΔ,ΔΑΒγωνίαιἴσαιἀλλήλαις taken up in any (possible way).
εἰσίν,φανερόν,ὅτιδύοὁποιαιοῦντῆςλοιπῆςμείζονέςεἰσιν. For if the angles BAC, CAD, and DAB are equal to
εἰδὲοὔ,ἔστωμείζωνἡὑπὸΒΑΓ,καὶσυνεστάτωπρὸςτῇ one another then (it is) clear that (the sum of) any two
ΑΒεὐθείᾳκαὶτῷπρὸςαὐτῇσημείῳτῷΑτῇὑπὸΔΑΒ is greater than the remaining (one). But, if not, let BAC
γωνίᾳἐντῷδιὰτῶνΒΑΓἐπιπέδῳἴσηἡὑπὸΒΑΕ,καὶ be greater (than CAD or DAB). And let (angle) BAE,
κείσθωτῇΑΔἴσηἡΑΕ,καὶδιὰτοῦΕσημείουδιαχθεῖσα equal to the angle DAB, have been constructed in the
ἡΒΕΓτεμνέτωτὰςΑΒ,ΑΓεὐθείαςκατὰτὰΒ,Γσημεῖα, plane through BAC, on the straight-line AB, at the point
καὶἐπεζεύχθωσαναἱΔΒ,ΔΓ.
A on it. And let AE be made equal to AD. And BEC be-
ΚαὶἐπεὶἴσηἐστὶνἡΔΑτῇΑΕ,κοινὴδὲἡΑΒ,δύο ing drawn across through point E, let it cut the straightδυσὶνἴσαι·καὶγωνίαἡὑπὸΔΑΒγωνίᾳτῇὑπὸΒΑΕἴση·
lines AB and AC at points B and C (respectively). And
βάσιςἄραἡΔΒβάσειτῇΒΕἐστινἴση.καὶἐπεὶδύοαἱΒΔ, let DB and DC have been joined.
ΔΓτῆςΒΓμείζονέςεἰσιν,ὧνἡΔΒτῇΒΕἐδείχθηἴση, And since DA is equal to AE, and AB (is) common,
443

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
λοιπὴἄραἡΔΓλοιπῆςτῆςΕΓμείζωνἐστίν. καὶἐπεὶἴση the two (straight-lines AD and AB are) equal to the
ἐστὶνἡΔΑτῇΑΕ,κοινὴδὲἡΑΓ,καὶβάσιςἡΔΓβάσεως two (straight-lines EA and AB, respectively). And anτῆςΕΓμείζωνἐστίν,γωνίαἄραἡὑπὸΔΑΓγωνάιςτῆςὑπὸ
gle DAB (is) equal to angle BAE. Thus, the base DB
ΕΑΓμείζωνἐστίν.ἐδείχθηδὲκαὶἡὑπὸΔΑΒτῇὑπὸΒΑΕ is equal to the base BE [Prop. 1.4]. And since the (sum
ἴση·αἱἄραὑπὸΔΑΒ,ΔΑΓτῆςὑπὸΒΑΓμείζονέςεἰσιν. of the) two (straight-lines) BD and DC is greater than
ὁμοίωςδὴδείξομεν,ὅτικαὶαἱλοιπαὶσύνδυολαμβανόμεναι BC [Prop. 1.20], of which DB was shown (to be) equal
τῆςλοιπῆςμείζονέςεἰσιν.
to BE, the remainder DC is thus greater than the re-
᾿Εὰν ἄρα στερεὰ γωνία ὑπὸ τριῶν γωνιῶν ἐπιπέδων mainder EC. And since DA is equal to AE, but AC
περιέχηται,δύοὁποιαιοῦντῆςλοιπῆςμείζονέςεἰσιπάντῃ (is) common, and the base DC is greater than the base
μεταλαμβανόμεναι·ὅπερἔδειδεῖξαι.
EC, the angle DAC is thus greater than the angle EAC
[Prop. 1.25]. And DAB was also shown (to be) equal to
BAE. Thus, (the sum of) DAB and DAC is greater than
BAC. So, similarly, we can also show that the remaining
(angles), being taken in pairs, are greater than the
remaining (one).
Thus, if a solid angle is contained by three plane angles
then (the sum of) any two (angles) is greater than
the remaining (one), (the angles) being taken up in any
(possible way). (Which is) the very thing it was required
to show.
. Proposition 21
῞Απασα στερεὰ γωνία ὑπὸ ἐλασσόνων [ἢ] τεσσάρων Any solid angle is contained by plane angles (whose
ὀρθῶνγωνιῶνἐπιπέδωνπεριέχεται. sum is) less [than] four right-angles. †
C
A
D
B
῎Εστω στερεὰ γωνία ἡ πρὸς τῷ Α περιεχομένη ὑπὸ Let the solid angle A be contained by the plane angles
ἐπιπέδωνγωνιῶντῶνὑπὸΒΑΓ,ΓΑΔ,ΔΑΒ·λέγω,ὅτιαἱ BAC, CAD, and DAB. I say that (the sum of) BAC,
ὑπὸΒΑΓ,ΓΑΔ,ΔΑΒτεσσάρωνὀρθῶνἐλάσσονέςεἰσιν. CAD, and DAB is less than four right-angles.
Εἰλήφθωγὰρἐφ᾿ἑκάστηςτῶνΑΒ,ΑΓ,ΑΔτυχόντα For let the random points B, C, and D have been
σημεῖατὰΒ,Γ,Δ,καὶἐπεζεύχθωσαναἱΒΓ,ΓΔ,ΔΒ.καὶ taken on each of (the straight-lines) AB, AC, and AD
ἐπεὶστερεὰγωνίαἡπρὸςτῷΒὑπὸτριῶνγωνιῶνἐπιπέδων (respectively). And let BC, CD, and DB have been
περιέχεταιτῶνὑπὸΓΒΑ,ΑΒΔ,ΓΒΔ,δύοὁποιαιοῦντῆς joined. And since the solid angle at B is contained
λοιπῆςμείζονέςεἰσιν·αἱἄραὑπὸΓΒΑ,ΑΒΔτῆςὑπὸΓΒΔ by the three plane angles CBA, ABD, and CBD, (the
μείζονέςεἰσιν. διὰτὰαὐτὰδὴκαὶαἱμὲνὑπὸΒΓΑ,ΑΓΔ sum of) any two is greater than the remaining (one)
τῆςὑπὸΒΓΔμείζονέςεἰσιν,αἱδὲὑπὸΓΔΑ,ΑΔΒτῆςὑπὸ [Prop. 11.20]. Thus, (the sum of) CBA and ABD is
ΓΔΒμείζονέςεἰσιν·αἱἓξἄραγωνίαιαἱὑπὸΓΒΑ,ΑΒΔ, greater than CBD. So, for the same (reasons), (the sum
ΒΓΑ,ΑΓΔ,ΓΔΑ,ΑΔΒτριῶντῶνὑπὸΓΒΔ,ΒΓΑ,ΓΔΒ of) BCA and ACD is also greater than BCD, and (the
μείζονέςεἰσιν.ἀλλὰαἱτρεῖςαἱὑπὸΓΒΔ,ΒΔΓ,ΒΓΔδυσὶν sum of) CDA and ADB is greater than CDB. Thus,
ὀρθαῖςἴσαιεἰσίν·αἱἓξἄρααἱὑπὸΓΒΑ,ΑΒΔ,ΒΓΑ,ΑΓΔ, the (sum of the) six angles CBA, ABD, BCA, ACD,
ΓΔΑ,ΑΔΒδύοὀρθῶνμείζονέςεἰσιν.καὶἐπεὶἑκάστουτῶν CDA, and ADB is greater than the (sum of the) three
ΑΒΓ,ΑΓΔ,ΑΔΒτριγώνωναἱτρεῖςγωνίαιδυσὶνὀρθαῖς (angles) CBD, BCD, and CDB. But, the (sum of the)
ἴσαιεἰσίν,αἱἄρατῶντριῶντριγώνωνἐννέαγωνίαιαἱὑπὸ three (angles) CBD, BDC, and BCD is equal to two
444

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
ΓΒΑ,ΑΓΒ,ΒΑΓ,ΑΓΔ,ΓΔΑ,ΓΑΔ,ΑΔΒ,ΔΒΑ,ΒΑΔἓξ right-angles [Prop. 1.32]. Thus, the (sum of the) six an-
ὀρθαῖςἴσαιεἰσίν,ὧναἱὑπὸΑΒΓ,ΒΓΑ,ΑΓΔ,ΓΔΑ,ΑΔΒ, gles CBA, ABD, BCA, ACD, CDA, and ADB is greater
ΔΒΑἓξγωνίαιδύοὀρθῶνεἰσιμείζονες·λοιπαὶἄρααἱὑπὸ than two right-angles. And since the (sum of the) three
ΒΑΓ,ΓΑΔ,ΔΑΒτρεῖς[γωνίαι]περιέχουσαιτὴνστερεὰν angles of each of the triangles ABC, ACD, and ADB
γωνίαντεσσάρωνὀρθῶνἐλάσσονέςεἰσιν.
is equal to two right-angles, the (sum of the) nine angles
῞Απασαἄραστερεὰγωνίαὑπὸἐλασσόνων[ἤ]τεσσάρων CBA, ACB, BAC, ACD, CDA, CAD, ADB, DBA, and
ὀρθῶνγωνιῶνἐπιπέδωνπεριέχεται·ὅπερἔδειδεῖξαι. BAD of the three triangles is equal to six right-angles, of
which the (sum of the) six angles ABC, BCA, ACD,
CDA, ADB, and DBA is greater than two right-angles.
Thus, the (sum of the) remaining three [angles] BAC,
CAD, and DAB, containing the solid angle, is less than
four right-angles.
Thus, any solid angle is contained by plane angles
(whose sum is) less [than] four right-angles. (Which is)
the very thing it was required to show.
† This proposition is only proved for the case of a solid angle contained by three plane angles. However, the generalization to a solid angle
contained by more than three plane angles is straightforward.
. Proposition 22
᾿Εὰνὧσιτρεῖςγωνίαιἐπίπεδοι,ὧναἱδύοτῆςλοιπῆς If there are three plane angles, of which (the sum of
μείζονέςεἰσιπάντῃμεταλαμβανόμεναι,περιέχωσιδὲαὐτὰς any) two is greater than the remaining (one), (the an-
ἴσαιεὐθεῖαι,δυνατόνἐστινἐκτῶνἐπιζευγνυουσῶντὰςἴσας gles) being taken up in any (possible way), and if equal
εὐθείαςτρίγωνονσυστήσασθαι.
straight-lines contain them, then it is possible to construct
a triangle from (the straight-lines created by) joining the
(ends of the) equal straight-lines.
À
Â
Ã
Ä
῎Εστωσαν τρεῖς γωνίαι ἐπίπεδοι αἱ ὑπὸ ΑΒΓ, ΔΕΖ, Let ABC, DEF , and GHK be three plane angles, of
ΗΘΚ, ὧν αἱ δύο τῆς λοιπῆς μείζονές εἰσι πάντῃ μετα- which the sum of any) two is greater than the remainλαμβανόμεναι,αἱμὲνὑπὸΑΒΓ,ΔΕΖτῆςὑπὸΗΘΚ,αἱ
ing (one), (the angles) being taken up in any (possible
δὲὑπὸΔΕΖ,ΗΘΚτῆςὑπὸΑΒΓ,καὶἔτιαἱὑπὸΗΘΚ, way)—(that is), ABC and DEF (greater) than GHK,
ΑΒΓ τῆςὑπὸΔΕΖ, καὶἔστωσανἴσαιαἱΑΒ, ΒΓ,ΔΕ, DEF and GHK (greater) than ABC, and, further, GHK
ΕΖ,ΗΘ,ΘΚεὐθεῖαι,καὶἐπεζεύχθωσαναἱΑΓ,ΔΖ,ΗΚ· and ABC (greater) than DEF . And let AB, BC, DE,
λέγω,ὅτιδυνατόνἐστινἐκτῶνἴσωνταῖςΑΓ,ΔΖ,ΗΚ EF , GH, and HK be equal straight-lines. And let AC,
τρίγωνονσυστήσασθαι, τουτέστινὅτιτῶνΑΓ,ΔΖ,ΗΚ DF , and GK have been joined. I say that that it is possiδύοὁποιαιοῦντῆςλοιπῆςμείζονέςεἰσιν.
ble to construct a triangle out of (straight-lines) equal to
Εἰ μὲν οὖν αἱ ὑπὸ ΑΒΓ, ΔΕΖ, ΗΘΚ γωνίαι ἴσαι AC, DF , and GK—that is to say, that (the sum of) any
ἀλλήλαιςεἰσίν, φανερόν, ὅτικαὶτῶνΑΓ,ΔΖ,ΗΚἴσων two of AC, DF , and GK is greater than the remaining
γινομένωνδυνατόνἐστινἐκτῶνἴσωνταῖςΑΓ,ΔΖ,ΗΚ (one).
τρίγωνονσυστήσασθαι. εἰδὲοὔ,ἔστωσανἄνισοι,καὶσυ- Now, if the angles ABC, DEF , and GHK are equal
νεστάτωπρὸςτῇΘΚεὐθείᾳκαὶτῷπρὸςαὐτῇσημείῳτῷ to one another then (it is) clear that, (with) AC, DF ,
ΘτῇὑπὸΑΒΓγωνίᾳἴσηἡὑπὸΚΘΛ·καὶκείσθωμιᾷτῶν and GK also becoming equal, it is possible to construct a
ΑΒ,ΒΓ,ΔΕ,ΕΖ,ΗΘ,ΘΚἴσηἡΘΛ,καὶἐπεζεύχθωσαν triangle from (straight-lines) equal to AC, DF , and GK.
αἱΚΛ,ΗΛ.καὶἐπεὶδύοαἱΑΒ,ΒΓδυσὶταῖςΚΘ,ΘΛἴσαι And if not, let them be unequal, and let KHL, equal to
εἰσίν,καὶγωνίαἡπρὸςτῷΒγωνίᾳτῇὑπὸΚΘΛἴση,βάσις angle ABC, have been constructed on the straight-line
ἄραἡΑΓβάσειτῇΚΛἴση.καὶἐπεὶαἱὑπὸΑΒΓ,ΗΘΚτῆς HK, at the point H on it. And let HL be made equal to
A
B
C
D
E
F
G
H
K
L
445

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
ὑπὸΔΕΖμείζονέςεἰσιν,ἴσηδὲἡὑπὸΑΒΓτῇὑπὸΚΘΛ, one of AB, BC, DE, EF , GH, and HK. And let KL
ἡἄραὑπὸΗΘΛτῆςὑπὸΔΕΖμείζωνἐστίν. καὶἐπεὶδύο and GL have been joined. And since the two (straightαἱΗΘ,ΘΛδύοταῖςΔΕ,ΕΖἴσαιεἰσίν,καὶγωνίαἡὑπὸ
lines) AB and BC are equal to the two (straight-lines)
ΗΘΛγωνίαςτῆςὑπὸΔΕΖμείζων,βάσιςἄραἡΗΛβάσεως KH and HL (respectively), and the angle at B (is) equal
τῆςΔΖμείζωνἐστίν. ἀλλὰαἱΗΚ,ΚΛτῆςΗΛμείζονές to KHL, the base AC is thus equal to the base KL
εἰσιν.πολλῷἄρααἱΗΚ,ΚΛτῆςΔΖμείζονέςεἰσιν.ἴσηδὲ [Prop. 1.4]. And since (the sum of) ABC and GHK
ἡΚΛτῇΑΓ·αἱΑΓ,ΗΚἄρατῆςλοιπῆςτῆςΔΖμείζονές is greater than DEF , and ABC equal to KHL, GHL
εἰσιν. ὁμοίωςδὴδείξομεν,ὅτικαὶαἱμὲνΑΓ,ΔΖτῆςΗΚ is thus greater than DEF . And since the two (straightμείζονέςεἰσιν,καὶἔτιαἱΔΖ,ΗΚτῆςΑΓμείζονέςεἰσιν.
lines) GH and HL are equal to the two (straight-lines)
δυνατὸνἄραἐστὶνἐκτῶνἴσωνταῖςΑΓ,ΔΖ,ΗΚτρίγωνον DE and EF (respectively), and angle GHL (is) greater
συστήσασθαι·ὅπερἔδειδεῖξαι.
than DEF , the base GL is thus greater than the base DF
[Prop. 1.24]. But, (the sum of) GK and KL is greater
than GL [Prop. 1.20]. Thus, (the sum of) GK and KL is
much greater than DF . And KL (is) equal to AC. Thus,
(the sum of) AC and GK is greater than the remaining
(straight-line) DF . So, similarly, we can show that (the
sum of) AC and DF is greater than GK, and, further,
that (the sum of) DF and GK is greater than AC. Thus,
it is possible to construct a triangle from (straight-lines)
equal to AC, DF , and GK. (Which is) the very thing it
was required to show.
. Proposition 23
᾿Εκ τριῶν γωνιῶν ἐπιπέδων, ὧν αἱ δύο τῆς λοιπῆς To construct a solid angle from three (given) plane
μείζονές εἰσι πάντῃ μεταλαμβανόμεναι, στερεὰν γωνίαν angles, (the sum of) two of which is greater than the reσυστήσασθαι·
δεῖ δὴ τὰς τρεῖς τεσσάρων ὀρθῶν ἐλάσς- maining (one, the angles) being taken up in any (possible
οναςεἶναι.
way). So, it is necessary for the (sum of the) three (angles)
to be less than four right-angles [Prop. 11.21].
Â
H
B
E
À Ã
G
῎Εστωσαν αἱ δοθεῖσαι τρεῖς γωνίαι ἐπίπεδοι αἱ ὑπὸ Let ABC, DEF , and GHK be the three given plane
ΑΒΓ,ΔΕΖ,ΗΘΚ,ὧναἱδύοτῆςλοιπῆςμείζονεςἔστωσαν angles, of which let (the sum of) two be greater than the
πάντῃμεταλαμβανόμεναι,ἔτιδὲαἱτρεῖςτεσσάρωνὀρθῶν remaining (one, the angles) being taken up in any (pos-
A
C
D
F
K
ἐλάσσονες·δεῖδὴἐκτῶνἴσωνταῖςὑπὸΑΒΓ,ΔΕΖ,ΗΘΚ sible way), and, further, (let) the (sum of the) three (be)
στερεὰνγωνίανσυστήσασθαι.
less than four right-angles. So, it is necessary to construct
ἈπειλήφθωσανἴσαιαἱΑΒ,ΒΓ,ΔΕ,ΕΖ,ΗΘ,ΘΚ,καὶ a solid angle from (plane angles) equal to ABC, DEF ,
ἐπεζεύχθωσαναἱΑΓ,ΔΖ,ΗΚ·δυνατὸνἄραἐστὶνἐκτῶν and GHK.
ἴσωνταῖςΑΓ,ΔΖ,ΗΚτρίγωνονσυστήσασθαι.συνεστάτω Let AB, BC, DE, EF , GH, and HK be cut off (so
τὸΛΜΝ,ὥστεἴσηνεἶναιτὴνμὲνΑΓτῇΛΜ,τὴνδὲΔΖ as to be) equal (to one another). And let AC, DF , and
τῇΜΝ,καὶἔτιτὴνΗΚτῇΝΛ,καὶπεριγεγράφθωπερὶ GK have been joined. It is, thus, possible to construct a
τὸΛΜΝτρίγωνονκύκλοςὁΛΜΝ,καὶεἰλήφθωαὐτοῦτὸ triangle from (straight-lines) equal to AC, DF , and GK
κέντρονκαὶἔστωτὸΞ,καὶἐπεζεύχθωσαναἱΛΞ,ΜΞ,ΝΞ· [Prop. 11.22]. Let (such a triangle), LMN, have be constructed,
such that AC is equal to LM, DF to MN, and,
further, GK to NL. And let the circle LMN have been
circumscribed about triangle LMN [Prop. 4.5]. And let
446

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
Ê
Å
È
its center have been found, and let it be (at) O. And let
LO, MO, and NO have been joined.
R
M
Q
Ä Ç
Æ
L
P
O
N
Λέγω,ὅτιἡΑΒμείζωνἐστὶτῆςΛΞ.εἰγὰρμή,ἤτοι I say that AB is greater than LO. For, if not, AB is
ἴσηἐστὶνἡΑΒτῇΛΞἢἐλάττων.ἔστωπρότερονἴση.καὶ either equal to, or less than, LO. Let it, first of all, be
ἐπεὶἴσηἐστὶνἡΑΒτῇΛΞ,ἀλλὰἡμὲνΑΒτῇΒΓἐστιν equal. And since AB is equal to LO, but AB is equal to
ἴση,ἡδὲΞΛτῇΞΜ,δύοδὴαἱΑΒ,ΒΓδύοταῖςΛΞ,ΞΜ BC, and OL to OM, so the two (straight-lines) AB and
ἴσαιεἰσὶνἑκατέραἑκατέρᾳ·καὶβάσιςἡΑΓβάσειτῇΛΜ BC are equal to the two (straight-lines) LO and OM, re-
ὑπόκειταιἴση·γωνίαἄραἡὑπὸΑΒΓγωνίᾳτῇὑπὸΛΞΜ spectively. And the base AC was assumed (to be) equal
ἐστινἴση.διὰτὰαὐτὰδὴκαὶἡμὲνὑπὸΔΕΖτῇὑπὸΜΞΝ to the base LM. Thus, angle ABC is equal to angle
ἐστινἴση,καὶἔτιἡὑπὸΗΘΚτῇὑπὸΝΞΛ·αἱἄρατρεῖςαἱ LOM [Prop. 1.8]. So, for the same (reasons), DEF is
ὑπὸΑΒΓ,ΔΕΖ,ΗΘΚγωνίαιτρισὶταῖςὑπὸΛΞΜ,ΜΞΝ, also equal to MON, and, further, GHK to NOL. Thus,
ΝΞΛεἰσινἴσαι. ἀλλὰαἱτρεῖςαἱὑπὸΛΞΜ,ΜΞΝ,ΝΞΛ the three angles ABC, DEF , and GHK are equal to the
τέτταρσινὀρθαῖςεἰσινἴσαι·καὶαἱτρεῖςἄρααἱὑπὸΑΒΓ, three angles LOM, MON, and NOL, respectively. But,
ΔΕΖ,ΗΘΚτέτταρσινὀρθαῖςἴσαιεἰσίν. ὑπόκεινταιδὲκαὶ the (sum of the) three angles LOM, MON, and NOL is
τεσσάρωνὀρθῶνἐλάσσονες·ὅπερἄτοπον. οὐκἄραἡΑΒ equal to four right-angles. Thus, the (sum of the) three
τῇΛΞἴσηἐστίν.λέγωδή,ὅτιοὑδὲἐλάττωνἐστὶνἡΑΒτῆς angles ABC, DEF , and GHK is also equal to four right-
ΛΞ.εἰγὰρδυνατόν,ἔστω·καὶκείσθωτῇμὲνΑΒἴσηἡΞΟ, angles. And it was also assumed (to be) less than four
τῇδὲΒΓἴσηἡΞΠ,καὶἐπεζεύχθωἡΟΠ.καὶἐπεὶἴσηἐστὶν right-angles. The very thing (is) absurd. Thus, AB is
ἡΑΒτῇΒΓ,ἴσηἐστὶκαὶἡΞΟτῇΞΠ·ὥστεκαὶλοιπὴἡ not equal to LO. So, I say that AB is not less than LO
ΛΟτῇΠΜἐστινἴση.παράλληλοςἄραἐστὶνἡΛΜτῇΟΠ, either. For, if possible, let it be (less). And let OP be
καὶἰσογώνιοντὸΛΜΞτῷΟΠΞ·ἔστινἄραὡςἡΞΛπρὸς made equal to AB, and OQ equal to BC, and let PQ
ΛΜ,οὕτωςἡΞΟπρὸςΟΠ·ἐναλλὰξὡςἡΛΞπρὸςΞΟ, have been joined. And since AB is equal to BC, OP
οὕτωςἡΛΜπρὸςΟΠ.μείζωνδὲἡΛΞτῆςΞΟ·μείζωνἄρα is also equal to OQ. Hence, the remainder LP is also
καὶἡΛΜτῆςΟΠ.ἀλλὰἡΛΜκεῖταιτῇΑΓἴση·καὶἡΑΓ equal to (the remainder) QM. LM is thus parallel to PQ
ἄρατῆςΟΠμείζωνἐστίν.ἐπεὶοὖνδύοαἱΑΒ,ΒΓδυσὶταῖς [Prop. 6.2], and (triangle) LMO (is) equiangular with
ΟΞ,ΞΠἴσαιεἰσίν,καὶβάσιςἡΑΓβάσεωςτῆςΟΠμείζων (triangle) PQO [Prop. 1.29]. Thus, as OL is to LM, so
ἐστίν,γωνίαἄραἡὑπὸΑΒΓγωνίαςτῆςὑπὸΟΞΠμεῖζων OP (is) to PQ [Prop. 6.4]. Alternately, as LO (is) to OP ,
ἐστίν.ὁμοίωςδὴδείξομεν,ὅτικαὶἡμὲνὑπὸΔΕΖτῆςὑπὸ so LM (is) to PQ [Prop. 5.16]. And LO (is) greater than
ΜΞΝμείζωνἐστίν, ἡδὲὑπὸΗΘΚτῆςὑπὸΝΞΛ.αἱἄρα OP . Thus, LM (is) also greater than PQ [Prop. 5.14].
τρεῖςγωνίαιαἱὑπὸΑΒΓ,ΔΕΖ,ΗΘΚτριῶντῶνὑπὸΛΞΜ, But LM was made equal to AC. Thus, AC is also greater
ΜΞΝ,ΝΞΛμείζονέςεἰσιν.ἀλλὰαἱὑπὸΑΒΓ,ΔΕΖ,ΗΘΚ than PQ. Therefore, since the two (straight-lines) AB
τεσσάρωνὀρθῶνἐλάσσονεςὑπόκεινται·πολλῷἄρααἱὑπὸ and BC are equal to the two (straight-lines) PO and OQ
ΛΞΜ,ΜΞΝ,ΝΞΛτεσσάρωνὀρθῶνἐλάσσονέςεἰσιν.ἀλλὰ (respectively), and the base AC is greater than the base
καὶἴσαι·ὅπερἐστὶνἄτοπον. οὐκἄραἡΑΒἐλάσσωνἐστὶ PQ, the angle ABC is thus greater than the angle POQ
τῆςΛΞ.ἐδείχθηδέ,ὅτιοὐδὲἴση·μείζωνἄραἡΑΒτῆςΛΞ. [Prop. 1.25]. So, similarly, we can show that DEF is
ἈνεστάτωδὴἀπὸτοῦΞσημείουτῷτοῦΛΜΝκύκλου also greater than MON, and GHK than NOL. Thus,
ἐπιπέδῳπρὸςὀρθὰςἡΞΡ,καὶᾧμεῖζόνἐστιτὸἀπὸτῆς the (sum of the) three angles ABC, DEF , and GHK is
ΑΒτετράγωνοντοῦἀπὸτῆςΛΞ,ἐκείνῳἴσονἔστωτὸἀπὸ greater than the (sum of the) three angles LOM, MON,
447

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
τῆςΞΡ,καὶἐπεζεύχθωσαναἱΡΛ,ΡΜ,ΡΝ.
and NOL. But, (the sum of) ABC, DEF , and GHK was
ΚαὶἐπεὶἡΡΞὀρθὴἐστιπρὸςτὸτοῦΛΜΝκύκλου assumed (to be) less than four right-angles. Thus, (the
ἐπίπεδον,καὶπρὸςἑκάστηνἄρατῶνΛΞ,ΜΞ,ΝΞὀρθή sum of) LOM, MON, and NOL is much less than four
ἐστινἡΡΞ.καὶἐπεὶἴσηἐστὶνἡΛΞτῇΞΜ,κοινὴδὲκαὶ right-angles. But, (it is) also equal (to four right-angles).
πρὸςὀρθὰςἡΞΡ,βάσιςἄραἡΡΛβάσειτῂΡΜἐστινἴση. The very thing is absurd. Thus, AB is not less than LO.
διὰτὰαὐτὰδὴκαὶἡΡΝἑκατέρᾳτῶνΡΛ,ΡΜἐστινἴση· And it was shown (to be) not equal either. Thus, AB (is)
αἱτρεῖςἄρααἱΡΛ,ΡΜ,ΡΝἴσαιἀλλήλαιςεἰσίν. καὶἐπεὶ greater than LO.
ᾧμεῖζόνἐστιτὸἀπὸτῆςΑΒτοῦἀπὸτῆςΛΞ,ἐκείνῳἴσον So let OR have been set up at point O at right-
ὑπόκειταιτὸἀπὸτῆςΞΡ,τὸἄραἀπὸτῆςΑΒἴσονἐστὶτοῖς angles to the plane of circle LMN [Prop. 11.12]. And
ἀπὸτῶνΛΞ,ΞΡ.τοῖςδὲἀπὸτῶνΛΞ,ΞΡἴσονἐστὶτὸἀπὸ let the (square) on OR be equal to that (area) by which
τῆςΛΡ·ὀρθὴγὰρἡὑπὸΛΞΡ·τὸἄραἀπὸτῆςΑΒἴσον the square on AB is greater than the (square) on LO
ἐστὶτῷἀπὸτῆςΡΛ·ἴσηἄραἡΑΒτῇΡΛ.ἀλλὰτῇμὲνΑΒ [Prop. 11.23 lem.]. And let RL, RM, and RN have been
ἴσηἐστὶνἑκάστητῶνΒΓ,ΔΕ,ΕΖ,ΗΘ,ΘΚ,τῇδὲΡΛἴση joined.
ἑκατέρατῶνΡΜ,ΡΝ·ἑκάστηἄρατῶνΑΒ,ΒΓ,ΔΕ,ΕΖ, And since RO is at right-angles to the plane of cir-
ΗΘ,ΘΚἑκάστῃτῶνΡΛ,ΡΜ,ΡΝἴσηἐστίν.καὶἐπεὶδύο cle LMN, RO is thus also at right-angles to each of LO,
αἱΛΡ,ΡΜδυσὶταῖςΑΒ,ΒΓἴσαιεἰσίν,καὶβάσιςἡΛΜ MO, and NO. And since LO is equal to OM, and OR
βάσειτῇΑΓὑπόκειταιἴση,γωνίαἄραἡὑπὸΛΡΜγωνίᾳ is common and at right-angles, the base RL is thus equal
τῇὑπὸΑΒΓἐστινἴση.διὰτὰαὐτὰδὴκαὶἡμὲνὑπὸΜΡΝ to the base RM [Prop. 1.4]. So, for the same (reasons),
τῇὑπὸΔΕΖἐστινἴση,ἡδὲὑπὸΛΡΝτῇὑπὸΗΘΚ. RN is also equal to each of RL and RM. Thus, the three
᾿ΕκτριῶνἄραγωνιῶνἐπιπέδωντῶνὑπὸΛΡΜ,ΜΡΝ, (straight-lines) RL, RM, and RN are equal to one an-
ΛΡΝ,αἵεἰσινἴσαιτρισὶταῖςδοθείσαιςταῖςὑπὸΑΒΓ,ΔΕΖ, other. And since the (square) on OR was assumed to
ΗΘΚ,στερεὰγωνίασυνέσταταιἡπρὸςτῷΡπεριεχομένη be equal to that (area) by which the (square) on AB is
ὑπὸτῶνΛΡΜ,ΜΡΝ,ΛΡΝγωνιῶν·ὅπερἔδειποιῆσαι. greater than the (square) on LO, the (square) on AB
is thus equal to the (sum of the squares) on LO and
OR. And the (square) on LR is equal to the (sum of
the squares) on LO and OR. For LOR (is) a right-angle
[Prop. 1.47]. Thus, the (square) on AB is equal to the
(square) on RL. Thus, AB (is) equal to RL. But, each
of BC, DE, EF , GH, and HK is equal to AB, and each
of RM and RN equal to RL. Thus, each of AB, BC,
DE, EF , GH, and HK is equal to each of RL, RM,
and RN. And since the two (straight-lines) LR and RM
are equal to the two (straight-lines) AB and BC (respectively),
and the base LM was assumed (to be) equal to
the base AC, the angle LRM is thus equal to the angle
ABC [Prop. 1.8]. So, for the same (reasons), MRN is
also equal to DEF , and LRN to GHK.
Thus, the solid angle R, contained by the angles
LRM, MRN, and LRN, has been constructed out of
the three plane angles LRM, MRN, and LRN, which
are equal to the three given (plane angles) ABC, DEF ,
and GHK (respectively). (Which is) the very thing it was
required to do.
448

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
C
ÄÑÑ.
῝Ονδὲτρόπον,ᾧμεῖζόνἐστιτὸἀπὸτῆςΑΒτοῦἀπὸ And we can demonstrate, thusly, in which manner to
τῆςΛΞ,ἐκείνῳἴσονλαβεῖνἔστιτὸἀπὸτῆςΞΡ,δείξομεν take the (square) on OR equal to that (area) by which
οὕτως.ἐκκείσθωσαναἱΑΒ,ΛΞεὐθεῖαι,καὶἔστωμείζωνἡ the (square) on AB is greater than the (square) on LO.
ΑΒ,καὶγεγράφθωἐπ᾿αὐτῆςἡμικύκλιοντὸΑΒΓ,καὶεἰςτὸ Let the straight-lines AB and LO be set out, and let AB
ΑΒΓἡμικύκλιονἐνηρμόσθωτῇΛΞεὐθείᾳμὴμείζονιοὔσῃ be greater, and let the semicircle ABC have been drawn
τῆςΑΒδιαμέτρουἴσηἡΑΓ,καὶἐπεζεύχθωἡΓΒ.ἐπεὶοὖν around it. And let AC, equal to the straight-line LO,
ἐνἡμικυκλίῳτῷΑΓΒγωνίαἐστὶνἡὑπὸΑΓΒ,ὀρθὴἄρα which is not greater than the diameter AB, have been
ἐστὶνἡὑπὸΑΓΒ.τὸἄραἀπὸτῆςΑΒἴσονἐστὶτοῖςἀπὸ inserted into the semicircle ABC [Prop. 4.1]. And let
τῶνΑΓ,ΓΒ.ὥστετὸἀπὸτῆςΑΒτοῦἀπὸτῆςΑΓμεῖζόν CB have been joined. Therefore, since the angle ACB
ἐστιτῷἀπὸτῆςΓΒ.ἴσηδὲἡΑΓτῇΛΞ.τὸἄραἀπὸτῆς is in the semicircle ACB, ACB is thus a right-angle
ΑΒτοῦἀπὸτῆςΛΞμεῖζόνἐστιτῷἀπὸτῆςΓΒ.ἐὰνοὖν [Prop. 3.31]. Thus, the (square) on AB is equal to the
τῇΒΓἴσηντὴνΞΡἀπολάβωμεν,ἔσταιτὸἀπὸτῆςΑΒτοῦ (sum of the) squares on AC and CB [Prop. 1.47]. Hence,
ἀπὸτῆςΛΞμεῖζοντῷἀπὸτῆςΞΡ·ὅπερπροέκειτοποιῆσαι. the (square) on AB is greater than the (square) on AC
by the (square) on CB. And AC (is) equal to LO. Thus,
the (square) on AB is greater than the (square) on LO
by the (square) on CB. Therefore, if we take OR equal
to BC then the (square) on AB will be greater than the
(square) on LO by the (square) on OR. (Which is) the
very thing it was prescribed to do.
A
Lemma
. Proposition 24
᾿Εὰνστερεὸνὑπὸπαραλλήλωνἐπιπέδωνπεριέχηται,τὰ
Â
If a solid (figure) is contained by (six) parallel planes
ἀπεναντίον αὐτοῦ ἐπίπεδα ἴσα τε καὶ παραλληλόγραμμά then its opposite planes are both equal and parallelo-
ἐστιν.
grammic.
À
B
H
A
G
C
F
D
E
ΣτερεὸνγὰρτὸΓΔΘΗὑπὸπαραλλήλωνἐπιπέδωνπε- For let the solid (figure) CDHG have been contained
ριεχέσθωτῶνΑΓ,ΗΖ,ΑΘ,ΔΖ,ΒΖ,ΑΕ·λέγω,ὅτιτὰἀπε- by the parallel planes AC, GF , and AH, DF , and BF ,
ναντίοναὐτοῦἐπίπεδαἴσατεκαὶπαραλληλόγραμμάἐστιν. AE. I say that its opposite planes are both equal and
᾿Επεὶ γὰρ δύο ἐπίπεδα παράλληλα τὰ ΒΗ, ΓΕ ὑπὸ parallelogrammic.
ἐπιπέδουτοῦΑΓτέμνεται,αἱκοιναὶαὐτῶντομαὶπαράλληλοί For since the two parallel planes BG and CE are
εἰσιν. παράλληλοςἄρα ἐστὶν ἡ ΑΒ τῇ ΔΓ. πάλιν, ἑπεὶ cut by the plane AC, their common sections are parallel
δύο ἐπίπεδα παράλληλα τὰ ΒΖ, ΑΕ ὑπὸ ἐπιπέδου τοῦ [Prop. 11.16]. Thus, AB is parallel to DC. Again, since
ΑΓ τέμνεται, αἱ κοιναὶ αὐτῶν τομαὶ παράλληλοί εἰσιν. the two parallel planes BF and AE are cut by the plane
B
449

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
παράλληλοςἄραἐστὶνἡΒΓτῇΑΔ.ἐδείχθηδὲκαὶἡΑΒ AC, their common sections are parallel [Prop. 11.16].
τῇΔΓπαράλληλος·παραλληλόγραμμονἄραἐστὶτὸΑΓ. Thus, BC is parallel to AD. And AB was also shown (to
ὁμοίωςδὴδείξομεν,ὅτικαὶἕκαστοντῶνΔΖ,ΖΗ,ΗΒ,ΒΖ, be) parallel to DC. Thus, AC is a parallelogram. So, sim-
ΑΕπαραλληλόγραμμόνἐστιν.
ilarly, we can also show that DF , FG, GB, BF , and AE
᾿ΕπεζεύχθωσαναἱΑΘ,ΔΖ.καὶἐπεὶπαράλληλόςἐστινἡ are each parallelograms.
μὲνΑΒτῇΔΓ,ἡδὲΒΘτῇΓΖ,δύοδὴαἱΑΒ,ΒΘἁπτόμεναι Let AH and DF have been joined. And since AB is
ἀλλήλωνπαρὰδύοεὐθείαςτὰςΔΓ,ΓΖἁπτομέναςἀλλήλων parallel to DC, and BH to CF , so the two (straight-lines)
εἰσὶνοὐκἐντῷαὐτῷἐπιπέδῳ·ἴσαςἄραγωνίαςπεριέξουσιν· joining one another, AB and BH, are parallel to the two
ἴσηἄραἡὑπὸΑΒΘγωνίατῇὑπὸΔΓΖ.καὶἐπεὶδύοαἱΑΒ, straight-lines joining one another, DC and CF (respec-
ΒΘδυσὶταῖςΔΓ,ΓΖἴσαιεἰσίν,καὶγωνίαἡὑπὸΑΒΘγωνίᾳ tively), not (being) in the same plane. Thus, they will
τῇὑπὸΔΓΖἐστινἴση,βάσιςἄραἡΑΘβάσειτῇΔΖἐστιν contain equal angles [Prop. 11.10]. Thus, angle ABH
ἴση,καὶτὸΑΒΘτρίγωνοντῷΔΓΖτριγώνῳἴσονἐστίν.καί (is) equal to (angle) DCF . And since the two (straight-
ἐστιτοῦμὲνΑΒΘδιπλάσιοντὸΒΗπαραλληλόγραμμον, lines) AB and BH are equal to the two (straight-lines)
τοῦ δὲ ΔΓΖ διπλάσιον τὸ ΓΕ παραλληλόγραμμον· ἴσον DC and CF (respectively) [Prop. 1.34], and angle ABH
ἄρατὸΒΗπαραλληλόγραμμοντῷΓΕπαραλληλογράμμῳ· is equal to angle DCF , the base AH is thus equal to the
ὁμοίωςδὴδείξομεν,ὅτικαὶτὸμὲνΑΓτῷΗΖἐστινἴσον, base DF , and triangle ABH is equal to triangle DCF
τὸδὲΑΕτῷΒΖ.
[Prop. 1.4]. And parallelogram BG is double (triangle)
᾿Εὰνἄραστερεὸνὑπὸπαραλλήλωνἐπιπέδωνπεριέχηται, ABH, and parallelogram CE double (triangle) DCF
τὰἀπεναντίοναὐτοῦἐπίπεδαἴσατεκαὶπαραλληλόγραμμά [Prop. 1.34]. Thus, parallelogram BG (is) equal to paral-
ἐστιν·ὅπερἔδειδεῖξαι.
lelogram CE. So, similarly, we can show that AC is also
equal to GF , and AE to BF .
Thus, if a solid (figure) is contained by (six) parallel
planes then its opposite planes are both equal and parallelogrammic.
(Which is) the very thing it was required to
show.
. Proposition 25
᾿Εὰν στερεὸν παραλληλεπίπεδον ἐπιπέδῳ τμηθῇ πα- If a parallelipiped solid is cut by a plane which is parραλλήλῳὄντιτοῖςἀπεναντίονἐπιπέδοις,ἔσταιὡςἡβάσις
È
allel to the opposite planes (of the parallelipiped) then as
πρὸςτὴνβάσιν,οὕτωςτὸστερεὸνπρὸςτὸστερεόν. the base (is) to the base, so the solid will be to the solid.
X Q R U D Y T
Ç Ë
Ä Ã Å É
O
B G I
Æ
P V F C W S
L K A E H M N
ΣτερεὸνγὰρπαραλληλεπίπεδοντὸΑΒΓΔἐπιπέδῳτῷ For let the parallelipiped solid ABCD have been cut
ΖΗτετμήσθωπαραλλήλῳὄντιτοῖςἀπεναντίονἐπιπέδοις by the plane FG which is parallel to the opposite planes
τοῖςΡΑ,ΔΘ·λέγω,ὅτιἐστὶνὡςἡΑΕΖΦβάσιςπρὸςτὴν RA and DH. I say that as the base AEFV (is) to the base
ΕΘΓΖ βάσιν, οὕτως τὸ ΑΒΖΥ στερεὸν πρὸς τὸ ΕΗΓΔ EHCF , so the solid ABFU (is) to the solid EGCD.
Ê Í Ï Ì
À Á
Â
στερεόν.
For let AH have been produced in each direction. And
᾿Εκβεβλήσθω γὰρ ἡ ΑΘ ἐφ᾿ ἑκάτερα τὰ μέρη, καὶ let any number whatsoever (of lengths), AK and KL,
κείσθωσαντῇμὲνΑΕἴσαιὁσαιδηποτοῦναἱΑΚ,ΚΛ,τῇδὲ be made equal to AE, and any number whatsoever (of
ΕΘἴσαιὁσαιδηποτοῦναἱΘΜ,ΜΝ,καὶσυμπεπληρώσθω lengths), HM and MN, equal to EH. And let the paralτὰΛΟ,ΚΦ,ΘΧ,ΜΣπαραλληλόγραμμακαὶτὰΛΠ,ΚΡ,
lelograms LP , KV , HW , and MS have been completed,
450

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
ΔΜ,ΜΤστερεά. and the solids LQ, KR, DM, and MT .
ΚαὶἐπεὶἴσαιεἰσὶναἱΛΚ,ΚΑ,ΑΕεὐθεῖαιἀλλήλαις,ἴσα And since the straight-lines LK, KA, and AE are
ἐστὶκαὶτὰμὲνΛΟ,ΚΦ,ΑΖπαραλληλόγραμμαἀλλήλοις,τὰ equal to one another, the parallelograms LP , KV , and
δὲΚΞ,ΚΒ,ΑΗἀλλήλοιςκαὶἔτιτὰΛΨ,ΚΠ,ΑΡἀλλήλοις· AF are also equal to one another, and KO, KB, and AG
ἀπεναντίονγάρ. διὰτὰαὐτὰδὴκαὶτὰμὲνΕΓ,ΘΧ,ΜΣ (are equal) to one another, and, further, LX, KQ, and
παραλληλόγραμμαἴσαεἰσὶνἀλλήλοις,τὰδὲΘΗ,ΘΙ,ΙΝἴσα AR (are equal) to one another. For (they are) opposite
εἰσὶνἀλλήλοις,καὶἔτιτὰΔΘ,ΜΩ,ΝΤ·τρίαἄραἐπίπεδα [Prop. 11.24]. So, for the same (reasons), the paralleloτῶνΛΠ,ΚΡ,ΑΥστερεῶντρισὶνἐπιπέδοιςἐστὶνἴσα.ἀλλὰ
grams EC, HW , and MS are also equal to one another,
τὰτρίατρισὶτοῖςἀπεναντίονἐστὶνἴσα·τὰἄρατρίαστερεὰ and HG, HI, and IN are equal to one another, and,
τὰΛΠ,ΚΡ,ΑΥἴσαἀλλήλοιςἐστίν.διὰτὰαὐτὰδὴκαὶτὰ further, DH, MY , and NT (are equal to one another).
τρίαστερεὰτὰΕΔ, ΔΜ,ΜΤἴσαἀλλήλοιςἐστίν· ὁσα- Thus, three planes of (one of) the solids LQ, KR, and
πλασίων ἄρα ἐστὶν ἡ ΛΖ βάσιςτῆς ΑΖ βάσεως, τοσαυ- AU are equal to the (corresponding) three planes (of the
ταπλάσιόνἐστικαὶτὸΛΥστερεὸντοῦΑΥστερεοῦ. διὰ others). But, the three planes (in one of the soilds) are
τὰαὐτὰδὴὁσαπλασίωνἐστὶνἡΝΖβάσιςτῆςΖΘβάσεως, equal to the three opposite planes [Prop. 11.24]. Thus,
τοσαυταπλάσιόνἐστικαὶτὸΝΥστερεὸντοῦΘΥστερεοῦ. the three solids LQ, KR, and AU are equal to one anκαὶεἰἴσηἐστὶνἡΛΖβάσιςτῇΝΖβάσει,ἴσονἐστὶκαὶτὸ
other [Def. 11.10]. So, for the same (reasons), the three
ΛΥστερεὸντῷΝΥστερεῷ,καὶεἰὑπερέχειἡΛΖβάσιςτῆς solids ED, DM, and MT are also equal to one another.
ΝΖβάσεως,ὑπερέχεικαὶτὸΛΥστερεὸντοῦΝΥστερεοῦ, Thus, as many multiples as the base LF is of the base AF ,
καὶεἰἐλλείπει,ἐλλείπει.τεσσάρωνδὴὄντωνμεγεθῶν,δύο so many multiples is the solid LU also of the the solid AU.
μὲνβάσεωντῶνΑΖ,ΖΘ,δύοδὲστερεῶντῶνΑΥ,ΥΘ, So, for the same (reasons), as many multiples as the base
εἴληπταιἰσάκιςπολλαπλάσιατῆςμὲνΑΖβάσεωςκαὶτοῦ NF is of the base FH, so many multiples is the solid NU
ΑΥστερεοῦἥτεΛΖβάσιςκαὶτὸΛΥστερεόν,τῆςδὲΘΖ also of the solid HU. And if the base LF is equal to the
βάσεωςκαὶτοῦΘΥστερεοῦἥτεΝΖβάσιςκαὶτὸΝΥ base NF then the solid LU is also equal to the solid NU. †
στερεόν,καὶδέδεικται,ὅτιεἱὑπερέχειἡΛΖβάσιςτῆςΖΝ And if the base LF exceeds the base NF then the solid
βάσεως,ὑπερέχεικαὶτὸΛΥστερεὸντοῦΝΥ[στερεοῦ],καὶ LU also exceeds the solid NU. And if (LF ) is less than
εἰἴση,ἴσον,καὶεἰἐλλείπει,ἐλλείπει. ἔστινἄραὡςἡΑΖ (NF ) then (LU) is (also) less than (NU). So, there are
βάσιςπρὸςτὴνΖΘβάσιν,οὕτωςτὸΑΥστερεὸνπρὸςτὸ four magnitudes, the two bases AF and FH, and the two
ΥΘστερεόν·ὅπερἔδειδεῖξαι.
solids AU and UH, and equal multiples have been taken
of the base AF and the solid AU— (namely), the base
LF and the solid LU—and of the base HF and the solid
HU—(namely), the base NF and the solid NU. And it
has been shown that if the base LF exceeds the base FN
then the solid LU also exceeds the [solid] NU, and if
(LF is) equal (to FN) then (LU is) equal (to NU), and
if (LF is) less than (FN) then (LU is) less than (NU).
Thus, as the base AF is to the base FH, so the solid AU
(is) to the solid UH [Def. 5.5]. (Which is) the very thing
it was required to show.
† Here, Euclid assumes that LF NF implies LU NU. This is easily demonstrated.
¦. Proposition 26
Πρὸςτῇδοθείσῃεὐθείᾳκαὶτῷπρὸςαὐτῇσημείῳτῇ To construct a solid angle equal to a given solid angle
δοθείσῃστερεᾷγωνίᾳἴσηνστερεὰνγωνίανσυστήσασθαι. on a given straight-line, and at a given point on it.
῎ΕστωἡμὲνδοθεῖσαεὐθεῖαἡΑΒ,τὸδὲπρὸςαὐτῇ Let AB be the given straight-line, and A the given
δοθὲνσημεῖοντὸΑ,ἡδὲδοθεῖσαστερεὰγωνίαἡπρὸς point on it, and D the given solid angle, contained by the
τῷΔπεριεχομένηὑπὸτῶνὑπὸΕΔΓ,ΕΔΖ,ΖΔΓγωνιῶν plane angles EDC, EDF , and FDC. So, it is necessary
ἐπιπέδων· δεῖ δὴ πρὸς τῇ ΑΒ εὐθείᾳκαὶ τῷ πρὸς αὐτῇ to construct a solid angle equal to the solid angle D on
σημείῳτῷΑτῇπρὸςτῷΔστερεᾷγωνίᾳἴσηνστερεὰν the straight-line AB, and at the point A on it.
γωνίανσυστήσασθαι.
451

ËÌÇÁÉÁÏƺ ELEMENTS
Â
F
BOOK 11
H
À
Ä
Ã
D
C
E
G
L
A
K
B
ΕἰλήφθωγὰρἐπὶτῆςΔΖτυχὸνσημεῖοντὸΖ,καὶἤχθω For let some random point F have been taken on DF ,
ἀπὸτοῦΖἐπὶτὸδιὰτῶνΕΔ,ΔΓἐπίπεδονκάθετοςἡΖΗ, and let FG have been drawn from F perpendicular to
καὶσυμβαλλέτωτῷἐπιπέδῳκατὰτὸΗ,καὶἐπεζεύχθωἡ the plane through ED and DC [Prop. 11.11], and let it
ΔΗ,καὶσυνεστάτωπρὸςτῇΑΒεὐθείᾳκαὶτῷπρὸςαὐτῇ meet the plane at G, and let DG have been joined. And
σημείῳτῷΑτῇμὲνὑπὸΕΔΓγωνίᾳἴσηἡὑπὸΒΑΛ,τῇδὲ let BAL, equal to the angle EDC, and BAK, equal to
ὑπὸΕΔΗἴσηἡὑπὸΒΑΚ,καὶκείσθωτῇΔΗἴσηἡΑΚ,καὶ EDG, have been constructed on the straight-line AB at
ἀνεστάτωἀπὸτοῦΚσημείουτῷδιὰτῶνΒΑΛἐπιπέδῳπρὸς the point A on it [Prop. 1.23]. And let AK be made equal
ὀρθὰςἡΚΘ,καὶκείσθωἴσητῇΗΖἡΚΘ,καὶἐπεζεύχθω to DG. And let KH have been set up at the point K
ἡΘΑ·λέγω,ὅτιἡπρὸςτῷΑστερεὰγωνίαπεριεχομένη at right-angles to the plane through BAL [Prop. 11.12].
ὑπὸτῶνΒΑΛ,ΒΑΘ,ΘΑΛγωνιῶνἴσηἐστὶτῇπρὸςτῷΔ And let KH be made equal to GF . And let HA have
στερεᾷγωνίᾳτῇπεριεχομένῃὑπὸτῶνΕΔΓ,ΕΔΖ,ΖΔΓ been joined. I say that the solid angle at A, contained by
γωνιῶν.
the (plane) angles BAL, BAH, and HAL, is equal to the
ἈπειλήφθωσανγὰρἴσαιαἱΑΒ,ΔΕ,καὶἐπεζεύχθωσαν solid angle at D, contained by the (plane) angles EDC,
αἱΘΒ,ΚΒ,ΖΕ,ΗΕ.καὶἐπεὶἡΖΗὀρθήἐστιπρὸςτὸ EDF , and FDC.
ὑποκείμενονἐπίπεδον,καὶπρὸςπάσαςἄρατὰςἁπτομένας For let AB and DE have been cut off (so as to be)
αὐτῆςεὐθείαςκαὶοὔσαςἐντῷὑποκειμένῳἐπιπέδῳὀρθὰς equal, and let HB, KB, FE, and GE have been joined.
ποιήσειγωνίας· ὀρθὴἄραἐστὶνἑκατέρατῶνὑπὸΖΗΔ, And since FG is at right-angles to the reference plane
ΖΗΕγωνιῶν. διὰτὰαὐτὰδὴκαὶἑκατέρατῶνὑπὸΘΚΑ, (EDC), it will also make right-angles with all of the
ΘΚΒγωνιῶνὀρθήἐστιν. καὶἐπεὶδύοαἱΚΑ,ΑΒδύο straight-lines joined to it which are also in the reference
ταῖςΗΔ,ΔΕἴσαιεἰσὶνἑκατέραἑκατέρᾳ,καὶγωνίαςἴσας plane [Def. 11.3]. Thus, the angles FGD and FGE
περιέχουσιν,βάσιςἄραἡΚΒβάσειτῇΗΕἴσηἐστίν. ἔστι are right-angles. So, for the same (reasons), the anδὲκαὶἡΚΘτῇΗΖἴση·καὶγωνίαςὀρθὰςπεριέχουσιν·ἴση
gles HKA and HKB are also right-angles. And since
ἄρακαὶἡΘΒτῇΖΕ.πάλινἐπεὶδύοαἱΑΚ,ΚΘδυσὶταῖς the two (straight-lines) KA and AB are equal to the two
ΔΗ,ΗΖἴσαιεἰσίν,καὶγωνίαςὀρθὰςπεριέχουσιν,βάσιςἄρα (straight-lines) GD and DE, respectively, and they con-
ἡΑΘβάσειτῇΖΔἴσηἐστίν.ἔστιδὲκαὶἡΑΒτῇΔΕἴση· tain equal angles, the base KB is thus equal to the base
δύοδὴαἱΘΑ,ΑΒδύοταῖςΔΖ,ΔΕἴσαιεἰσίν.καὶβάσιςἡ GE [Prop. 1.4]. And KH is also equal to GF . And they
ΘΒβάσειτῇΖΕἴση·γωνίαἄραἡὑπὸΒΑΘγωνίᾳτῇὑπὸ contain right-angles (with the respective bases). Thus,
ΕΔΖἐστινἴση.διὰτὰαὐτὰδὴκαὶἡὑπὸΘΑΛτῇὑπὸΖΔΓ HB (is) also equal to FE [Prop. 1.4]. Again, since the
ἐστινἴση.ἔστιδὲκαὶἡὑπὸΒΑΛτῇὑπὸΕΔΓἴση. two (straight-lines) AK and KH are equal to the two
ΠρὸςἄρατῇδοθείσῃεὐθείᾳτῇΑΒκαὶτῷπρὸςαὐτῇ (straight-lines) DG and GF (respectively), and they conσημείῳτῷΑτῇδοθείσῃστερεᾷγωνίᾳτῇπρὸςτῷΔἴση
tain right-angles, the base AH is thus equal to the base
συνέσταται·ὅπερἔδειποιῆσαι. FD [Prop. 1.4]. And AB (is) also equal to DE. So,
the two (straight-lines) HA and AB are equal to the two
(straight-lines) DF and DE (respectively). And the base
HB (is) equal to the base FE. Thus, the angle BAH is
equal to the angle EDF [Prop. 1.8]. So, for the same
(reasons), HAL is also equal to FDC. And BAL is also
equal to EDC.
Thus, (a solid angle) has been constructed, equal to
the given solid angle at D, on the given straight-line AB,
452

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
at the given point A on it. (Which is) the very thing it
was required to do.
Þ. Proposition 27
Ἀπὸτῆςδοθείσηςεὐθείαςτῷδοθέντιστερεῷπαραλλη- To describe a parallelepiped solid similar, and simiλεπιπέδῳὅμοιόντεκαὶὁμοίωςκείμενονστερεὸνπαραλλη-
larly laid out, to a given parallelepiped solid on a given
λεπίπεδονἀναγράψαι.
straight-line.
῎ΕστωἡμὲνδοθεῖσαεὐθεῖαἡΑΒ,τὸδὲδοθὲνστερεὸν Let the given straight-line be AB, and the given parπαραλληλεπίπεδοντὸΓΔ·δεῖδὴἀπὸτῆςδοθείσηςεὐθείας
allelepiped solid CD. So, it is necessary to describe a
τῆς ΑΒ τῷ δοθέντι στερεῷ παραλληλεπιπέδῳ τῷ ΓΔ parallelepiped solid similar, and similarly laid out, to the
ὅμοιόντεκαὶὁμοίωςκείμενονστερεὸνπαραλληλεπίπεδον given parallelepiped solid CD on the given straight-line
ἀναγράψαι.
AB.
ΣυνεστάτωγὰρπρὸςτῇΑΒεὐθείᾳκαὶτῷπρὸςαὐτῇ For, let a (solid angle) contained by the (plane angles)
σημείῳτῷΑτῇπρὸςτῷΓστερεᾷγωνίᾳἴσηἡπεριεχομένη BAH, HAK, and KAB have been constructed, equal to
ὑπὸτῶνΒΑΘ,ΘΑΚ,ΚΑΒ,ὥστεἴσηνεἶναιτὴνμὲνὑπὸ solid angle at C, on the straight-line AB at the point A on
ΒΑΘγωνίαντῇὑπὸΕΓΖ,τὴνδὲὑπὸΒΑΚτῇὑπὸΕΓΗ, it [Prop. 11.26], such that angle BAH is equal to ECF ,
τὴνδὲὑπὸΚΑΘτῇὑπὸΗΓΖ·καὶγεγονέτωὡςμὲνἡΕΓ and BAK to ECG, and KAH to GCF . And let it have
πρὸςτὴνΓΗ,οὕτωςἡΒΑπρὸςτὴνΑΚ,ὡςδὲἡΗΓπρὸς been contrived that as EC (is) to CG, so BA (is) to AK,
τὴνΓΖ,οὕτωςἡΚΑπρὸςτὴνΑΘ.καὶδι᾿ἴσουἄραἐστὶν and as GC (is) to CF , so KA (is) to AH [Prop. 6.12].
ὡςἡΕΓπρὸςτὴνΓΖ,οὕτωςἡΒΑπρὸςτὴνΑΘ.καὶσυμ-
And thus, via equality, as EC is to CF , so BA (is) to AH
πεπληρώσθωτὸΘΒπαραλληλόγραμμονκαὶτὸΑΛστερεόν. [Prop. 5.22]. And let the parallelogram HB have been
Ä
completed, and the solid AL.
D
À ÃÅ
L
F
H
M
G
A B
C
E
ΚαὶἐπείἐστινὡςἡΕΓπρὸςτὴνΓΗ,οὕτωςἡΒΑπρὸς And since as EC is to CG, so BA (is) to AK, and
τὴνΑΚ,καὶπερὶἴσαςγωνίαςτὰςὑπὸΕΓΗ,ΒΑΚαἱπλευραὶ the sides about the equal angles ECG and BAK are
ἀνάλογόνεἰσιν,ὅμοιονἄραἐστὶτὸΗΕπαραλληλόγραμμον (thus) proportional, the parallelogram GE is thus simiτῷΚΒπαραλληλογράμμῳ.
διὰτὰαὐτὰδὴκαὶτὸμὲνΚΘ lar to the parallelogram KB. So, for the same (reasons),
παραλληλόγραμμοντῷΗΖπαραλληλογράμμῳὅμοιόνἐστι the parallelogram KH is also similar to the parallelogram
καὶἔτιτὸΖΕτῷΘΒ·τρίαἄραπαραλληλόγραμματοῦΓΔ GF , and, further, FE (is similar) to HB. Thus, three
στερεοῦτρισὶπαραλληλογράμμοιςτοῦΑΛστερεοῦὅμοιά of the parallelograms of solid CD are similar to three of
ἐστιν. ἀλλὰτὰμὲντρίατρισὶτοῖςἀπεναντίονἴσατέἐστι the parallelograms of solid AL. But, the (former) three
καὶὅμοια,τὰδὲτρίατρισὶτοῖςἀπεναντίονἴσατέἐστικαὶ are equal and similar to the three opposite, and the (lat-
ὅμοια·ὅλονἄρατὸΓΔστερεὸνὅλῳτῷΑΛστερεῷὅμοιόν ter) three are equal and similar to the three opposite.
ἐστιν.
Thus, the whole solid CD is similar to the whole solid
Ἀπὸ τῆς δοθείσης ἄρα εὐθείας τῆς ΑΒ τῷ δοθέντι AL [Def. 11.9].
στερεῷ παραλληλεπιπέδῳ τῷ ΓΔ ὅμοιόν τε καὶ ὁμοίως Thus, AL, similar, and similarly laid out, to the given
κείμενονἀναγέγραπταιτὸΑΛ·ὅπερἔδειποιῆσαι. parallelepiped solid CD, has been described on the given
straight-lines AB. (Which is) the very thing it was required
to do.
. Proposition 28
᾿Εὰν στερεὸν παραλληλεπίπεδονἐπιπέδῳ τμηθῇ κατὰ If a parallelepiped solid is cut by a plane (passing)
Â
K
453

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
τὰςδιαγωνίουςτῶνἀπεναντίονἐπιπέδων,δίχατμηθήσεται through the diagonals of (a pair of) opposite planes then
τὸστερεὸνὑπὸτοῦἐπιπέδου.
the solid will be cut in half by the plane.
Â
À
Στερεὸν γὰρ παραλληλεπίπεδον τὸ ΑΒ ἐπιπέδῳ τῷ For let the parallelepiped solid AB have been cut by
ΓΔΕΖ τετμήσθω κατὰ τὰς διαγωνίους τῶν ἀπεναντίον the plane CDEF (passing) through the diagonals of the
ἐπιπέδωντὰςΓΖ,ΔΕ·λέγω,ὅτιδίχατμηθήσεταιτὸΑΒ opposite planes CF and DE. † I say that the solid AB will
στερεὸνὑπὸτοῦΓΔΕΖἐπιπέδου. be cut in half by the plane CDEF .
᾿Επεὶ γὰρ ἴσον ἐστὶ τὸ μὲν ΓΗΖ τρίγωνον τῷ ΓΖΒ For since triangle CGF is equal to triangle CFB, and
τριγώνῳ,τὸδὲΑΔΕτῷΔΕΘ,ἔστιδὲκαὶτὸμὲνΓΑπα- ADE (is equal) to DEH [Prop. 1.34], and paralleloραλληλόγραμμοντῷΕΒἴσον·ἀπεναντίονγάρ·τὸδὲΗΕ
gram CA is also equal to EB—for (they are) opposite
τῷΓΘ,καὶτὸπρίσμαἄρατὸπεριεχόμενονὑπὸδύομὲν [Prop. 11.24]—and GE (equal) to CH, thus the prism
τριγώνων τῶν ΓΗΖ,ΑΔΕ, τριῶν δὲπαραλληλογράμμων contained by the two triangles CGF and ADE, and the
τῶνΗΕ,ΑΓ,ΓΕἴσονἐστὶτῷπρίσματιτῷπεριεχομένῳ three parallelograms GE, AC, and CE, is also equal to
ὑπὸδύομὲντριγώνωντῶνΓΖΒ,ΔΕΘ,τριῶνδὲπαραλ- the prism contained by the two triangles CFB and DEH,
ληλογράμμωντῶνΓΘ,ΒΕ,ΓΕ·ὑπὸγὰρἴσωνἐπιπέδων and the three parallelograms CH, BE, and CE. For they
περιέχονταιτῷτεπλήθεικαὶτῷμεγέθει. ὥστεὅλοντὸ are contained by planes (which are) equal in number and
ΑΒστερεὸνδίχατέτμηταιὑπὸτοῦΓΔΕΖἐπιπέδου·ὅπερ in magnitude [Def. 11.10]. ‡ Thus, the whole of solid AB
ἔδειδεῖξαι.
† Here, it is assumed that the two diagonals lie in the same plane. The proof is easily supplied.
‡ However, strictly speaking, the prisms are not similarly arranged, being mirror images of one another.
H
B
D
C
is cut in half by the plane CDEF . (Which is) the very
thing it was required to show.
. Proposition 29
Τὰἐπὶτῆςαὐτῆςβάσεωςὄνταστερεὰπαραλληλεπίπεδα Parallelepiped solids which are on the same base, and
καὶὑπὸτὸαὐτὸὕψος,ὧναἱἐφεστῶσαιἐπὶτῶναὐτῶνεἰσιν
Â
(have) the same height, and in which the (ends of the
εὐθειῶν,ἴσαἀλλήλοιςἐστίν.
straight-lines) standing up are on the same straight-lines,
are equal to one another.
À Å Æ
D E
H K
G
F
N
M
Ä
C
B
A
L
῎ΕστωἐπὶτῆςαὐτῆςβάσεωςτῆςΑΒστερεὰπαραλλη- For let the parallelepiped solids CM and CN be on
E
F
A
G
454

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
λεπίπεδατὰΓΜ,ΓΝὑπὸτὸαὐτὸὕψος,ὧναἱἐφεστῶσαι the same base AB, and (have) the same height, and let
αἱΑΗ,ΑΖ,ΛΜ,ΛΝ,ΓΔ,ΓΕ,ΒΘ,ΒΚἐπὶτῶναὐτῶν the (ends of the straight-lines) standing up in them, AG,
εὐθειῶνἔστωσαντῶνΖΝ,ΔΚ·λέγω,ὅτιἴσονἐστὶτὸΓΜ AF , LM, LN, CD, CE, BH, and BK, be on the same
στερεὸντῷΓΝστερεῷ.
straight-lines, FN and DK. I say that solid CM is equal
᾿ΕπεὶγὰρπαραλληλόγραμμόνἐστινἑκάτεροντῶνΓΘ, to solid CN.
ΓΚ,ἴσηἐστὶνἡΓΒἑκατέρᾳτῶνΔΘ,ΕΚ·ὥστεκαὶἡ For since CH and CK are each parallelograms, CB
ΔΘτῇΕΚἐστινἴση. κοινὴἀφῃρήσθωἡΕΘ·λοιπὴἄρα is equal to each of DH and EK [Prop. 1.34]. Hence,
ἡ ΔΕ λοιπῇ τῇ ΘΚ ἐστιν ἴση. ὥστε καὶ τὸ μὲν ΔΓΕ DH is also equal to EK. Let EH have been subtracted
τρίγωνοντῷΘΒΚτριγώνῳἴσονἐστίν,τὸδὲΔΗπαραλ- from both. Thus, the remainder DE is equal to the reληλόγραμμοντῷΘΝπαραλληλογράμμῳ.διὰτὰαὐτὰδὴκαὶ
mainder HK. Hence, triangle DCE is also equal to triτὸΑΖΗτρίγωνοντῷΜΛΝτριγώνῳἴσονἐστίν.ἔστιδὲκαὶ
angle HBK [Props. 1.4, 1.8], and parallelogram DG to
τὸμὲνΓΖπαραλληλόγραμμοντῷΒΜπαραλληλογράμμῳ parallelogram HN [Prop. 1.36]. So, for the same (rea-
ἴσον,τὸδὲΓΗτῷΒΝ·ἀπεναντίονγάρ·καὶτὸπρίσμαἄρα sons), traingle AFG is also equal to triangle MLN. And
τὸπεριεχόμενονὑπὸδύομὲντριγώνωντῶνΑΖΗ,ΔΓΕ, parallelogram CF is also equal to parallelogram BM,
τριῶνδὲπαραλληλογράμμωντῶνΑΔ,ΔΗ,ΓΗἴσονἐστὶ and CG to BN [Prop. 11.24]. For they are opposite.
τῷπρίσματιτῷπεριεχομένῳὑπὸδύομὲντριγώνωντῶν Thus, the prism contained by the two triangles AFG and
ΜΛΝ,ΘΒΚ,τριῶνδὲπαραλληλογράμμωντῶνΒΜ,ΘΝ, DCE, and the three parallelograms AD, DG, and CG, is
ΒΝ.κοινὸνπροσκείσθωτὸστερεὸν,οὗβάσιςμὲντὸΑΒ equal to the prism contained by the two triangles MLN
παραλληλόγραμμον,ἀπεναντίονδὲτὸΗΕΘΜ·ὅλονἄρατὸ and HBK, and the three parallelograms BM, HN, and
ΓΜστερεὸνπαραλληλεπίπεδονὅλῳτῷΓΝστερεῷπαραλ- BN. Let the solid whose base (is) parallelogram AB, and
ληλεπιπέδῳἴσονἐστίν.
(whose) opposite (face is) GEHM, have been added to
Τὰἄραἐπὶτῆςαὐτῆςβάσεωςὄνταστερεὰπαραλλη- both (prisms). Thus, the whole parallelepiped solid CM
λεπίπεδακαὶὑπὸτὸαὐτὸὕψος,ὧναἱἐφεστῶσαιἐπὶτῶν is equal to the whole parallelepiped solid CN.
αὐτῶνεἰσινεὐθειῶν,ἴσαἀλλήλοιςἐστίν·ὅπερἔδειδεῖξαι. Thus, parallelepiped solids which are on the same
base, and (have) the same height, and in which the
(ends of the straight-lines) standing up (are) on the same
straight-lines, are equal to one another. (Which is) the
very thing it was required to show.
Ð. Proposition 30
Τὰἐπὶτῆςαὐτῆςβάσεωςὄνταστερεὰπαραλληλεπίπεδα Parallelepiped solids which are on the same base, and
Æ Ç Ã Ê
καὶὑπὸτὸαὐτὸὕψος,ὧναἱἐφεστῶσαιοὐκεἰσὶνἐπὶτῶν (have) the same height, and in which the (ends of the
αὐτῶνεὐθειῶν,ἴσαἀλλήλοιςἐστίν.
straight-lines) standing up are not on the same straightlines,
are equal to one another.
À Å
N P K R
M
H
G O E Q
F
D
Ä
L
B
A
C
῎ΕστωἐπὶτῆςαὐτῆςβάσεωςτῆςΑΒστερεὰπαραλλη- Let the parallelepiped solids CM and CN be on the
λεπίπεδατὰΓΜ,ΓΝὑπὸτὸαὐτὸὕψος,ὧναἱἐφεστῶσαιαἱ same base, AB, and (have) the same height, and let the
ΑΖ,ΑΗ,ΛΜ,ΛΝ,ΓΔ,ΓΕ,ΒΘ,ΒΚμὴἔστωσανἐπὶτῶν (ends of the straight-lines) standing up in them, AF , AG,
ÂÈ
455

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
αὐτῶνεὐθειῶν·λέγω,ὅτιἴσονἐστὶτὸΓΜστερεὸντῷΓΝ LM, LN, CD, CE, BH, and BK, not be on the same
στερεῷ.
straight-lines. I say that the solid CM is equal to the
᾿ΕκβεβλήσθωσανγὰραἱΝΚ,ΔΘκαὶσυμπιπτέτωσαν solid CN.
ἀλλήλαιςκατὰτὸΡ,καὶἔτιἐκβεβλήσθωσαναἱΖΜ,ΗΕ For let NK and DH have been produced, and let
ἐπὶτὰΟ,Π,καὶἐπεζεύχθωσαναἱΑΞ,ΛΟ,ΓΠ,ΒΡ.ἴσον them have joined one another at R. And, further, let FM
δήἐστιτὸΓΜστερεόν, οὗβάσιςμὲντὸΑΓΒΛπαραλ- and GE have been produced to P and Q (respectively).
ληλόγραμμον,ἀπεναντίονδὲτὸΖΔΘΜ,τῷΓΟστερεῷ,οὗ And let AO, LP , CQ, and BR have been joined. So, solid
βάσιςμὲντὸΑΓΒΛπαραλληλόγραμμον,ἀπεναντίονδὲτὸ CM, whose base (is) parallelogram ACBL, and oppo-
ΞΠΡΟ·ἐπίτεγὰρτῆςαὐτῆςβάσεώςεἰσιτῆςΑΓΒΛκαὶ site (face) FDHM, is equal to solid CP , whose base (is)
ὑπὸτὸαὐτὸὕψος,ὧναἱἐφεστῶσαιαἱΑΖ,ΑΞ,ΛΜ,ΛΟ, parallelogram ACBL, and opposite (face) OQRP . For
ΓΔ,ΓΠ,ΒΘ,ΒΡἐπὶτῶναὐτῶνεἰσινεὐθειῶντῶνΖΟ,ΔΡ. they are on the same base, ACBL, and (have) the same
ἀλλὰτὸΓΟστερεόν,οὗβάσιςμένἐστιτὸΑΓΒΛπαραλ- height, and the (ends of the straight-lines) standing up in
ληλόγραμμον,ἀπεναντίονδὲτὸΞΠΡΟ,ἴσονἐστὶτῷΓΝ them, AF , AO, LM, LP , CD, CQ, BH, and BR, are on
στερεῷ,οὗβάσιςμὲντὸΑΓΒΛπαραλληλόγραμμον,ἀπε- the same straight-lines, FP and DR [Prop. 11.29]. But,
ναντίονδὲτὸΗΕΚΝ·ἐπίτεγὰρπάλιντῆςαὐτῆςβάσεώς solid CP , whose base is parallelogram ACBL, and oppoεἰσιτῆςΑΓΒΛκαὶὑπὸτὸαὐτὸὕψος,ὧναἱἐφεστῶσαιαἱ
site (face) OQRP , is equal to solid CN, whose base (is)
ΑΗ,ΑΞ,ΓΕ,ΓΠ,ΛΝ,ΛΟ,ΒΚ,ΒΡἐπὶτῶναὐτῶνεἰσιν parallelogram ACBL, and opposite (face) GEKN. For,
εὐθειῶντῶνΗΠ,ΝΡ.ὥστεκαὶτὸΓΜστερεὸνἴσονἐστὶ again, they are on the same base, ACBL, and (have)
τῷΓΝστερεῷ.
the same height, and the (ends of the straight-lines)
Τὰἄραἐπὶτῆςαὐτῆςβάσεωςστερεὰπαραλληλεπίπεδα standing up in them, AG, AO, CE, CQ, LN, LP , BK,
καὶὑπὸτὸαὐτὸὕψος,ὧναἱἐφεστῶσαιοὐκεἰσὶνἐπὶτῶν and BR, are on the same straight-lines, GQ and NR
αὐτῶνεὐθειῶν,ἴσαἀλλήλοιςἐστίν·ὅπερἔδειδεῖξαι. [Prop. 11.29]. Hence, solid CM is also equal to solid
CN.
Thus, parallelepiped solids (which are) on the same
base, and (have) the same height, and in which the (ends
of the straight-lines) standing up are not on the same
straight-lines, are equal to one another. (Which is) the
very thing it was required to show.
Ð. Proposition 31
Τὰἐπὶἴσωνβάσεωνὄνταστερεὰπαραλληλεπίπεδακαὶ Parallelepiped solids which are on equal bases, and
ὑπὸτὸαὐτὸὕψοςἴσαἀλλήλοιςἐστίν.
(have) the same height, are equal to one another.
῎ΕστωἐπὶἴσωνβάσεωντῶνΑΒ,ΓΔστερεὰπαραλλη- Let the parallelepiped solids AE and CF be on the
λεπίπεδατὰΑΕ,ΓΖὑπὸτὸαὐτὸὕψος.λέγω,ὅτιἴσονἐστὶ equal bases AB and CD (respectively), and (have) the
τὸΑΕστερεὸντῷΓΖστερεῷ. same height. I say that solid AE is equal to solid CF .
῎ΕστωσανδὴπρότεροναἱἐφεστηκυῖαιαἱΘΚ,ΒΕ,ΑΗ, So, let the (straight-lines) standing up, HK, BE, AG,
ΛΜ,ΟΠ,ΔΖ,ΓΞ,ΡΣπρὸςὀρθὰςταῖςΑΒ,ΓΔβάσεσιν, LM, PQ, DF , CO, and RS, first of all, be at right-angles
καὶἐκβεβλήσθωἐπ᾿εὐθείαςτῇΓΡεὐθεῖαἡΡΤ,καὶσυ- to the bases AB and CD. And let RT have been produced
νεστάτωπρὸςτῇΡΤεὐθείᾳκαὶτῷπρὸςαὐτῇσημείῳτῷ in a straight-line with CR. And let (angle) TRU, equal
ΡτῇὑπὸΑΛΒγωνίᾳἴσηἡὑπὸΤΡΥ,καὶκείσθωτῇμὲν to angle ALB, have been constructed on the straight-line
ΑΛἴσηἡΡΤ,τῇδὲΛΒἴσηἡΡΥ,καὶσυμπεπληρώσθωἥ RT , at the point R on it [Prop. 1.23]. And let RT be
τεΡΧβάσιςκαὶτὸΨΥστερεόν. made equal to AL, and RU to LB. And let the base RW ,
and the solid XU, have been completed.
456

ËÌÇÁÉÁÏƺ ELEMENTS
Ç À Á
Ë
Ê Ì
¦
Ï Ã
Â Í Ó
É
À
Å
C
Q
O
D
R
e
F
S
K
d
T
a
Y U b W
BOOK 11
H
B
Ä
G M
A L
ΚαὶἐπεὶδύοαἱΤΡ,ΡΥδυσὶταῖςΑΛ,ΛΒἴσαιεἰσίν, And since the two (straight-lines) TR and RU are
καὶγωνίαςἴσαςπεριέχουσιν,ἴσονἄρακαὶὅμοιοντὸΡΧ equal to the two (straight-lines) AL and LB (respecπαραλληλόγραμμον
τῷ ΘΛ παραλληλογράμμῳ. καὶ ἐπεὶ tively), and they contain equal angles, parallelogram
πάλινἴσημὲνἡΑΛτῇΡΤ,ἡδὲΛΜτῇΡΣ,καὶγωνίας RW is thus equal and similar to parallelogram HL
ὀρθὰςπεριέχουσιν,ἴσονἄρακαὶὅμοιόνἐστιτὸΡΨπαραλ- [Prop. 6.14]. And, again, since AL is equal to RT ,
ληλόγραμμοντῷΑΜπαραλληλογράμμῳ. διὰτὰαὐτὰδὴ and LM to RS, and they contain right-angles, paralκαὶτὸΛΕτῷΣΥἴσοντέἐστικαὶὅμοιον·τρίαἄραπα-
lelogram RX is thus equal and similar to parallelogram
ραλληλόγραμματοῦΑΕστερεοῦτρισὶπαραλληλογράμμοις AM [Prop. 6.14]. So, for the same (reasons), LE is also
τοῦΨΥστερεοῦἴσατέἐστικαὶὅμοια. ἀλλὰτὰμὲντρία equal and similar to SU. Thus, three parallelograms of
τρισὶ τοῖς ἀπεναντίον ἴσα τέ ἐστι καὶ ὅμοια, τὰ δὲ τρία solid AE are equal and similar to three parallelograms
τρισὶτοῖςἀπεναντίον·ὅλονἄρατὸΑΕστερεὸνπαραλλη- of solid XU. But, the three (faces of the former solid)
λεπίπεδονὅλῳτῷΨΥστερεῷπαραλληλεπιπέδῳἴσονἐστίν. are equal and similar to the three opposite (faces), and
διήχθωσαναἱΔΡ,ΧΥκαὶσυμπιπτέτωσανἀλλήλαιςκατὰ the three (faces of the latter solid) to the three opposite
τὸΩ,καὶδιὰτοῦΤτῇΔΩπαράλληλοςἤχθωἡαΤϡ,καὶ (faces) [Prop. 11.24]. Thus, the whole parallelepiped
ἐκβεβλήσθωἡΟΔκατὰτὸα,καὶσυμπεπληρώσθωτὰΩΨ, solid AE is equal to the whole parallelepiped solid XU
ΡΙ στερεά. ἴσονδήἐστιτὸΨΩστερεόν, οὗβάσιςμέν [Def. 11.10]. Let DR and WU have been drawn across,
ἐστιτὸΡΨπαραλληλόγραμμον,ἀπεναντίονδὲτὸΩϟ,τῷ and let them have met one another at Y . And let aTb
ΨΥστερεῷ,οὗβάσιςμὲντὸΡΨπαραλληλόγραμμον,ἀπε- have been drawn through T parallel to DY . And let PD
ναντίονδὲτὸΥΦ·ἐπίτεγὰρτῆςαὐτῆςβάσεώςεἰσιτῆς have been produced to a. And let the solids Y X and
ΡΨκαὶὑπὸτὸαὐτὸὕψος,ὧναἱἐφεστῶσαιαἱΡΩ,ΡΥ, RI have been completed. So, solid XY , whose base is
Τϡ,ΤΧ,Σϛ,Σ˜ο,Ψϟ,ΨΦἐπὶτῶναὐτῶνεἰσινεὐθειῶντῶν parallelogram RX, and opposite (face) Y c, is equal to
ΩΧ,ϛΦ.ἀλλὰτὸΨΥστερεὸντῷΑΕἐστινἴσον·καὶτὸΨΩ solid XU, whose base (is) parallelogram RX, and oppo-
ἄραστερεὸντῷΑΕστερεῷἐστινἴσον. καὶἐπεὶἴσονἐστὶ site (face) UV . For they are on the same base RX, and
τὸΡΥΧΤπαραλληλόγραμμοντῷΩΤπαραλληλογράμμῳ· (have) the same height, and the (ends of the straight-
ἐπίτεγὰρτῆςαὐτῆςβάσεώςεἰσιτῆςΡΤκαὶἐνταῖςαὐταῖς lines) standing up in them, RY , RU, Tb, TW, Se, Sd,
παραλλήλοιςταῖςΡΤ,ΩΧ·ἀλλὰτὸΡΥΧΤτῷΓΔἐστιν Xc and XV , are on the same straight-lines, Y W and
ἴσον,ἐπεὶκαὶτῷΑΒ,καὶτὸΩΤἄραπαραλληλόγραμμον eV [Prop. 11.29]. But, solid XU is equal to AE. Thus,
P
c
I
X
E
V
457

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
τῷΓΔἐστινἴσον. ἄλλοδὲτὸΔΤ·ἔστινἄραὡςἡΓΔ solid XY is also equal to solid AE. And since parallelβάσιςπρὸςτὴνΔΤ,οὕτωςἡΩΤπρὸςτὴνΔΤ.καὶἐπεὶ
ogram RUWT is equal to parallelogram Y T . For they
στερεὸνπαραλληλεπίπεδοντὸΓΙἐπιπέδῳτῷΡΖτέτμηται are on the same base RT , and between the same parπαραλλήλῳὄντιτοῖςἀπεναντίονἐπιπέδοις,ἔστινὡςἡΓΔ
allels RT and Y W [Prop. 1.35]. But, RUWT is equal
βάσιςπρὸςτὴνΔΤβάσιν,οὕτωςτὸΓΖστερεὸνπρὸςτὸΡΙ to CD, since (it is) also (equal) to AB. Parallelogram
στερεόν. διὰτὰαὐτὰδή,ἐπεὶστερεὸνπαραλληλεπίπεδον Y T is thus also equal to CD. And DT is another (parτὸΩΙἐπιπέδῳτῷΡΨτέτμηταιπαραλλήλῳὄντιτοῖςἀπε-
allelogram). Thus, as base CD is to DT , so Y T (is) to
ναντίονἐπιπέδοις,ἔστινὡςἡΩΤβάσιςπρὸςτὴνΤΛβάσιν, DT [Prop. 5.7]. And since the parallelepiped solid CI
οὕτωςτὸΩΨστερεὸνπρὸςτὸΡΙ.ἀλλ᾿ὡςἡΓΔβάσις has been cut by the plane RF , which is parallel to the
πρὸςτὴνΔΤ,οὕτωςἡΩΤπρὸςτὴνΔΤ·καὶὡςἄρατὸ opposite planes (of CI), as base CD is to base DT , so
ΓΖστερεὸνπρὸςτὸΡΙστερεόν, οὕτωςτὸΩΨστερεὸν solid CF (is) to solid RI [Prop. 11.25]. So, for the same
πρὸςτὸΡΙ.ἑκάτερονἄρατῶνΓΖ,ΩΨστερεῶνπρὸςτὸ (reasons), since the parallelepiped solid Y I has been cut
ΡΙτὸναὐτὸνἔχειλόγον·ἴσονἄραἐστὶτὸΓΖστερεὸντῷ by the plane RX, which is parallel to the opposite planes
ΩΨστερεῷ. ἀλλὰτὸΩΨτῷΑΕἐδείχθηἴσον·καὶτὸΑΕ (of Y I), as base Y T is to base TD, so solid Y X (is) to
ἄρατῷΓΖἐστινἴσον.
Å
À Ã È
É
Ç
Í Æ Ë
Â
Ä Ì Ï ÊÁ
solid RI [Prop. 11.25]. But, as base CD (is) to DT , so
Y T (is) to DT . And, thus, as solid CF (is) to solid RI,
so solid Y X (is) to solid RI. Thus, solids CF and Y X
each have the same ratio to RI [Prop. 5.11]. Thus, solid
CF is equal to solid Y X [Prop. 5.9]. But, Y X was show
(to be) equal to AE. Thus, AE is also equal to CF .
Q F
V
T
L
B C R
ΜὴἔστωσανδὴαἱἐφεστηκυῖαιαἱΑΗ,ΘΚ,ΒΕ,ΛΜ, And so let the (straight-lines) standing up, AG, HK,
ΓΞ,ΟΠ,ΔΖ,ΡΣπρὸςὀρθὰςταῖςΑΒ,ΓΔβάσεσιν·λέγω BE, LM, CO, PQ, DF , and RS, not be at right-angles
πάλιν,ὅτιἵσοντὸΑΕστερεὸντῷΓΖστερεῷ.ἤχθωσανγὰρ to the bases AB and CD. Again, I say that solid AE
ἀπὸτῶνΚ,Ε,Η,Μ,Π,Ζ,Ξ,Σσημείωνἐπὶτὸὑποκείμενον (is) equal to solid CF . For let KN, ET , GU, MV , QW ,
ἐπίπεδονκάθετοιαἱΚΝ,ΕΤ,ΗΥ,ΜΦ,ΠΧ,ΖΨ,ΞΩ,ΣΙ, FX, OY , and SI have been drawn from points K, E,
καὶσυμβαλλέτωσαντῷἐπιπέδῳκατὰτὰΝ,Τ,Υ,Φ,Χ,Ψ, G, M, Q, F , O, and S (respectively) perpendicular to
Ω,Ισημεῖα,καὶἐπεζεύχθωσαναἱΝΤ,ΝΥ,ΥΦ,ΤΦ,ΧΨ, the reference plane (i.e., the plane of the bases AB and
ΧΩ,ΩΙ,ΙΨ.ἴσονδήἐστιτὸΚΦστερεὸντῷΠΙστερεῷ· CD), and let them have met the plane at points N, T ,
ἐπίτεγὰρἴσωνβάσεώνεἰσιτῶνΚΜ,ΠΣκαὶὑπὸτὸαὐτὸ U, V , W , X, Y , and I (respectively). And let NT , NU,
ὕψος,ὧναἱἐφεστῶσαιπρὸςὀρθάςεἰσιταῖςβάσεσιν.ἀλλὰ UV , TV , WX, WY , Y I, and IX have been joined. So
τὸμὲνΚΦστερεὸντῷΑΕστερεῷἐστινἴσον,τὸδὲΠΙτῷ solid KV is equal to solid QI. For they are on the equal
ΓΖ·ἐπίτεγὰρτῆςαὐτῆςβάσεώςεἰσικαὶὑπὸτὸαὐτὸὕψος, bases KM and QS, and (have) the same height, and the
ὧναἱἐφεστῶσαιοὔκεἰσινἐπὶτῶναὐτῶνεὐθειῶν. καὶτὸ (straight-lines) standing up in them are at right-angles
ΑΕἄραστερεὸντῷΓΖστερεῷἐστινἴσον.
to their bases (see first part of proposition). But, solid
Τὰἄραἐπὶἴσωνβάσεωνὄνταστερεὰπαραλληλεπίπεδα KV is equal to solid AE, and QI to CF . For they are
καὶὑπὸτὸαὐτὸὕψοςἴσαἀλλήλοιςἐστίν·ὅπερἔδειδεῖξαι. on the same base, and (have) the same height, and the
(straight-lines) standing up in them are not on the same
straight-lines [Prop. 11.30]. Thus, solid AE is also equal
to solid CF .
Thus, parallelepiped solids which are on equal bases,
M
G
U
A
E
K
N
H
P
W
O
Y
D
X
I
S
458

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
and (have) the same height, are equal to one another.
(Which is) the very thing it was required to show.
Ð. Proposition 32
Τὰ ὑπὸ τὸ αὐτὸ ὕψος ὄντα στερεὰ παραλληλεπίπεδα Parallelepiped solids which (have) the same height
πρὸςἄλληλάἐστινὡςαἱβάσεις.
are to one another as their bases.
Ã
Ä
À Â
῎Εστωὑπὸτὸαὐτὸὕψοςστερεὰπαραλληλεπίπεδατὰ Let AB and CD be parallelepiped solids (having) the
ΑΒ,ΓΔ·λέγω, ὅτιτὰΑΒ,ΓΔστερεὰπαραλληλεπίπεδα same height. I say that the parallelepiped solids AB and
πρὸςἄλληλάἐστινὡςαἱβάσεις,τουτέστινὅτιἐστὶνὡςἡ CD are to one another as their bases. That is to say, as
ΑΕβάσιςπρὸςτὴνΓΖβάσιν,οὕτωςτὸΑΒστερεὸνπρὸς base AE is to base CF , so solid AB (is) to solid CD.
τὸΓΔστερεόν.
For let FH, equal to AE, have been applied to FG (in
ΠαραβεβλήσθωγὰρπαρὰτὴνΖΗτῷΑΕἴσοντὸΖΘ, the angle FGH equal to angle LCG) [Prop. 1.45]. And
καὶἀπὸβάσεωςμὲντῆςΖΘ,ὕψουςδὲτοῦαὐτοῦτῷΓΔ let the parallelepiped solid GK, (having) the same height
στερεὸνπαραλληλεπίπεδονσυμπεπληρώσθωτὸΗΚ.ἴσον as CD, have been completed on the base FH. So solid
δήἐστιτὸΑΒστερεὸντῷΗΚστερεῷ·ἐπίτεγὰρἴσων AB is equal to solid GK. For they are on the equal bases
βάσεώνεἰσιτῶνΑΕ,ΖΘκαὶὑπὸτὸαὐτοὕψος. καὶἐπεὶ AE and FH, and (have) the same height [Prop. 11.31].
στερεὸνπαραλληλεπίπεδοντὸΓΚἐπιπέδῳτῷΔΗτέτμηται And since the parallelepiped solid CK has been cut by
παραλλήλῳὄντιτοῖςἀπεναντίονἐπιπέδοις,ἔστινἄραὡςἡ the plane DG, which is parallel to the opposite planes (of
ΓΖβάσιςπρὸςτὴνΖΘβάσιν,οὕτωςτὸΓΔστερεὸνπρὸςτὸ CK), thus as the base CF is to the base FH, so the solid
ΔΘστερεόν.ἴσηδὲἡμὲνΖΘβάσιςτῇΑΕβάσει,τὸδὲΗΚ CD (is) to the solid DH [Prop. 11.25]. And base FH (is)
στερεὸντῷΑΒστερεῷ·ἔστινἄρακαὶὡςἡΑΕβάσιςπρὸς equal to base AE, and solid GK to solid AB. And thus
τὴνΓΖβάσιν,οὕτωςτὸΑΒστερεὸνπρὸςτὸΓΔστερεόν. as base AE is to base CF , so solid AB (is) to solid CD.
Τὰἄραὑπὸτὸαὐτὸὕψοςὄνταστερεὰπαραλληλεπίπεδα Thus, parallelepiped solids which (have) the same
πρὸςἄλληλάἐστινὡςαἱβάσεις·ὅπερἔδειδεῖξαι.
A
B
E
L
height are to one another as their bases. (Which is) the
very thing it was required to show.
Ð. Proposition 33
Τὰὅμοιαστερεὰπαραλληλεπίπεδαπρὸςἄλληλαἐντρι- Similar parallelepiped solids are to one another as the
πλασίονιλόγῳεἰσὶτῶνὁμολόγωνπλευρῶν.
cubed ratio of their corresponding sides.
῎Εστω ὅμοια στερεὰ παραλληλεπίπεδα τὰ ΑΒ, ΓΔ, Let AB and CD be similar parallelepiped solids, and
ὁμόλογοςδὲἔστωἡΑΕτῇΓΖ·λέγω,ὅτιτὸΑΒστερεὸν let AE correspond to CF . I say that solid AB has to solid
πρὸςτὸΓΔστερεὸντριπλασίοναλόγονἔχει,ἤπερἡΑΕ CD the cubed ratio that AE (has) to CF .
πρὸςτὴνΓΖ.
For let EK, EL, and EM have been produced in a
᾿Εκβεβλήσθωσανγὰρἐπ᾿εὐθείαςταῖςΑΕ,ΗΕ,ΘΕαἱ straight-line with AE, GE, and HE (respectively). And
ΕΚ,ΕΛ,ΕΜ,καὶκείσθωτῇμὲνΓΖἴσηἡΕΚ,τῇδὲΖΝ let EK be made equal to CF , and EL equal to FN, and,
ἴσηἡΕΛ,καὶἔτιτῇΖΡἴσηἡΕΜ,καὶσυμπεπληρώσθωτὸ further, EM equal to FR. And let the parallelogram KL
ΚΛπαραλληλόγραμμονκαὶτὸΚΟστερεόν. have been completed, and the solid KP .
C
F
D
G
K
H
459

ËÌÇÁÉÁÏƺ ELEMENTS
Æ
Ê
ΚαὶἐπεὶδύοαἱΚΕ,ΕΛδυσὶταῖςΓΖ,ΖΝἴσαιεἰσίν, And since the two (straight-lines) KE and EL are
ἀλλὰκαὶγωνίαἡὑπὸΚΕΛγωνίᾳτῇὑπὸΓΖΝἐστινἴση, equal to the two (straight-lines) CF and FN, but angle
ἐπειδήπερκαὶἡὑπὸΑΕΗτῇὑπὸΓΖΝἐστινἴσηδιὰτὴν KEL is also equal to angle CFN, inasmuch as AEG is
ὁμοιότητατῶνΑΒ,ΓΔστερεῶν,ἴσονἄραἐστὶ[καὶὅμοιον] also equal to CFN, on account of the similarity of the
τὸΚΛπαραλληλόγραμμοντῷΓΝπαραλληλογράμμῳ. διὰ solids AB and CD, parallelogram KL is thus equal [and
τὰαὐτὰδὴκαὶτὸμὲνΚΜπαραλληλόγραμμονἴσονἐστὶκαὶ similar] to parallelogram CN. So, for the same (reasons),
ὅμοιοντῷΓΡ[παραλληλογράμμῳ]καὶἔτιτὸΕΟτῷΔΖ· parallelogram KM is also equal and similar to [parallelτρίαἄραπαραλληλόγραμματοῦΚΟστερεοῦτρισὶπαραλ-
ogram] CR, and, further, EP to DF . Thus, three parληλογράμμοιςτοῦΓΔστερεοῦἴσαἐστὶκαὶὅμοια.ἀλλὰτὰ
allelograms of solid KP are equal and similar to three
μὲντρίατρισὶτοῖςἀπεναντίονἴσαἐστὶκαὶὅμοια,τὰδὲτρία parallelograms of solid CD. But the three (former parτρισὶτοῖςἀπεναντίονἴσαἐστὶκαὶὅμοια·ὅλονἄρατὸΚΟ
allelograms) are equal and similar to the three opposite
στερεὸνὅλῳτῷΓΔστερεῷἴσονἐστὶκαὶὅμοιον. συμπε- (parallelograms), and the three (latter parallelograms)
πληρώσθωτὸΗΚπαραλληλόγραμμον,καὶἀπὸβάσεωνμὲν are equal and similar to the three opposite (paralleloτῶνΗΚ,ΚΛπαραλληλόγραμμων,ὕψουςδὲτοῦαὐτοῦτῷ
grams) [Prop. 11.24]. Thus, the whole of solid KP is
ΑΒστερεὰσυμπεπληρώσθωτὰΕΞ,ΛΠ.καὶἐπεὶδιὰτὴν equal and similar to the whole of solid CD [Def. 11.10].
ὁμοιότητατῶνΑΒ,ΓΔστερεῶνἐστινὡςἡΑΕπρὸςτὴν Let parallelogram GK have been completed. And let the
ΓΖ,οὕτωςἡΕΗπρὸςτὴνΖΝ,καὶἡΕΘπρὸςτὴνΖΡ,ἴση the solids EO and LQ, with bases the parallelograms GK
δὲἡμὲνΓΖτῇΕΚ,ἡδὲΖΝτῇΕΛ,ἡδὲΖΡτῇΕΜ,ἔστιν and KL (respectively), and with the same height as AB,
ἄραὡςἡΑΕπρὸςτὴνΕΚ,οὕτωςἡΗΕπρὸςτὴνΕΛκαὶἡ have been completed. And since, on account of the sim-
ΘΕπρὸςτὴνΕΜ.ἀλλ᾿ὡςμὲνἡΑΕπρὸςτὴνΕΚ,οὕτωςτὸ ilarity of solids AB and CD, as AE is to CF , so EG (is)
ΑΗ[παραλληλόγραμμον]πρὸςτὸΗΚπαραλληλόγραμμον, to FN, and EH to FR [Defs. 6.1, 11.9], and CF (is)
ὡςδὲἡΗΕπρὸςτὴνΕΛ,οὕτωςτὸΗΚπρὸςτὸΚΛ,ὡς equal to EK, and FN to EL, and FR to EM, thus as
δὲἡΘΕπρὸςΕΜ,οὕτωςτὸΠΕπρὸςτὸΚΜ·καὶὡςἄρα AE is to EK, so GE (is) to EL, and HE to EM. But,
τὸΑΗπαραλληλόγραμμονπρὸςτὸΗΚ,οὕτωςτὸΗΚπρὸς as AE (is) to EK, so [parallelogram] AG (is) to paralτὸΚΛκαὶτὸΠΕπρὸςτὸΚΜ.ἀλλ᾿ὡςμὲντὸΑΗπρὸς
lelogram GK, and as GE (is) to EL, so GK (is) to KL,
τὸΗΚ,οὕτωςτὸΑΒστερεὸνπρὸςτὸΕΞστερεόν,ὡςδὲ and as HE (is) to EM, so QE (is) to KM [Prop. 6.1].
τὸΗΚπρὸςτὸΚΛ,οὕτωςτὸΞΕστερεὸνπρὸςτὸΠΛ And thus as parallelogram AG (is) to GK, so GK (is)
στερεόν,ὡςδὲτὸΠΕπρὸςτὸΚΜ,οὕτωςτὸΠΛστερεὸν to KL, and QE (is) to KM. But, as AG (is) to GK, so
πρὸςτὸΚΟστερεόν·καὶὡςἄρατὸΑΒστερεὸνπρὸςτὸ solid AB (is) to solid EO, and as GK (is) to KL, so solid
ΕΞ,οὕτωςτὸΕΞπρὸςτὸΠΛκαὶτὸΠΛπρὸςτὸΚΟ. OE (is) to solid QL, and as QE (is) to KM, so solid QL
ἐὰνδὲτέσσαραμεγέθηκατὰτὸσυνεχὲςἀνάλογονᾖ,τὸ (is) to solid KP [Prop. 11.32]. And, thus, as solid AB
πρῶτονπρὸςτὸτέταρτοντριπλασίοναλόγονἔχειἤπερπρὸς is to EO, so EO (is) to QL, and QL to KP . And if four
τὸδεύτερον·τὸΑΒἄραστερεὸνπρὸςτὸΚΟτριπλασίονα magnitudes are continuously proportional then the first
λόγονἔχειἤπερτὸΑΒπρὸςτὸΕΞ.ἀλλ᾿ὡςτὸΑΒπρὸς has to the fourth the cubed ratio that (it has) to the secτὸΕΞ,οὕτωςτὸΑΗπαραλληλόγραμμονπρὸςτὸΗΚκαὶἡ
ond [Def. 5.10]. Thus, solid AB has to KP the cubed
ΑΕεὐθεῖαπρὸςτὴνΕΚ·ὥστεκαὶτὸΑΒστερεὸνπρὸςτὸ ratio which AB (has) to EO. But, as AB (is) to EO, so
ΚΟτριπλασίοναλόγονἔχειἤπερἡΑΕπρὸςτὴνΕΚ.ἴσον parallelogram AG (is) to GK, and the straight-line AE
δὲτὸ[μὲν]ΚΟστερεὸντῷΓΔστερεῷ,ἡδὲΕΚεὐθεῖα to EK [Prop. 6.1]. Hence, solid AB also has to KP the
τῇΓΖ·καὶτὸΑΒἄραστερεὸνπρὸςτὸΓΔστερεὸντρι- cubed ratio that AE (has) to EK. And solid KP (is)
C
BOOK 11
N
D
F
R
460

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
πλασίοναλόγονἔχειἤπερἡὁμόλογοςαὐτοῦπλευρὰἡΑΕ equal to solid CD, and straight-line EK to CF . Thus,
πρὸςτὴνὁμόλογονπλευρὰντὴνΓΖ.
solid AB also has to solid CD the cubed ratio which its
 È
corresponding side AE (has) to the corresponding side
CF .
B O
À
H Q
Å Ä Ã
G
K
A
E
L
Ç
M
P
Τὰἄραὅμοιαστερεὰπαραλληλεπίπεδαἐντριπλασίονι Thus, similar parallelepiped solids are to one another
λόγῳἐστὶτῶνὁμολόγωνπλευρῶν·ὅπερἔδειδεῖξαι. as the cubed ratio of their corresponding sides. (Which
is) the very thing it was required to show.
ÈÖ×Ñ.
Corollary
᾿Εκ δὴ τούτου φανερόν, ὅτι ἐὰν τέσσαρες εὐθεῖαι So, (it is) clear, from this, that if four straight-lines are
ἀνάλογονὦσιν,ἔσταιὡςἡπρώτηπρὸςτὴντετάρτην,οὕτω (continuously) proportional then as the first is to the
τὸἀπὸτῆςπρώτηςστερεὸνπαραλληλεπίπεδονπρὸςτὸἀπὸ fourth, so the parallelepiped solid on the first will be to
τῆςδευτέραςτὸὅμοιονκαὶὁμοίωςἀναγραφόμενον,ἐπείπερ the similar, and similarly described, parallelepiped solid
καὶἡπρώτηπρὸςτὴντετάρτηντριπλασίοναλόγονἔχειἤπερ on the second, since the first also has to the fourth the
πρὸςτὴνδευτέραν.
cubed ratio that (it has) to the second.
Ð. Proposition 34 †
Τῶνἴσωνστερεῶνπαραλληλεπιπέδωνἀντιπεπόνθασιν The bases of equal parallelepiped solids are recipαἱβάσειςτοῖςὕψεσιν·καὶὧνστερεῶνπαραλληλεπιπέδων
rocally proportional to their heights. And those paral-
ἀντιπεπόνθασιναἱβάσειςτοῖςὕψεσιν,ἵσαἐστὶνἐκεῖνα. lelepiped solids whose bases are reciprocally proportional
῎ΕστωἴσαστερεὰπαραλληλεπίπεδατὰΑΒ,ΓΔ·λέγω, to their heights are equal.
ὅτιτῶνΑΒ,ΓΔστερεῶνπαραλληλεπιπέδωνἀντιπεπόνθασιν Let AB and CD be equal parallelepiped solids. I say
αἱβάσειςτοῖςὕψεσιν,καίἐστινὡςἡΕΘβάσιςπρὸςτὴν that the bases of the parallelepiped solids AB and CD
ΝΠβάσιν,οὕτωςτὸτοῦΓΔστερεοῦὕψοςπρὸςτὸτοῦΑΒ are reciprocally proportional to their heights, and (so) as
στερεοῦὕψος.
base EH is to base NQ, so the height of solid CD (is) to
῎ΕστωσανγὰρπρότεροναἱἐφεστηκυῖαιαἱΑΗ,ΕΖ,ΛΒ, the height of solid AB.
ΘΚ,ΓΜ,ΝΞ,ΟΔ,ΠΡπρὸςὀρθὰςταῖςβάσεσιναὐτῶν· For, first of all, let the (straight-lines) standing up,
λέγω,ὅτιἐστὶνὡςἡΕΘβάσιςπρὸςτὴνΝΠβάσιν,οὕτως AG, EF , LB, HK, CM, NO, PD, and QR, be at right-
ἡΓΜπρὸςτὴνΑΗ.
angles to their bases. I say that as base EH is to base
ΕἰμὲνοὖνἴσηἐστὶνἡΕΘβάσιντῇΝΠβάσει,ἔστιδὲ NQ, so CM (is) to AG.
καὶτὸΑΒστερεὸντῷΓΔστερεῷἴσον,ἔσταικαὶἡΓΜτῇ Therefore, if base EH is equal to base NQ, and solid
ΑΗἴση.τὰγὰρὑπὸτὸαὐτὸὕψοςστερεὰπαραλληλεπίπεδα AB is also equal to solid CD, CM will also be equal to
πρὸςἄλληλάἐστινὡςαἱβάσεις. καὶἔσταιὡςἡΕΘβάσις AG. For parallelepiped solids of the same height are to
πρὸςτὴνΝΠ,οὕτωςἡΓΜπρὸςτὴνΑΗ,καὶφανερόν,ὅτι one another as their bases [Prop. 11.32]. And as base
461

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
τῶνΑΒ,ΓΔστερεῶνπαραλληλεπιπέδωνἀντιπεπόνθασιναἱ EH (is) to NQ, so CM will be to AG. And (so it is) clear
βάσειςτοῖςὕψεσιν.
that the bases of the parallelepiped solids AB and CD
Ã
are reciprocally proportional to their heights.
ÊÅ
R D
À Ì
M O
K B
V
Â
Ä
È Ç
G F
T
Æ
H L Q P
A E C N
ΜὴἔστωδὴἴσηἡΕΘβάσιςτῇΝΠβάσει,ἀλλ᾿ἔστω So let base EH not be equal to base NQ, but let EH
μείζωνἡΕΘ.ἔστιδὲκαὶτὸΑΒστερεὸντῷΓΔστερεῷ be greater. And solid AB is also equal to solid CD. Thus,
ἴσον·μείζωνἄραἐστὶκαὶἡΓΜτῆςΑΗ.κείσθωοὖντῇ CM is also greater than AG. Therefore, let CT be made
ΑΗἴσηἡΓΤ,καὶσυμπεπληρώσθωἀπὸβάσεωςμὲντῆς equal to AG. And let the parallelepiped solid V C have
ΝΠ,ὕψουςδὲτοῦΓΤ,στερεὸνπαραλληλεπίπεδοντὸΦΓ. been completed on the base NQ, with height CT . And
καὶἐπεὶἴσονἐστὶτὸΑΒστερεὸντῷΓΔστερεῷ,ἔξωθενδὲ since solid AB is equal to solid CD, and CV (is) extrinsic
τὸΓΦ,τὰδὲἴσαπρὸςτὸαὐτὸτὸναὐτὸνἔχειλόγον,ἔστιν (to them), and equal (magnitudes) have the same ratio to
ἄραὡςτὸΑΒστερεὸνπρὸςτὸΓΦστερεόν,οὕτωςτὸΓΔ the same (magnitude) [Prop. 5.7], thus as solid AB is to
στερεὸνπρὸςτὸΓΦστερεόν.ἀλλ᾿ὡςμὲντὸΑΒστερεὸν solid CV , so solid CD (is) to solid CV . But, as solid AB
πρὸςτὸΓΦστερεόν,οὕτωςἡΕΘβάσιςπρὸςτὴνΝΠβάσιν· (is) to solid CV , so base EH (is) to base NQ. For the
ἰσοϋψῆγὰρτὰΑΒ,ΓΦστερεά·ὡςδὲτὸΓΔστερεὸνπρὸς solids AB and CV (are) of equal height [Prop. 11.32].
τὸΓΦστερεόν, οὕτωςἡΜΠβάσιςπρὸςτὴνΤΠβάσιν And as solid CD (is) to solid CV , so base MQ (is) to base
καὶἡΓΜπρὸςτὴνΓΤ·καὶὡςἄραἡΕΘβάσιςπρὸςτὴν TQ [Prop. 11.25], and CM to CT [Prop. 6.1]. And, thus,
ΝΠβάσιν,οὕτωςἡΜΓπρὸςτὴνΓΤ.ἴσηδὲἡΓΤτῇΑΗ· as base EH is to base NQ, so MC (is) to AG. And CT
καὶὡςἄραἡΕΘβάσιςπρὸςτὴνΝΠβάσιν,οὕτωςἡΜΓ (is) equal to AG. And thus as base EH (is) to base NQ,
πρὸςτὴνΑΗ.τῶνΑΒ,ΓΔἄραστερεῶνπαραλληλεπιπέδων so MC (is) to AG. Thus, the bases of the parallelepiped
ἀντιπεπόνθασιναἱβάσειςτοῖςὕψεσιν.
solids AB and CD are reciprocally proportional to their
ΠάλινδὴτῶνΑΒ,ΓΔστερεῶνπαραλληλεπιπέδωνἀντι- heights.
πεπονθέτωσαναἱβάσειςτοῖςὕψεσιν, καὶἔστωὡςἡΕΘ So, again, let the bases of the parallelepipid solids AB
βάσιςπρὸςτὴνΝΠβάσιν,οὕτωςτὸτοῦΓΔστερεοῦὕψος and CD be reciprocally proportional to their heights, and
πρὸςτὸτοῦΑΒστερεοῦὕψος·λέγω,ὅτιἴσονἐστὶτὸΑΒ let base EH be to base NQ, as the height of solid CD (is)
στερεὸντῷΓΔστερεῷ.
to the height of solid AB. I say that solid AB is equal to
῎Εστωσαν[γὰρ]πάλιναἱἐφεστηκυῖαιπρὸςὀρθὰςταῖς solid CD. [For] let the (straight-lines) standing up again
βάσεσιν.καὶεἰμὲνἴσηἐστὶνἡΕΘβάσιςτῇΝΠβάσει,καί be at right-angles to the bases. And if base EH is equal
ἐστινὡςἡΕΘβάσιςπρὸςτὴνΝΠβάσιν,οὕτωςτὸτοῦ to base NQ, and as base EH is to base NQ, so the height
ΓΔστερεοῦὕψοςπρὸςτὸτοῦΑΒστερεοῦὕψος,ἴσονἄρα of solid CD (is) to the height of solid AB, the height of
ἐστὶκαὶτὸτοῦΓΔστερεοῦὕψοςτῷτοῦΑΒστερεοῦὕψει. solid CD is thus also equal to the height of solid AB.
τὰδὲἐπὶἴσωνβάσεωνστερεάπαραλληλεπίπεδακαὶὑπὸτὸ And parallelepiped solids on equal bases, and also with
αὐτὸὕψοςἴσαἀλλήλοιςἐστίν·ἴσονἄραἐστὶτὸΑΒστερεὸν the same height, are equal to one another [Prop. 11.31].
τῷΓΔστερεῷ.
Thus, solid AB is equal to solid CD.
ΜὴἔστωδὴἡΕΘβάσιςτῇΝΠ[βάσει]ἴση,ἀλλ᾿ἔστω So, let base EH not be equal to [base] NQ, but let
μείζωνἡΕΘ·μεῖζονἄραἐστὶκαὶτὸτοῦΓΔστερεοῦὕψος EH be greater. Thus, the height of solid CD is also
τοῦ τοῦ ΑΒ στερεοῦ ὕψους, τουτέστιν ἡ ΓΜ τῆς ΑΗ. greater than the height of solid AB, that is to say CM
κείσθωτῇΑΗἴσηπάλινἡΓΤ,καὶσυμπεπληρώσθωὁμοίως (greater) than AG. Let CT again be made equal to AG,
τὸΓΦστερεόν. ἐπείἐστινὡςἡΕΘβάσιςπρὸςτὴνΝΠ and let the solid CV have been similarly completed.
βάσιν, οὕτωςἡΜΓπρὸςτὴνΑΗ,ἴσηδὲἡΑΗτῇΓΤ, Since as base EH is to base NQ, so MC (is) to AG,
462

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
ἔστινἄραὡςἡΕΘβάσιςπρὸςτὴνΝΠβάσιν,οὕτωςἡΓΜ and AG (is) equal to CT , thus as base EH (is) to base
πρὸςτὴνΓΤ.ἀλλ᾿ὡςμὲνἡΕΘ[βάσις]πρὸςτὴνΝΠβάσιν, NQ, so CM (is) to CT . But, as [base] EH (is) to base
οὕτωςτὸΑΒστερεὸνπρὸςτὸΓΦστερεόν· ἰσοϋψῆγάρ NQ, so solid AB (is) to solid CV . For solids AB and CV
ἐστιτὰΑΒ,ΓΦστερεά·ὡςδὲἡΓΜπρὸςτὴνΓΤ,οὕτως are of equal heights [Prop. 11.32]. And as CM (is) to
ἥτεΜΠ βάσιςπρὸςτὴνΠΤ βάσινκαὶτὸΓΔστερεὸν CT , so (is) base MQ to base QT [Prop. 6.1], and solid
πρὸςτὸΓΦστερεόν. καὶὡςἄρατὸΑΒστερεὸνπρὸςτὸ CD to solid CV [Prop. 11.25]. And thus as solid AB (is)
ΓΦστερεόν,οὕτωςτὸΓΔστερεὸνπρὸςτὸΓΦστερεόν· to solid CV , so solid CD (is) to solid CV . Thus, AB and
ἑκάτερονἄρατῶν
Ã
ΑΒ, ΓΔ πρὸςτὸΓΦ τὸν αὐτὸνἔχει CD each have the same ratio to CV . Thus, solid AB is
λόγον.ἴσονἄραἐστὶτὸΑΒστερεὸντῷΓΔστερεῷ. equal to solid CD [Prop. 5.9].
À
Â
Ä
K
B
Í G
F
Ì Ë
L
ÇÏ
H V
U
A
Ê È Æ
E
T
S
a
D
P W
N
O
Y
Å
R
Q X
M
C
ΜὴἔστωσανδὴαἱἐφεστηκυῖαιαἱΖΕ,ΒΛ,ΗΑ,ΚΘ, So, let the (straight-lines) standing up, FE, BL, GA,
ΞΝ, ΔΟ,ΜΓ, ΡΠ πρὸςὀρθὰςταῖςβάσεσιναὐτῶν, καὶ KH, ON, DP , MC, and RQ, not be at right-angles to
ἤχθωσανἀπὸτῶνΖ,Η,Β,Κ,Ξ,Μ,Ρ,Δσημείωνἐπὶ their bases. And let perpendiculars have been drawn to
τὰδιὰτῶνΕΘ,ΝΠἐπίπεδακάθετοικαὶσυμβαλλέτωσαν the planes through EH and NQ from points F , G, B, K,
τοῖςἐπιπέδοιςκατὰτὰΣ,Τ,Υ,Φ,Χ,Ψ,Ω,ς,καὶσυμ- O, M, R, and D, and let them have joined the planes at
πεπληρώσθωτὰΖΦ,ΞΩστερεά·λέγω,ὅτικαὶοὕτωςἴσων (points) S, T , U, V , W , X, Y , and a (respectively). And
ὄντωντῶνΑΒ,ΓΔστερεῶνἀντιπεπόνθασιναἱβάσειςτοῖς let the solids FV and OY have been completed. In this
ὕψεσιν,καίἐστινὡςἡΕΘβἁσινπρὸςτὴνΝΠβάσιν,οὕτως case, also, I say that the solids AB and CD being equal,
τὸτοῦΓΔστερεοῦὕψοςπρὸςτὸτοῦΑΒστερεοῦὕψος. their bases are reciprocally proportional to their heights,
᾿ΕπεὶἴσονἐστὶτὸΑΒστερεὸντῷΓΔστερεῷ,ἀλλὰτὸ and (so) as base EH is to base NQ, so the height of solid
μὲνΑΒτῷΒΤἐστινἴσον·ἐπίτεγὰρτῆςαὐτῆςβάσεώς CD (is) to the height of solid AB.
εἰσιτῆςΖΚκαὶὑπὸτὸαὐτὸὕψος·τὸδὲΓΔστερεὸντῷ Since solid AB is equal to solid CD, but AB is equal
ΔΨἐστινἴσον·ἐπίτεγὰρπάλιντῆςαὐτῆςβάσεώςεἰσιτῆς to BT . For they are on the same base FK, and (have) the
ΡΞκαὶὑπὸτὸαὐτὸὕψος·καὶτὸΒΤἄραστερεὸντῷΔΨ same height [Props. 11.29, 11.30]. And solid CD is equal
στερεῷἴσονἐστίν. ἔστινἄραὡςἡΖΚβάσιςπρὸςτὴνΞΡ is equal to DX. For, again, they are on the same base RO,
βάσιν,οὕτωςτὸτοῦΔΨστερεοῦὕψοςπρὸςτὸτοῦΒΤ and (have) the same height [Props. 11.29, 11.30]. Solid
στερεοῦὕψος. ἴσηδὲἡμὲνΖΚβάσιςτῇΕΘβάσει,ἡδὲ BT is thus also equal to solid DX. Thus, as base FK (is)
ΞΡβάσιςτῇΝΠβάσει·ἔστινἄραὡςἡΕΘβάσιςπρὸςτὴν to base OR, so the height of solid DX (is) to the height
ΝΠβάσιν,οὕτωςτὸτοῦΔΨστερεοῦὕψοςπρὸςτὸτοῦΒΤ of solid BT (see first part of proposition). And base FK
στερεοῦὕψος. τὰδ᾿αὐτὰὕψηἐστὶτῶνΔΨ,ΒΤστερεῶν (is) equal to base EH, and base OR to NQ. Thus, as
καὶτῶνΔΓ,ΒΑ·ἔστινἄραὡςἡΕΘβάσιςπρὸςτὴνΝΠ base EH is to base NQ, so the height of solid DX (is) to
463

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
βάσιν,οὕτωςτὸτοῦΔΓστερεοῦὕψοςπρὸςτὸτοῦΑΒστε- the height of solid BT . And solids DX, BT are the same
ρεοῦὕψος. τῶνΑΒ,ΓΔἄραστερεῶνπαραλληλεπιπέδων height as (solids) DC, BA (respectively). Thus, as base
ἀντιπεπόνθασιναἱβάσειςτοῖςὕψεσιν.
EH is to base NQ, so the height of solid DC (is) to the
ΠάλινδὴτῶνΑΒ,ΓΔστερεῶνπαραλληλεπιπέδωνἀντι- height of solid AB. Thus, the bases of the parallelepiped
πεπονθέτωσαναἱβάσειςτοῖςὕψεσιν, καὶἔστωὡςἡΕΘ solids AB and CD are reciprocally proportional to their
βάσιςπρὸςτὴνΝΠβάσιν,οὕτωςτὸτοῦΓΔστερεοῦὕψος heights.
πρὸςτὸτοῦΑΒστερεοῦὕψος·λέγω,ὅτιἴσονἐστὶτὸΑΒ So, again, let the bases of the parallelepiped solids
στερεὸντῷΓΔστερεῷ.
AB and CD be reciprocally proportional to their heights,
Τῶνγὰραὐτῶνκατασκευασθέντων,ἐπείἐστινὡςἡΕΘ and (so) let base EH be to base NQ, as the height of
βάσιςπρὸςτὴνΝΠβάσιν,οὕτωςτὸτοῦΓΔστερεοῦὕψος solid CD (is) to the height of solid AB. I say that solid
πρὸςτὸτοῦΑΒστερεοῦὕψος,ἴσηδὲἡμὲνΕΘβάσιςτῇ AB is equal to solid CD.
ΖΚβάσει,ἡδὲΝΠτῇΞΡ,ἔστινἄραὡςἡΖΚβάσιςπρὸς For, with the same construction (as before), since as
τὴνΞΡβάσιν, οὕτωςτὸτοῦΓΔστερεοῦὕψοςπρὸςτὸ base EH is to base NQ, so the height of solid CD (is) to
τοῦΑΒστερεοῦὕψος. τὰδ᾿αὐτὰὕψηἐστὶτῶνΑΒ,ΓΔ the height of solid AB, and base EH (is) equal to base
στερεῶνκαὶτῶνΒΤ,ΔΨ·ἔστινἄραὡςἡΖΚβάσιςπρὸς FK, and NQ to OR, thus as base FK is to base OR,
τὴνΞΡβάσιν,οὕτωςτὸτοῦΔΨστερεοῦὕψοςπρὸςτὸτοῦ so the height of solid CD (is) to the height of solid AB.
ΒΤστερεοῦὕψος. τῶνΒΤ,ΔΨἄραστερεῶνπαραλληλε- And solids AB, CD are the same height as (solids) BT ,
πιπέδωνἀντιπεπόνθασιναἱβάσειςτοῖςὕψεσιν·ἴσονἄραἐστὶ DX (respectively). Thus, as base FK is to base OR, so
τὸΒΤστερεὸντῷΔΨστερεῷ. ἀλλὰτὸμὲνΒΤτῷΒΑ the height of solid DX (is) to the height of solid BT .
ἴσονἐστίν·ἐπίτεγὰρτῆςαὐτῆςβάσεως[εἰσι]τῆςΖΚκαὶ Thus, the bases of the parallelepiped solids BT and DX
ὑπὸτὸαὐτὸὕψος.τὸδὲΔΨστερεὸντῷΔΓστερεῷἴσον are reciprocally proportional to their heights. Thus, solid
ἐστίν. καὶτὸΑΒἄραστερεὸντῷΓΔστερεῷἐστινἴσον· BT is equal to solid DX (see first part of proposition).
ὅπερἔδειδεῖξαι.
But, BT is equal to BA. For [they are] on the same base
FK, and (have) the same height [Props. 11.29, 11.30].
And solid DX is equal to solid DC [Props. 11.29, 11.30].
Thus, solid AB is also equal to solid CD. (Which is) the
very thing it was required to show.
† This proposition assumes that (a) if two parallelepipeds are equal, and have equal bases, then their heights are equal, and (b) if the bases of
two equal parallelepipeds are unequal, then that solid which has the lesser base has the greater height.
Ð. Proposition 35
᾿Εὰν ὦσι δύο γωνίαι ἐπίπεδοι ἴσαι, ἐπὶ δὲ τῶν κο- If there are two equal plane angles, and raised
ρυφῶν αὐτῶν μετέωροι εὐθεῖαιἐπισταθῶσιν ἴσαςγωνίας straight-lines are stood on the apexes of them, containing
περιέχουσαιμετὰτῶνἐξἀρχῆςεὐθειῶνἑκατέρανἑκατέρᾳ, equal angles respectively with the original straight-lines
ἐπὶδὲτῶνμετεώρωνληφθῇτυχόντασημεῖα,καὶἀπ᾿αὐτῶν (forming the angles), and random points are taken on
ἐπὶτὰἐπίπεδα, ἐνοἷςεἰσιναἱἐξἀρχῆςγωνίαι, κάθετοι the raised (straight-lines), and perpendiculars are drawn
ἀχθῶσιν,ἀπὸδὲτῶνγενομένωνσημείωνἐντοῖςἐπιπέδοις from them to the planes in which the original angles are,
ἐπὶτὰςἐξἀρχῆςγωνίαςἐπιζευχθῶσινεὐθεῖαι,ἴσαςγωνίας and straight-lines are joined from the points created in
περιέξουσιμετὰτῶνμετεώρων.
the planes to the (vertices of the) original angles, then
῎ΕστωσανδύογωνίαιεὐθύγραμμοιἴσαιαἱὑπὸΒΑΓ, they will enclose equal angles with the raised (straight-
ΕΔΖ,ἀπὸδὲτῶνΑ,Δσημείωνμετέωροιεὐθεῖαιἐφεστάτ- lines).
ωσαναἱΑΗ,ΔΜἴσαςγωνίαςπεριέχουσινμετὰτῶνἐξ Let BAC and EDF be two equal rectilinear angles.
ἀρχῆςεὐθειῶνἑκατέρανἑκατέρᾳ,τὴνμὲνὑπὸΜΔΕτῇ And let the raised straight-lines AG and DM have been
ὑπὸΗΑΒ,τὴνδὲὑπὸΜΔΖτῇὑπὸΗΑΓ,καὶεἰλήφθω stood on points A and D, containing equal angles respec-
ἐπὶτῶνΑΗ,ΔΜτυχόντασημεῖατὰΗ,Μ,καὶἤχθωσαν tively with the original straight-lines. (That is) MDE
ἀπὸτῶνΗ,ΜσημείωνἐπὶτὰδιὰτῶνΒΑΓ,ΕΔΖἐπίπεδα (equal) to GAB, and MDF (to) GAC. And let the ranκάθετοι
αἱ ΗΛ, ΜΝ, καὶ συμβαλλέτωσαν τοῖς ἐπιπέδοις dom points G and M have been taken on AG and DM
κατὰτὰΛ,Ν,καὶἐπεζεύχθωσαναἱΛΑ,ΝΔ·λέγω,ὅτιἴση (respectively). And let the GL and MN have been drawn
ἐστὶνἡὑπὸΗΑΛγωνίατῇὑπὸΜΔΝγωνίᾳ.
from points G and M perpendicular to the planes through
464

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
 Ã
Ä
Å
Æ
BAC and EDF (respectively). And let them have joined
the planes at points L and N (respectively). And let LA
and ND have been joined. I say that angle GAL is equal
to angle MDN.
ΚείσθωτῇΔΜἴσηἡΑΘ,καὶἤχθωδιὰτοῦΘσημείου Let AH be made equal to DM. And let HK have been
τῇΗΛπαράλληλοςἡΘΚ.ἡδὲΗΛκάθετόςἐστινἐπὶτὸ drawn through point H parallel to GL. And GL is perδιὰτῶνΒΑΓἐπίπεδον·καὶἡΘΚἄρακάθετόςἐστινἐπὶτὸ
pendicular to the plane through BAC. Thus, HK is also
διὰτῶνΒΑΓἐπίπεδον. ἤχθωσανἀπὸτῶνΚ,Νσημείων perpendicular to the plane through BAC [Prop. 11.8].
ἐπὶτὰςΑΓ,ΔΖ,ΑΒ,ΔΕεὐθείαςκάθετοιαἱΚΓ,ΝΖ,ΚΒ, And let KC, NF , KB, and NE have been drawn from
ΝΕ,καὶἐπεζεύχθωσαναἱΘΓ,ΓΒ,ΜΖ,ΖΕ.ἐπεὶτὸἀπὸ points K and N perpendicular to the straight-lines AC,
τῆςΘΑἴσονἐστὶτοῖςἀπὸτῶνΘΚ,ΚΑ,τῷδὲἀπὸτῆςΚΑ DF , AB, and DE. And let HC, CB, MF , and FE have
ἴσαἐστὶτὰἀπὸτῶνΚΓ,ΓΑ,καὶτὸἀπὸτῆςΘΑἄραἴσον been joined. Since the (square) on HA is equal to the
ἐστὶτοῖςἀπὸτῶνΘΚ,ΚΓ,ΓΑ.τοῖςδὲἀπὸτῶνΘΚ,ΚΓ (sum of the squares) on HK and KA [Prop. 1.47], and
ἴσονἐστὶτὸἀπὸτῆςΘΓ·τὸἄραἀπὸτῆςΘΑἴσονἐστὶτοῖς the (sum of the squares) on KC and CA is equal to the
ἀπὸτῶνΘΓ,ΓΑ.ὀρθὴἄραἐστὶνἡὑπὸΘΓΑγωνία.διὰτὰ (square) on KA [Prop. 1.47], thus the (square) on HA
αὐτὰδὴκαὶἡὑπὸΔΖΜγωνίαὀρθήἐστιν. ἴσηἄραἐστὶν is equal to the (sum of the squares) on HK, KC, and
ἡὑπὸΑΓΘγωνίατῇὑπὸΔΖΜ.ἔστιδὲκαὶἡὑπὸΘΑΓ CA. And the (square) on HC is equal to the (sum of
τῇὑπὸΜΔΖἴση.δύοδὴτρίγωνάἐστιτὰΜΔΖ,ΘΑΓδύο the squares) on HK and KC [Prop. 1.47]. Thus, the
γωνίαςδυσὶγωνίαιςἴσαςἔχονταἑκατέρανἑκατέρᾳκαὶμίαν (square) on HA is equal to the (sum of the squares)
πλευρὰνμιᾷπλευρᾷἴσηντὴνὑποτείνουσανὑπὸμίαντῶν on HC and CA. Thus, angle HCA is a right-angle
ἴσωνγωνιῶντὴνΘΑτῇΜΔ·καὶτὰςλοιπὰςἄραπλευρὰς [Prop. 1.48]. So, for the same (reasons), angle DFM
ταῖςλοιπαῖςπλευραῖςἴσαςἕξειἑκατέρανἑκαρέρᾳ. ἴσηἄρα is also a right-angle. Thus, angle ACH is equal to (an-
ἐστὶνἡΑΓτῇΔΖ.ὁμοίωςδὴδείξομεν,ὅτικαὶἡΑΒτῇ gle) DFM. And HAC is also equal to MDF . So, MDF
ΔΕἐστινἴση. ἐπεὶοὖνἴσηἐστὶνἡμὲνΑΓτῇΔΖ,ἡδὲ and HAC are two triangles having two angles equal to
ΑΒτῇΔΕ,δύοδὴαἱΓΑ,ΑΒδυσὶταῖςΖΔ,ΔΕἴσαιεἰσίν. two angles, respectively, and one side equal to one side—
ἀλλὰκαὶγωνίαἡὑπὸΓΑΒγωνίᾳτῇὑπὸΖΔΕἐστινἴση· (namely), that subtending one of the equal angles —(that
βάσιςἄραἡΒΓβάσειτῇΕΖἴσηἐστὶκαὶτὸτρίγωνοντῷ is), HA (equal) to MD. Thus, they will also have the reτριγώνῳκαὶαἱλοιπαὶγωνίαιταῖςλοιπαῖςγωνίαις·ἴσηἄραἡ
maining sides equal to the remaining sides, respectively
ὑπὸΑΓΒγωνίατῇὑπὸΔΖΕ.ἔστιδὲκαὶὀρθὴἡὑπὸΑΓΚ [Prop. 1.26]. Thus, AC is equal to DF . So, similarly, we
ὀρθῇτῇὑπὸΔΖΝἴση·καὶλοιπὴἄραἡὑπὸΒΓΚλοιπῇτῇ can show that AB is also equal to DE. Therefore, since
ὑπὸΕΖΝἐστινἴση.διὰτὰαὐτὰδὴκαὶἡὑπὸΓΒΚτῇὑπὸ AC is equal to DF , and AB to DE, so the two (straight-
ΖΕΝἐστινἴση. δύοδὴτρίγωνάἐστιτὰΒΓΚ,ΕΖΝ[τὰς] lines) CA and AB are equal to the two (straight-lines)
δύογωνίαςδυσὶγωνίαιςἴσαςἔχονταἑκατέρανἑκατέρᾳκαὶ FD and DE (respectively). But, angle CAB is also equal
μίανπλευρὰνμιᾷπλευρᾷἴσηντὴνπρὸςταῖςἴσαιςγωνίαις to angle FDE. Thus, base BC is equal to base EF , and
τὴνΒΓτῇΕΖ·καὶτὰςλοιπὰςἄραπλευρὰςταῖςλοιπαῖς triangle (ACB) to triangle (DFE), and the remaining
πλευραῖςἴσαςἕξουσιν.ἴσηἄραἐστὶνἡΓΚτῇΖΝ.ἔστιδὲ angles to the remaining angles (respectively) [Prop. 1.4].
A
B
H
C
K
G
L
D
E
F
M
N
465

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
καὶἡΑΓτῇΔΖἴση·δύοδὴαἱΑΓ,ΓΚδυσὶταῖςΔΖ,ΖΝ Thus, angle ACB (is) equal to DFE. And the right-angle
ἴσαιεἰσίν·καὶὀρθὰςγωνίαςπεριέχουσιν. βάσιςἄραἡΑΚ ACK is also equal to the right-angle DFN. Thus, the
βάσειτῇΔΝἴσηἐστίν. καὶἐπεὶἴσηἐστὶνἡΑΘτῇΔΜ, remainder BCK is equal to the remainder EFN. So,
ἴσονἐστὶκαὶτὸἀπὸτῆςΑΘτῷἀπὸτῆςΔΜ.ἀλλὰτῷμὲν for the same (reasons), CBK is also equal to FEN.
ἀπὸτῆςΑΘἴσαἐστὶτὰἀπὸτῶνΑΚ,ΚΘ·ὀρθὴγὰρἡὑπὸ So, BCK and EFN are two triangles having two an-
ΑΚΘ·τῷδὲἀπὸτῆςΔΜἴσατὰἀπὸτῶνΔΝ,ΝΜ·ὀρθὴ gles equal to two angles, respectively, and one side equal
γὰρἡὑπὸΔΝΜ·τὰἄραἀπὸτῶνΑΚ,ΚΘἴσαἐστὶτοῖς to one side—(namely), that by the equal angles—(that
ἀπὸτῶνΔΝ,ΝΜ,ὧντὸἀπὸτῆςΑΚἴσονἐστὶτῷἀπὸτῆς is), BC (equal) to EF . Thus, they will also have the re-
ΔΝ·λοιπὸνἄρατὸἀπὸτῆςΚΘἴσονἐστὶτῷἀπὸτῆςΝΜ· maining sides equal to the remaining sides (respectively)
ἴσηἄραἡΘΚτῇΜΝ.καὶἐπεὶδύοαἱΘΑ,ΑΚδυσὶταῖς [Prop. 1.26]. Thus, CK is equal to FN. And AC (is) also
ΜΔ,ΔΝἴσαιεἰσὶνἑκατέραἑκατέρᾳ,καὶβάσιςἡΘΚβάσει equal to DF . So, the two (straight-lines) AC and CK are
τῇΜΝἐδείχθηἴση,γωνίαἄραἡὑπὸΘΑΚγωνίᾳτῇὑπὸ equal to the two (straight-lines) DF and FN (respec-
ΜΔΝἐστινἴση.
tively). And they enclose right-angles. Thus, base AK is
᾿Εὰνἄραὦσιδύογωνίαιἐπίπεδοιἴσαικαὶτὰἑξῆςτῆς equal to base DN [Prop. 1.4]. And since AH is equal to
προτάσεως[ὅπερἔδειδεῖξαι].
DM, the (square) on AH is also equal to the (square) on
DM. But, the the (sum of the squares) on AK and KH
is equal to the (square) on AH. For angle AKH (is) a
right-angle [Prop. 1.47]. And the (sum of the squares)
on DN and NM (is) equal to the square on DM. For angle
DNM (is) a right-angle [Prop. 1.47]. Thus, the (sum
of the squares) on AK and KH is equal to the (sum of
the squares) on DN and NM, of which the (square) on
AK is equal to the (square) on DN. Thus, the remaining
(square) on KH is equal to the (square) on NM. Thus,
HK (is) equal to MN. And since the two (straight-lines)
HA and AK are equal to the two (straight-lines) MD
and DN, respectively, and base HK was shown (to be)
equal to base MN, angle HAK is thus equal to angle
MDN [Prop. 1.8].
Thus, if there are two equal plane angles, and so on
of the proposition. [(Which is) the very thing it was required
to show].
ÈÖ×Ñ.
᾿Εκδὴτούτουφανερόν,ὅτι,ἑὰνὦσιδύογωνίαιἐπίπεδοι So, it is clear, from this, that if there are two equal
Corollary
ἴσαι,ἐπισταθῶσιδὲἐπ᾿αὐτῶνμετέωροιεὐθεῖαιἴσαιἴσας plane angles, and equal raised straight-lines are stood
γωνίαςπεριέχουσαιμετὰτῶνἐξἀρχῆςεὐθειῶνἑκατέραν on them (at their apexes), containing equal angles re-
ἑκατέρᾳ,αἱἀπ᾿αὐτῶνκάθετοιἀγόμεναιἐπὶτὰἐπίπεδα,ἐν spectively with the original straight-lines (forming the
οἷςεἰσιναἱἐξἀρχῆςγωνίαι,ἴσαιἀλλήλαιςεἰσίν. ὅπερἔδει angles), then the perpendiculars drawn from (the raised
δεῖξαι.
ends of) them to the planes in which the original angles
lie are equal to one another. (Which is) the very thing it
was required to show.
Ц. Proposition 36
᾿Εὰν τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, τὸ ἐκ τῶν τριῶν If three straight-lines are (continuously) proportional
στερεὸν παραλληλεπίπεδονἴσον ἐστὶ τῷ ἀπὸ τῆς μέσης then the parallelepiped solid (formed) from the three
στερεῷπαραλληλεπιπέδῳἰσοπλεύρῳμέν,ἰσογωνίῳδὲτῷ (straight-lines) is equal to the equilateral parallelepiped
προειρημένῳ.
solid on the middle (straight-line which is) equiangular
to the aforementioned (parallelepiped solid).
466

ËÌÇÁÉÁÏƺ ELEMENTS
Â
Å
Ä
Æ
Ã
À
H
A
B
M
O
L
N
BOOK 11
C
῎ΕστωσαντρεῖςεὐθεῖαιἀνάλογοναἱΑ,Β,Γ,ὡςἡΑ Let A, B, and C be three (continuously) proportional
πρὸςτὴνΒ,οὕτωςἡΒπρὸςτὴνΓ·λέγω,ὅτιτὸἐκτῶν straight-lines, (such that) as A (is) to B, so B (is) to C.
Α,Β,ΓστερεὸνἴσονἐστὶτῷἀπὸτῆςΒστερεῷἰσοπλεύρῳ I say that the (parallelepiped) solid (formed) from A, B,
μέν,ἰσογωνίῳδὲτῷπροειρημένῳ.
and C is equal to the equilateral solid on B (which is)
᾿ΕκκείσθωστερεὰγωνίαἡπρὸςτῷΕπεριεχομένηὑπὸ equiangular with the aforementioned (solid).
τῶνὑπὸΔΕΗ,ΗΕΖ,ΖΕΔ,καὶκείσθωτῇμὲνΒἴσηἑκάστη Let the solid angle at E, contained by DEG, GEF ,
τῶνΔΕ,ΗΕ,ΕΖ,καὶσυμπεπληρώσθωτὸΕΚστερεὸνπα- and FED, be set out. And let DE, GE, and EF each
ραλληλεπίπεδον,τῇδὲΑἴσηἡΛΜ,καὶσυνεστάτωπρὸς be made equal to B. And let the parallelepiped solid
τῇΛΜεὐθείᾳκαὶτῷπρὸςαὐτῇσημείῳτῷΛτῇπρὸςτῷ EK have been completed. And (let) LM (be made)
Εστερεᾷγωνίᾳἴσηστερεὰγωνίαἡπερειχομένηὑπὸτῶν equal to A. And let the solid angle contained by NLO,
ΝΛΞ,ΞΛΜ,ΜΛΝ,καὶκείσθωτῇμὲνΒἴσηἡΛΞ,τῇδὲ OLM, and MLN have been constructed on the straight-
ΓἴσηἡΛΝ.καὶἐπείἐστινὡςἡΑπρὸςτὴνΒ,οὕτωςἡΒ line LM, and at the point L on it, (so as to be) equal
πρὸςτὴνΓ,ἴσηδὲἡμὲνΑτῇΛΜ,ἡδὲΒἑκατέρᾳτῶνΛΞ, to the solid angle E [Prop. 11.23]. And let LO be made
ΕΔ,ἡδὲΓτῇΛΝ,ἔστινἄραὡςἡΛΜπρὸςτὴνΕΖ,οὕτως equal to B, and LN equal to C. And since as A (is)
ἡΔΕπρὸςτὴνΛΝ.καὶπερὶἴσαςγωνίαςτὰςὑπὸΝΛΜ, to B, so B (is) to C, and A (is) equal to LM, and B
ΔΕΖαἱπλευραὶἀντιπεπόνθασιν·ἴσονἄραἐστὶτὸΜΝπα- to each of LO and ED, and C to LN, thus as LM (is)
ραλληλόγραμμοντῷΔΖπαραλληλογραμάμμῳ.καὶἐπεὶδύο to EF , so DE (is) to LN. And (so) the sides around
γωνίαιἐπίπεδοιεὐθύγραμμοιἴσαιεἰσὶναἱὑπὸΔΕΖ,ΝΛΜ, the equal angles NLM and DEF are reciprocally proκαὶἐπ᾿αὐτῶνμετέωροιεὐθεῖαιἐφεστᾶσιναἱΛΞ,ΕΗἴσαι
portional. Thus, parallelogram MN is equal to parallelτεἀλλήλαιςκαὶἴσαςγωνίαςπεριέχουσαιμετὰτῶνἐξἀρχῆς
ogram DF [Prop. 6.14]. And since the two plane rectiεὐθειῶνἑκατέρανἑκατέρᾳ,αἱἄραἀπὸτῶνΗ,Ξσημείων
linear angles DEF and NLM are equal, and the raised
κάθετοιἀγόμεναιἐπὶτὰδιὰτῶνΝΛΜ,ΔΕΖἐπίπεδαἴσαι straight-lines stood on them (at their apexes), LO and
ἀλλήλαιςεἰσίν·ὥστετὰΛΘ,ΕΚστερεὰὑπὸτὸαὐτὸὕψος EG, are equal to one another, and contain equal angles
ἐστίν. τὰδὲἐπὶἴσωνβάσεωνστερεὰπαραλληλεπίπεδακαὶ respectively with the original straight-lines (forming the
ὑπὸτὸαὐτὸὕψοςἴσαἀλλήλοιςἐστίν·ἴσονἄραἐστὶτὸΘΛ angles), the perpendiculars drawn from points G and O
στερεὸντῷΕΚστερεῷ.καίἐστιτὸμὲνΛΘτὸἐκτῶνΑ, to the planes through NLM and DEF (respectively) are
Β,Γστερεόν,τὸδὲΕΚτὸἀπὸτῆςΒστερεόν·τὸἄραἐκ thus equal to one another [Prop. 11.35 corr.]. Thus, the
τῶνΑ,Β,Γστερεὸνπαραλληλεπίπεδονἴσονἐστὶτῷἀπὸ solids LH and EK (have) the same height. And paralτῆςΒστερεῷἰσοπλεύρῳμέν,ἰσογωνίῳδὲτῷπροειρημένῳ·
lelepiped solids on equal bases, and with the same height,
ὅπερἔδειδεῖξαι.
are equal to one another [Prop. 11.31]. Thus, solid HL
is equal to solid EK. And LH is the solid (formed) from
A, B, and C, and EK the solid on B. Thus, the parallelepiped
solid (formed) from A, B, and C is equal to
the equilateral solid on B (which is) equiangular with the
aforementioned (solid). (Which is) the very thing it was
required to show.
ÐÞ. Proposition 37 †
᾿Εὰντέσσαρεςεὐθεῖαιἀνάλογονὦσιν,καὶτὰἀπ᾿αὐτῶν If four straight-lines are proportional then the similar,
K
F
G
E
D
467

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
στερεὰπαραλληλεπίπεδαὅμοιάτεκαὶὁμοίωςἀναγραφόμενα and similarly described, parallelepiped solids on them
ἀνάλογονἔσται· καὶ ἐὰν τὰ ἀπ᾿ αὐτῶν στερεὰ παραλλη- will also be proportional. And if the similar, and similarly
λεπίπεδαὅμοιάτεκαὶὁμοίωςἀναγραφόμεναἀνάλογονᾖ, described, parallelepiped solids on them are proportional
καὶαὐταὶαἱεὐθεῖαιἀνάλογονἔσονται.
then the straight-lines themselves will be proportional.
à Ä
L
K
Å Æ
A B
C
D
N
M
À Â
῎ΕστωσαντέσσαρεςεὐθεῖαιἀνάλογοναἱΑΒ,ΓΔ,ΕΖ, Let AB, CD, EF , and GH, be four proportional
ΗΘ,ὡςἡΑΒπρὸςτὴνΓΔ,οὕτωςἡΕΖπρὸςτὴνΗΘ,καὶ straight-lines, (such that) as AB (is) to CD, so EF (is)
ἀναγεγράφθωσανἀπὸτῶνΑΒ,ΓΔ,ΕΖ,ΗΘὅμοιάτεκαὶ to GH. And let the similar, and similarly laid out, par-
ὁμοίωςκείμεναστερεὰπαραλληλεπίπεδατὰΚΑ,ΛΓ,ΜΕ, allelepiped solids KA, LC, ME and NG have been de-
ΝΗ·λέγω,ὅτιἐστὶνὡςτὸΚΑπρὸςτὸΛΓ,οὕτωςτὸΜΕ scribed on AB, CD, EF , and GH (respectively). I say
πρὸςτὸΝΗ.
that as KA is to LC, so ME (is) to NG.
᾿ΕπεὶγὰρὅμοιόνἐστιτὸΚΑστερεὸνπαραλληλεπίπεδον For since the parallelepiped solid KA is similar to LC,
τῷΛΓ,τὸΚΑἄραπρὸςτὸΛΓτριπλασίοναλόγονἔχειἤπερ KA thus has to LC the cubed ratio that AB (has) to CD
ἡΑΒπρὸςτὴνΓΔ.διὰτὰαὐτὰδὴκαὶτὸΜΕπρὸςτὸΝΗ [Prop. 11.33]. So, for the same (reasons), ME also has to
τριπλασίοναλόγονἔχειἤπερἡΕΖπρὸςτὴνΗΘ.καίἐστιν NG the cubed ratio that EF (has) to GH [Prop. 11.33].
ὡςἡΑΒπρὸςτὴνΓΔ,οὕτωςἡΕΖπρὸςτὴνΗΘ.καὶὡς And since as AB is to CD, so EF (is) to GH, thus, also,
ἄρατὸΑΚπρὸςτὸΛΓ,οὕτωςτὸΜΕπρὸςτὸΝΗ. as AK (is) to LC, so ME (is) to NG.
ἈλλὰδὴἔστωὡςτὸΑΚστερεὸνπρὸςτὸΛΓστερεόν, And so let solid AK be to solid LC, as solid ME (is)
οὕτωςτὸΜΕστερεὸνπρὸςτὸΝΗ·λέγω,ὅτιἐστὶνὡςἡ to NG. I say that as straight-line AB is to CD, so EF (is)
ΑΒεὐθεῖαπρὸςτὴνΓΔ,οὕτωςἡΕΖπρὸςτὴνΗΘ. to GH.
᾿ΕπεὶγὰρπάλιντὸΚΑπρὸςτὸΛΓτριπλασίοναλόγον For, again, since KA has to LC the cubed ratio that
ἔχειἤπερἡΑΒπρὸςτὴνΓΔ,ἔχειδὲκαὶτὸΜΕπρὸςτὸ AB (has) to CD [Prop. 11.33], and ME also has to NG
ΝΗτριπλασίοναλόγονἤπερἡΕΖπρὸςτὴνΗΘ,καίἐστιν the cubed ratio that EF (has) to GH [Prop. 11.33], and
ὡςτὸΚΑπρὸςτὸΛΓ,οὕτωςτὸΜΕπρὸςτὸΝΗ,καὶὡς as KA is to LC, so ME (is) to NG, thus, also, as AB (is)
ἄραἡΑΒπρὸςτὴνΓΔ,οὕτωςἡΕΖπρὸςτὴνΗΘ. to CD, so EF (is) to GH.
᾿Εὰνἄρατέσσαρεςεὐθεῖαιἀνάλογονὦσικαὶτὰἑξῆς Thus, if four straight-lines are proportional, and so
τῆςπροτάσεως·ὅπερἔδειδεῖξαι.
on of the proposition. (Which is) the very thing it was
required to show.
† This proposition assumes that if two ratios are equal then the cube of the former is also equal to the cube of the latter, and vice versa.
Ð. Proposition 38
᾿Εὰνκύβουτῶνἀπεναντίονἐπιπέδωναἱπλευραὶδίχα If the sides of the opposite planes of a cube are cut
τμηθῶσιν,διὰδὲτῶντομῶνἐπίπεδαἐκβληθῇ,ἡκοινὴτομὴ in half, and planes are produced through the pieces, then
τῶνἐπιπέδωνκαὶἡτοῦκύβουδιάμετροςδίχατέμνουσιν the common section of the (latter) planes and the diam-
ἀλλήλας.
eter of the cube cut one another in half.
E
F
G
H
468

ËÌÇÁÉÁÏƺ ELEMENTS
ÃË
Ç
Å ÌÍ Ä
Â
È Ê Æ À
D
O
C
A
K
S
U
T
BOOK 11
ΚύβουγὰρτοῦΑΖτῶνἀπεναντίονἐπιπέδωντῶνΓΖ, For let the opposite planes CF and AH of the cube
ΑΘαἱπλευραὶδίχατετμήσθωσανκατὰτὰΚ,Λ,Μ,Ν,Ξ, AF have been cut in half at the points K, L, M, N, O,
Π,Ο,Ρσημεῖα,διὰδὲτῶντομῶνἐπίπεδαἐκβεβλήσθωτὰ Q, P , and R. And let the planes KN and OR have been
ΚΝ,ΞΡ,κοινὴδὲτομὴτῶνἐπιπέδωνἔστωἡΥΣ,τοῦδὲ produced through the pieces. And let US be the common
ΑΖκύβουδιαγώνιοςἡΔΗ.λέγω,ὅτιἴσηἐστὶνἡμὲνΥΤ section of the planes, and DG the diameter of cube AF .
τῇΤΣ,ἡδὲΔΤτῇΤΗ.
I say that UT is equal to TS, and DT to TG.
᾿Επεζεύχθωσαν γὰρ αἱ ΔΥ, ΥΕ, ΒΣ, ΣΗ. καὶ ἐπεὶ For let DU, UE, BS, and SG have been joined. And
παράλληλόςἐστινἡΔΞτῇΟΕ,αἱἐναλλὰξγωνίαιαἱὑπὸ since DO is parallel to PE, the alternate angles DOU and
ΔΞΥ,ΥΟΕἴσαιἀλλήλαιςεἰσίν. καὶἐπεὶἴσηἐστὶνἡμὲν UPE are equal to one another [Prop. 1.29]. And since
ΔΞτῇΟΕ,ἡδὲΞΥτῇΥΟ,καὶγωνίαςἴσαςπεριέχουσιν, DO is equal to PE, and OU to UP , and they contain
βάσιςἄραἡΔΥτῇΥΕἐστινἴση,καὶτὸΔΞΥτρίγωνοντῷ equal angles, base DU is thus equal to base UE, and tri-
ΟΥΕτριγώνῳἐστὶνἴσονκαὶαἱλοιπαὶγωνίαιταῖςλοιπαῖς angle DOU is equal to triangle PUE, and the remaining
γωνίαιςἴσαι·ἴσηἄραἡὑπὸΞΥΔγωνίατῇὑπὸΟΥΕγωνίᾳ. angles (are) equal to the remaining angles [Prop. 1.4].
διὰδὴτοῦτοεὐθεῖάἐστινἡΔΥΕ.διὰτὰαὐτὰδὴκαὶΒΣΗ Thus, angle OUD (is) equal to angle PUE. So, for this
εὐθεῖάἐστιν,καὶἴσηἡΒΣτῇΣΗ.καὶἐπεὶἡΓΑτῇΔΒ (reason), DUE is a straight-line [Prop. 1.14]. So, for
ἴση ἐστὶ καὶπαράλληλος, ἀλλὰἡΓΑ καὶ τῇ ΕΗ ἴση τέ the same (reason), BSG is also a straight-line, and BS
ἐστι καὶπαράλληλος,καὶἡΔΒἄρατῇΕΗ ἴσητέἐστι equal to SG. And since CA is equal and parallel to DB,
καὶπαράλληλος. καὶἐπιζευγνύουσιναὐτὰςεὐθεῖαιαἱΔΕ, but CA is also equal and parallel to EG, DB is thus also
ΒΗ·παράλληλοςἄραἐστὶνἡΔΕτῇΒΗ.ἴσηἄραἡμὲνὑπὸ equal and parallel to EG [Prop. 11.9]. And the straight-
ΕΔΤγωνίατῇὑπὸΒΗΤ·ἐναλλὰξγάρ·ἡδὲὑπὸΔΤΥτῇ lines DE and BG join them. DE is thus parallel to BG
ὑπὸΗΤΣ.δύοδὴτρίγωνάἐστιτὰΔΤΥ,ΗΤΣτὰςδύο [Prop. 1.33]. Thus, angle EDT (is) equal to BGT . For
γωνίαςταῖςδυσὶγωνίαιςἴσαςἔχοντακαὶμίανπλευρὰνμιᾷ (they are) alternate [Prop. 1.29]. And (angle) DTU (is
πλευρᾷἴσηντὴνὑποτείνουσανὑπὸμίαντῶνἴσωνγωνιῶν equal) to GTS [Prop. 1.15]. So, DTU and GTS are two
τὴνΔΥτῇΗΣ·ἡμίσειαιγάρεἰσιτῶνΔΕ,ΒΗ·καὶτὰς triangles having two angles equal to two angles, and one
λοιπὰςπλευρὰςταῖςλοιπαῖςπλευραῖςἴσαςἕξει. ἴσηἄραἡ side equal to one side—(namely), that subtended by one
μὲνΔΤτῇΤΗ,ἡδὲΥΤτῇΤΣ.
of the equal angles—(that is), DU (equal) to GS. For
᾿Εὰνἄρακύβουτῶνἀπεναντίονἐπιπέδωναἱπλευραὶδίχα they are halves of DE and BG (respectively). (Thus),
τμηθῶσιν,διὰδὲτῶντομῶνἐπίπεδαἐκβληθῇ,ἡκοινὴτομὴ they will also have the remaining sides equal to the reτῶνἐπιπέδωνκαὶἡτοῦκύβουδιάμετροςδίχατέμνουσιν
maining sides [Prop. 1.26]. Thus, DT (is) equal to TG,
ἀλλήλας·ὅπερἔδειδεῖξαι.
B
Q
M
and UT to TS.
Thus, if the sides of the opposite planes of a cube are
cut in half, and planes are produced through the pieces,
then the common section of the (latter) planes and the
diameter of the cube cut one another in half. (Which is)
the very thing it was required to show.
L
N
F
H
P
R
E
G
469

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 11
Ð. Proposition 39
᾿Εὰνᾖδύοπρίσματαἰσοϋψῆ,καὶτὸμὲνἔχῇβάσινπα- If there are two equal height prisms, and one has a
ραλληλόγραμμον,τὸδὲτρίγωνον,διπλάσιονδὲᾖτὸπαραλ-
Ç
parallelogram, and the other a triangle, (as a) base, and
ληλόγραμμοντοῦτριγώνου,ἴσαἔσταιτὰπρίσματα. the parallelogram is double the triangle, then the prisms
will be equal.
B D M P
O L N
À
A C
H
E F G K
῎ΕστωδύοπρίσματαἰσοϋψῆτὰΑΒΓΔΕΖ,ΗΘΚΛΜΝ, Let ABCDEF and GHKLMN be two equal height
καὶ τὸ μὲν ἐχέτω βάσιν τὸ ΑΖ παραλληλόγραμμον, τὸ prisms, and let the former have the parallelogram AF ,
δὲτὸΗΘΚτρίγωνον, διπλάσιονδὲἔστωτὸΑΖπαραλ- and the latter the triangle GHK, as a base. And let parληλόγραμμοντοῦΗΘΚτριγώνου·λέγω,ὅτιἴσονἐστὶτὸ
allelogram AF be twice triangle GHK. I say that prism
Å Ä
Â
Æ
Ã
ΑΒΓΔΕΖπρίσματῷΗΘΚΛΜΝπρίσματι.
ABCDEF is equal to prism GHKLMN.
ΣυμπεπληρώσθωγὰρτὰΑΞ,ΗΟστερεά.ἐπεὶδιπλάσιόν For let the solids AO and GP have been com-
ἐστιτὸΑΖπαραλληλόγραμμοντοῦΗΘΚτριγώνου,ἔστι pleted. Since parallelogram AF is double triangle GHK,
δὲ καὶ τὸ ΘΚ παραλληλόγραμμον διπλάσιον τοῦ ΗΘΚ and parallelogram HK is also double triangle GHK
τριγώνου,ἴσονἄραἐστὶτὸΑΖπαραλληλόγραμμοντῷΘΚ [Prop. 1.34], parallelogram AF is thus equal to paralπαραλληλογράμμῳ.
τὰδὲἐπὶἴσωνβάσεωνὄνταστερεὰ lelogram HK. And parallelepiped solids which are on
παραλληλεπίπεδακαὶὑπὸτὸαὐτὸὕψοςἴσαἀλλήλοιςἐστίν· equal bases, and (have) the same height, are equal to
ἴσονἄραἐστὶτὸΑΞστερεὸντῷΗΟστερεῷ. καίἐστι one another [Prop. 11.31]. Thus, solid AO is equal to
τοῦμὲνΑΞστερεοῦἥμισυτὸΑΒΓΔΕΖπρίσμα, τοῦδὲ solid GP . And prism ABCDEF is half of solid AO, and
ΗΟστερεοῦἥμισυτὸΗΘΚΛΜΝπρίσμα·ἴσονἄραἐστὶτὸ prism GHKLMN half of solid GP [Prop. 11.28]. Prism
ΑΒΓΔΕΖπρίσματῷΗΘΚΛΜΝπρίσματι.
ABCDEF is thus equal to prism GHKLMN.
᾿Εὰνἄραᾖδύοπρίσματαἰσοϋψῆ,καὶτὸμὲνἔχῇβάσιν Thus, if there are two equal height prisms, and one
παραλληλόγραμμον,τὸδὲτρίγωνον,διπλάσιονδὲᾖτὸπα- has a parallelogram, and the other a triangle, (as a) base,
ραλληλόγραμμοντοῦτριγώνου,ἴσαἔστὶτὰπρίσματα·ὅπερ and the parallelogram is double the triangle, then the
ἔδειδεῖξαι.
prisms are equal. (Which is) the very thing it was required
to show.
470

ELEMENTS BOOK 12
Proportional Stereometry †
† The novel feature of this book is the use of the so-called method of exhaustion (see Prop. 10.1), a precursor to integration which is generally
attributed to Eudoxus of Cnidus.
471

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
. Proposition 1
Τὰἐντοῖςκύκλοιςὅμοιαπολύγωναπρὸςἄλληλάἐστιν Similar polygons (inscribed) in circles are to one an-
ὡςτὰἀπὸτῶνδιαμέτρωντετράγωνα.
other as the squares on the diameters (of the circles).
Å
A
À Ä
F
B
Æ
E G
L
 Ã
M
N
C
H K
D
῎ΕστωσανκύκλοιοἱΑΒΓ,ΖΗΘ,καὶἐναὐτοῖςὅμοια Let ABC and FGH be circles, and let ABCDE and
πολύγωναἔστω τὰΑΒΓΔΕ, ΖΗΘΚΛ,διάμετροιδὲτῶν FGHKL be similar polygons (inscribed) in them (reκύκλωνἔστωσανΒΜ,ΗΝ·λέγω,ὅτιἐστὶνὡςτὸἀπὸτῆς
spectively), and let BM and GN be the diameters of the
ΒΜτετράγωνονπρὸςτὸἀπὸτῆςΗΝτετράγωνον,οὕτως circles (respectively). I say that as the square on BM is to
τὸΑΒΓΔΕπολύγωνονπρὸςτὸΖΗΘΚΛπολύγωνον. the square on GN, so polygon ABCDE (is) to polygon
᾿Επεζεύχθωσαν γὰρ αἱ ΒΕ, ΑΜ, ΗΛ, ΖΝ. καὶ ἐπεὶ FGHKL.
ὅμοιοντὸΑΒΓΔΕπολύγωνοντῷΖΗΘΚΛπολυγώνῳ,ἴση For let BE, AM, GL, and FN have been joined. And
ἐστὶκαὶἡὑπὸΒΑΕγωνίατῇὑπὸΗΖΛ,καίἐστινὡςἡΒΑ since polygon ABCDE (is) similar to polygon FGHKL,
πρὸςτὴνΑΕ,οὕτωςἡΗΖπρὸςτὴνΖΛ.δύοδὴτρίγωνά angle BAE is also equal to (angle) GFL, and as BA
ἐστιτὰΒΑΕ,ΗΖΛμίανγωνίανμιᾷγωνίᾳἴσηνἔχοντατὴν is to AE, so GF (is) to FL [Def. 6.1]. So, BAE and
ὑπὸΒΑΕτῇὑπὸΗΖΛ,περὶδὲτὰςἴσαςγωνίαςτὰςπλευρὰς GFL are two triangles having one angle equal to one
ἀνάλογον·ἰσογώνιονἄραἐστὶτὸΑΒΕτρίγωνοντῷΖΗΛ angle, (namely), BAE (equal) to GFL, and the sides
τριγώνῳ.ἴσηἄραἐστὶνἡὑπὸΑΕΒγωνίατῇὑπὸΖΛΗ.ἀλλ᾿ around the equal angles proportional. Triangle ABE is
ἡμὲνὑπὸΑΕΒτῇὑπὸΑΜΒἐστινἴση·ἐπὶγὰρτῆςαὐτῆς thus equiangular with triangle FGL [Prop. 6.6]. Thus,
περιφερείαςβεβήκασιν·ἡδὲὑπὸΖΛΗτῇὑπὸΖΝΗ·καὶἡ angle AEB is equal to (angle) FLG. But, AEB is equal
ὑπὸΑΜΒἄρατῇὑπὸΖΝΗἐστινἴση. ἔστιδὲκαὶὀρθὴ to AMB, and FLG to FNG, for they stand on the same
ἡὑπὸΒΑΜὀρθῇτῇὑπὸΗΖΝἴση· καὶἡλοιπὴἄρατῇ circumference [Prop. 3.27]. Thus, AMB is also equal
λοιπῇἐστινἴση.ἰσογώνιονἄραἐστὶτὸΑΒΜτρίγωνοντῷ to FNG. And the right-angle BAM is also equal to the
ΖΗΝτρίγωνῳ. ἀνάλογονἄραἐστὶνὡςἡΒΜπρὸςτὴν right-angle GFN [Prop. 3.31]. Thus, the remaining (an-
ΗΝ, οὕτωςἡΒΑ πρὸςτὴν ΗΖ. ἀλλὰτοῦμὲν τῆςΒΜ gle) is also equal to the remaining (angle) [Prop. 1.32].
πρὸςτὴνΗΝλόγονδιπλασίωνἐστὶνὁτοῦἀπὸτῆςΒΜ Thus, triangle ABM is equiangular with triangle FGN.
τετραγώνουπρὸςτὸἀπὸτῆςΗΝτετράγωνον,τοῦδὲτῆς Thus, proportionally, as BM is to GN, so BA (is) to GF
ΒΑπρὸςτὴνΗΖδιπλασίωνἐστὶνὁτοῦΑΒΓΔΕπολυγώνου [Prop. 6.4]. But, the (ratio) of the square on BM to the
πρὸςτὸΖΗΘΚΛπολύγωνον·καὶὡςἄρατὸἁπὸτῆςΒΜ square on GN is the square of the ratio of BM to GN,
τετράγωνονπρὸςτὸἀπὸτῆςΗΝτετράγωνον,οὕτωςτὸ and the (ratio) of polygon ABCDE to polygon FGHKL
ΑΒΓΔΕπολύγωνονπρὸςτὸΖΗΘΚΛπολύγωνον. is the square of the (ratio) of BA to GF [Prop. 6.20].
Τὰἄραἐντοῖςκύκλοιςὅμοιαπολύγωναπρὸςἄλληλά And, thus, as the square on BM (is) to the square on
ἐστινὡςτὰἀπὸτῶνδιαμέτρωντετράγωνα·ὅπερἔδειδεῖξαι. GN, so polygon ABCDE (is) to polygon FGHKL.
Thus, similar polygons (inscribed) in circles are to one
another as the squares on the diameters (of the circles).
(Which is) the very thing it was required to show.
. Proposition 2
Οἱκύκλοιπρὸςἀλλήλουςεἰσὶνὡςτὰἀπὸτῶνδιαμέτρων Circles are to one another as the squares on (their)
τετράγωνα.
diameters.
῎ΕστωσανκύκλοιοἱΑΒΓΔ,ΕΖΗΘ,διάμετροιδὲαὐτῶν Let ABCD and EFGH be circles, and [let] BD and
472

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
[ἔστωσαν]αἱΒΔ,ΖΘ·λέγω,ὅτιἐστὶνὡςὁΑΒΓΔκύκλος FH [be] their diameters. I say that as circle ABCD is to
πρὸςτὸνΕΖΗΘκύκλον,οὕτωςτὸἀπὸτῆςΒΔτετράγωνον circle EFGH, so the square on BD (is) to the square on
πρὸςτὸἀπὸτῆςΖΘτετράγωνον.
à Æ
FH.
A
E
Å Â
O
R K N
Ç Ä
B
D F
H
À
L
M
P
Q
G
C
Ê
È
Ë
Ì
S
T
ΕἰγὰρμήἐστινὡςὁΑΒΓΔκύκλοςπρὸςτὸνΕΖΗΘ, For if the circle ABCD is not to the (circle) EFGH,
οὕτωςτὸἀπὸτῆςΒΔτετράγωνονπρὸςτὸἀπὸτῆςΖΘ, as the square on BD (is) to the (square) on FH, then as
ἔσταιὡςτὸἀπὸτῆςΒΔπρὸςτὸἀπὸτῆςΖΘ,οὕτωςὁ the (square) on BD (is) to the (square) on FH, so circle
ΑΒΓΔκύκλοςἤτοιπρὸςἔλασσόντιτοῦΕΖΗΘκύκλου ABCD will be to some area either less than, or greater
χωρίονἢπρὸςμεῖζον. ἔστωπρότερονπρὸςἔλασσοντὸ than, circle EFGH. Let it, first of all, be (in that ratio) to
Σ.καιἐγγεγράφθωεἰςτὸνΕΖΗΘκύκλοντετράγωνοντὸ (some) lesser (area), S. And let the square EFGH have
ΕΖΗΘ.τὸδὴἐγγεγραμμένοντετράγωνονμεῖζόνἐστινἢτὸ been inscribed in circle EFGH [Prop. 4.6]. So the in-
ἥμισυτοῦΕΖΗΘκύκλου,ἐπειδήπερἐὰνδιὰτῶνΕ,Ζ,Η,Θ scribed square is greater than half of circle EFGH, inasσημείωνἐφαπτομένας[εὐθείας]τοῦκύκλουἀγάγωμεν,τοῦ
much as if we draw tangents to the circle through the
περιγραφομένουπερὶτὸνκύκλοντετραγώνουἥμισύἐστι points E, F , G, and H, then square EFGH is half of the
τὸΕΖΗΘτετράγωνον,τοῦδὲπεριγραφέντοςτετραγώνου square circumscribed about the circle [Prop. 1.47], and
ἐλάττωνἐστὶνὁκύκλος·ὥστετὸΕΖΗΘἐγγεγραμμένον the circle is less than the circumscribed square. Hence,
τετράγωνονμεῖζόνἐστιτοῦἡμίσεωςτοῦΕΖΗΘκύκλου. the inscribed square EFGH is greater than half of cirτετμήσθωσανδίχααἱΕΖ,ΖΗ,ΗΘ,ΘΕπεριφέρειαικατὰ
cle EFGH. Let the circumferences EF , FG, GH, and
τὰ Κ, Λ, Μ, Ν σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΕΚ, ΚΖ, HE have been cut in half at points K, L, M, and N
ΖΛ,ΛΗ,ΗΜ,ΜΘ,ΘΝ,ΝΕ·καὶἕκαστονἄρατῶνΕΚΖ, (respectively), and let EK, KF , FL, LG, GM, MH,
ΖΛΗ,ΗΜΘ,ΘΝΕτριγώνωνμεῖζόνἐστινἢτὸἥμισυτοῦ HN, and NE have been joined. And, thus, each of
καθ᾿ἑαυτὸτμήματοςτοῦκύκλου,ἐπειδήπερἐὰνδιὰτῶν the triangles EKF , FLG, GMH, and HNE is greater
Κ,Λ,Μ,Νσημείωνἐφαπτομέναςτοῦκύκλουἀγάγωμεν than half of the segment of the circle about it, inasmuch
καὶἀναπληρώσωμεντὰἐπὶτῶνΕΖ,ΖΗ,ΗΘ,ΘΕεὐθειῶν as if we draw tangents to the circle through points K,
παραλληλόγραμμα,ἕκαστοντῶνΕΚΖ,ΖΛΗ,ΗΜΘ,ΘΝΕ L, M, and N, and complete the parallelograms on the
τριγώνωνἥμισυἔσταιτοῦκαθ᾿ἑαυτὸπαραλληλογράμμου, straight-lines EF , FG, GH, and HE, then each of the
ἀλλὰ τὸ καθ᾿ ἑαυτὸ τμῆμα ἔλαττόν ἐστι τοῦ παραλλη- triangles EKF , FLG, GMH, and HNE will be half
λογράμμου·ὥστεἕκαστοντῶνΕΚΖ,ΖΛΗ,ΗΜΘ,ΘΝΕ of the parallelogram about it, but the segment about it
τριγώνωνμεῖζόνἐστιτοῦἡμίσεωςτοῦκαθ᾿ἑαυτὸτμήματος is less than the parallelogram. Hence, each of the triτοῦκύκλου.
τέμνοντεςδὴτὰςὑπολειπομέναςπεριφερείας angles EKF , FLG, GMH, and HNE is greater than
δίχακαὶἐπιζευγνύντεςεὐθείαςκαὶτοῦτοἀεὶποιοῦντεςκα- half of the segment of the circle about it. So, by cutting
ταλείψομέντιναἀποτμήματατοῦκύκλου,ἃἔσταιἐλάσσονα the circumferences remaining behind in half, and joining
τῆςὑπεροχῆς,ᾗὑπερέχειὁΕΖΗΘκύκλοςτοῦΣχωρίου. straight-lines, and doing this continually, we will (even-
473

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
ἐδείχθηγὰρἐντῷπρώτῳθεωρήματιτοῦδεκάτουβιβλίου, tually) leave behind some segments of the circle whose
ὅτιδύομεγεθῶνἀνίσωνἐκκειμένων,ἐὰνἀπὸτοῦμείζονος (sum) will be less than the excess by which circle EFGH
ἀφαιρεθῇμεῖζονἢτὸἥμισυκαὶτοῦκαταλειπομένουμεῖζον exceeds the area S. For we showed in the first theo-
ἢτὸἥμισυ,καὶτοῦτοἀεὶγίγνηται,λειφθήσεταίτιμέγεθος, rem of the tenth book that if two unequal magnitudes
ὃ ἔσται ἔλασσον τοῦ ἐκκειμένου ἐλάσσονος μεγέθους. are laid out, and if (a part) greater than a half is subλελείφθω
οὖν, καὶ ἔστω τὰ ἐπὶ τῶν ΕΚ, ΚΖ, ΖΛ, ΛΗ, tracted from the greater, and (if from) the remainder (a
ΗΜ,ΜΘ,ΘΝ,ΝΕτμήματατοῦΕΖΗΘκύκλουἐλάττονα part) greater than a half (is subtracted), and this hapτῆςὑπεροχῆς,ᾗὑπερέχειὁΕΖΗΘκύκλοςτοῦΣχωρίου.
pens continually, then some magnitude will (eventually)
λοιπὸνἄρατὸΕΚΖΛΗΜΘΝπολύγωνονμεῖζόνἐστιτοῦΣ be left which will be less than the lesser laid out magχωρίου.
ἐγγεγράφθωκαὶεἰςτὸνΑΒΓΔκύκλοντῷΕΚΖ- nitude [Prop. 10.1]. Therefore, let the (segments) have
ΛΗΜΘΝπολυγώνῳὅμοιονπολύγωνοντὸΑΞΒΟΓΠΔΡ· been left, and let the (sum of the) segments of the circle
ἔστινἄραὡςτὸἀπὸτῆςΒΔτετράγωνονπρὸςτὸἀπὸτῆς EFGH on EK, KF , FL, LG, GM, MH, HN, and NE
ΖΘτετράγωνον,οὕτωςτὸΑΞΒΟΓΠΔΡπολύγωνονπρὸς be less than the excess by which circle EFGH exceeds
τὸΕΚΖΛΗΜΘΝπολύγωνον.ἀλλὰκαὶὡςτὸἀπὸτῆςΒΔ area S. Thus, the remaining polygon EKFLGMHN is
τετράγωνονπρὸςτὸἀπὸτῆςΖΘ,οὕτωςὁΑΒΓΔκύκλος greater than area S. And let the polygon AOBPCQDR,
πρὸςτὸΣχωρίον·καὶὡςἄραὁΑΒΓΔκύκλοςπρὸςτὸΣ similar to the polygon EKFLGMHN, have been inχωρίον,οὕτωςτὸΑΞΒΟΓΠΔΡπολύγωνονπρὸςτὸΕΚΖ-
scribed in circle ABCD. Thus, as the square on BD is
ΛΗΜΘΝ πολύγωνον· ἐναλλὰξἄραὡς ὁ ΑΒΓΔ κύκλος to the square on FH, so polygon AOBPCQDR (is) to
πρὸς τὸ ἐν αὐτῷ πολύγωνον, οὕτως τὸ Σ χωρίον πρὸς polygon EKFLGMHN [Prop. 12.1]. But, also, as the
τὸΕΚΖΛΗΜΘΝπολύγωνον. μείζωνδὲὁΑΒΓΔκύκλος square on BD (is) to the square on FH, so circle ABCD
τοῦἐναὐτῷπολυγώνου·μεῖζονἄρακαὶτὸΣχωρίοντοῦ (is) to area S. And, thus, as circle ABCD (is) to area S,
ΕΚΖΛΗΜΘΝπολυγώνου. ἀλλὰκαὶἔλαττον·ὅπερἐστὶν so polygon AOBPGQDR (is) to polygon EKFLGMHN
ἀδύνατον. οὐκἄραἐστὶνὡςτὸἀπὸτῆςΒΔτετράγωνον [Prop. 5.11]. Thus, alternately, as circle ABCD (is) to
πρὸςτὸἀπὸτῆςΖΘ,οὕτωςὁΑΒΓΔκύκλοςπρὸςἔλασσόν the polygon (inscribed) within it, so area S (is) to polyτιτοῦΕΖΗΘκύκλουχωρίον.ὁμοίωςδὴδείξομεν,ὅτιοὐδὲ
gon EKFLGMHN [Prop. 5.16]. And circle ABCD (is)
ὡςτὸἀπὸΖΘπρὸςτὸἀπὸΒΔ,οὕτωςὁΕΖΗΘκύκλος greater than the polygon (inscribed) within it. Thus, area
πρὸςἔλασσόντιτοῦΑΒΓΔκύκλουχωρίον.
S is also greater than polygon EKFLGMHN. But, (it is)
Λέγωδή, ὅτιοὐδὲὡς τὸἀπὸτῆςΒΔπρὸςτὸἀπὸ also less. The very thing is impossible. Thus, the square
τῆςΖΘ,οὕτωςὁΑΒΓΔκύκλοςπρὸςμεῖζόντιτοῦΕΖΗΘ on BD is not to the (square) on FH, as circle ABCD (is)
κύκλουχωρίον.
to some area less than circle EFGH. So, similarly, we can
Εἰγὰρδυνατόν,ἔστωπρὸςμεῖζοντὸΣ.ἀνάπαλινἄρα show that the (square) on FH (is) not to the (square) on
[ἐστὶν]ὡςτὸἀπὸτῆςΖΘτετράγωνονπρὸςτὸἀπὸτῆςΔΒ, BD as circle EFGH (is) to some area less than circle
οὕτωςτὸΣχωρίονπρὸςτὸνΑΒΓΔκύκλον. ἀλλ᾿ὡςτὸ ABCD either.
ΣχωρίονπρὸςτὸνΑΒΓΔκύκλον,οὕτωςὁΕΖΗΘκύκλος So, I say that neither (is) the (square) on BD to
πρὸςἔλαττόντιτοῦΑΒΓΔκύκλουχωρίον·καὶὡςἄρατὸ the (square) on FH, as circle ABCD (is) to some area
ἀπὸτῆςΖΘπρὸςτὸἀπὸτῆςΒΔ,οὕτωςὁΕΖΗΘκύκλος greater than circle EFGH.
πρὸςἔλασσόντιτοῦΑΒΓΔκύκλουχωρίον·ὅπερἀδύνατον For, if possible, let it be (in that ratio) to (some)
ἐδείχθη.οὐκἄραἐστὶνὡςτὸἀπὸτῆςΒΔτετράγωνονπρὸς greater (area), S. Thus, inversely, as the square on FH
τὸἀπὸτῆςΖΘ,οὕτωςὁΑΒΓΔκύκλοςπρὸςμεῖζόντιτοῦ [is] to the (square) on DB, so area S (is) to circle ABCD
ΕΖΗΘκύκλουχωρίον.ἐδείχθηδέ,ὅτιοὐδὲπρὸςἔλασσον· [Prop. 5.7 corr.]. But, as area S (is) to circle ABCD, so
ἔστινἄραὡςτὸἀπὸτῆςΒΔτετράγωνονπρὸςτὸἀπὸτῆς circle EFGH (is) to some area less than circle ABCD
ΖΘ,οὕτωςὁΑΒΓΔκύκλοςπρὸςτὸνΕΖΗΘκύκλον. (see lemma). And, thus, as the (square) on FH (is) to
Οἱ ἄρα κύκλοι πρὸς ἀλλήλους εἰσὶν ὡς τὰ ἀπὸ τῶν the (square) on BD, so circle EFGH (is) to some area
διαμέτρωντετράγωνα·ὅπερἔδειδεῖξαι.
less than circle ABCD [Prop. 5.11]. The very thing was
shown (to be) impossible. Thus, as the square on BD is
to the (square) on FH, so circle ABCD (is) not to some
area greater than circle EFGH. And it was shown that
neither (is it in that ratio) to (some) lesser (area). Thus,
as the square on BD is to the (square) on FH, so circle
ABCD (is) to circle EFGH.
Thus, circles are to one another as the squares on
474

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
(their) diameters. (Which is) the very thing it was required
to show.
ÄÑÑ.
Λέγωδή,ὅτιτοῦΣχωρίουμείζονοςὄντοςτοῦΕΖΗΘ So, I say that, area S being greater than circle EFGH,
Lemma
κύκλουἐστὶνὡςτὸΣχωρίονπρὸςτὸνΑΒΓΔκύκλον, as area S is to circle ABCD, so circle EFGH (is) to some
οὕτωςὁΕΖΗΘκύκλοςπρὸςἔλαττόντιτοῦΑΒΓΔκύκλου area less than circle ABCD.
χωρίον.
For let it have been contrived that as area S (is) to
ΓεγονέτωγὰρὡςτὸΣχωρίονπρὸςτὸνΑΒΓΔκύκλον, circle ABCD, so circle EFGH (is) to area T . I say that
οὕτως ὁ ΕΖΗΘ κύκλος πρὸς τὸ Τ χωρίον. λέγω, ὅτι area T is less than circle ABCD. For since as area S is
ἔλαττόνἐστιτὸΤχωρίοντοῦΑΒΓΔκύκλου. ἐπεὶγάρ to circle ABCD, so circle EFGH (is) to area T , alter-
ἐστινὡςτὸΣχωρίονπρὸςτὸνΑΒΓΔκύκλον,οὕτωςὁ nately, as area S is to circle EFGH, so circle ABCD (is)
ΕΖΗΘκύκλοςπρὸςτὸΤχωρίον,ἐναλλάξἐστινὡςτὸΣ to area T [Prop. 5.16]. And area S (is) greater than circle
χωρίονπρὸςτὸνΕΖΗΘκύκλον,οὕτωςὁΑΒΓΔκύκλος EFGH. Thus, circle ABCD (is) also greater than area
πρὸς τὸ Τ χωρίον. μεῖζον δὲ τὸΣχωρίον τοῦΕΖΗΘ T [Prop. 5.14]. Hence, as area S is to circle ABCD, so
κύκλου·μείζωνἄρακαὶὁΑΒΓΔκύκλοςτοῦΤχωρίου. circle EFGH (is) to some area less than circle ABCD.
ὥστε ἐστὶν ὡς τὸ Σ χωρίον πρὸς τὸν ΑΒΓΔ κύκλον, (Which is) the very thing it was required to show.
οὕτωςὁΕΖΗΘκύκλοςπρὸςἔλαττόντιτοῦΑΒΓΔκύκλου
χωρίον·ὅπερἔδειδεῖξαι.. Proposition 3
Πᾶσαπυραμὶςτρίγωνονἔχουσαβάσινδιαιρεῖταιεἰςδύο Any pyramid having a triangular base is divided into
πυραμίδαςἴσαςτεκαὶὁμοίαςἀλλήλαιςκαὶ[ὁμοίας]τῇὅλῇ
two pyramids having triangular bases (which are) equal,
τριγώνουςἐχουσαςβάσειςκαὶεἰςδύοπρίσματαἴσα·καὶτὰ similar to one another, and [similar] to the whole, and
δύοπρίσματαμείζονάἐστινἢτὸἥμισυτῆςὅληςπυραμίδος. into two equal prisms. And the (sum of the) two prisms
is greater than half of the whole pyramid.
D
 Ä
À Ã
L
H K
C
G F
A E B
῎Εστωπυραμίς,ἧςβάσιςμένἐστιτὸΑΒΓτρίγωνον,κο- Let there be a pyramid whose base is triangle ABC,
ρυφὴδὲτὸΔσημεῖον·λέγω,ὅτιἡΑΒΓΔπυραμὶςδιαιρεῖται and (whose) apex (is) point D. I say that pyramid
εἰςδύοπυραμίδαςἴσαςἀλλήλαιςτριγώνουςβάσειςἐχούσας ABCD is divided into two pyramids having triangular
καὶὁμοίαςτῇὅλῇκαὶεἰςδύοπρίσματαἴσα· καὶτὰδύο bases (which are) equal to one another, and similar to
πρίσματαμείζονάἐστινἢτὸἥμισυτῆςὅληςπυραμίδος. the whole, and into two equal prisms. And the (sum of
ΤετμήσθωσανγὰραἱΑΒ,ΒΓ,ΓΑ,ΑΔ,ΔΒ,ΔΓδίχα the) two prisms is greater than half of the whole pyramid.
κατὰτὰΕ,Ζ,Η,Θ,Κ,Λσημεῖα,καὶἐπεζεύχθωσαναἱ For let AB, BC, CA, AD, DB, and DC have been
ΘΕ,ΕΗ,ΗΘ,ΘΚ,ΚΛ,ΛΘ,ΚΖ,ΖΗ.ἐπεὶἴσηἐστὶνἡμὲν cut in half at points E, F , G, H, K, and L (respectively).
ΑΕτῇΕΒ,ἡδὲΑΘτῇΔΘ,παράλληλοςἄραἐστὶνἡΕΘ And let HE, EG, GH, HK, KL, LH, KF , and FG have
τῇΔΒ.διὰτὰαὐτὰδὴκαὶἡΘΚτῇΑΒπαράλληλόςἐστιν. been joined. Since AE is equal to EB, and AH to DH,
475

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
παραλληλόγραμμονἄραἐστὶτὸΘΕΒΚ·ἴσηἄραἐστὶνἡΘΚ EH is thus parallel to DB [Prop. 6.2]. So, for the same
τῇΕΒ.ἀλλὰἡΕΒτῇΕΑἐστινἴση·καὶἡΑΕἄρατῇΘΚ (reasons), HK is also parallel to AB. Thus, HEBK is
ἐστινἴση.ἔστιδὲκαὶἡΑΘτῇΘΔἴση·δύοδὴαἱΕΑ,ΑΘ a parallelogram. Thus, HK is equal to EB [Prop. 1.34].
δυσὶταῖςΚΘ,ΘΔἴσαιεἰσὶνἑκατέραἑκατέρᾳ·καὶγωνίαἡ But, EB is equal to EA. Thus, AE is also equal to HK.
ὑπὸΕΑΘγωνίᾳτῇὑπὸΚΘΔἴση·βάσιςἄραἡΕΘβάσειτῇ And AH is also equal to HD. So the two (straight-lines)
ΚΔἐστινἴση.ἴσονἄρακαὶὅμοιόνἐστιτὸΑΕΘτρίγωνον EA and AH are equal to the two (straight-lines) KH
τῷΘΚΔτριγώνῳ. διὰτὰαὐτὰδὴκαὶτὸΑΘΗτρίγωνον and HD, respectively. And angle EAH (is) equal to anτῷΘΛΔτριγώνῳἴσοντέἐστικαὶὅμοιον.
καὶἐπεὶδύο gle KHD [Prop. 1.29]. Thus, base EH is equal to base
εὐθεῖαιἁπτόμεναιἀλλήλωναἱΕΘ,ΘΗπαρὰδύοεὐθείας KD [Prop. 1.4]. Thus, triangle AEH is equal and simi-
ἁπτομέναςἀλλήλωντὰςΚΔ,ΔΛεἰσινοὐκἐντῷαὐτῷ lar to triangle HKD [Prop. 1.4]. So, for the same (rea-
ἐπιπέδῳοὖσαι,ἴσαςγωνίαςπεριέξουσιν.ἴσηἄραἐστὶνἡὑπὸ sons), triangle AHG is also equal and similar to trian-
ΕΘΗγωνίατῇὑπὸΚΔΛγωνίᾳ.καὶἐπεὶδύοεὐθεῖαιαἱΕΘ, gle HLD. And since EH and HG are two straight-lines
ΘΗδυσὶταῖςΚΔ,ΔΛἴσαιεἰσὶνἑκατέραεκατέρᾳ,καὶγωνία joining one another (which are respectively) parallel to
ἡὑπὸΕΘΗγωνίᾳτῇὑπὸΚΔΛἐστινἴση,βάσιςἄραἡΕΗ two straight-lines joining one another, KD and DL, not
βάσειτῇΚΛ[ἐστιν]ἴση·ἴσονἄρακαὶὅμοιόνἐστιτὸΕΘΗ being in the same plane, they will contain equal angles
τρίγωνοντῷΚΔΛτριγώνῳ. διὰτὰαὐτὰδὴκαὶτὸΑΕΗ [Prop. 11.10]. Thus, angle EHG is equal to angle KDL.
τρίγωνοντῷΘΚΛτριγώνῳἴσοντεκαὶὅμοιόνἐστιν.ἡἄρα And since the two straight-lines EH and HG are equal
πυραμίς,ἧςβάσιςμένἐστιτὸΑΕΗτρίγωνον,κορυφὴδὲτὸ to the two straight-lines KD and DL, respectively, and
Θσημεῖον,ἴσηκαὶὁμοίαἐστὶπυραμίδι,ἧςβάσιςμένἐστιτὸ angle EHG is equal to angle KDL, base EG [is] thus
ΘΚΛτρίγωνον,κορυφὴδὲτὸΔσημεῖον.καὶἐπεὶτριγώνου equal to base KL [Prop. 1.4]. Thus, triangle EHG is
τοῦΑΔΒπαρὰμίαντῶνπλευρῶντὴνΑΒἦκταιἡΘΚ, equal and similar to triangle KDL. So, for the same (rea-
ἰσογώνιόνἐστιτὸΑΔΒτρίγωνοντῷΔΘΚτριγώνῳ,καὶ sons), triangle AEG is also equal and similar to triangle
τὰςπλευρὰςἀνάλογονἔχουσιν·ὅμοιονἄραἐστὶτὸΑΔΒ HKL. Thus, the pyramid whose base is triangle AEG,
τρίγωνοντῷΔΘΚτριγώνῳ. διὰτὰαὐτὰδὴκαὶτὸμὲν and apex the point H, is equal and similar to the pyra-
ΔΒΓτρίγωνοντῷΔΚΛτριγώνῳὅμοιόνἐστιν,τὸδὲΑΔΓ mid whose base is triangle HKL, and apex the point D
τῷΔΛΘ.καὶἐπεὶδύοεὐθεῖαιἁπτόμεναιἀλλήλωναἱΒΑ, [Def. 11.10]. And since HK has been drawn parallel to
ΑΓπαρὰδύοεὐθείαςἁπτομέναςἀλλήλωντὰςΚΘ,ΘΛεἰσιν one of the sides, AB, of triangle ADB, triangle ADB
οὐκἐντῷαὐτῷἐπιπέδῳ,ἴσαςγωνίαςπεριέξουσιν.ἴσηἄρα is equiangular to triangle DHK [Prop. 1.29], and they
ἐστὶνἡὑπὸΒΑΓγωνίατῇὑπὸΚΘΛ.καίἐστινὡςἡΒΑ have proportional sides. Thus, triangle ADB is similar to
πρὸςτὴνΑΓ,οὕτωςἡΚΘπρὸςτὴνΘΛ·ὅμοιονἄραἐστὶ triangle DHK [Def. 6.1]. So, for the same (reasons), triτὸΑΒΓτρίγωνοντῷΘΚΛτριγώνῳ.
καὶπυραμὶςἄρα,ἧς angle DBC is also similar to triangle DKL, and ADC to
βάσιςμένἐστιτὸΑΒΓτρίγωνον,κορυφὴδὲτὸΔσημεῖον, DLH. And since two straight-lines joining one another,
ὁμοίαἐστὶπυραμίδι,ἧςβάσιςμένἐστιτὸΘΚΛτρίγωνον, BA and AC, are parallel to two straight-lines joining one
κορυφὴδὲτὸΔσημεῖον.ἀλλὰπυραμίς,ἧςβάσιςμέν[ἐστι] another, KH and HL, not in the same plane, they will
τὸΘΚΛτρίγωνον,κορυφὴδὲτὸΔσημεῖον,ὁμοίαἐδείχθη contain equal angles [Prop. 11.10]. Thus, angle BAC is
πυραμίδι,ἧςβάσιςμένἐστιτὸΑΕΗτρίγωνον,κορυφὴδὲ equal to (angle) KHL. And as BA is to AC, so KH (is)
τὸΘσημεῖον.ἑκατέραἄρατῶνΑΕΗΘ,ΘΚΛΔπυραμίδων to HL. Thus, triangle ABC is similar to triangle HKL
ὁμοίαἐστὶτῇὅλῃτῇΑΒΓΔπυραμίδι.
[Prop. 6.6]. And, thus, the pyramid whose base is trian-
Καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΖ τῇ ΖΓ, διπλάσιόν ἐστι τὸ gle ABC, and apex the point D, is similar to the pyra-
ΕΒΖΗπαραλληλόγραμμοντοῦΗΖΓτριγώνου. καὶἐπεὶ, mid whose base is triangle HKL, and apex the point D
ἐὰνᾖδύοπρίσματαἰσοϋψῆ,καὶτὸμὲνἔχῃβάσινπαραλ- [Def. 11.9]. But, the pyramid whose base [is] triangle
ληλόγραμμον,τὸδὲτρίγωνον,διπλάσιονδὲᾖτὸπαραλ- HKL, and apex the point D, was shown (to be) similar
ληλόγραμμον τοῦ τριγώνου, ἴσα ἐστὶ τὰ πρίσματα, ἴσον to the pyramid whose base is triangle AEG, and apex the
ἄραἐστὶτὸπρίσματὸπεριεχόμενονὑπὸδύομὲντριγώνων point H. Thus, each of the pyramids AEGH and HKLD
τῶνΒΚΖ,ΕΘΗ,τριῶνδὲπαραλληλογράμμωντῶνΕΒΖΗ, is similar to the whole pyramid ABCD.
ΕΒΚΘ,ΘΚΖΗτῷπρισματιτῷπεριεχομένῳὑπὸδύομὲν And since BF is equal to FC, parallelogram EBFG
τριγώνων τῶν ΗΖΓ, ΘΚΛ,τριῶν δὲ παραλληλογράμμων is double triangle GFC [Prop. 1.41]. And since, if two
τῶνΚΖΓΛ,ΛΓΗΘ,ΘΚΖΗ.καὶφανερόν,ὅτιἑκάτροντῶν prisms (have) equal heights, and the former has a parπρισμάτων,οὗτεβάσιςτὸΕΒΖΗπαραλληλόγραμμον,ἀπε-
allelogram as a base, and the latter a triangle, and the
ναντίονδὲἡΘΚεὐθεῖα,καὶοὗβάσιςτὸΗΖΓτρίγωνον, parallelogram (is) double the triangle, then the prisms
ἀπεναντίονδὲτὸΘΚΛτρίγωνον, μεῖζόνἐστιν ἑκατέρας are equal [Prop. 11.39], the prism contained by the two
476

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
τῶνπυραμίδων,ὧνβάσειςμὲντὰΑΕΗ,ΘΚΛτρίγωνα,κο- triangles BKF and EHG, and the three parallelograms
ρυφαὶ,δὲτὰΘ,Δσημεῖα,ἐπειδήπερ[καί]ἐὰνἐπιζεύξωμεν EBFG, EBKH, and HKFG, is thus equal to the prism
τὰςΕΖ,ΕΚεὐθείας,τὸμὲνπρίσμα,οὗβάσιςτὸΕΒΖΗπα- contained by the two triangles GFC and HKL, and
ραλληλόγραμμον,ἀπεναντίονδὲἡΘΚεὐθεῖα,μεῖζόνἐστι the three parallelograms KFCL, LCGH, and HKFG.
τῆςπυραμίδος,ἧςβάσιςτὸΕΒΖτρίγωνον,κορυφὴδὲτὸ And (it is) clear that each of the prisms whose base (is)
Κσημεῖον. ἀλλ᾿ἡπυραμίς,ἧςβάσιςτὸΕΒΖτρίγωνον, parallelogram EBFG, and opposite (side) straight-line
κορυφὴδὲτὸΚσημεῖον,ἴσηἐστὶπυραμίδι,ἧςβάσιςτὸ HK, and whose base (is) triangle GFC, and opposite
ΑΕΗτρίγωνον,κορυφὴδὲτὸΘσημεῖον· ὑπὸγὰρἴσων (plane) triangle HKL, is greater than each of the pyraκαὶὁμοίωνἐπιπέδωνπεριέχονται.
ὥστεκαὶτὸπρίσμα,οὗ mids whose bases are triangles AEG and HKL, and
βάσιςμὲντὸΕΒΖΗπαραλληλόγραμμον,ἀπεναντίονδὲἡ apexes the points H and D (respectively), inasmuch as,
ΘΚεὐθεῖα,μεῖζόνἐστιπυραμίδος,ἧςβάσιςμὲντὸΑΕΗ if we [also] join the straight-lines EF and EK then the
τρίγωνον,κορυφὴδὲτὸΘσημεῖον.ἴσονδὲτὸμὲνπρίσμα, prism whose base (is) parallelogram EBFG, and oppoοὗβάσιςτὸΕΒΖΗπαραλληλόγραμμον,ἀπεναντίονδὲἡΘΚ
site (side) straight-line HK, is greater than the pyramid
εὐθεῖα,τῷπρίσματι,οὗβάσιςμὲντὸΗΖΓτρίγωνον,ἀπε- whose base (is) triangle EBF , and apex the point K. But
ναντίονδὲτὸΘΚΛτρίγωνον·ἡδὲπυραμίς,ἧςβάσιςτὸ the pyramid whose base (is) triangle EBF , and apex the
ΑΕΗτρίγωνον,κορυφὴδὲτὸΘσημεῖον,ἴσηἐστὶπυραμίδι, point K, is equal to the pyramid whose base is triangle
ἧςβάσιςτὸΘΚΛτρίγωνον,κορυφὴδὲτὸΔσημεῖον. τὰ AEG, and apex point H. For they are contained by equal
ἄραεἰρημέναδύοπρίσματαμείζονάἐστιτῶνεἰρημένωνδύο and similar planes. And, hence, the prism whose base
πυραμίδων,ὧνβάσειςμὲντὰΑΕΗ,ΘΚΛτρίγωνα,κορυφαὶ (is) parallelogram EBFG, and opposite (side) straightδὲτὰΘ,Δσημεῖα.
line HK, is greater than the pyramid whose base (is)
῾Ηἄραὅληπυραμίς,ἧςβάσιςτὸΑΒΓτρίγωνον,κο- triangle AEG, and apex the point H. And the prism
ρυφὴδὲτὸΔσημεῖον,διῄρηταιεἴςτεδύοπυραμίδαςἴσας whose base is parallelogram EBFG, and opposite (side)
ἀλλήλαις[καὶὁμοίαςτῇὅλῃ]καὶεἰςδύοπρίσματαἴσα,καὶτὰ straight-line HK, (is) equal to the prism whose base (is)
δύοπρίσματαμείζονάἐστινἢτὸἥμισυτῆςὅληςπυραμίδος· triangle GFC, and opposite (plane) triangle HKL. And
ὅπερἔδειδεῖξαι.
the pyramid whose base (is) triangle AEG, and apex the
point H, is equal to the pyramid whose base (is) triangle
HKL, and apex the point D. Thus, the (sum of the)
aforementioned two prisms is greater than the (sum of
the) aforementioned two pyramids, whose bases (are) triangles
AEG and HKL, and apexes the points H and D
(respectively).
Thus, the whole pyramid, whose base (is) triangle
ABC, and apex the point D, has been divided into two
pyramids (which are) equal to one another [and similar
to the whole], and into two equal prisms. And the (sum
of the) two prisms is greater than half of the whole pyramid.
(Which is) the very thing it was required to show.
. Proposition 4
᾿Εὰνὦσιδύοπυραμίδεςὑπὸτὸαὐτὸὕψοςτριγώνους If there are two pyramids with the same height, hav-
ἔχουσαι βάσεις, διαιρεθῇ δὲ ἑκατέρα αὐτῶν εἴς τε δύο ing trianglular bases, and each of them is divided into two
πυραμίδας ἴσας ἀλλήλαιςκαὶ ὁμοίας τῇ ὅλῃ καὶ εἰς δύο pyramids equal to one another, and similar to the whole,
πρίσματαἴσα,ἔσταιὡςἡτῆςμιᾶςπυραμίδοςβάσιςπρὸςτὴν and into two equal prisms then as the base of one pyraτῆςἑτέραςπυραμίδοςβάσιν,οὕτωςτὰἐντῇμιᾷπυραμίδι
mid (is) to the base of the other pyramid, so (the sum of)
πρίσματα πάντα πρὸς τὰ ἐν τῇ ἑτάρᾳ πυραμίδι πρίσματα all the prisms in one pyramid will be to (the sum of all)
πάνταἰσοπληθῇ.
the equal number of prisms in the other pyramid.
῎Εστωσανδύοπυραμίδεςὑπὸτὸαὐτὸὕψοςτριγώνους Let there be two pyramids with the same height, hav-
ἔχουσαιβάσειςτὰςΑΒΓ,ΔΕΖ,κορυφὰςδὲτὰΗ,Θσημεῖα, ing the triangular bases ABC and DEF , (with) apexes
καὶ διῃρήσθω ἑκατέρα αὐτῶν εἴς τε δύο πυραμίδας ἴσας the points G and H (respectively). And let each of them
ἀλλήλαιςκαὶὁμοίαςτῇὅλῃκαὶεἰςδύοπρίσματαἴσα·λέγω, have been divided into two pyramids equal to one an-
477

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
ὅτιἐστὶνὡςἡΑΒΓβάσιςπρὸςτὴνΔΕΖβάσιν,οὕτωςτὰ other, and similar to the whole, and into two equal prisms
ἐντῇΑΒΓΗπυραμίδιπρίσματαπάνταπρὸςτὰἐντῇΔΕΖΘ [Prop. 12.3]. I say that as base ABC is to base DEF ,
πυραμίδιπρίσματαἰσοπληθῆ.
À Â
so (the sum of) all the prisms in pyramid ABCG (is) to
(the sum of) all the equal number of prisms in pyramid
DEFH.
G
H
Ç Æ
Ë Í P
S
N
U
Ã È Ê
M
T
D
F
A L
R
K
Q V
C
B O
E
᾿ΕπεὶγὰρἴσηἐστὶνἡμὲνΒΞτῇΞΓ,ἡδὲΑΛτῇΛΓ, For since BO is equal to OC, and AL to LC, LO is
παράλληλοςἄραἐστὶνἡΛΞτῇΑΒκαὶὅμοιοντὸΑΒΓ thus parallel to AB, and triangle ABC similar to triangle
τρίγωνοντῷΛΞΓτριγώνῳ. διὰτὰαὐτὰδὴκαὶτὸΔΕΖ LOC [Prop. 12.3]. So, for the same (reasons), triangle
τρίγωνον τῷ ΡΦΖ τριγώνῳ ὅμοιόν ἐστιν. καὶ ἐπεὶ δι- DEF is also similar to triangle RV F . And since BC is
πλασίωνἐστὶνἡμὲνΒΓτῆςΓΞ,ἡδὲΕΖτῆςΖΦ,ἔστιν double CO, and EF (double) FV , thus as BC (is) to
ἄραὡςἡΒΓπρὸςτὴνΓΞ,οὕτωςἡΕΖπρὸςτὴνΖΦ.καὶ CO, so EF (is) to FV . And the similar, and similarly
ἀναγέγραπταιἀπὸμὲντῶνΒΓ,ΓΞὅμοιάτεκαὶὁμοίως laid out, rectilinear (figures) ABC and LOC have been
κείμεναεὐθύγραμματὰΑΒΓ,ΛΞΓ,ἀπὸδὲτῶνΕΖ,ΖΦ described on BC and CO (respectively), and the sim-
ὅμοιάτεκαὶὁμοίωςκείμενα[εὐθύγραμμα]τὰΔΕΖ,ΡΦΖ· ilar, and similarly laid out, [rectilinear] (figures) DEF
ἔστινἄραὡςτὸΑΒΓτρίγωνονπρὸςτὸΛΞΓτρίγωνον, and RV F on EF and FV (respectively). Thus, as triοὕτωςτὸΔΕΖτρίγωνονπρὸςτὸΡΦΖτρίγωνον·ἐναλλὰξ
angle ABC is to triangle LOC, so triangle DEF (is) to
ἄραἐστὶνὡςτὸΑΒΓτρίγωνονπρὸςτὸΔΕΖ[τρίγωνον], triangle RV F [Prop. 6.22]. Thus, alternately, as trianοὕτωςτὸΛΞΓ[τρίγωνον]πρὸςτὸΡΦΖτρίγωνον.
ἀλλ᾿ gle ABC is to [triangle] DEF , so [triangle] LOC (is)
Å
Ä
Ì
ὡςτὸΛΞΓτρίγωνονπρὸςτὸΡΦΖτρίγωνον, οὕτωςτὸ to triangle RV F [Prop. 5.16]. But, as triangle LOC
πρίσμα,οὗβάσιςμέν[ἐστι]τὸΛΞΓτρίγωνον,ἀπεναντίονδὲ (is) to triangle RV F , so the prism whose base [is] trianτὸΟΜΝ,πρὸςτὸπρίσμα,οὗβάσιςμὲντὸΡΦΖτρίγωνον,
gle LOC, and opposite (plane) PMN, (is) to the prism
ἀπεναντίονδὲτὸΣΤΥ·καὶὡςἄρατὸΑΒΓτρίγωνονπρὸς whose base (is) triangle RV F , and opposite (plane) STU
τὸΔΕΖτρίγωνον,οὕτωςτὸπρίσμα,οὗβάσιςμὲντὸΛΞΓ (see lemma). And, thus, as triangle ABC (is) to trianτρίγωνον,ἀπεναντίονδὲτὸΟΜΝ,πρὸςτὸπρίσμα,οὗβάσις
gle DEF , so the prism whose base (is) triangle LOC,
μὲντὸΡΦΖτρίγωνον,ἀπεναντίονδὲτὸΣΤΥ.ὡςδὲτὰ and opposite (plane) PMN, (is) to the prism whose base
εἰρημέναπρίσματαπρὸςἄλληλα,οὕτωςτὸπρίσμα,οὗβάσις (is) triangle RV F , and opposite (plane) STU. And as
μὲν τὸ ΚΒΞΛπαραλληλόγραμμον, ἀπεναντίονδὲ ἡΟΜ the aforementioned prisms (are) to one another, so the
εὐθεῖα, πρὸςτὸπρίσμα, οὗβάσιςμὲντὸΠΕΦΡ παραλ- prism whose base (is) parallelogram KBOL, and oppoληλόγραμμον,ἀπεναντίονδὲἡΣΤεὐθεῖα.
καὶτὰδύοἄρα site (side) straight-line PM, (is) to the prism whose base
πρίσματα,οὗτεβάσιςμὲντὸΚΒΞΛπαραλληλόγραμμον, (is) parallelogram QEV R, and opposite (side) straight-
ἀπεναντίονδὲἡΟΜ,καὶοὗβάσιςμὲντὸΛΞΓ,ἀπεναντίον line ST [Props. 11.39, 12.3]. Thus, also, (is) the (sum
δὲτὸΟΜΝ,πρὸςτὰπρίσματα,οὗτεβάσιςμὲντὸΠΕΦΡ, of the) two prisms—that whose base (is) parallelogram
ἀπεναντίον δὲ ἡ ΣΤ εὐθεῖα, καὶ οὗ βάσις μὲν τὸ ΡΦΖ KBOL, and opposite (side) PM, and that whose base
τρίγωνον,ἀπεναντίονδὲτὸΣΤΥ.καὶὡςἄραἡΑΒΓβάσις (is) LOC, and opposite (plane) PMN—to (the sum of)
πρὸςτὴνΔΕΖβάσιν,οὕτωςτὰεἰρημέναδύοπρίσματαπρὸς the (two) prisms—that whose base (is) QEV R, and opτὰεἰρημέναδύοπρίσματα.
posite (side) straight-line ST , and that whose base (is)
Καὶ ὁμοίως, ἐὰν διαιρεθῶσιν αἱ ΟΜΝΗ, ΣΤΥΘ πυ- triangle RV F , and opposite (plane) STU [Prop. 5.12].
ραμίδεςεἴςτεδύοπρίσματακαὶδύοπυραμίδας,ἔσταιὡςἡ And, thus, as base ABC (is) to base DEF , so the (sum
478

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
ΟΜΝβάσιςπρὸςτὴνΣΤΥβάσιν,οὕτωςτὰἐντῇΟΜΝΗ of the first) aforementioned two prisms (is) to the (sum
πυραμίδιδύοπρίσματαπρὸςτὰἐντῇΣΤΥΘπυραμίδιδύο of the second) aforementioned two prisms.
πρίσματα. ἀλλ᾿ὡςἡΟΜΝβάσιςπρὸςτὴνΣΤΥβάσιν, And, similarly, if pyramids PMNG and STUH are diοὕτωςἡΑΒΓβάσιςπρὸςτὴνΔΕΖβάσιν·ἴσονγὰρἑκάτερον
vided into two prisms, and two pyramids, as base PMN
τῶνΟΜΝ,ΣΤΥτριγώνωνἑκατέρῳτῶνΛΞΓ,ΡΦΖ.καὶὡς (is) to base STU, so (the sum of) the two prisms in pyra-
ἄραἡΑΒΓβάσιςπρὸςτὴνΔΕΖβάσιν,οὕτωςτὰτέσσαρα mid PMNG will be to (the sum of) the two prisms in
πρίσματαπρὸςτὰτέσσαραπρίσματα. ὁμοίωςδὲκἂντὰς pyramid STUH. But, as base PMN (is) to base STU, so
ὑπολειπομέναςπυραμίδας διέλωμεν εἴς τε δύο πυραμίδας base ABC (is) to base DEF . For the triangles PMN
καὶεἰςδύοπρίσματα,ἔσταιὡςἡΑΒΓβάσιςπρὸςτὴνΔΕΖ and STU (are) equal to LOC and RV F , respectively.
βάσιν,οὕτωςτὰἐντῇΑΒΓΗπυραμίδιπρίσματαπάνταπρὸς And, thus, as base ABC (is) to base DEF , so (the sum
τὰἐντῇΔΕΖΘπυραμίδιπρίσματαπάνταἰσοπληθῆ·ὅπερ of) the four prisms (is) to (the sum of) the four prisms
ἔδειδεῖξαι.
[Prop. 5.12]. So, similarly, even if we divide the pyramids
left behind into two pyramids and into two prisms,
as base ABC (is) to base DEF , so (the sum of) all the
prisms in pyramid ABCG will be to (the sum of) all the
equal number of prisms in pyramid DEFH. (Which is)
the very thing it was required to show.
ÄÑÑ.
῞Οτι δέ ἐστιν ὡς τὸ ΛΞΓ τρίγωνον πρὸς τὸ ΡΦΖ And one may show, as follows, that as triangle LOC
Lemma
τρίγωνον, οὕτωςτὸπρίσμα,οὗβάσιςτὸΛΞΓτρίγωνον, is to triangle RV F , so the prism whose base (is) trian-
ἀπεναντίονδὲτὸΟΜΝ,πρὸςτὸπρίσμα,οὗβάσιςμὲντὸ gle LOC, and opposite (plane) PMN, (is) to the prism
ΡΦΖ[τρίγωνον],ἀπεναντίονδὲτὸΣΤΥ,οὕτωδεικτέον. whose base (is) [triangle] RV F , and opposite (plane)
᾿Επὶγὰρτῆςαὐτῆςκαταγραφῆςνενοήσθωσανἀπὸτῶν STU.
Η, Θ κάθετοι ἐπὶ τὰ ΑΒΓ, ΔΕΖ ἐπίπεδα, ἴσαι δηλαδὴ For, in the same figure, let perpendiculars have been
τυγχάνουσαι διὰ τὸ ἰσοϋψεῖς ὑποκεῖσθαι τὰς πυραμίδας. conceived (drawn) from (points) G and H to the planes
καὶἐπεὶδύοεὐθεῖαιἥτεΗΓκαὶἡἀπὸτοῦΗκάθετος ABC and DEF (respectively). These clearly turn out to
ὑπὸπαραλλήλωνἐπιπέδωντῶνΑΒΓ,ΟΜΝτέμνονται,εἰς be equal, on account of the pyramids being assumed (to
τοὺςαὐτοὺςλόγουςτμηθήσονται.καὶτέτμηταιἡΗΓδίχα be) of equal height. And since two straight-lines, GC
ὑπὸτοῦΟΜΝἐπιπέδουκατὰτὸΝ·καὶἡἀπὸτοῦΗἄρα and the perpendicular from G, are cut by the parallel
κάθετοςἐπὶτὸ ΑΒΓ ἐπίπεδονδίχατμηθήσεταιὑπὸτοῦ planes ABC and PMN they will be cut in the same ra-
ΟΜΝἐπιπέδου.διὰτὰαὐτὰδὴκαὶἡἀπὸτοῦΘκάθετοςἐπὶ tios [Prop. 11.17]. And GC was cut in half by the plane
τὸΔΕΖἐπίπεδονδίχατμηθήσεταιὑπὸτοῦΣΤΥἐπιπέδου. PMN at N. Thus, the perpendicular from G to the plane
καίεἰσινἴσαιαἱἀπὸτῶνΗ,ΘκάθετοιἐπὶτὰΑΒΓ,ΔΕΖ ABC will also be cut in half by the plane PMN. So,
ἐπίπεδα· ἴσαιἄρακαὶαἱἀπὸτῶνΟΜΝ,ΣΤΥτριγώνων for the same (reasons), the perpendicular from H to the
ἐπὶτὰΑΒΓ,ΔΕΖκάθετοι.ἰσοϋψῆἄρα[ἐστὶ]τὰπρίσματα, plane DEF will also be cut in half by the plane STU. And
ὧνβάσειςμένεἰσιτὰΛΞΓ,ΡΦΖτρίγωνα,ἀπεναντίονδὲ the perpendiculars from G and H to the planes ABC and
τὰΟΜΝ,ΣΤΥ.ὥστεκαὶτὰστερεὰπαραλληλεπίπεδατὰ DEF (respectively) are equal. Thus, the perpendiculars
ἀπὸτῶνεἰρημένωνπρισμάτωνἀναγραφόμεναἰσοϋψῆκαὶ from the triangles PMN and STU to ABC and DEF
πρὸςἄλληλά[εἰσιν]ὡςαἱβάσεις·καὶτὰἡμίσηἄραἐστὶν (respectively, are) also equal. Thus, the prisms whose
ὡςἡΛΞΓβάσιςπρὸςτὴνΡΦΖβάσιν,οὕτωςτὰεἰρημένα bases are triangles LOC and RV F , and opposite (sides)
πρίσματαπρὸςἄλληλα·ὅπερἔδειδεῖξαι.
PMN and STU (respectively), [are] of equal height.
And, hence, the parallelepiped solids described on the
aforementioned prisms [are] of equal height and (are) to
one another as their bases [Prop. 11.32]. Likewise, the
halves (of the solids) [Prop. 11.28]. Thus, as base LOC
is to base RV F , so the aforementioned prisms (are) to
one another. (Which is) the very thing it was required to
show.
479

ËÌÇÁÉÁÏƺ ELEMENTS BOOK 12
. Proposition 5
Αἱὑπὸτὸαὐτὸὕψοςοὖσαιπυραμίδεςκαὶτριγώνους Pyramids which are of the same height, and have tri-
ἔχουσαιβάσειςπρὸςἀλλήλαςεἰσὶνὡςαἱβάσεις. angular bases, are to one another as their bases.
À
Ç Å
Ã
Ä
Æ
Ë
È
Â
Í
Ê
Ì
É
B
E
῎Εστωσανὑπὸτὸαὐτὸὕψοςπυραμίδες,ὧνβάσειςμὲν Let there be pyramids of the same height whose bases
τὰΑΒΓ,ΔΕΖτρίγωνα,κορυφαὶδὲτὰΗ,Θσημεῖα·λέγω, (are) the triangles ABC and DEF , and apexes the points
ὅτιἐστὶνὡςἡΑΒΓβάσιςπρὸςτὴνΔΕΖβάσιν,οὕτωςἡ G and H (respectively). I say that as base ABC is to base
ΑΒΓΗπυραμὶςπρὸςτὴνΔΕΖΘπυραμίδα.
DEF , so pyramid ABCG (is) to pyramid DEFH.
ΕἰγὰρμήἐστινὡςἡΑΒΓβάσιςπρὸςτὴνΔΕΖβάσιν, For if base ABC is not to base DEF , as pyramid
οὕτωςἡΑΒΓΗπυραμὶςπρὸςτὴνΔΕΖΘπυραμίδα,ἔσταιὡς ABCG (is) to pyramid DEFH, then base ABC will be
ἡΑΒΓβάσιςπρὸςτὴνΔΕΖβάσιν,οὕτωςἡΑΒΓΗπυραμὶς to base DEF , as pyramid ABCG (is) to some solid ei-
ἤτοιπρὸςἔλασσόντιτῆςΔΕΖΘπυραμίδοςστερεὸνἢπρὸς ther less than, or greater than, pyramid DEFH. Let it,
μεῖζον. ἔστωπρότερονπρὸςἔλασσοντὸΧ,καὶδιῃρὴσθω first of all, be (in this ratio) to (some) lesser (solid), W .
ἡΔΕΖΘπυραμὶςεἴςτεδύοπυραμίδαςἴσαςἀλλήλαιςκαὶ And let pyramid DEFH have been divided into two pyra-
ὁμοίαςτῇὅλῃκαὶεἰςδύοπρίσματαἴσα·τὰδὴδύοπρίσαμτα mids equal to one another, and similar to the whole, and
μείζονάἐστινἢτὸἥμισυτῆςὅληςπυραμίδος.καὶπάλιναἱἐκ into two equal prisms. So, the (sum of the) two prisms
τῆςδιαιρέσεωςγινόμεναιπυραμίδεςὁμοίωςδιῃρήσθωσαν, is greater than half of the whole pyramid [Prop. 12.3].
καὶτοῦτοἀεὶγινέσθω,ἕωςοὗλειφθῶσίτινεςπυραμίδεςἀπὸ And, again, let the pyramids generated by the division
τῆςΔΕΖΘπυραμίδος,αἵεἰσινἐλάττονεςτῆςὑπεροχῆς,ᾗ have been similarly divided, and let this be done contin-
ὑπερέχειἡΔΕΖΘπυραμὶςτοῦΧστερεοῦ.λελείφθωσανκαὶ ually until some pyramids are left from pyramid DEFH
ἔστωσανλόγουἕνεκεναἱΔΠΡΣ,ΣΤΥΘ·λοιπὰἄρατὰἐν which (when added together) are less than the excess by
τῇΔΕΖΘπυραμίδιπρίσματαμείζονάἐστιτοῦΧστερεοῦ. which pyramid DEFH exceeds the solid W [Prop. 10.1].
διῃρήσθωκαὶἡΑΒΓΗπυραμὶςὁμοίωςκαὶἰσοπληθῶςτῇ Let them have been left, and, for the sake of argument,
ΔΕΖΘπυραμίδι·ἔστινἄραὡςἡΑΒΓβάσιςπρὸςτὴνΔΕΖ let them be DQRS and STUH. Thus, the (sum of the)
βάσιν,οὕτωςτὰἐντῇΑΒΓΗπυραμίδιπρίσματαπρὸςτὰἐν remaining prisms within pyramid DEFH is greater than
τῇΔΕΖΘπυραμίδιπρίσματα,ἀλλὰκαὶὡςἡΑΒΓβάσιςπρὸς solid W . Let pyramid ABCG also have been divided simτὴνΔΕΖβάσιν,οὕτωςἡΑΒΓΗπυραμὶςπρὸςτὸΧστερεόν·
ilarly, and a similar number of times, as pyramid DEFH.
καὶὡςἄραἡΑΒΓΗπυραμὶςπρὸςτὸΧστερεόν,οὕτως Thus, as base ABC is to base DEF , so the (sum of the)
τὰἐντῇΑΒΓΗπυραμίδιπρίσματαπρὸςτὰἐντῇΔΕΖΘ prisms within pyramid ABCG (is) to the (sum of the)
πυραμίδιπρίσματα·ἐναλλὰξἄραὡςἡΑΒΓΗπυραμὶςπρὸς prisms within pyramid DEFH [Prop. 12.4]. But, also,
τὰἐναὐτῇπρίσματα,οὕτωςτὸΧστερεὸνπρὸςτὰἐντῇ as base ABC (is) to base DEF , so pyramid ABCG (is)
ΔΕΖΘπυραμίδιπρίσματα.μείζωνδὲἡΑΒΓΗπυραμὶςτῶν to solid W . And, thus, as pyramid ABCG (is) to solid
ἐναὐτῇπρισμάτων·μεῖζονἄρακαὶτὸΧστερεὸντῶνἐντῇ W , so the (sum of the) prisms within pyramid ABCG
ΔΕΖΘπυραμίδιπρισμάτων. ἀλλὰκαὶἔλαττον·ὅπερἐστὶν (is) to the (sum of the) prisms within pyramid DEFH
ἀδύνατον. οὐκἄραἐστὶνὡςἡΑΒΓβάσιςπρὸςτὴνΔΕΖ [Prop. 5.11]. Thus, alternately, as pyramid ABCG (is) to
βάσιν,οὕτωςἡΑΒΓΗπυραμὶςπρὸςἔλασσόντιτῆςΔΕΖΘ the (sum of the) prisms within it, so solid W (is) to the
πυραμίδοςστερεόν. ὁμοίωςδὴδειχθήσεται,ὅτιοὐδὲὡςἡ (sum of the) prisms within pyramid DEFH [Prop. 5.16].
ΔΕΖβάσιςπρὸςτὴνΑΒΓβάσιν,οὕτωςἡΔΕΖΘπυραμὶς And pyramid ABCG (is) greater than the (sum of the)
πρὸςἔλαττόντιτῆςΑΒΓΗπυραμίδοςστερεόν.
prisms within it. Thus, solid W (is) also greater than the
Λέγωδή,ὅτιοὐκἔστινοὐδὲὡςἡΑΒΓβάσιςπρὸςτὴν (sum of the) prisms within pyramid DEFH [Prop. 5.14].
ΔΕΖβάσιν, οὕτωςἡΑΒΓΗπυραμὶςπρὸςμεῖζόντιτῆς But, (it is) also less. This very thing is impossible. Thus,
ΔΕΖΘπυραμίδοςστερεόν.
as base ABC is to base DEF , so pyramid ABCG (is)
A
K
P
G
M
L
O
N
C
D
S
Q
H
T
R
U
V
F
W
480

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
Εἰγὰρδυνατόν,ἔστωπρὸςμεῖζοντὸΧ·ἀνάπαλινἄρα not to some solid less than pyramid DEFH. So, simi-
ἐστὶνὡςἡΔΕΖβάσιςπρὸςτὴνΑΒΓβάσιν,οὕτωςτὸΧ larly, we can show that base DEF is not to base ABC,
στερεὸνπρὸςτὴνΑΒΓΗπυραμίδα. ὡςδὲτὸΧστερεὸν as pyramid DEFH (is) to some solid less than pyramid
πρὸςτὴνΑΒΓΗπυραμίδα,οὕτωςἡΔΕΖΘπυραμὶςπρὸς ABCG either.
ἔλασσόντιτῆςΑΒΓΗπυραμίδος,ὡςἔμπροσθενἐδείχθη· So, I say that neither is base ABC to base DEF , as
καὶὡςἄραἡΔΕΖβάσιςπρὸςτὴνΑΒΓβάσιν,οὕτωςἡ pyramid ABCG (is) to some solid greater than pyramid
ΔΕΖΘ πυραμὶς πρὸς ἔλασσόν τι τῆς ΑΒΓΗ πυραμίδος· DEFH.
ὅπερἄτοπονἐδείχθη.οὐκἄραἐστὶνὡςἡΑΒΓβάσιςπρὸς For, if possible, let it be (in this ratio) to some
τὴνΔΕΖβάσιν, οὕτωςἡΑΒΓΗπυραμὶςπρὸςμεῖζόντι greater (solid), W . Thus, inversely, as base DEF
τῆςΔΕΖΘπυραμίδοςστερεόν. ἐδείχθηδέ,ὅτιοὐδὲπρὸς (is) to base ABC, so solid W (is) to pyramid ABCG
ἔλασσον.ἔστινἄραὡςἡΑΒΓβάσιςπρὸςτὴνΔΕΖβάσιν, [Prop. 5.7. corr.]. And as solid W (is) to pyramid ABCG,
οὕτωςἡΑΒΓΗπυραμὶςπρὸςτὴνΔΕΖΘπυραμίδα·ὅπερ so pyramid DEFH (is) to some (solid) less than pyramid
ἔδειδεῖξαι.
ABCG, as shown before [Prop. 12.2 lem.]. And, thus,
as base DEF (is) to base ABC, so pyramid DEFH (is)
to some (solid) less than pyramid ABCG [Prop. 5.11].
The very thing was shown (to be) absurd. Thus, base
ABC is not to base DEF , as pyramid ABCG (is) to
some solid greater than pyramid DEFH. And, it was
shown that neither (is it in this ratio) to a lesser (solid).
Thus, as base ABC is to base DEF , so pyramid ABCG
(is) to pyramid DEFH. (Which is) the very thing it was
required to show.
¦. Proposition 6
Αἱὐπὸτὸαὐτὸὕψοςοὖσαιπυραμίδεςκαὶπολυγώνους Pyramids which are of the same height, and have
Å Æ
ἔχουσαιβάσειςπρὸςἀλλήλαςεἰσὶνὡςαἱβάσεις. polygonal bases, are to one another as their bases.
M
N
Ã
Ä
Â
À
L
G
A
E
F
῎Εστωσανὑπὸτὸαὐτὸὕψοςπυραμίδες,ὧν[αἱ]βάσεις Let there be pyramids of the same height whose bases
μὲν τὰ ΑΒΓΔΕ, ΖΗΘΚΛπολύγωνα, κορυφαὶδὲ τὰ Μ, (are) the polygons ABCDE and FGHKL, and apexes
Νσημεῖα·λέγω,ὅτιἐστὶνὡςἡΑΒΓΔΕβάσιςπρὸςτὴν the points M and N (respectively). I say that as base
ΖΗΘΚΛβάσιν,οὕτωςἡΑΒΓΔΕΜπυραμὶςπρὸςτὴνΖΗΘ- ABCDE is to base FGHKL, so pyramid ABCDEM (is)
ΚΛΝπυραμίδα.
to pyramid FGHKLN.
᾿Επεζεύχθωσαν γὰρ αἱ ΑΓ, ΑΔ, ΖΘ, ΖΚ. ἐπεὶ οὖν For let AC, AD, FH, and FK have been joined.
δύο πυραμίδεςεἰσὶν αἱ ΑΒΓΜ, ΑΓΔΜ τριγώνους ἔχου- Therefore, since ABCM and ACDM are two pyramids
σαιβάσειςκαὶὕψοςἴσον,πρὸςἀλλήλαςεἰσὶνὡςαἱβάσεις· having triangular bases and equal height, they are to
ἔστινἄραὡςἡΑΒΓβάσιςπρὸςτὴνΑΓΔβάσιν,οὕτωςἡ one another as their bases [Prop. 12.5]. Thus, as base
ΑΒΓΜπυραμὶςπρὸςτὴνΑΓΔΜπυραμίδα. καὶσυνθέντι ABC is to base ACD, so pyramid ABCM (is) to pyra-
ὡςἡΑΒΓΔβάσιςπρὸςτὴνΑΓΔβάσιν,οὕτωςἡΑΒΓΔΜ mid ACDM. And, via composition, as base ABCD
D
C
B
K
H
481

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
πυραμὶςπρὸςτὴνΑΓΔΜπυραμίδα. ἀλλὰκαὶὡςἡΑΓΔ (is) to base ACD, so pyramid ABCDM (is) to pyraβάσιςπρὸςτὴνΑΔΕβάσιν,οὕτωςἡΑΓΔΜπυραμὶςπρὸς
mid ACDM [Prop. 5.18]. But, as base ACD (is) to
τὴνΑΔΕΜπυραμίδα.δι᾿ἴσουἄραὡςἡΑΒΓΔβάσιςπρὸς base ADE, so pyramid ACDM (is) also to pyramid
τὴνΑΔΕβάσιν,οὕτωςἡΑΒΓΔΜπυραμὶςπρὸςτὴνΑΔΕΜ ADEM [Prop. 12.5]. Thus, via equality, as base ABCD
πυραμίδα.καὶσυνθέντιπάλιν,ὡςἡΑΒΓΔΕβάσιςπρὸςτὴν (is) to base ADE, so pyramid ABCDM (is) to pyramid
ΑΔΕβάσιν,οὕτωςἡΑΒΓΔΕΜπυραμὶςπρὸςτὴνΑΔΕΜ ADEM [Prop. 5.22]. And, again, via composition, as
πυραμίδα. ὁμοίωςδὴδειχθήσεται,ὅτικαὶὡςἡΖΗΘΚΛ base ABCDE (is) to base ADE, so pyramid ABCDEM
βάσιςπρὸςτὴνΖΗΘβάσιν,οὕτωςκαὶἡΖΗΘΚΛΝπυραμὶς (is) to pyramid ADEM [Prop. 5.18]. So, similarly, it can
πρὸςτὴνΖΗΘΝπυραμίδα. καὶἐπεὶδύοπυραμίδεςεἱσὶναἱ also be shown that as base FGHKL (is) to base FGH,
ΑΔΕΜ,ΖΗΘΝτριγώνουςἔχουσαιβάσειςκαὶὕψοςἴσον, so pyramid FGHKLN (is) also to pyramid FGHN. And
ἔστινἄραὡςἡΑΔΕβάσιςπρὸςτὴνΖΗΘβάσιν,οὕτως since ADEM and FGHN are two pyramids having tri-
ἡΑΔΕΜπυραμὶςπρὸςτὴνΖΗΘΝπυραμίδα. ἀλλ᾿ὡςἡ angular bases and equal height, thus as base ADE (is) to
ΑΔΕβάσιςπρὸςτὴνΑΒΓΔΕβάσιν,οὕτωςἦνἡΑΔΕΜ base FGH, so pyramid ADEM (is) to pyramid FGHN
πυραμὶςπρὸςτὴνΑΒΓΔΕΜπυραμίδα.καὶδι᾿ἴσουἄραὡς [Prop. 12.5]. But, as base ADE (is) to base ABCDE, so
ἡΑΒΓΔΕβάσιςπρὸςτὴνΖΗΘβάσιν,οὕτωςἡΑΒΓΔΕΜ pyramid ADEM (was) to pyramid ABCDEM. Thus, via
πυραμὶςπρὸςτὴνΖΗΘΝπυραμίδα.ἀλλὰμὴνκαὶὡςἡΖΗΘ equality, as base ABCDE (is) to base FGH, so pyramid
βάσιςπρὸςτὴνΖΗΘΚΛβάσιν,οὕτωςἦνκαὶἡΖΗΘΝπυ- ABCDEM (is) also to pyramid FGHN [Prop. 5.22].
ραμὶςπρὸςτὴνΖΗΘΚΛΝπυραμίδα,καὶδι᾿ἴσουἄραὡςἡ But, furthermore, as base FGH (is) to base FGHKL,
ΑΒΓΔΕβάσιςπρὸςτὴνΖΗΘΚΛβάσιν,οὕτωςἡΑΒΓΔΕΜ so pyramid FGHN was also to pyramid FGHKLN.
πυραμὶςπρὸςτὴνΖΗΘΚΛΝπυραμίδα·ὅπερἔδειδεῖξαι. Thus, via equality, as base ABCDE (is) to base FGHKL,
so pyramid ABCDEM (is) also to pyramid FGHKLN
[Prop. 5.22]. (Which is) the very thing it was required to
show.
Þ. Proposition 7
Πᾶνπρίσματρίγωνονἔχονβάσινδιαιρεῖταιεἰςτρεῖςπυ- Any prism having a triangular base is divided into
ραμίδαςἴσαςἀλλήλαιςτριγώνουςβάσειςἐχούσας. three pyramids having triangular bases (which are) equal
to one another.
F
E
D
C
B
A
῎Εστωπρίσμα, οὗβάσιςμὲντὸΑΒΓτρίγωνον, ἀπε- Let there be a prism whose base (is) triangle ABC,
ναντίονδὲτὸΔΕΖ·λέγω,ὅτιτὸΑΒΓΔΕΖπρίσμαδιαιρεῖται and opposite (plane) DEF . I say that prism ABCDEF
εἰς τρεῖς πυραμίδας ἴσας ἀλλήλαις τριγώνους ἐχούσας is divided into three pyramids having triangular bases
βάσεις.
(which are) equal to one another.
᾿Επεζεύχθωσαν γὰρ αἱ ΒΔ, ΕΓ, ΓΔ. ἐπεὶ παραλ- For let BD, EC, and CD have been joined. Since
ληλόγραμμόν ἐστι τὸ ΑΒΕΔ, διάμετρος δὲ αὐτὸῦ ἐστιν ABED is a parallelogram, and BD is its diagonal, tri-
ἡΒΔ,ἴσονἄραἐστιτὸΑΒΔτρίγωνοντῷΕΒΔτρίγωνῳ· angle ABD is thus equal to triangle EBD [Prop. 1.34].
482

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
καὶἡπυραμὶςἄρα,ἧςβάσιςμὲντὸΑΒΔτρίγωνον, κο- And, thus, the pyramid whose base (is) triangle ABD,
ρυφὴ δὲ τὸ Γ σημεῖον, ἴση ἐστὶ πυραμίδι, ἧς βάσις μέν and apex the point C, is equal to the pyramid whose base
ἐστιτὸΔΕΒτρίγωνον, κορυφὴδὲτὸΓσημεῖον. ἀλλὰ is triangle DEB, and apex the point C [Prop. 12.5]. But,
ἡπυραμίς,ἧςβάσιςμένἐστιτὸΔΕΒτρίγωνον,κορυφὴ the pyramid whose base is triangle DEB, and apex the
δὲτὸΓσημεῖον,ἡαὐτήἐστιπυραμίδι,ἧςβάσιςμένἐστι point C, is the same as the pyramid whose base is trianτὸΕΒΓτρίγωνον,κορυφὴδὲτὸΔσημεῖον·ὑπὸγὰρτῶν
gle EBC, and apex the point D. For they are contained
αὐτῶνἐπιπέδωνπεριέχεται. καὶπυραμὶςἄρα,ἧςβάσιςμέν by the same planes. And, thus, the pyramid whose base
ἐστιτὸΑΒΔτρίγωνον,κορυφὴδὲτὸΓσημεῖον,ἴσηἐστὶ is ABD, and apex the point C, is equal to the pyramid
πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΕΒΓ τρίγωνον, κορυφὴ whose base is EBC and apex the point D. Again, since
δὲτὸΔσημεῖον. πάλιν,ἐπεὶπαραλληλόγραμμόνἐστιτὸ FCBE is a parallelogram, and CE is its diagonal, trian-
ΖΓΒΕ,διάμετροςδέἐστιναὐτοῦἡΓΕ,ἴσονἐστὶτὸΓΕΖ gle CEF is equal to triangle CBE [Prop. 1.34]. And,
τρίγωνον τῷ ΓΒΕ τριγώνῳ. καὶ πυραμὶςἄρα, ἧς βάσις thus, the pyramid whose base is triangle BCE, and apex
μένἐστιτὸΒΓΕτρίγωνον,κορυφὴδὲτὸΔσημεῖον,ἴση the point D, is equal to the pyramid whose base is trian-
ἐστὶπυραμίδι,ἧςβάσιςμένἐστιτὸΕΓΖτρίγωνον,κορυφὴ gle ECF , and apex the point D [Prop. 12.5]. And the
δὲτὸΔσημεῖον.ἡδὲπυραμίς,ἧςβάσιςμένἐστιτὸΒΓΕ pyramid whose base is triangle BCE, and apex the point
τρίγωνον,κορυφὴδὲτὸΔσημεῖον,ἴσηἐδείχθηπυραμίδι,ἧς D, was shown (to be) equal to the pyramid whose base is
βάσιςμένἐστιτὸΑΒΔτρίγωνον,κορυφὴδὲτὸΓσημεῖον· triangle ABD, and apex the point C. Thus, the pyramid
καὶπυραμὶςἄρα,ἧςβάσιςμένἐστιτὸΓΕΖτρίγωνον,κο- whose base is triangle CEF , and apex the point D, is
ρυφὴ δὲ τὸ Δ σημεῖον, ἴσηἐστὶ πυραμίδι, ἧςβάσιςμέν also equal to the pyramid whose base [is] triangle ABD,
[ἐστι]τὸΑΒΔτρίγωνον,κορυφὴδὲτὸΓσημεῖον·διῄρηται and apex the point C. Thus, the prism ABCDEF has
ἄρατὸΑΒΓΔΕΖπρίσμαεἰςτρεῖςπυραμίδαςἴσαςἀλλήλαις been divided into three pyramids having triangular bases
τριγώνουςἐχούσαςβάσεις.
(which are) equal to one another.
Καὶἐπεὶπυραμίς,ἧςβάσιςμένἐστιτὸΑΒΔτρίγωνον, And since the pyramid whose base is triangle ABD,
κορυφὴδὲτὸΓσημεῖον,ἡαὐτήἐστιπυραμίδι,ἧςβάσις and apex the point C, is the same as the pyramid whose
τὸΓΑΒτρίγωνον,κορυφὴδὲτὸΔσημεῖον·ὑπὸγὰρτῶν base is triangle CAB, and apex the point D. For they are
αὐτῶν ἐπιπέδων περιέχονται· ἡ δὲ πυραμίς, ἧς βάσις τὸ contained by the same planes. And the pyramid whose
ΑΒΔτρίγωνον,κορυφὴδὲτὸΓσημεῖον,τρίτονἐδείχθη base (is) triangle ABD, and apex the point C, was shown
τοῦπρίσματος,οὗβάσιςτὸΑΒΓτρίγωνον,ἀπεναντίονδὲ (to be) a third of the prism whose base is triangle ABC,
τὸΔΕΖ,καὶἡπυραμὶςἄρα,ἧςβάσιςτὸΑΒΓτρίγωνον, and opposite (plane) DEF , thus the pyramid whose base
κορυφὴδὲτὸΔσημεῖον,τρίτονἐστὶτοῦπρίσματοςτοῦ is triangle ABC, and apex the point D, is also a third of
ἔχοντοςβάσιςτὴναὐτὴντὸΑΒΓτρίγωνον,ἀπεναντίονδὲ the pyramid having the same base, triangle ABC, and
τὸΔΕΖ. opposite (plane) DEF .
ÈÖ×Ñ.
Corollary
᾿Εκδὴτούτουφανερόν,ὅτιπᾶσαπυραμὶςτρίτονμέρος And, from this, (it is) clear that any pyramid is the
ἐστὶτοῦπρίσματοςτοῦτὴναὐτὴνβάσινἔχοντοςαὐτῇκαὶ third part of the prism having the same base as it, and an
ὕψοςἴσον·ὅπερἔδειδεῖξαι.
equal height. (Which is) the very thing it was required to
show.
. Proposition 8
Αἱὅμοιαιπυραμίδεςκαὶτριγώνουςἔχουσαιβάσειςἐν Similar pyramids which also have triangular bases are
τριπλασίονιλόγῳεἰσὶτῶνὁμολόγωνπλευρῶν.
in the cubed ratio of their corresponding sides.
῎Εστωσαν ὅμοιαι καὶ ὁμοίως κείμεναι πυραμίδες, ὧν Let there be similar, and similarly laid out, pyramids
βάσειςμένεἰσιτὰΑΒΓ,ΔΕΖτρίγωνα,κορυφαὶδὲτὰΗ, whose bases are triangles ABC and DEF , and apexes
Θσημεῖα· λέγω, ὅτιἡΑΒΓΗπυραμὶςπρὸςτὴνΔΕΖΘ the points G and H (respectively). I say that pyramid
πυραμίδατριπλασίοναλόγονἔχειἤπερἡΒΓπρὸςτὴνΕΖ. ABCG has to pyramid DEFH the cubed ratio of that
BC (has) to EF .
483

ËÌÇÁÉÁÏƺ ELEMENTS
Ã
À
Ä
Å
Æ
Â
È
Ç
Ê
K
M
L
D
BOOK 12
G N
H
B C E F
ΣυμπεπληρώσθωγὰρτὰΒΗΜΛ,ΕΘΠΟστερεὰπαραλ- For let the parallelepiped solids BGML and EHQP
ληλεπίπεδα.καὶἐπεὶὁμοίαἐστὶνἡΑΒΓΗπυραμὶςτῇΔΕΖΘ have been completed. And since pyramid ABCG is simiπυραμίδι,ἵσηἄραἐστὶνἡμὲνὑπὸΑΒΓγωνίατῇὑπὸΔΕΖ
lar to pyramid DEFH, angle ABC is thus equal to angle
γωνίᾳ,ἡδὲὑπὸΗΒΓτῇὑπὸΘΕΖ,ἡδὲὑπὸΑΒΗτῇὑπὸ DEF , and GBC to HEF , and ABG to DEH. And as AB
ΔΕΘ,καίἐστινὡςἡΑΒπρὸςτὴνΔΕ,οὕτωςἡΒΓπρὸς is to DE, so BC (is) to EF , and BG to EH [Def. 11.9].
τὴνΕΖ,καὶἡΒΗπρὸςτὴνΕΘ.καὶἐπείἐστινὡςἡΑΒ And since as AB is to DE, so BC (is) to EF , and (so)
πρὸςτὴνΔΕ,οὕτωςἡΒΓπρὸςτὴνΕΖ,καὶπερὶἴσαςγωνίας the sides around equal angles are proportional, parallelαἱπλευραὶἀνάλογόνεἰσιν,ὅμοιονἄραἐστὶτὸΒΜπαραλ-
ogram BM is thus similar to paralleleogram EQ. So,
ληλόγραμμοντῷΕΠπαραλληλογράμμῳ.διὰτὰαὐτὰδὴκαὶ for the same (reasons), BN is also similar to ER, and
τὸμὲνΒΝτῷΕΡὅμοιόνἐστι,τὸδὲΒΚτῷΕΞ·τὰτρία BK to EO. Thus, the three (parallelograms) MB, BK,
ἄρατὰΜΒ,ΒΚ,ΒΝτρισὶτοῖςΕΠ,ΕΞ,ΕΡὅμοιάἐστιν. and BN are similar to the three (parallelograms) EQ,
ἀλλὰτὰμὲντρίατὰΜΒ,ΒΚ,ΒΝτρισὶτοῖςἀπεναντίονἴσα EO, ER (respectively). But, the three (parallelograms)
τεκαὶὅμοιάἐστιν,τὰδὲτρίατὰΕΠ,ΕΞ,ΕΡτρισὶτοῖς MB, BK, and BN are (both) equal and similar to the
ἀπεναντίονἴσατεκαὶὅμοιάἐστιν.τὰΒΗΜΛ,ΕΘΠΟἄρα three opposite (parallelograms), and the three (parallelστερεὰὑπὸὁμοίωνἐπιπέδωνἴσωντὸπλῆθοςπεριέχεται.
ograms) EQ, EO, and ER are (both) equal and simi-
ὅμοιονἄραἐστὶτὸΒΗΜΛστερεὸντῷΕΘΠΟστερεῷ.τὰ lar to the three opposite (parallelograms) [Prop. 11.24].
δὲὅμοιαστερεὰπαραλληλεπίπεδαἐντριπλασίονιλόγῳἐστὶ Thus, the solids BGML and EHQP are contained by
τῶνὁμολόγωνπλευρῶν. τὸΒΗΜΛἄραστερεὸνπρὸςτὸ equal numbers of similar (and similarly laid out) planes.
ΕΘΠΟστερεὸντριπλασίοναλόγονἔχειἤπερἡὁμόλογος Thus, solid BGML is similar to solid EHQP [Def. 11.9].
πλευρὰἡΒΓπρὸςτὴνὁμόλογονπλευρὰντὴνΕΖ.ὡςδὲτὸ And similar parallelepiped solids are in the cubed ratio of
ΒΗΜΛστερεὸνπρὸςτὸΕΘΠΟστερεόν,οὕτωςἡΑΒΓΗ corresponding sides [Prop. 11.33]. Thus, solid BGML
πυραμὶςπρὸς τὴν ΔΕΖΘ πυραμίδα, ἐπειδήπερἡπυραμὶς has to solid EHQP the cubed ratio that the correspond-
ἕκτονμέροςἐστὶτοῦστερεοῦδιὰτὸκαὶτὸπρίσμαἥμισυ ing side BC (has) to the corresponding side EF . And as
ὂντοῦστερεοῦπαραλληλεπιπέδουτριπλάσιονεἶναιτῆςπυ- solid BGML (is) to solid EHQP , so pyramid ABCG (is)
ραμίδος. καὶἡΑΒΓΗἄραπυραμὶςπρὸςτὴνΔΕΖΘπυ- to pyramid DEFH, inasmuch as the pyramid is the sixth
ραμίδατριπλασίοναλόγονἔχειἤπερἡΒΓπρὸςτὴνΕΖ· part of the solid, on account of the prism, being half of the
ὅπερἔδειδεῖξαι.
A
parallelepiped solid [Prop. 11.28], also being three times
the pyramid [Prop. 12.7]. Thus, pyramid ABCG also has
to pyramid DEFH the cubed ratio that BC (has) to EF .
(Which is) the very thing it was required to show.
Corollary
ÈÖ×Ñ.
᾿Εκδὴτούτουφανερόν,ὅτικαὶαἱπολυγώνουςἔχου- So, from this, (it is) also clear that similar pyraσαιβάσειςὅμοιαιπυραμίδεςπρὸςἀλλήλαςἐντριπλασίονι
mids having polygonal bases (are) to one another as the
λόγῳεἰσὶτῶνὁμολόγωνπλευρῶν.διαιρεθεισῶνγὰραὐτῶν cubed ratio of their corresponding sides. For, dividing
εἰςτὰςἐναὐταῖςπυραμίδαςτριγώνουςβάσειςἐχούσαςτῷ them into the pyramids (contained) within them which
καὶ τὰ ὅμοια πολύγωνα τῶν βάσεων εἰς ὅμοια τρίγωνα have triangular bases, with the similar polygons of the
διαιρεῖσθαικαὶἴσατῷπλήθεικαὶὁμόλογατοῖςὅλοιςἔσται bases also being divided into similar triangles (which are)
O
Q
P
R
484

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
ὡς [ἡ] ἐν τῇ ἑτέρᾳ μία πυραμὶς τρίγωνον ἔχουσα βάσιν both equal in number, and corresponding, to the wholes
πρὸς τὴν ἐν τῇ ἑτέρᾳ μίαν πυραμίδα τρίγωνον ἔχουσαν [Prop. 6.20]. As one pyramid having a triangular base in
βάσιν,οὕτωςκαὶἅπασαιαἱἐντῇἑτέρᾳπυραμίδιπυραμίδες the former (pyramid having a polygonal base is) to one
τριγώνουςἔχουσαιβάσειςπρὸςτὰςἐντῇἑτέρᾳπυραμίδι pyramid having a triangular base in the latter (pyramid
πυραμίδας τριγώνους βάσεις ἐχούσας, τουτέστιν αὐτὴ ἡ having a polygonal base), so (the sum of) all the pyraπολύγωνον
βάσιν ἔχουσα πυραμὶς πρὸς τὴν πολύγωνον mids having triangular bases in the former pyramid will
βάσινἔχουσανπυραμίδα. ἡδὲτρίγωνονβάσινἔχουσαπυ- also be to (the sum of) all the pyramids having trianguραμὶςπρὸςτὴντρίγωνονβάσινἔχουσανἐντριπλασίονιλόγῳ
lar bases in the latter pyramid [Prop. 5.12]—that is to
ἐστὶτῶνὁμολόγονπλευρῶν·καὶἡπολύγωνονἄραβάσιν say, the (former) pyramid itself having a polygonal base
ἔχουσαπρὸςτὴνὁμοίανβάσινἔχουσαντριπλασίοναλόγον to the (latter) pyramid having a polygonal base. And a
ἔχειἤπερἡπλευρὰπρὸςτὴνπλευράν.
pyramid having a triangular base is to a (pyramid) having
a triangular base in the cubed ratio of corresponding
sides [Prop. 12.8]. Thus, a (pyramid) having a polygonal
base also has to to a (pyramid) having a similar base the
cubed ratio of a (corresponding) side to a (corresponding)
side.
. Proposition 9
Τῶν ἴσων πυραμίδων καὶ τριγώνους βάσεις ἐχουσῶν The bases of equal pyramids which also have trian-
ἀντιπεπόνθασιναἱβάσειςτοῖςὕψεσιν· καὶὧν πυραμίδων gular bases are reciprocally proportional to their heights.
τριγώνουςβάσειςἐχουσῶνἀντιπεπόνθασιναἱβάσειςτοῖς
Â
And those pyramids which have triangular bases whose
ὕψεσιν,ἴσαιεἰσὶνἐκεῖναι.
Ä
bases are reciprocally proportional to their heights are
Ê Ç
equal.
O
À
H
Å
L
R P
G
M
A
B C
È
E D
F Q
῎Εστωσανγὰρἴσαιπυραμίδεςτριγώνουςβάσειςἔχουσαι For let there be (two) equal pyramids having the triτὰςΑΒΓ,ΔΕΖ,κορυφὰςδὲτὰΗ,Θσημεῖα·λέγω,ὅτιτῶν
angular bases ABC and DEF , and apexes the points G
ΑΒΓΗ, ΔΕΖΘπυραμίδωνἀντιπεπόνθασιναἱβάσειςτοῖς and H (respectively). I say that the bases of the pyramids
ὕψεσιν,καίἐστινὡςἡΑΒΓβάσιςπρὸςτὴνΔΕΖβάσιν, ABCG and DEFH are reciprocally proportional to their
οὕτωςτὸτῆςΔΕΖΘπυραμίδοςὕψοςπρὸςτὸτῆςΑΒΓΗ heights, and (so) that as base ABC is to base DEF , so
πυραμίδοςὕψος.
the height of pyramid DEFH (is) to the height of pyra-
ΣυμπεπληρώσθωγὰρτὰΒΗΜΛ,ΕΘΠΟστερεὰπαραλ- mid ABCG.
ληλεπίπεδα.καὶἐπεὶἴσηἐστὶνἡΑΒΓΗπυραμὶςτῇΔΕΖΘ For let the parallelepiped solids BGML and EHQP
πυραμίδι, καίἐστιτῆςμὲνΑΒΓΗπυραμίδοςἑξαπλάσιον have been completed. And since pyramid ABCG is
τὸΒΗΜΛστερεόν,τῆςδὲΔΕΖΘπυραμίδοςἑξαπλάσιον equal to pyramid DEFH, and solid BGML is six times
τὸΕΘΠΟστερεόν,ἴσονἄραἐστὶτὸΒΗΜΛστερεὸντῷ pyramid ABCG (see previous proposition), and solid
ΕΘΠΟστερεῷ. τῶνδὲἴσωνστερεῶνπαραλληλεπιπώδων EHQP (is) six times pyramid DEFH, solid BGML is
485

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
ἀντιπεπόνθασιναἱβάσειςτοῖςὕψεσιν·ἔστινἄραὡςἡΒΜ thus equal to solid EHQP . And the bases of equal parβάσιςπρὸςτὴνΕΠβάσιν,οὕτωςτὸτοῦΕΘΠΟστερεοῦ
allelepiped solids are reciprocally proportional to their
ὕψοςπρὸςτὸτοῦΒΗΜΛστερεοῦὕψος. ἀλλ᾿ὡςἡΒΜ heights [Prop. 11.34]. Thus, as base BM is to base EQ,
βάσιςπρὸςτὴνΕΠ,οὕτωςτὸΑΒΓτρίγωνονπρὸςτὸΔΕΖ so the height of solid EHQP (is) to the height of solid
τρίγωνον. καὶὡςἄρατὸΑΒΓτρίγωνονπρὸςτὸΔΕΖ BGML. But, as base BM (is) to base EQ, so triangle
τρίγωνον,οὕτωςτὸτοῦΕΘΠΟστερεοῦὕψοςπρὸςτὸτοῦ ABC (is) to triangle DEF [Prop. 1.34]. And, thus, as
ΒΗΜΛστερεοῦὕψος. ἀλλὰτὸμὲντοῦΕΘΠΟστερεοῦ triangle ABC (is) to triangle DEF , so the height of solid
ὕψοςτὸαὐτὸἐστιτῷτῆςΔΕΖΘπυραμίδοςὕψει,τὸδὲ EHQP (is) to the height of solid BGML [Prop. 5.11].
τοῦΒΗΜΛστερεοῦὕψοςτὸαὐτόἐστιτῷτῆςΑΒΓΗπυ- But, the height of solid EHQP is the same as the height
ραμίδοςὕψει·ἔστινἄραὡςἡΑΒΓβάσιςπρὸςτὴνΔΕΖ of pyramid DEFH, and the height of solid BGML is
βάσιν,οὕτωςτὸτῆςΔΕΖΘπυραμίδοςὕψοςπρὸςτὸτῆς the same as the height of pyramid ABCG. Thus, as base
ΑΒΓΗπυραμίδοςὕψος.τῶνΑΒΓΗ,ΔΕΖΘἄραπυραμίδων ABC is to base DEF , so the height of pyramid DEFH
ἀντιπεπόνθασιναἱβάσειςτοῖςὕψεσιν.
(is) to the height of pyramid ABCG. Thus, the bases
ἈλλὰδὴτῶνΑΒΓΗ,ΔΕΖΘπυραμίδωνἀντιπεπονθέτ- of pyramids ABCG and DEFH are reciprocally proporωσαναἱβάσειςτοῖςὕψεσιν,καὶἔστωὡςἡΑΒΓβάσιςπρὸς
tional to their heights.
τὴνΔΕΖβάσιν,οὕτωςτὸτῆςΔΕΖΘπυραμίδοςὕψοςπρὸς And so, let the bases of pyramids ABCG and DEFH
τὸτῆςΑΒΓΗπυραμίδοςὕψος·λέγω,ὅτιἴσηἐστὶνἡΑΒΓΗ be reciprocally proportional to their heights, and (thus)
πυραμὶςτῇΔΕΖΘπυραμίδι.
let base ABC be to base DEF , as the height of pyramid
Τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπείἐστιν ὡς ἡ DEFH (is) to the height of pyramid ABCG. I say that
ΑΒΓβάσιςπρὸςτὴνΔΕΖβάσιν,οὕτωςτὸτῆςΔΕΖΘπυ- pyramid ABCG is equal to pyramid DEFH.
ραμίδοςὕψοςπρὸςτὸτῆςΑΒΓΗπυραμίδοςὕψος,ἀλλ᾿ὡς For, with the same construction, since as base ABC
ἡΑΒΓβάσιςπρὸςτὴνΔΕΖβάσιν,οὕτωςτὸΒΜπαραλ- is to base DEF , so the height of pyramid DEFH (is) to
ληλόγραμμονπρὸςτὸΕΠπαραλληλόγραμμον,καὶὡςἄρα the height of pyramid ABCG, but as base ABC (is) to
τὸΒΜπαραλληλόγραμμονπρὸςτὸΕΠπαραλληλόγραμμον, base DEF , so parallelogram BM (is) to parallelogram
οὕτωςτὸτῆςΔΕΖΘπυραμίδοςὕψοςπρὸςτὸτῆςΑΒΓΗ EQ [Prop. 1.34], thus as parallelogram BM (is) to paralπυραμίδοςὕψος.ἀλλὰτὸ[μὲν]τῆςΔΕΖΘπυραμίδοςὕψος
lelogram EQ, so the height of pyramid DEFH (is) also
τὸαὐτόἐστιτῷτοῦΕΘΠΟπαραλληλεπιπέδουὕψει,τὸδὲ to the height of pyramid ABCG [Prop. 5.11]. But, the
τῆςΑΒΓΗπυραμίδοςὕψοςτὸαὐτόἐστιτῷτοῦΒΗΜΛ height of pyramid DEFH is the same as the height of
παραλληλεπιπέδουὕψει· ἔστιν ἄραὡς ἡΒΜ βάσιςπρὸς parallelepiped EHQP , and the height of pyramid ABCG
τὴν ΕΠ βάσιν, οὕτως τὸ τοῦ ΕΘΠΟ παραλληλεπιπέδου is the same as the height of parallelepiped BGML. Thus,
ὕψος πρὸς τὸ τοῦ ΒΗΜΛ παραλληλεπιπέδουὕψος. ὧν as base BM is to base EQ, so the height of parallelepiped
δὲ στερεῶν παραλληλεπιπέδων ἀντιπεπόνθασιν αἱ βάσεις EHQP (is) to the height of parallelepiped BGML. And
τοῖς ὕψεσιν, ἴσα ἐστὶν ἐκεῖνα· ἴσον ἄρα ἐστὶ τὸ ΒΗΜΛ those parallelepiped solids whose bases are reciprocally
στερεὸνπαραλληλεπίπεδοντῷΕΘΠΟστερεῷπαραλληλε- proportional to their heights are equal [Prop. 11.34].
πιπέδῳ. καίἐστιτοῦμὲνΒΗΜΛἕκτονμέροςἡΑΒΓΗ Thus, the parallelepiped solid BGML is equal to the parπυραμίς,τοῦδὲΕΘΠΟπαραλληλεπιπέδουἕκτονμέροςἡ
allelepiped solid EHQP . And pyramid ABCG is a sixth
ΔΕΖΘπυραμίς·ἴσηἄραἡΑΒΓΗπυραμὶςτῇΔΕΖΘπυ- part of BGML, and pyramid DEFH a sixth part of parραμίδι.
allelepiped EHQP . Thus, pyramid ABCG is equal to
Τῶνἄραἴσωνπυραμίδωνκαὶτριγώνουςβάσειςἐχουσῶν pyramid DEFH.
ἀντιπεπόνθασιναἱβάσειςτοῖςὕψεσιν· καὶὧν πυραμίδων Thus, the bases of equal pyramids which also have
τριγώνουςβάσειςἐχουσῶνἀντιπεπόνθασιναἱβάσειςτοῖς triangular bases are reciprocally proportional to their
ὕψεσιν,ἴσαιεἰσὶνἐκεῖναι·ὅπερἔδειδεῖξαι. heights. And those pyramids having triangular bases
whose bases are reciprocally proportional to their heights
are equal. (Which is) the very thing it was required to
show.
. Proposition 10
Πᾶςκῶνοςκυλίνδρουτρίτονμέροςἐστὶτοῦτὴναὐτὴν Every cone is the third part of the cylinder which has
βάσινἔχοντοςαὐτῷκαὶὕψοςἴσον.
the same base as it, and an equal height.
᾿Εχέτω γὰρκῶνοςκυλίνδρῷβάσιντετὴναὐτὴντὸν For let there be a cone (with) the same base as a cylin-
486

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
ΑΒΓΔ κύκλον καὶ ὕψος ἴσον· λέγω, ὅτι ὁ κῶνος τοῦ der, (namely) the circle ABCD, and an equal height. I
κυλίνδρουτρίτονἐστὶμέρος,τουτέστινὅτιὁκύλινδροςτοῦ
Â
say that the cone is the third part of the cylinder—that is
κώνουτριπλασίωνἐστίν.
to say, that the cylinder is three times the cone.
A
À
E
H
B
D
Εἰγὰρ μήἐστιν ὁκύλινδροςτοῦκώνουτριπλασίων, For if the cylinder is not three times the cone then the
ἔσται ὁ κύλινδρος τοῦ κώνου ἤτοι μείζων ἢ τριπλασίων cylinder will be either more than three times, or less than
ἢἐλάσσωνἢτριπλασίων. ἔστωπρότερονμείζων ἢτρι- three times, (the cone). Let it, first of all, be more than
πλασίων,καὶἐγγεγράφθωεἰςτὸνΑΒΓΔκύκλοντετράγων- three times (the cone). And let the square ABCD have
οντὸΑΒΓΔ·τὸδὴΑΒΓΔτετράγωνονμείζόνἐστινἢτὸ been inscribed in circle ABCD [Prop. 4.6]. So, square
ἥμισυτοῦΑΒΓΔκύκλου.καὶἀνεστάτωἀπὸτοῦΑΒΓΔτε- ABCD is more than half of circle ABCD [Prop. 12.2].
τραγώνουπρίσμαἰσοϋψὲςτῷκυλίνδρῳ.τὸδὴἀνιστάμενον And let a prism of equal height to the cylinder have been
πρίσμαμεῖζόνἐστιν ἢτὸἥμισυτοῦκυλίνδου, ἐπειδήπερ set up on square ABCD. So, the prism set up is more
κἂν περὶ τὸν ΑΒΓΔ κύκλον τετράγωνον περιγράψωμεν, than half of the cylinder, inasmuch as if we also circumτὸ
ἐγγεγραμμένον εἰς τὸν ΑΒΓΔ κύκλον τετράγωνον scribe a square around circle ABCD [Prop. 4.7] then the
ἥμισύἐστι τοῦπεριγεγραμμένου· καίἐστι τὰ ἀπ᾿αὐτῶν square inscribed in circle ABCD is half of the circum-
ἀνιστάμεναστερεὰπαραλληλεπίπεδαπρίσματαἰσοϋψῆ·τὰ scribed (square). And the solids set up on them are parδὲὑπὸτὸαὐτὸὕψοςὄνταστερεὰπαραλληλεπίπεδαπρὸς
allelepiped prisms of equal height. And parallelepiped
ἄλληλάἐστινὡςαἱβάσεις·καὶτὸἐπὶτοῦΑΒΓΔἄρατε- solids having the same height are to one another as their
τραγώνουἀνασταθὲνπρίσμαἥμισύἐστιτοῦἀνασταθέντος bases [Prop. 11.32]. And, thus, the prism set up on
πρίσματοςἀπὸτοῦπερὶτὸνΑΒΓΔκύκλονπεριγραφέντος square ABCD is half of the prism set up on the square
τετραγώνου·καίἐστινὁκύλινδροςἐλάττωντοῦπρίσματος circumscribed about circle ABCD. And the cylinder is
τοῦἀνατραθέντοςἀπὸτοῦπερὶτὸνΑΒΓΔκύκλονπερι- less than the prism set up on the square circumscribed
γραφέντοςτετραγώνου·τὸἄραπρίσματὸἀνασταθὲνἀπὸ about circle ABCD. Thus, the prism set up on square
τοῦΑΒΓΔτετραγώνουἰσοϋψὲςτῷκυλίνδρῳμεῖζόνἐστι ABCD of the same height as the cylinder is more than
τοῦ ἡμίσεως τοῦ κυλίνδρου. τετμήσθωσαν αἱ ΑΒ, ΒΓ, half of the cylinder. Let the circumferences AB, BC,
ΓΔ,ΔΑπεριφέρειαιδίχακατὰτὰΕ,Ζ,Η,Θσημεῖα,καὶ CD, and DA have been cut in half at points E, F , G,
ἐπεζεύχθωσαναἱΑΕ,ΕΒ,ΒΖ,ΖΓ,ΓΗ,ΗΔ,ΔΘ,ΘΑ· and H. And let AE, EB, BF , FC, CG, GD, DH, and
καὶἕκαστονἄρατῶνΑΕΒ,ΒΖΓ,ΓΗΔ,ΔΘΑτριγώνων HA have been joined. And thus each of the triangles
μειζόν ἐστιν ἢ τὸ ἥμισυ τοῦ καθ᾿ ἑαυτὸ τηήματος τοῦ AEB, BFC, CGD, and DHA is more than half of the
ΑΒΓΔκύκλου,ὡςἔμπροσθενἐδείκνυμεν. ἀνεστάτωἐφ᾿ segment of circle ABCD about it, as was shown pre-
ἑκάστουτῶνΑΕΒ,ΒΖΓ,ΓΗΔ,ΔΘΑτριγώνωνπρίσματα viously [Prop. 12.2]. Let prisms of equal height to the
ἰσοϋψῆτῷκυλίνδρῳ·καὶἕκαστονἄρατῶνἀνασταθέντων cylinder have been set up on each of the triangles AEB,
πρισμάτωνμεῖζόνἐστινἢτὸἥμισυμέροςτοῦκαθ᾿ἑαυτὸ BFC, CGD, and DHA. And each of the prisms set up is
τμήματοςτοῦκυλίνδρου,ἐπειδήπερἐὰνδιὰτῶνΕ,Ζ,Η, greater than the half part of the segment of the cylinder
ΘσημείωνπαραλλήλουςταῖςΑΒ,ΒΓ,ΓΔ,ΔΑἀγάγωμεν, about it—inasmuch as if we draw (straight-lines) parallel
καὶσυμπληρώσωμεντὰἐπὶτῶνΑΒ,ΒΓ,ΓΔ,ΔΑπαραλ- to AB, BC, CD, and DA through points E, F , G, and H
F
C
G
487

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
ληλόγραμμα,καὶἀπ᾿αὐτῶνἀναστήσωμενστερεὰπαραλλη- (respectively), and complete the parallelograms on AB,
λεπίπεδαἰσοϋψῆτῷκυλίνδρῳ,ἑκάσουτῶνἀνασταθέντων BC, CD, and DA, and set up parallelepiped solids of
ἡμίσηἐστὶτὰπρίσματατὰἐπὶτῶνΑΕΒ,ΒΖΓ,ΓΗΔ,ΔΘΑ equal height to the cylinder on them, then the prisms on
τριγώνων· καί ἐστι τὰ τοῦ κυλίνδρου τμήματα ἐλάττονα triangles AEB, BFC, CGD, and DHA are each half of
τῶνἀνασταθέντωνστερεῶνπαραλληλεπιπέδων·ὥστεκαὶ the set up (parallelepipeds). And the segments of the
τὰἐπὶτῶνΑΕΒ, ΒΖΓ,ΓΗΔ,ΔΘΑτριγώνων πρίσματα cylinder are less than the set up parallelepiped solids.
μείζονάἐστινἢτὸἥμισυτῶνκαθ᾿ἑαυτὰτοῦκυλίνδρου Hence, the prisms on triangles AEB, BFC, CGD, and
τμημάτων. τέμνοντες δὴ τὰς ὑπολειπομέναςπεριφερείας DHA are also greater than half of the segments of the
δίχακαὶἐπιζευγνύντεςεὐθείαςκαὶἀνιστάντεςἐφ᾿ἑκάσου cylinder about them. So (if) the remaining circumferτῶντριγώνωνπρίσματαἰσοϋψῆτῷκυλίνδρῳκαὶτοῦτοἀεὶ
ences are cut in half, and straight-lines are joined, and
ποιοῦντεςκαταλείψομέντινα ἀποτμήματατοῦκυλίνδρου, prisms of equal height to the cylinder are set up on each
ἃἔσταιἐλάττονατῆςὑπεροχῆς, ᾗὑπερέχειὁκυλίνδρος of the triangles, and this is done continually, then we will
τοῦτριπλασίουτοῦκώνου. λελείφθω,καὶἔστωτὰΑΕ, (eventually) leave some segments of the cylinder whose
ΕΒ,ΒΖ,ΖΓ,ΓΗ,ΗΔ,ΔΘ,ΘΑ·λοιπὸνἄρατὸπρίσμα,οὗ (sum) is less than the excess by which the cylinder exβάσιςμὲντὸΑΕΒΖΓΗΔΘπολύγωνον,ὕψοςδὲτὸαὐτὸ
ceeds three times the cone [Prop. 10.1]. Let them have
τῷκυλίνδρῷ,μεῖζόνἐστὶνἢτριπλάσιοντοῦκώνου. ἀλλὰ been left, and let them be AE, EB, BF , FC, CG, GD,
τὸπρίσμα,οὗβάσιςμὲνἐστὶτὸΑΕΒΖΓΗΔΘπολύγωνον, DH, and HA. Thus, the remaining prism whose base
ὕψος δὲ τὸ αὐτὸ τῷ κυλίνδρῳ, τριπλάσιόν ἐστι τῆς πυ- (is) polygon AEBFCGDH, and height the same as the
ραμίδος, ἧςβάσιςμένἐστιτὸΑΕΒΖΓΗΔΘπολύγωνον, cylinder, is greater than three times the cone. But, the
κορυφὴδὲἡαὐτὴτῷκώνῳ·καὶἡπυραμὶςἄρα,ἧςβάσις prism whose base is polygon AEBFCGDH, and height
μέν[ἐστι]τὸΑΕΒΖΓΗΔΘπολύγωνον,κορυφὴδὲἡαὐτὴ the same as the cylinder, is three times the pyramid whose
τῷκώνῳ,μείζωνἐστὶτοῦκώνουτοῦβάσινἔχοντεςτὸν base is polygon AEBFCGDH, and apex the same as the
ΑΒΓΔκύκλον. ἀλλὰκαὶἐλάττων· ἐμπεριέχεταιγὰρὑπ᾿ cone [Prop. 12.7 corr.]. And thus the pyramid whose
αὐτοῦ· ὅπερἐστὶνἀδύνατον. οὐκἄραἐστὶνὁκύλινδρος base [is] polygon AEBFCGDH, and apex the same as
τοῦκώνουμεῖζωνἢτριπλάσιος.
the cone, is greater than the cone having (as) base circle
Λέγω δή, ὅτι οὐδὲ ἐλάττων ἐστὶν ἢ τριπλάσιος ὁ ABCD. But (it is) also less. For it is encompassed by it.
κύλινδροςτοῦκώνου.
The very thing (is) impossible. Thus, the cylinder is not
Εἰγὰρδυνατόν,ἔστωἐλάττωνἢτριπλάσιοςὁκύλινδρος more than three times the cone.
τοῦκώνου·ἀνάπαλινἄραὁκῶνοςτοῦκυλίνδρουμεῖζων So, I say that neither (is) the cylinder less than three
ἐστὶνἢτρίτονμέρος.ἐγγεγράφθωδὴεἰςτὸνΑΒΓΔκύκλον times the cone.
τετράγωνοντὸΑΒΓΔ·τὸΑΒΓΔἄρατετράγωνονμεῖζόν For, if possible, let the cylinder be less than three times
ἐστινἢτὸἥμισυτοῦΑΒΓΔκύκλου.καὶἀνεστάτωἀπὸτοῦ the cone. Thus, inversely, the cone is greater than the
ΑΒΓΔτετραγώνουπυραμὶςτὴναὐτὴνκορυφὴνἔχουσατῷ third part of the cylinder. So, let the square ABCD have
κώνῳ·ἠἄραἀνασταθεῖσαπυραμὶςμείζωνἐστὶνἢτὸἥμισυ been inscribed in circle ABCD [Prop. 4.6]. Thus, square
μέρος τοῦ κώνου, ἐπειδήπερ, ὡς ἕμπροσθεν ἐδείκνυμεν, ABCD is greater than half of circle ABCD. And let a
ὅτιἐὰνπερὶτὸνκύκλοντετράγωνονπεριγράψωμεν,ἔσται pyramid having the same apex as the cone have been set
τὸ ΑΒΓΔ τετράγωνον ἥμισυ τοῦ περὶ τὸν κύκλονπερι- up on square ABCD. Thus, the pyramid set up is greater
γεγραμμένουτετραγώνου· καὶἐὰνἀπὸτῶντετραγώνων than the half part of the cone, inasmuch as we showed
στερεὰπαραλληλεπίπεδαἀναστήσωμενἰσοϋψῆτῷκώνῳ,ἂ previously that if we circumscribe a square about the cirκαὶκαλεῖταιπρίσματα,ἔσταιτὸἀνασταθὲνἀπὸτοῦΑΒΓΔ
cle [Prop. 4.7] then the square ABCD will be half of the
τετραγώνου ἥμισυ τοῦ ἀνασταθέντος ἀπὸ τοῦ περὶ τὸν square circumscribed about the circle [Prop. 12.2]. And
κύκλονπεριγραφέντοςτετραγώνου·πρὸςἄλληλαγάρεἰσιν if we set up on the squares parallelepiped solids—which
ὡς αἱ βάσεις. ὥστε καὶ τὰ τρίτα· καὶ πυραμὶς ἄρα, ἧς are also called prisms—of the same height as the cone,
βάσιςτὸΑΒΓΔτετράγωνον,ἥμισύἐστιτῆςπυραμίδοςτῆς then the (prism) set up on square ABCD will be half
ἀνασταθείσηςἀπὸτοῦπερὶτὸνκύκλονπεριγραφέντοςτε- of the (prism) set up on the square circumscribed about
τραγώνου. καίἐστιμείζωνἡπυραμὶςἡἀνασταθεῖσαἀπὸ the circle. For they are to one another as their bases
τοῦπερὶτὸνκύκλοντετραγώνουτοῦκώνου· ἐμπεριέχει [Prop. 11.32]. Hence, (the same) also (goes for) the
γὰραὐτόν.ἡἄραπυραμὶς,ἧςβάσιςτὸΑΒΓΔτετράγωνον, thirds. Thus, the pyramid whose base is square ABCD
κορυφὴδὲἡαὐτὴτῷκώνῳ,μείζωνἐστὶνἢτὸἥμισυτοῦ is half of the pyramid set up on the square circumscribed
κώνου. τετμήσθωσαν αἱ ΑΒ, ΒΓ, ΓΔ, ΔΑ περιφέρειαι about the circle [Prop. 12.7 corr.]. And the pyramid set
δίχα κατὰ τὰ Ε, Ζ, Η, Θ σημεῖα, καὶ ἐπεζεύχθωσαν αἱ up on the square circumscribed about the circle is greater
488

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
ΑΕ, ΕΒ, ΒΖ,ΖΓ,ΓΗ, ΗΔ, ΔΘ,ΘΑ·καὶἕκαστονἄρα than the cone. For it encompasses it. Thus, the pyramid
τῶνΑΕΒ,ΒΖΓ,ΓΗΔ,ΔΘΑτριγώνωνμεῖζόνἐστινἢτὸ whose base is square ABCD, and apex the same as the
ἥμισυμέροςτουκαθ᾿ἑαυτὸτμήματοςτοῦΑΒΓΔκύκλου. cone, is greater than half of the cone. Let the circumκαὶἀνεστάτωσανἐφ᾿ἑκάστουτῶνΑΕΒ,ΒΖΓ,ΓΗΔ,ΔΘΑ
ferences AB, BC, CD, and DA have been cut in half
τριγώνωνπυραμίδεςτὴναὐτὴνκορυφὴνἔχουσαιτῷκώνῳ· at points E, F , G, and H (respectively). And let AE,
καὶ ἑκάστη ἄρα τῶν ἀνασταθεισῶν πυραμίδων κατὰ τὸν EB, BF , FC, CG, GD, DH, and HA have been joined.
αὐτὸν τρόπον μείζων ἐστὶν ἢ τὸ ἥμισυ μέρος τοῦ καθ᾿ And, thus, each of the triangles AEB, BFC, CGD, and
ἑαυτὴντμήματοςτοῦκώνου. τέμνοντες δὴ τὰς ὑπολει- DHA is greater than the half part of the segment of cirπομένας
περιφερείας δίχα καὶ ἐπιζευγνύντες εὐθείας καὶ cle ABCD about it [Prop. 12.2]. And let pyramids having
ἀνιστάντεςἐφ᾿ἑκάστουτῶντριγώνωνπυραμίδατὴναὐτὴν the same apex as the cone have been set up on each of the
κορυφὴν ἔχουσαν τῷ κώνῳ καὶ τοῦτο ἀεὶ ποιοῦτες κα- triangles AEB, BFC, CGD, and DHA. And, thus, in the
ταλείψομέντιναἀποτμήματατοῦκώνου,ἃἔσταιἐλάττονα same way, each of the pyramids set up is more than the
τῆςὑπεροχῆς,ᾗὑπερέχειὁκῶνοςτοῦτρίτουμέρουςτοῦ half part of the segment of the cone about it. So, (if) the
κυλίνδρου.λελείφθω,καὶἔστωτὰἐπὶτῶνΑΕ,ΕΒ,ΒΖ,ΖΓ, remaining circumferences are cut in half, and straight-
ΓΗ,ΗΔ,ΔΘ,ΘΑ·λοιπὴἄραἡπυραμίς,ἧςβάσιςμένἐστι lines are joined, and pyramids having the same apex as
τὸΑΕΒΖΓΗΔΘπολύγωνον,κορυφὴδὲἡαὐτὴτῷκώνῳ, the cone are set up on each of the triangles, and this is
μείζωνἐστὶνἢτρίτονμέροςτοῦκυλίνδρου.ἀλλ᾿ἡπυραμίς, done continually, then we will (eventually) leave some
ἧςβάσιςμένἐστιτὸΑΕΒΖΓΗΔΘπολύγωνον,κορυφὴδὲ segments of the cone whose (sum) is less than the excess
ἡαυτὴτῷκώνῳ,τρίτονἐστὶμέροςτοῦπρίσματος,οὗβάσις by which the cone exceeds the third part of the cylinder
μένἐστιτὸΑΕΒΖΓΗΔΘπολύγωνον,ὕψοςδὲτὸαὐτὸτῷ [Prop. 10.1]. Let them have been left, and let them be
κυλίνδρῳ·τὸἄραπρίσμα,οὗβάσιςμένἐστιτὸΑΕΒΖΓΗΔΘ the (segments) on AE, EB, BF , FC, CG, GD, DH, and
πολύγωνον,ὕψοςδὲτὸαὐτὸτῷκυλίνδρῳ,μεῖζόνἐστιτοῦ HA. Thus, the remaining pyramid whose base is polyκυλίνδρου,οὗβάσιςἐστὶνὁΑΒΓΔκύκλος.ἀλλὰκαὶἔλατ-
gon AEBFCGDH, and apex the same as the cone, is
τον·ἐμπεριέχεταιγὰρὑπ᾿αὐτοῦ·ὅπερἐστὶνἀδύνατον.οὐκ greater than the third part of the cylinder. But, the pyra-
ἄραὁκύλινδροςτοῦκώνουἐλάττωνἐστὶνἢτριπλάσιος. mid whose base is polygon AEBFCGDH, and apex the
ἐδείχθηδέ,ὅτιοὐδὲμείζωνἢτριπλάσιος·τριπλάσιοςἄραὁ same as the cone, is the third part of the prism whose
κύλινδροςτοῦκώνου·ὥστεὁκῶνοςτρίτονἐστὶμέροςτοῦ base is polygon AEBFCGDH, and height the same as
κυλίνδρου.
the cylinder [Prop. 12.7 corr.]. Thus, the prism whose
Πᾶςἄρακῶνοςκυλίνδρουτρίτονμέροςἐστὶτοῦτὴν base is polygon AEBFCGDH, and height the same as
αὐτὴνβάσινἔχοντοςαὐτῷκαὶὕψοςἴσον·ὅπερἔδειδεῖξαι. the cylinder, is greater than the cylinder whose base is
circle ABCD. But, (it is) also less. For it is encompassed
by it. The very thing is impossible. Thus, the cylinder is
not less than three times the cone. And it was shown that
neither (is it) greater than three times (the cone). Thus,
the cylinder (is) three times the cone. Hence, the cone is
the third part of the cylinder.
Thus, every cone is the third part of the cylinder which
has the same base as it, and an equal height. (Which is)
the very thing it was required to show.
. Proposition 11
Οἱὑποτὸαὐτὸὕψοςὄντεςκῶνοικαὶκύλινδροιπρὸς Cones and cylinders having the same height are to one
ἀλλήλουςεἰσὶνὡςαἱβάσεις.
another as their bases.
῎Εστωσανὑπὸτὸαὐτὸὕψοςκῶνοικαὶκύλινδροι,ὧν Let there be cones and cylinders of the same height
βάσειςμὲν[εἰσιν]οἱΑΒΓΔ,ΕΖΗΘκύκλοι,ἄξονεςδὲοἱ whose bases [are] the circles ABCD and EFGH, axes
ΚΛ,ΜΝ,διάμετροιδὲτῶνβάσεωναἱΑΓ,ΕΗ·λέγω,ὅτι KL and MN, and diameters of the bases AC and EG (re-
ἐστὶνὡςὁΑΒΓΔκύκλοςπρὸςτὸνΕΖΗΘκύκλον,οὕτως spectively). I say that as circle ABCD is to circle EFGH,
ὁΑΛκῶνοςπρὸςτὸνΕΝκῶνον.
so cone AL (is) to cone EN.
489

ËÌÇÁÉÁÏƺ ELEMENTS
Ì É Ä
Ç Â Ë
à Æ
Å À Í È Ê
BOOK 12
Εἰγὰρμή,ἔσταιὡςὁΑΒΓΔκύκλοςπρὸςτὸνΕΖΗΘ For if not, then as circle ABCD (is) to circle EFGH,
κύκλον,οὕτωςὁΑΛκῶνοςἤτοιπρὸςἔλασσόντιτοῦΕΝ so cone AL will be to some solid either less than, or
κώνουστερεὸνἢπρὸςμεῖζον. ἔστωπρότερονπρὸςἔλασ- greater than, cone EN. Let it, first of all, be (in this raσοντὸΞ,καὶᾧἔλασσόνἐστιτὸΞστερεὸντοῦΕΝκώνου,
tio) to (some) lesser (solid), O. And let solid X be equal
ἐκείνῳἴσονἔστωτὸΨστερεόν·ὁΕΝκῶνοςἄραἴσοςἐστὶ to that (magnitude) by which solid O is less than cone
τοῖςΞ,Ψστερεοῖς. ἐγγεγράφθωεἰςτὸνΕΖΗΘκύκλον EN. Thus, cone EN is equal to (the sum of) solids O
τετράγωνον τὸ ΕΖΗΘ·τὸ ἄρατετράγωνον μεῖζόν ἐστιν and X. Let the square EFGH have been inscribed in cir-
ἢ τὸ ἥμισυ τοῦ κύκλου. ἀνεστάτω ἀπὸ τοῦ ΕΖΗΘ τε- cle EFGH [Prop. 4.6]. Thus, the square is greater than
τραγώνουπυραμὶςἰσοϋψὴςτῷκώνῳ· ἡἄραἀνασταθεῖσα half of the circle [Prop. 12.2]. Let a pyramid of the same
πυραμὶςμείζωνἐστὶνἢτὸἥμισυτοῦκώνου,ἐπειδήπερἐὰν height as the cone have been set up on square EFGH.
περιγράψωμενπερὶτὸνκύκλοντετράγωνον,καὶἀπ᾿αὐτοῦ Thus, the pyramid set up is greater than half of the cone,
ἀναστήσωμενπυραμίδαἰσοϋψῆτῷκώνῳ,ἡἐγγραφεῖσαπυ- inasmuch as, if we circumscribe a square about the cirραμὶςἥμισύἐστιτῆςπεριγραφείσης·πρὸςἀλλήλαςγάρεἰσιν
cle [Prop. 4.7], and set up on it a pyramid of the same
ὡςαἱβάσεις·ἐλάττωνδὲὁκῶνοςτῆςπεριγραφείσηςπυ- height as the cone, then the inscribed pyramid is half
ραμίδος. τετμήσθωσαναἱΕΖ,ΖΗ,ΗΘ,ΘΕπεριφέρειαι of the circumscribed pyramid. For they are to one anδίχα
κατὰτὰ Ο, Π, Ρ, Σ σημεῖα, καὶἐπεζεύχθωσαν αἱ other as their bases [Prop. 12.6]. And the cone (is) less
ΘΟ,ΟΕ,ΕΠ,ΠΖ,ΖΡ,ΡΗ,ΗΣ,ΣΘ.ἕκαστονἄρατῶν than the circumscribed pyramid. Let the circumferences
ΘΟΕ,ΕΠΖ,ΖΡΗ,ΗΖΘτριγώνωνμεῖζόνἐστινἢτὸἥμισυ EF , FG, GH, and HE have been cut in half at points
τοῦ καθ᾿ ἑαυτὸ τμήματος τοῦ κύκλου. ἀνεστάτω ἐφ᾿ P , Q, R, and S. And let HP , PE, EQ, QF , FR, RG,
ἑκάστουτῶνΘΟΕ,ΕΠΖ,ΖΡΗ,ΗΣΘτριγώνωνπυραμὶς GS, and SH have been joined. Thus, each of the trian-
ἰσοϋψὴςτῷκώνῳ·καὶἑκάστηἄρατῶνἀνασταθεισῶνπυ- gles HPE, EQF , FRG, and GSH is greater than half
ραμίδωνμείζωνἐστὶνἢτὸἥμισυτοῦκαθ᾿ἑαυτὴντμήματος of the segment of the circle about it [Prop. 12.2]. Let
τοῦκώνου. τέμνοντεςδὴτὰςὑπολειπομέναςπεριφερείας pyramids of the same height as the cone have been set up
δίχακαὶἐπιζευγνύντεςεὐθείαςκαὶἀνιστάντεςἐπὶἑκάστου on each of the triangles HPE, EQF , FRG, and GSH.
τῶντριγώνωνπυραμίδαςἰσοϋψεῖςτῷκώνῳκαὶἀεὶτοῦτο And, thus, each of the pyramids set up is greater than
ποιοῦντες καταλείψομέν τινα ἀποτμήματα τοῦ κώνου, ἃ half of the segment of the cone about it [Prop. 12.10].
ἔσταιἐλάσσονατοῦΨστερεοῦ. λελείφθω,καὶἔστωτὰ So, (if) the remaining circumferences are cut in half, and
ἐπὶτῶνΘΟΕ,ΕΠΖ,ΖΡΗ,ΗΣΘ·λοιπὴἄραἡπυραμίς,ἧς straight-lines are joined, and pyramids of equal height
βάσιςτὸΘΟΕΠΖΡΗΣπολύγωνον,ὕψοςδὲτὸαὐτὸτῷ to the cone are set up on each of the triangles, and
κώνῳ,μείζωνἐστὶτοῦΞστερεοῦ.ἐγγεγράφθωκαὶεἰςτὸν this is done continually, then we will (eventually) leave
ΑΒΓΔκύκλοντῷΘΟΕΠΖΡΗΣπολυγώνῳὅμοιόντεκαὶ some segments of the cone (the sum of) which is less
ὁμοίωςκείμενονπολύγωνοντὸΔΤΑΥΒΦΓΧ,καὶἀνεστάτω than solid X [Prop. 10.1]. Let them have been left, and
ἐπ᾿αὐτοῦπυραμὶςἰσοϋψὴςτῷΑΛκώνῳ.ἐπεὶοὖνἐστινὡς let them be the (segments) on HPE, EQF , FRG, and
τὸἀπὸτῆςΑΓπρὸςτὸἀπὸτῆςΕΗ,οὕτωςτὸΔΤΑΥΒΦΓΧ GSH. Thus, the remaining pyramid whose base is polyπολύγωνονπρὸςτὸΘΟΕΠΖΡΗΣπολύγωνον,
ὡςδὲτὸ gon HPEQFRGS, and height the same as the cone, is
ἀπὸτῆςΑΓπρὸςτὸἀπὸτῆςΕΗ,οὕτωςὁΑΒΓΔκύκλος greater than solid O [Prop. 6.18]. And let the polygon
πρὸςτὸνΕΖΗΘκύκλον,καὶὡςἄραὁΑΒΓΔκύκλοςπρὸς DTAUBV CW , similar, and similarly laid out, to polygon
τὸν ΕΖΗΘ κύκλον, οὕτως τὸ ΔΤΑΥΒΦΓΧ πολύγωνον HPEQFRGS, have been inscribed in circle ABCD. And
πρὸςτὸΘΟΕΠΖΡΗΣπολύγωνον.ὡςδὲὁΑΒΓΔκύκλος on it let a pyramid of the same height as cone AL have
πρὸςτὸνΕΖΗΘκύκλον,οὕτωςὁΑΛκῶνοςπρὸςτὸΞ been set up. Therefore, since as the (square) on AC is
στερεόν,ὡςδὲτὸΔΤΑΥΒΦΓΧπολύγωνονπρὸςτὸΘΟ- to the (square) on EG, so polygon DTAUBV CW (is) to
ΕΠΖΡΗΣπολύγωνον,οὕτωςἡπυραμίς,ἧςβάσιςμὲντὸ polygon HPEQFRGS [Prop. 12.1], and as the (square)
ΔΤΑΥΒΦΓΧπολύγωνον,κορυφὴδὲτὸΛσημεῖον,πρὸς on AC (is) to the (square) on EG, so circle ABCD (is)
A
T
U
D
K
B
W
V
C
L
E
P
Q
H
M
F
S
R
G
N
X
O
490

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
τὴνπυραμίδα,ἧςβάσιςμὲντὸΘΟΕΠΖΡΗΣπολύγωνον, to circle EFGH [Prop. 12.2], thus as circle ABCD (is)
κορυφὴδὲτὸΝσημεῖον.καὶὡςἄραὁΑΛκῶνοςπρὸςτὸ to circle EFGH, so polygon DTAUBV CW also (is) to
Ξστερεόν,οὕτωςἡπυραμίς,ἧςβάσιςμὲντὸΔΤΑΥΒΦΓΧ polygon HPEQFRGS. And as circle ABCD (is) to cirπολύγωνον,κορυφὴδὲτὸΛσημεῖον,πρὸςτὴνπυραμίδα,
cle EFGH, so cone AL (is) to solid O. And as poly-
ἧςβάσιςμὲντὸΘΟΕΠΖΡΗΣπολύγωνον,κορυφὴδὲτὸΝ gon DTAUBV CW (is) to polygon HPEQFRGS, so the
σημεῖον·ἐναλλὰξἄραἐστὶνὡςὁΑΛκῶνοςπρὸςτὴνἐν pyramid whose base is polygon DTAUBV CW , and apex
αὐτῷπυραμίδα,οὕτωςτὸΞστερεὸνπρὸςτὴνἐντῷΕΝ the point L, (is) to the pyramid whose base is polygon
κώνῳπυραμίδα. μείζωνδὲὁΑΛκῶνοςτῆςἐναὐτῷπυ- HPEQFRGS, and apex the point N [Prop. 12.6]. And,
ραμίδος·μεῖζονἄρακαὶτὸΞστερεὸντῆςἐντῷΕΝκώνῳ thus, as cone AL (is) to solid O, so the pyramid whose
πυραμίδος.ἀλλὰκαὶἔλασσον·ὅπερἄτοπον.οὐκἄραἐστὶν base is DTAUBV CW , and apex the point L, (is) to the
ὡςὁΑΒΓΔκύκλοςπρὸςτὸνΕΖΗΘκύκλον,οὕτωςὁΑΛ pyramid whose base is polygon HPEQFRGS, and apex
κῶνοςπρὸςἔλασσόντιτοῦΕΝκώνουστερεόν. ὁμοίως the point N [Prop. 5.11]. Thus, alternately, as cone AL
δὲδείξομεν,ὅτιοὐδέἐστινὡςὁΕΖΗΘκύκλοςπρὸςτὸν is to the pyramid within it, so solid O (is) to the pyramid
ΑΒΓΔκύκλον,οὕτωςὁΕΝκῶνοςπρὸςἔλασσόντιτοῦ within cone EN [Prop. 5.16]. But, cone AL (is) greater
ΑΛκώνουστερεόν.
than the pyramid within it. Thus, solid O (is) also greater
Λέγωδή,ὅτιοὐδέἐστινὡςὁΑΒΓΔκύκλοςπρὸςτὸν than the pyramid within cone EN [Prop. 5.14]. But, (it
ΕΖΗΘκύκλον,οὕτωςὁΑΛκῶνοςπρὸςμεῖζόντιτοῦΕΝ is) also less. The very thing (is) absurd. Thus, circle
κώνουστερεόν.
ABCD is not to circle EFGH, as cone AL (is) to some
Εἰγὰρδυνατόν,ἕστωπρὸςμεῖζοντὸΞ·ἀνάπαλινἄρα solid less than cone EN. So, similarly, we can show that
ἐστὶνὡςὁΕΖΗΘκύκλοςπρὸςτὸνΑΒΓΔκύκλον,οὕτως neither is circle EFGH to circle ABCD, as cone EN (is)
τὸΞστερεὸνπρὸςτὸνΑΛκῶνον. ἀλλ᾿ὡςτὸΞστερεὸν to some solid less than cone AL.
πρὸςτὸνΑΛκῶνον,οὕτωςὁΕΝκῶνοςπρὸςἔλασσόντι So, I say that neither is circle ABCD to circle EFGH,
τοῦΑΛκώνουστερεόν·καὶὡςἄραὁΕΖΗΘκύκλοςπρὸς as cone AL (is) to some solid greater than cone EN.
τὸνΑΒΓΔκύκλον,οὕτωςὁΕΝκῶνοςπρὸςἔλασσόντι For, if possible, let it be (in this ratio) to (some)
τοῦΑΛκώνουστερεόν·ὅπερἀδύνατονἐδείχθη. οὐκἄρα greater (solid), O. Thus, inversely, as circle EFGH is to
ἐστὶνὡςὁΑΒΓΔκύκλοςπρὸςτὸνΕΖΗΘκύκλον,οὕτωςὁ circle ABCD, so solid O (is) to cone AL [Prop. 5.7 corr.].
ΑΛκῶνοςπρὸςμεῖζόντιτοῦΕΝκώνουστερεόν.ἐδείχθη But, as solid O (is) to cone AL, so cone EN (is) to some
δέ,ὅτιοὐδὲπρὸςἔλασσον·ἔστινἄραὡςὁΑΒΓΔκύκλος solid less than cone AL [Prop. 12.2 lem.]. And, thus, as
πρὸςτὸνΕΖΗΘκύκλον,οὕτωςὁΑΛκῶνοςπρὸςτὸνΕΝ circle EFGH (is) to circle ABCD, so cone EN (is) to
κῶνον.
some solid less than cone AL. The very thing was shown
Ἀλλ᾿ὡςὁκῶνοςπρὸςτὸνκῶνον,ὁκύλινδροςπρὸς (to be) impossible. Thus, circle ABCD is not to circle
τὸνκύλινδρον·τριπλασίωνγὰρἑκάτεροςἑκατέρου.καὶὡς EFGH, as cone AL (is) to some solid greater than cone
ἄραὁΑΒΓΔκύκλοςπρὸςτὸνΕΖΗΘκύκλον,οὕτωςοἱἐπ᾿ EN. And, it was shown that neither (is it in this ratio) to
αὐτῶνἰσοϋψεῖς.
(some) lesser (solid). Thus, as circle ABCD is to circle
Οἱἄραὑπὸτὸαὐτὸὕψοςὄντεςκῶνοικαὶκύλινδροι EFGH, so cone AL (is) to cone EN.
πρὸςἀλλήλουςεἰσὶνὡςαἱβάσεις·ὅπερἔδειδεῖξαι.
But, as the cone (is) to the cone, (so) the cylinder
(is) to the cylinder. For each (is) three times each
[Prop. 12.10]. Thus, circle ABCD (is) also to circle
EFGH, as (the ratio of the cylinders) on them (having)
the same height.
Thus, cones and cylinders having the same height are
to one another as their bases. (Which is) the very thing it
was required to show.
. Proposition 12
Οἱὅμοιοικῶνοικαὶκύλινδροιπρὸςἀλλήλουςἐντρι- Similar cones and cylinders are to one another in the
πλασίονιλόγῳεἰσὶτῶνἐνταῖςβάσεσιδιαμέτρων. cubed ratio of the diameters of their bases.
῎Εστωσανὅμοιοικῶνοικαὶκύλινδροι, ὧνβάσειςμὲν Let there be similar cones and cylinders of which the
οἱΑΒΓΔ,ΕΖΗΘκύκλοι,διάμετροιδὲτῶνβάσεωναἱΒΔ, bases (are) the circles ABCD and EFGH, the diameters
ΖΘ,ἄξονεςδὲτῶνκώνωνκαὶκυλίνδρωνοἱΚΛ,ΜΝ·λέγω, of the bases (are) BD and FH, and the axes of the cones
491

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
ὅτιὁκῶνος,οὗβάσιςμέν[ἐστιν]ὁΑΒΓΔκύκλος,κορυφὴ and cylinders (are) KL and MN (respectively). I say
δὲ τὸΛσημεῖον, πρὸςτὸνκῶνον, οὗβάσιςμέν[ἐστιν] that the cone whose base [is] circle ABCD, and apex the
ὁΕΖΗΘκύκλος, κορυφὴδὲτὸΝσημεῖον, τριπλασίονα point L, has to the cone whose base [is] circle EFGH,
λόγονἔχειἤπερἡΒΔπρὸςτὴνΖΘ.
and apex the point N, the cubed ratio that BD (has) to
FH.
Ä
Ì
à Í
É
Â
Ë
Å
Ê À
Æ
Ç
È
G
Εἰ γὰρ μὴἔχειὁΑΒΓΔΛ κῶνοςπρὸς τὸν ΕΖΗΘΝ For if cone ABCDL does not have to cone EFGHN
κῶνονπριπλασίοναλόγονἤπερἡΒΔπρὸςτὴνΖΘ,ἕξει the cubed ratio that BD (has) to FH then cone ABCDL
ὁΑΒΓΔΛκῶνοςἢπρὸςἔλασσόντιτοῦΕΖΗΘΝκώνου will have the cubed ratio to some solid either less than, or
στερεὸντριπλασίοναλόγονἢπρὸςμεῖζον.ἐχέτωπρότερον greater than, cone EFGHN. Let it, first of all, have (such
πρὸςἔλασσοντὸΞ,καὶἐγγεγράφθωεἰςτὸνΕΖΗΘκύκλον a ratio) to (some) lesser (solid), O. And let the square
τετράγωνοντὸΕΖΗΘ·τὸἄραΕΖΗΘτετράγωνονμεῖζόν EFGH have been inscribed in circle EFGH [Prop. 4.6].
ἐστινἢτὸἥμισυτοῦΕΖΗΘκύκλου. καὶἀνεστάτωἐπὶ Thus, square EFGH is greater than half of circle EFGH
τοῦΕΖΗΘτετραγώνουπυραμὶςτὴναὐτὴνκορυφὴνἔχουσα [Prop. 12.2]. And let a pyramid having the same apex
τῷκώνῳ· ἡἄραἀνασταθεῖσαπυραμὶςμείζωνἐστὶνἢτὸ as the cone have been set up on square EFGH. Thus,
ἥμισυ μέρος τοῦ κώνου. τετμήσθωσαν δὴ αἱ ΕΖ, ΖΗ, the pyramid set up is greater than the half part of the
ΗΘ,ΘΕπεριφέρειαιδίχακατὰτὰΟ,Π,Ρ,Σσημεῖα,καὶ cone [Prop. 12.10]. So, let the circumferences EF , FG,
ἐπεζεύχθωσαναἱΕΟ,ΟΖ,ΖΠ,ΠΗ,ΗΡ,ΡΘ,ΘΣ,ΣΕ.καὶ GH, and HE have been cut in half at points P , Q, R,
ἕκαστονἄρατῶνΕΟΖ,ΖΠΗ,ΗΡΘ,ΘΣΕτριγώνωνμεῖζόν and S (respectively). And let EP , PF, FQ, QG, GR,
ἐστινἢτὸἥμισυμέροςτοῦκαθ᾿ἑαυτὸτμήματοςτοῦΕΖΗΘ RH, HS, and SE have been joined. And, thus, each
κύκλου.καὶἀνεστάτωἐφ᾿ἑκάστουτῶνΕΟΖ,ΖΠΗ,ΗΡΘ, of the triangles EPF, FQG, GRH, and HSE is greater
ΘΣΕ τριγώνων πυραμὶς τὴν αὐτὴν κορυφὴν ἔχουσα τῷ than the half part of the segment of circle EFGH about it
κώνῳ·καὶἑκάστηἄρατῶνἀνασταθεισῶνπυραμίδωνμείζων [Prop. 12.2]. And let a pyramid having the same apex as
ἐστὶν ἢ τὸ ἥμισυ μέρος τοῦ καθ᾿ ἑαυτὴν τμήματος τοῦ the cone have been set up on each of the triangles EPF,
κώνου. τέμνοντεςδὴτὰςὑπολειπομέναςπεριφερείαςδίχα FQG, GRH, and HSE. And thus each of the pyramids
καὶἐπιζευγνύντεςεὐθείαςκαὶἀνιστάντεςἐφ᾿ἑκάστουτῶν set up is greater than the half part of the segment of the
τριγώνωνπυραμίδαςτὴναὐτὴνκορυφὴνἐχούσαςτῷκώνῳ cone about it [Prop. 12.10]. So, (if) the the remaining cirκαὶτοῦτοἀεὶποιοῦντεςκαταλείψομέντιναἀποτμήματατοῦ
cumferences are cut in half, and straight-lines are joined,
κώνου, ἃ ἔσται ἐλάσσονα τῆς ὑπεροχῆς, ᾗ ὑπερέχει ὁ and pyramids having the same apex as the cone are set
ΕΖΗΘΝκῶνοςτοῦΞστερεοῦ. λελείφθω, καὶἔστωτὰ up on each of the triangles, and this is done continu-
ἐπὶτῶνΕΟ,ΟΖ,ΖΠ,ΠΗ,ΗΡ,ΡΘ,ΘΣ,ΣΕ·λοιπὴἄραἡ ally, then we will (eventually) leave some segments of the
πυραμίς,ἧςβάσιςμένἐστιτὸΕΟΖΠΗΡΘΣπολύγωνον, cone whose (sum) is less than the excess by which cone
κορυφὴ δὲ τὸ Ν σημεῖον, μείζων ἐστὶ τοῦ Ξ στερεοῦ. EFGHN exceeds solid O [Prop. 10.1]. Let them have
ἐγγεγράφθωκαὶεἰςτὸνΑΒΓΔκύκλοντῷΕΟΖΠΗΡΘΣ been left, and let them be the (segments) on EP , PF,
πολυγώνῳὅμοιόντεκαὶὁμοίωςκείμενονπολύγωνοντὸ FQ, QG, GR, RH, HS, and SE. Thus, the remaining
ΑΤΒΥΓΦΔΧ, καὶ ἀνεστάτω ἐπὶ τοῦ ΑΤΒΥΓΦΔΧ πο- pyramid whose base is polygon EPFQGRHS, and apex
λυγώνου πυραμὶςτὴν αὐτὴν κορυφὴν ἔχουσα τῷ κώνῳ, the point N, is greater than solid O. And let the polygon
καὶτῶνμὲνπεριεχόντωντὴνπυραμίδα,ἧςβάσιςμένἐστι ATBUCV DW , similar, and similarly laid out, to polyτὸ
ΑΤΒΥΓΦΔΧ πολύγωνον, κορυφὴ δὲ τὸ Λ σημεῖον, gon EPFQGRHS, have been inscribed in circle ABCD
ἓντρίγωνονἔστωτὸΛΒΤ,τῶνδὲπερειχόντωντὴνπυ- [Prop. 6.18]. And let a pyramid having the same apex
ραμίδα, ἧς βάσις μέν ἐστι τὸ ΕΟΖΠΗΡΘΣ πολύγωνον, as the cone have been set up on polygon ATBUCV DW .
B
T
L
U
A
K
C
W
V
D
H
S
R
E
M
N
P
Q
F
O
492

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
κορυφὴδὲτὸΝσημεῖον,ἓντρίγωνονἔστωτὸΝΖΟ,καὶ And let LBT be one of the triangles containing the pyra-
ἐπεζεύχθωσαναἱΚΤ,ΜΟ.καὶἐπεὶὅμοιόςἐστινὁΑΒΓΔΛ mid whose base is polygon ATBUCV DW , and apex the
κῶνοςτῷΕΖΗΘΝκώνῳ, ἔστινἄραὡςἡΒΔπρὸςτὴν point L. And let NFP be one of the triangles containing
ΖΘ,οὕτωςὁΚΛἄξωνπρὸςτὸνΜΝἄξονα. ὡςδὲἡΒΔ the pyramid whose base is triangle EPFQGRHS, and
πρὸςτὴνΖΘ,οὕτωςἡΒΚπρὸςτὴνΖΜ·καὶὡςἄραἡΒΚ apex the point N. And let KT and MP have been joined.
πρὸςτὴνΖΜ,οὕτωςἡΚΛπρὸςτὴνΜΝ.καὶἐναλλὰξὡς And since cone ABCDL is similar to cone EFGHN, thus
ἡΒΚπρὸςτὴνΚΛ,οὕτωςἡΖΜπρὸςτὴνΜΝ.καὶπερὶ as BD is to FH, so axis KL (is) to axis MN [Def. 11.24].
ἴσαςγωνίαςτὰςὑπὸΒΚΛ,ΖΜΝαἱπλευραὶἀνάλογόνεἰσιν· And as BD (is) to FH, so BK (is) to FM. And, thus, as
ὅμοιονἄραἐστὶτὸΒΚΛτρίγωνοντῷΖΜΝτριγώνῳ.πάλιν, BK (is) to FM, so KL (is) to MN. And, alternately, as
ἐπείἐστινὡςἡΒΚπρὸςτὴνΚΤ,οὕτωςἡΖΜπρὸςτὴν BK (is) to KL, so FM (is) to MN [Prop. 5.16]. And
ΜΟ,καὶπερὶἴσαςγωνίαςτὰςὑπὸΒΚΤ,ΖΜΟ,ἐπειδήπερ, the sides around the equal angles BKL and FMN are
ὃμέροςἐστὶνἡὑπὸΒΚΤγωνίατῶνπρὸςτῷΚκέντρῳ proportional. Thus, triangle BKL is similar to triangle
τεσσάρωνὀρθῶν,τὸαὐτὸμέροςἐστὶκαὶἡὑπὸΖΜΟγωνία FMN [Prop. 6.6]. Again, since as BK (is) to KT , so
τῶνπρὸςτῷΜκέντρῳτεσσάρωνὀρθῶν·ἐπεὶοὖνπερὶἴσας FM (is) to MP , and (they are) about the equal angles
γωνίαςαἱπλευραὶἀνάλογόνεἰσιν,ὅμοιονἄραἐστιτὸΒΚΤ BKT and FMP , inasmuch as whatever part angle BKT
τρίγωνοντῷΖΜΟτριγώνῳ. πάλιν,ἐπεὶἐδείχθηὡςἡΒΚ is of the four right-angles at the center K, angle FMP is
πρὸςτὴνΚΛ,οὕτωςἡΖΜπρὸςτὴνΜΝ,ἴσηδὲἡμὲν also the same part of the four right-angles at the cen-
ΒΚτῇΚΤ,ἡδὲΖΜτῇΟΜ,ἔστινἄραὡςἡΤΚπρὸς ter M. Therefore, since the sides about equal angles
τὴνΚΛ,οὕτωςἡΟΜπρὸςτὴνΜΝ.καὶπερὶἴσαςγωνίας are proportional, triangle BKT is thus similar to trainτὰςὑπὸΤΚΛ,ΟΜΝ·ὀρθαὶγάρ·αἱπλευραὶἀνάλογόνεἰσιν·
gle FMP [Prop. 6.6]. Again, since it was shown that
ὅμοιονἄραἐστὶτὸΛΚΤτρίγωνοντῷΝΜΟτριγώνῳ.καὶ as BK (is) to KL, so FM (is) to MN, and BK (is)
ἐπεὶδιὰτὴνὁμοιότητατῶνΛΚΒ,ΝΜΖτριγώνωνἐστὶν equal to KT , and FM to PM, thus as TK (is) to KL,
ὡςἡΛΒπρὸςτὴνΒΚ,οὕτωςἡΝΖπρὸςτὴνΖΜ,διὰδὲ so PM (is) to MN. And the sides about the equal angles
τὴνὁμοιότητατῶνΒΚΤ,ΖΜΟτριγώνωνἐστὶνὡςἡΚΒ TKL and PMN—for (they are both) right-angles—are
πρὸςτὴνΒΤ,οὕτωςἡΜΖπρὸςτὴνΖΟ,δι᾿ἴσουἄραὡς proportional. Thus, triangle LKT (is) similar to triangle
ἡΛΒπρὸςτὴνΒΤ,οὕτωςἡΝΖπρὸςτὴνΖΟ.πάλιν,ἐπεὶ NMP [Prop. 6.6]. And since, on account of the similarity
διὰτὴνομοιότητατῶνΛΤΚ,ΝΟΜτριγώνωνἐστὶνὡςἡ of triangles LKB and NMF , as LB (is) to BK, so NF
ΛΤπρὸςτὴνΤΚ,οὕτωςἡΝΟπρὸςτὴνΟΜ,διὰδὲτὴν (is) to FM, and, on account of the similarity of triangles
ὁμοιότητατῶνΤΚΒ,ΟΜΖτριγώνωνἐστὶνὡςἡΚΤπρὸς BKT and FMP , as KB (is) to BT , so MF (is) to FP
τὴνΤΒ,οὕτωςἡΜΟπρὸςτὴνΟΖ,δι᾿ἴσουἄραὡςἡΛΤ [Def. 6.1], thus, via equality, as LB (is) to BT , so NF
πρὸςτὴνΤΒ,οὕτωςἡΝΟπρὸςτὴνΟΖ.ἐδείχθηδὲκαὶ (is) to FP [Prop. 5.22]. Again, since, on account of the
ὡςἡΤΒπρὸςτὴνΒΛ,οὕτωςἡΟΖπρὸςτὴνΖΝ.δι᾿ἴσου similarity of triangles LTK and NPM, as LT (is) to TK,
ἄραὡςἡΤΛπρὸςτὴνΛΒ,οὕτωςἡΟΝπρὸςτὴνΝΖ. so NP (is) to PM, and, on account of the similarity of
τῶνΛΤΒ,ΝΟΖἄρατριγώνωνἀνάλογόνεἰσιναἱπλευραί· triangles TKB and PMF , as KT (is) to TB, so MP (is)
ἰσογώνιαἄραἐστὶτὰΛΤΒ,ΝΟΖτρίγωνα·ὥστεκαὶὅμοια. to PF, thus, via equality, as LT (is) to TB, so NP (is)
καὶπυραμὶςἄρα,ἧςβάσιςμὲντὸΒΚΤτρίγωνον,κορυφὴ to PF [Prop. 5.22]. And it was shown that as TB (is)
δὲ τὸ Λ σημεῖον, ὁμοία ἐστὶ πυραμίδι, ἧς βάσις μὲν τὸ to BL, so PF (is) to FN. Thus, via equality, as TL (is)
ΖΜΟτρίγωνον,κορυφὴδὲτὸΝσημεῖον·ὑπὸγὰρὅμοίων to LB, so PN (is) to NF [Prop. 5.22]. Thus, the sides
ἐπιπέδωνπεριέχονταιἴσωντὸπλῆθος. αἱδὲὅμοιαιπυ- of triangles LTB and NPF are proportional. Thus, triραμίδεςκαὶτριγώνουςἔχουσαιβάσειςἐντριπλασίονιλόγῳ
angles LTB and NPF are equiangular [Prop. 6.5]. And,
εἰσὶτῶνὁμολόγωνπλευρῶν.ἡἄραΒΚΤΛπυραμὶςπρὸςτὴν hence, (they are) similar [Def. 6.1]. And, thus, the pyra-
ΖΜΟΝπυραμίδατριπλασίοναλόγονἔχειἤπερἡΒΚπρὸς mid whose base is triangle BKT , and apex the point L,
τὴνΖΜ.ὁμοίωςδὴἐπιζευγνύντεςἀπὸτῶνΑ,Χ,Δ,Φ,Γ,Υ is similar to the pyramid whose base is triangle FMP ,
ἐπὶτὸΚεὐθείαςκαὶἀπὸτῶνΕ,Σ,Θ,Ρ,Η,ΠἐπὶτὸΜκαὶ and apex the point N. For they are contained by equal
ἀνιστάντεςἐφ᾿ἑκάστουτῶντριγώνωνπυραμίδαςτὴναὐτὴν numbers of similar planes [Def. 11.9]. And similar pyraκορυφὴνἐχούσαςτοῖςκώνοιςδείξομεν,ὅτικαὶἑκάστητῶν
mids which also have triangular bases are in the cubed
ὁμοταγῶνπυραμίδωνπρὸςἑκάστηνὁμοταγῆπυραμίδατρι- ratio of corresponding sides [Prop. 12.8]. Thus, pyramid
πλασίοναλόγονἕξειἤπερἡΒΚὁμόλογοςπλευρὰπρὸςτὴν BKTL has to pyramid FMPN the cubed ratio that BK
ΖΜὁμόλογονπλευράν,τουτέστινἤπερἡΒΔπρὸςτὴνΖΘ. (has) to FM. So, similarly, joining straight-lines from
καὶὡςἓντῶνἡγουμένωνπρὸςἓντῶνἑπομένων,οὕτως (points) A, W , D, V , C, and U to (center) K, and from
ἅπαντατὰἡγούμεναπρὸςἅπαντατὰἑπόμενα· ἔστινἄρα (points) E, S, H, R, G, and Q to (center) M, and set-
493

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
καὶὡςἡΒΚΤΛπυραμὶςπρὸςτὴνΖΜΟΝπυραμίδα,οὕτως ting up pyramids having the same apexes as the cones
ἡὅληπυραμίς,ἧςβάσιςτὸΑΤΒΥΓΦΔΧπολύγωνον,κο- on each of the triangles (so formed), we can also show
ρυφὴδὲτὸΛσημεῖον,πρὸςτὴνὅληνπυραμίδα,ἧςβάσις that each of the pyramids (on base ABCD taken) in orμὲντὸΕΟΖΠΗΡΘΣπολύγωνον,κορυφὴδὲτὸΝσημεῖον·
der will have to each of the pyramids (on base EFGH
ὥστεκαὶπυραμίς,ἧςβάσιςμὲντὸΑΤΒΥΓΦΔΧ,κορυφὴδὲ taken) in order the cubed ratio that the corresponding
τὸΛ,πρὸςτὴνπυραμίδα,ἧςβάσις[μὲν]τὸΕΟΖΠΗΡΘΣ side BK (has) to the corresponding side FM—that is to
πολύγωνον,κορυφὴδὲτὸΝσημεῖον,τριπλασίοναλόγον say, that BD (has) to FH. And (for two sets of propor-
ἔχειἤπερἡΒΔπρὸςτὴνΖΘ.ὑπόκειταιδὲκαὶὁκῶνος,οὗ tional magnitudes) as one of the leading (magnitudes is)
βάσις[μὲν]ὁΑΒΓΔκύκλος,κορυφὴδὲτὸΛσημεῖον,πρὸς to one of the following, so (the sum of) all of the leading
τὸΞστερεὸντριπλασίοναλόγονἔχωνἤπερἡΒΔπρὸςτὴν (magnitudes is) to (the sum of) all of the following (mag-
ΖΘ·ἔστινἄραὡςὁκῶνος,οὗβάσιςμένἐστινὁΑΒΓΔ nitudes) [Prop. 5.12]. And, thus, as pyramid BKTL (is)
κύκλος,κορυφὴδὲτὸΛ,πρὸςτὸΞστερεόν,οὕτωςἡπυ- to pyramid FMPN, so the whole pyramid whose base
ραμίς,ἧςβάσιςμὲντὸΑΤΒΥΓΦΔΧ[πολύγωνον],κορυφὴ is polygon ATBUCV DW , and apex the point L, (is) to
δὲτὸΛ,πρὸςτὴνπυραμίδα,ἧςβάσιςμένἐστιτὸΕΟΖ- the whole pyramid whose base is polygon EPFQGRHS,
ΠΗΡΘΣπολύγωνον,κορυφὴδὲτὸΝ·ἐναλλὰξἄρα,ὡςὁ and apex the point N. And, hence, the pyramid whose
κῶνος,οὗβάσιςμὲνὁΑΒΓΔκύκλος,κορυφὴδὲτὸΛ, base is polygon ATBUCV DW , and apex the point L,
πρὸςτὴνἐναὐτῷπυραμίδα,ἧςβάσιςμὲντὸΑΤΒΥΓΦΔΧ has to the pyramid whose base is polygon EPFQGRHS,
πολύγωνον,κορυφὴδὲτὸΛ,οὕτωςτὸΞ[στερεὸν]πρὸςτὴν and apex the point N, the cubed ratio that BD (has)
πυραμίδα,ἧςβάσιςμένἐστιτὸΕΟΖΠΗΡΘΣπολύγωνον, to FH. And it was also assumed that the cone whose
κορυφὴδὲτὸΝ.μείζωνδὲὁεἰρημένοςκῶνοςτῆςἐναὐτῷ base is circle ABCD, and apex the point L, has to solid
πυραμίδος· ἐμπεριέχει γὰρ αὐτὴν. μεῖζον ἄρα καὶ τὸ Ξ O the cubed ratio that BD (has) to FH. Thus, as the
στερεὸντῆςπυραμίδος,ἧςβάσιςμένἐστιτὸΕΟΖΠΗΡΘΣ cone whose base is circle ABCD, and apex the point L,
πολύγωνον,κορυφὴδὲτὸΝ.ἀλλὰκαὶἔλαττον·ὅπερἐστὶν is to solid O, so the pyramid whose base (is) [polygon]
ἀδύνατον.οὐκἄραὁκῶνος,οὗβάσιςὁΑΒΓΔκύκλος,κο- ATBUCV DW , and apex the point L, (is) to the pyramid
ρυφὴδὲτὸΛ[σημεῖον],πρὸςἔλαττόντιτοῦκώνουστερεόν, whose base is polygon EPFQGRHS, and apex the point
οὗβάσιςμὲνὁΕΖΗΘκύκλος,κορυφὴδὲτὸΝσημεῖον,τρι- N. Thus, alternately, as the cone whose base (is) circle
πλασίοναλόγονἔχειἤπερἡΒΔπρὸςτὴνΖΘ.ὁμοίωςδὴ ABCD, and apex the point L, (is) to the pyramid within
δείξομεν,ὅτιοὐδὲὁΕΖΗΘΝκῶνοςπρὸςἔλαττόντιτοῦ it whose base (is) the polygon ATBUCV DW , and apex
ΑΒΓΔΛκώνουστερεὸντριπλασίοναλόγονἔχειἤπερἡΖΘ the point L, so the [solid] O (is) to the pyramid whose
πρὸςτὴνΒΔ.
base is polygon EPFQGRHS, and apex the point N
Λέγωδή,ὅτιοὐδὲὁΑΒΓΔΛκῶνοςπρὸςμεῖζόντιτοῦ [Prop. 5.16]. And the aforementioned cone (is) greater
ΕΖΗΘΝκώνουστερεὸντριπλασίοναλόγονἔχειἤπερἡΒΔ than the pyramid within it. For it encompasses it. Thus,
πρὸςτὴνΖΘ.
solid O (is) also greater than the pyramid whose base is
Εἰγὰρδυνατόν,ἐχέτωπρὸςμεῖζοντὸΞ.ἀνάπαλινἄρα polygon EPFQGRHS, and apex the point N. But, (it
τὸΞστερεὸνπρὸςτὸνΑΒΓΔΛκῶνοντριπλασίοναλόγον is) also less. The very thing is impossible. Thus, the cone
ἔχειἤπερἡΖΘπρὸςτὴνΒΔ.ὡςδὲτὸΞστερεὸνπρὸς whose base (is) circle ABCD, and apex the [point] L,
τὸνΑΒΓΔΛκῶνον,οὕτωςὁΕΖΗΘΝκῶνοςπρὸςἔλαττόν does not have to some solid less than the cone whose
τιτοῦΑΒΓΔΛκώνουστερεόν. καὶὁΕΖΗΘΝἄρακῶνος base (is) circle EFGH, and apex the point N, the cubed
πρὸςἔλαττόντιτοῦΑΒΓΔΛκώνουστερεὸντριπλασίονα ratio that BD (has) to EH. So, similarly, we can show
λόγονἔχειἤπερἡΖΘπρὸςτὴνΒΔ·ὅπερἀδύνατονἐδείχθη. that neither does cone EFGHN have to some solid less
οὐκ ἄρα ὁ ΑΒΓΔΛ κῶνος πρὸς μεῖζόν τι τοῦ ΕΖΗΘΝ than cone ABCDL the cubed ratio that FH (has) to BD.
κώνουστερεὸντριπλασίοναλόγονἔχειἤπερἡΒΔπρὸς So, I say that neither does cone ABCDL have to some
τὴνΖΘ.ἐδείχθηδέ,ὅτιοὐδὲπρὸςἔλαττον.ὁΑΒΓΔΛἄρα solid greater than cone EFGHN the cubed ratio that BD
κῶνοςπρὸςτὸνΕΖΗΘΝκῶνοντριπλασίοναλόγονἔχει (has) to FH.
ἤπερἡΒΔπρὸςτὴνΖΘ.
For, if possible, let it have (such a ratio) to a greater
῾Ωςδὲὁκῶνοςπρὸςτὸνκῶνον,ὁκύλινδροςπρὸςτὸν (solid), O. Thus, inversely, solid O has to cone ABCDL
κύλινδρον·τριπλάσιοςγὰρὁκύλινδροςτοῦκώνουὁἐπὶτῆς the cubed ratio that FH (has) to BD [Prop. 5.7 corr.].
αὐτῆςβάσεωςτῷκώνῳκαὶἰσοϋψὴςαὐτῷ.καὶὁκύλινδρος And as solid O (is) to cone ABCDL, so cone EFGHN
ἄραπρὸςτὸνκύλινδροντριπλασίοναλόγονἔχειἤπερἡΒΔ (is) to some solid less than cone ABCDL [12.2 lem.].
πρὸςτὴνΖΘ.
Thus, cone EFGHN also has to some solid less than cone
Οἱἄραὅμοιοικῶνοικαὶκύλινδροιπρὸςἀλλήλουςἐν ABCDL the cubed ratio that FH (has) to BD. The very
494

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
τριπλασίονιλόγῳεἰσὶτῶνἐνταῖςβάσεσιδιαμέτρων·ὅπερ thing was shown (to be) impossible. Thus, cone ABCDL
ἔδειδεῖξαι.
does not have to some solid greater than cone EFGHN
the cubed ratio than BD (has) to FH. And it was shown
that neither (does it have such a ratio) to a lesser (solid).
Thus, cone ABCDL has to cone EFGHN the cubed ratio
that BD (has) to FG.
And as the cone (is) to the cone, so the cylinder (is)
to the cylinder. For a cylinder is three times a cone on
the same base as the cone, and of the same height as it
[Prop. 12.10]. Thus, the cylinder also has to the cylinder
the cubed ratio that BD (has) to FH.
Thus, similar cones and cylinders are in the cubed ratio
of the diameters of their bases. (Which is) the very
thing it was required to show.
. Proposition 13
᾿Εὰνκύλινδροςἐπιπέδῳτμηθῇπαραλλήλῳὄντιτοῖςἀπε- If a cylinder is cut by a plane which is parallel to the
Ç Ê À Ì
ναντίονἐπιπέδοις,ἔσταιὡςὁκύλινδροςπρὸςτὸνκύλινδρον, opposite planes (of the cylinder) then as the cylinder (is)
οὕτωςὁἄξωνπρὸςτὸνἄξονα.
to the cylinder, so the axis will be to the axis.
Ä Æ Ã Å
P R A G C T V
L N E K F O M
È Ë Â Í É Q S B H D U W
ΚύλινδροςγὰρὁΑΔἐπιπέδῳτῷΗΘτετμήσθωπα- For let the cylinder AD have been cut by the plane
ραλλήλῳὄντιτοῖςἀπεναντίονἐπιπέδοιςτοῖςΑΒ,ΓΔ,καὶ GH which is parallel to the opposite planes (of the cylinσυμβαλλέτωτῷἄξονιτὸΗΘἐπίπεδονκατὰτὸΚσημεῖον·
der), AB and CD. And let the plane GH have met the
λέγω,ὅτιἐστὶνὡςὁΒΗκύλινδροςπρὸςτὸνΗΔκύλινδρον, axis at point K. I say that as cylinder BG is to cylinder
οὕτωςὁΕΚἄξωνπρὸςτὸνΚΖἄξονα. GD, so axis EK (is) to axis KF .
᾿ΕκβεβλήσθωγὰρὁΕΖἄξωνἐφ᾿ἑκάτερατὰμέρηἐπὶ For let axis EF have been produced in each direction
τὰΛ,Μσημεῖα,καὶἐκκείσθωσαντῷΕΚἄξονιἴσοιὁσοι- to points L and M. And let any number whatsoever (of
δηποτοῦνοἱΕΝ,ΝΛ,τῷδὲΖΚἴσοιὁσοιδηποτοῦνοἱΖΞ, lengths), EN and NL, equal to axis EK, be set out (on
ΞΜ,καὶνοείσθωὁἐπὶτοῦΛΜἄξονοςκύλινδροςὁΟΧ, the axis EL), and any number whatsoever (of lengths),
οὗβάσειςοἱΟΠ,ΦΧκύκλοι. καὶἐκβεβλήσθωδιὰτῶν FO and OM, equal to (axis) FK, (on the axis KM).
Ν, Ξ σημείων ἐπίπεδαπαράλληλατοῖςΑΒ, ΓΔ καὶταῖς And let the cylinder PW, whose bases (are) the circles
βάσεσιτοῦΟΧκυλίνδρουκαὶποιείτωσαντοὺςΡΣ,ΤΥ PQ and V W , have been conceived on axis LM. And
κύκλουςπερὶτὰΝ,Ξκέντρα. καὶἐπεὶοἱΛΝ,ΝΕ,ΕΚ let planes parallel to AB, CD, and the bases of cylinder
ἄξονεςἴσοιεἰσὶνἀλλήλοις,οἱἄραΠΡ,ΡΒ,ΒΗκύλινδροι PW, have been produced through points N and O, and
πρὸςἀλλήλουςεἰσὶνὡςαἱβάσεις. ἴσαιδέεἰσιναἱβάσεις· let them have made the circles RS and TU around the
ἴσοιἄρακαὶοἱΠΡ,ΡΒ,ΒΗκύλινδροιἀλλήλοις. επεὶοὖν centers N and O (respectively). And since axes LN, NE,
οἱΛΝ,ΝΕ,ΕΚἄξονεςἴσοιεἰσὶνἀλλήλοις,εἰσὶδὲκαὶοἱ and EK are equal to one another, the cylinders QR, RB,
ΠΡ,ΡΒ,ΒΗκύλινδροιἴσοιἀλλήλοις,καίἐστινἴσοντὸ and BG are to one another as their bases [Prop. 12.11].
πλῆθοςτῷπλήθει,ὁσαπλασίωνἄραὁΚΛἄξωντοῦΕΚ But the bases are equal. Thus, the cylinders QR, RB,
ἄξονος,τοσαυταπλασίωνἔσταικαὶὁΠΗκύλινδροςτοῦΗΒ and BG (are) also equal to one another. Therefore, since
κυλίνδρου.διὰτὰαὐτὰδὴκαὶὁσαπλασίωνἐστὶνὁΜΚἄξων the axes LN, NE, and EK are equal to one another,
τοῦΚΖἄξονος,τοσαυταπλασίωνἐστὶκαὶὁΧΗκύλινδρος and the cylinders QR, RB, and BG are also equal to one
τοῦΗΔκυλίνδρου. καὶεἰμὲνἴσοςἐστὶνὁΚΛἄξωντῷ another, and the number (of the former) is equal to the
ΚΜἄξονι,ἴσοςἔσταικαὶὁΠΗκύλινδροςτῷΗΧκυλίνδρῳ, number (of the latter), thus as many multiples as axis KL
495

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
εἰδὲμείζωνὁἄξωντοῦἄξονος,μείζωνκαὶὁκύλινδρος is of axis EK, so many multiples is cylinder QG also of
τοῦκυλίνδρου,καὶεἰἐλάσσων,ἐλάσσων.τεσσάρωνδὴμε- cylinder GB. And so, for the same (reasons), as many
γεθῶνὄντων,ἀξόνωνμὲντῶνΕΚ,ΚΖ,κυλίνδρωνδὲτῶν multiples as axis MK is of axis KF , so many multiples
ΒΗ,ΗΔ,εἴληπταιἰσάκιςπολλαπλάσια,τοῦμὲνΕΚἄξονος is cylinder WG also of cylinder GD. And if axis KL is
καὶτοῦΒΗκυλίνδρουὅτεΛΚἄξωνκαὶὁΠΗκύλινδρος, equal to axis KM then cylinder QG will also be equal
τοῦδὲΚΖἄξονεςκαὶτοῦΗΔκυλίνδρουὅτεΚΜἄξων to cylinder GW , and if the axis (is) greater than the axis
καὶὁΗΧκύλινδρος,καὶδέδεικται,ὅτιεἰὑπερέχειὁΚΛ then the cylinder (will also be) greater than the cylinder,
ἄξωντοῦΚΜἄξονος,ὑπερέχεικαὶὁΠΗκύλινδροςτοῦ and if (the axis is) less then (the cylinder will also be)
ΗΧκυλίνδρου,καὶεἰἴσος,ἴσος,καὶεἰἐλάσσων,ἐλάσσων. less. So, there are four magnitudes—the axes EK and
ἔστινἄραὡςὁΕΚἄξωνπρὸςτὸνΚΖἄξονα,οὕτωςὁΒΗ KF , and the cylinders BG and GD—and equal multiples
κύλινδροςπρὸςτὸνΗΔκύλινδρον·ὅπερἔδειδεῖξαι. have been taken of axis EK and cylinder BG—(namely),
axis LK and cylinder QG—and of axis KF and cylinder
GD—(namely), axis KM and cylinder GW . And it has
been shown that if axis KL exceeds axis KM then cylinder
QG also exceeds cylinder GW , and if (the axes are)
equal then (the cylinders are) equal, and if (KL is) less
then (QG is) less. Thus, as axis EK is to axis KF , so
cylinder BG (is) to cylinder GD [Def. 5.5]. (Which is)
the very thing it was required to show.
. Proposition 14
Οἱ ἐπὶ ἴσων βάσεων ὄντες κῶνοι καὶ κύλινδροιπρὸς Cones and cylinders which are on equal bases are to
αλλήλουςεἰσὶνὡςτὰὕψη.
À Ã
one another as their heights.
Ä
F K
E G
Â
C L D
Æ Å
A H B
῎ΕστωσανγὰρἐπὶἴσωνβάσεωντῶνΑΒ,ΓΔκύκλων For let EB and FD be cylinders on equal bases,
κύλινδροιοἱΕΒ,ΖΔ·λέγω,ὅτιἐστὶνὡςὁΕΒκύλινδρος (namely) the circles AB and CD (respectively). I say
πρὸςτὸνΖΔκύλινδρον,οὕτωςὁΗΘἄξωνπρὸςτὸνΚΛ that as cylinder EB is to cylinder FD, so axis GH (is) to
ἄξονα.
axis KL.
᾿ΕκβεβλήσθωγὰρὁΚΛἄξωνἐπὶτὸΝσημεῖον, καὶ For let the axis KL have been produced to point N.
κείσθωτῷΗΘἄξονιἴσοςὁΛΝ,καὶπερὶἄξονατὸνΛΝ And let LN be made equal to axis GH. And let the cylinκύλινδροςνενοήσθωὁΓΜ.ἐπεὶοὖνοἱΕΒ,ΓΜκύλινδροι
der CM have been conceived about axis LN. Therefore,
ὑπὸτὸαὐτὸὕψοςεἰσίν,πρὸςἀλλήλουςεἰσὶνὡςαἱβάσεις. since cylinders EB and CM have the same height they
ἴσαιδέεἰσίναἱβάσειςἀλλήλαις·ἴσοιἄραεἰσὶκαὶοἱΕΒ,ΓΜ are to one another as their bases [Prop. 12.11]. And the
κύλινδροι. καὶἐπεὶκύλινδροςὁΖΜἐπιπέδῳτέτμηταιτῷ bases are equal to one another. Thus, cylinders EB and
ΓΔπαραλλήλῳὄντιτοῖςἀπεναντίονἐπιπέδοις,ἔστινἄραὡς CM are also equal to one another. And since cylinder
ὁΓΜκύλινδροςπρὸςτὸνΖΔκύλινδρον,οὕτωςὁΛΝἄξων FM has been cut by the plane CD, which is parallel to
πρὸςτὸνΚΛἄξονα.ἴσοςδέἐστινὁμὲνΓΜκύλινδροςτῷ its opposite planes, thus as cylinder CM is to cylinder
ΕΒκυλίνδρῳ,ὁδὲΛΝἄξωντῷΗΘἄξονι·ἔστινἄραὡςὁ FD, so axis LN (is) to axis KL [Prop. 12.13]. And cylin-
ΕΒκύλινδροςπρὸςτὸνΖΔκύλινδρον,οὕτωςὁΗΘἄξων der CM is equal to cylinder EB, and axis LN to axis GH.
πρὸςτὸνΚΛἄξονα. ὡςδὲὁΕΒκύλινδροςπρὸςτὸνΖΔ Thus, as cylinder EB is to cylinder FD, so axis GH (is)
N
M
496

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
Ä
Ã
κύλινδρον,οὑτωςὁΑΒΗκῶνοςπρὸςτὸνΓΔΚκῶνον.καὶ to axis KL. And as cylinder EB (is) to cylinder FD, so
ὡςἄραὁΗΘἄξωνπρὸςτὸνΚΛἄξονα,οὕτωςὁΑΒΗ cone ABG (is) to cone CDK [Prop. 12.10]. Thus, also,
κῶνοςπρὸςτὸνΓΔΚκῶνονκαὶὁΕΒκύλινδροςπρὸςτὸν as axis GH (is) to axis KL, so cone ABG (is) to cone
ΖΔκύλινδρον·ὅπερἔδειδεῖξαι.
CDK, and cylinder EB to cylinder FD. (Which is) the
very thing it was required to show.
. Proposition 15
Τῶν ἴσων κώνων καὶ κυλίνδρων ἀντιπεπόνθασιν αἱ The bases of equal cones and cylinders are reciproβάσεις
τοῖς ὕψεσιν· καὶ ὧν κώνων καὶ κυλίνδρων ἀντι- cally proportional to their heights. And, those cones and
πεπόνθασιναἱβάσειςτοῖςὕψεσιν,ἴσοιεἰσὶνἐκεῖνοι.
Ë À
cylinders whose bases (are) reciprocally proportional to
their heights are equal.
Å È
Í Ê Â Æ
L O
P S G
F
D
Q
M
N
A
H
K C
B R U E
῎Εστωσανἴσοικῶνοικαὶκύλινδροι,ὧνβάσειςμὲνοἱ Let there be equal cones and cylinders whose bases
ΑΒΓΔ, ΕΖΗΘ κύκλοι, διάμετροι δὲ αὐτῶν αἱ ΑΓ, ΕΗ, are the circles ABCD and EFGH, and the diameters
ἅξονεςδὲοἱΚΛ,ΜΝ,οἵτινεςκαὶὕψηεἰσὶτῶνκώνωνἢ of (the bases) AC and EG, and (whose) axes (are) KL
κυλίνδρων,καὶσυμπεπληρώσθωσανοἱΑΞ,ΕΟκύλινδροι. and MN, which are also the heights of the cones and
λέγω,ὅτιτῶνΑΞ,ΕΟκυλίνδρωνἀντιπεπόνθασιναἱβάσεις cylinders (respectively). And let the cylinders AO and
τοῖςὕψεσιν,καίἐστινὡςἡΑΒΓΔβάσιςπρὸςτὴνΕΖΗΘ EP have been completed. I say that the bases of cylinders
Ç Ì T
βάσιν,οὕτωςτὸΜΝὕψοςπρὸςτὸΚΛὕψος. AO and EP are reciprocally proportional to their heights,
ΤὸγὰρΛΚὕψοςτῷΜΝὕψειἤτοιἴσονἐστὶνἢοὔ. and (so) as base ABCD is to base EFGH, so height MN
ἔστωπρότερονἴσον. ἔστιδὲκαὶὁΑΞκύλινδροςτῷΕΟ (is) to height KL.
κυλίνδρῳἴσος. οἱδὲὑπὸτὸαὐτὸὕψοςὄντεςκῶνοικαὶ For height LK is either equal to height MN, or not.
κύλινδροιπρὸςἀλλήλουςεἰσὶνὡςαἱβάσεις· ἴσηἄρακαὶ Let it, first of all, be equal. And cylinder AO is also equal
ἡΑΒΓΔβάσιςτῇΕΖΗΘβάσει. ὥστεκαὶἀντιπέπονθεν, to cylinder EP . And cones and cylinders having the same
ὡςἡΑΒΓΔβάσιςπρὸςτὴνΕΖΗΘβάσιν,οὕτωςτὸΜΝ height are to one another as their bases [Prop. 12.11].
ὕψοςπρὸςτὸΚΛὕψος. ἀλλὰδὴμὴἔστωτὸΛΚὕψος Thus, base ABCD (is) also equal to base EFGH. And,
τῷΜΝἴσον,ἀλλ᾿ἔστωμεῖζοντὸΜΝ,καὶἀφῃρήσθωἀπὸ hence, reciprocally, as base ABCD (is) to base EFGH,
τοῦΜΝὕψουςτῷΚΛἴσοντὸΠΝ,καὶδιὰτοῦΠσημείου so height MN (is) to height KL. And so, let height LK
τετμήσθωὁΕΟκύλινδροςἐπιπέδῳτῷΤΥΣπαραλλήλῳ not be equal to MN, but let MN be greater. And let QN,
τοῖςτῶνΕΖΗΘ,ΡΟκύκλωνἐπιπέδοις,καὶἀπὸβάσεωςμὲν equal to KL, have been cut off from height MN. And
τοῦΕΖΗΘκύκλου,ὕψουςδὲτοῦΝΠκύλινδροςνενοήσθω let the cylinder EP have been cut, through point Q, by
ὁΕΣ.καίἐπεὶἴσοςἐστὶνὁΑΞκύλινδροςτῷΕΟκυλίνδρῳ, the plane TUS (which is) parallel to the planes of the
ἔστινἄραὡςὁΑΞκύλινδροςπρὸςτὸνΕΣκυλίνδρον,οὕτως circles EFGH and RP . And let cylinder ES have been
ὁΕΟκύλινδροςπρὸςτὸνΕΣκύλινδρον.ἀλλ᾿ὡςμὲνὁΑΞ conceived, with base the circle EFGH, and height NQ.
κύλινδροςπρὸςτὸνΕΣκύλινδρον,οὕτωςἡΑΒΓΔβάσις And since cylinder AO is equal to cylinder EP , thus, as
πρὸςτὴνΕΖΗΘ·ὑπὸγὰρτὸαὐτὸὕψοςεἰσὶνοἱΑΞ,ΕΣ cylinder AO (is) to cylinder ES, so cylinder EP (is) to
κύλινδροι·ὡςδὲὁΕΟκύλινδροςπρὸςτὸνΕΣ,οὕτωςτὸ cylinder ES [Prop. 5.7]. But, as cylinder AO (is) to cylin-
ΜΝὕψοςπρὸςτὸΠΝὕψος·ὁγὰρΕΟκύλινδροςἐπιπέδῳ der ES, so base ABCD (is) to base EFGH. For cylinders
τέτμηταιπαραλλήλῳὄντιτοῖςἀπεναντίονἐπιπέδοις. ἔστιν AO and ES (have) the same height [Prop. 12.11]. And
ἄρακαὶὡςἡΑΒΓΔβάσιςπρὸςτὴνΕΖΗΘβάσιν,οὕτωςτὸ as cylinder EP (is) to (cylinder) ES, so height MN (is)
ΜΝὕψοςπρὸςτὸΠΝὕψος. ἴσονδὲτὸΠΝὕψοςτῷΚΛ to height QN. For cylinder EP has been cut by a plane
ὕψει·ἔστινἄραὡςἡΑΒΓΔβάσιςπρὸςτὴνΕΖΗΘβάσιν, which is parallel to its opposite planes [Prop. 12.13].
οὕτωςτὸΜΝὕψοςπρὸςτὸΚΛὕψος. τῶνἄραΑΞ,ΕΟ And, thus, as base ABCD is to base EFGH, so height
κυλίνδρωνἀντιπεπόνθασιναἱβάσειςτοῖςὕψεσιν. MN (is) to height QN [Prop. 5.11]. And height QN
497

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
ἈλλὰδὴτῶνΑΞ,ΕΟκυλίνδρωνἀντιπεπονθέτωσαναἱ (is) equal to height KL. Thus, as base ABCD is to base
βάσειςτοῖςὕψεσιν,καὶἔστωὡςἡΑΒΓΔβάσιςπρὸςτὴν EFGH, so height MN (is) to height KL. Thus, the bases
ΕΖΗΘβάσιν,οὕτωςτὸΜΝὕψοςπρὸςτὸΚΛὕψος·λέγω, of cylinders AO and EP are reciprocally proportional to
ὅτιἴσοςἐστὶνὁΑΞκύλινδροςτῷΕΟκυλίνδρῳ. their heights.
Τῶν γὰρ αὐτῶν κατασκευασθέντων ἐπεί ἐστιν ὡς ἡ And, so, let the bases of cylinders AO and EP be
ΑΒΓΔβάσιςπρὸςτὴνΕΖΗΘβάσιν,οὕτωςτὸΜΝὕψος reciprocally proportional to their heights, and (thus) let
πρὸςτὸΚΛὕψος,ἴσονδὲτὸΚΛὕψοςτῷΠΝὕψει,ἔσται base ABCD be to base EFGH, as height MN (is) to
ἄραὡςἡΑΒΓΔβάσιςπρὸςτὴνΕΖΗΘβάσιν,οὕτωςτὸ height KL. I say that cylinder AO is equal to cylinder
ΜΝὕψοςπρὸςτὸΠΝὕψος. ἀλλ᾿ὡςμὲνἡΑΒΓΔβάσις EP .
πρὸςτὴνΕΖΗΘβάσιν,οὕτωςὁΑΞκύλινδροςπρὸςτὸν For, with the same construction, since as base ABCD
ΕΣκύλινδρον·ὑπὸγὰρτὸαὐτὸὕψοςεἰσίν·ὡςδὲτὸΜΝ is to base EFGH, so height MN (is) to height KL, and
ὕψοςπρὸςτὸΠΝ[ὕψος],οὕτωςὁΕΟκύλινδροςπρὸςτὸν height KL (is) equal to height QN, thus, as base ABCD
ΕΣκύλινδρον·ἔστινἄραὡςὁΑΞκύλινδροςπρὸςτὸνΕΣ (is) to base EFGH, so height MN will be to height
κύλινδρον,οὕτωςὁΕΟκύλινδροςπρὸςτὸνΕΣ.ἴσοςἄρα QN. But, as base ABCD (is) to base EFGH, so cylin-
ὁΑΞκύλινδροςτῷΕΟκυλίνδρῳ.ὡσαύτωςδὲκαὶἐπὶτῶν der AO (is) to cylinder ES. For they are the same height
κώνων·ὅπερἔδειδεῖξαι.
[Prop. 12.11]. And as height MN (is) to [height] QN,
so cylinder EP (is) to cylinder ES [Prop. 12.13]. Thus,
as cylinder AO is to cylinder ES, so cylinder EP (is) to
(cylinder) ES [Prop. 5.11]. Thus, cylinder AO (is) equal
to cylinder EP [Prop. 5.9]. In the same manner, (the
proposition can) also (be demonstrated) for the cones.
(Which is) the very thing it was required to show.
¦. Proposition 16
Δύο κύκλων περὶ τὸ αὐτὸ κέντρον ὄντων εἰς τὸν There being two circles about the same center, to
μείζονακύκλονπολύγωνονἰσόπλευρόντεκαὶἀρτιόπλευρον inscribe an equilateral and even-sided polygon in the
ἐγγράψαιμὴψαῦοντοῦἐλάσσονοςκύκλου.
greater circle, not touching the lesser circle.
Â
Ã
Ä
ÀÅ
Æ
B
E
H
K
F
G
A
L
M D
N
C
῎Εστωσαν οἱδοθέντες δύοκύκλοιοἱΑΒΓΔ, ΕΖΗΘ Let ABCD and EFGH be the given two circles, about
περὶτὸαὐτὸκέντροντὸΚ·δεῖδὴεἰςτὸνμείζονακύκλον the same center, K. So, it is necessary to inscribe an
τὸν ΑΒΓΔ πολύγωνον ἰσόπλευρόν τε καὶ ἀρτιόπλευρον equilateral and even-sided polygon in the greater circle
ἐγγράψαιμὴψαῦοντοῦΕΖΗΘκύκλου.
ABCD, not touching circle EFGH.
῎Ηχθω γὰρ διὰ τοῦ Κ κέντρου εὐθεῖα ἡ ΒΚΔ, καὶ Let the straight-line BKD have been drawn through
ἀπὸ τοῦ Η σημείου τῇ ΒΔ εὐθείᾳ πρὸς ὀρθὰς ἤχθω ἡ the center K. And let GA have been drawn, at right-
ΗΑκαὶδιήχθωἐπὶτὸΓ·ἡΑΓἄραἐφάπτεταιτοῦΕΖΗΘ angles to the straight-line BD, through point G, and let it
κύκλου. τέμνοντεςδὴτὴνΒΑΔπεριφέρειανδίχακαὶτὴν have been drawn through to C. Thus, AC touches circle
ἡμίσειαναὐτῆςδίχακαὶτοῦτοἀεὶποιοῦντεςκαταλείψομεν EFGH [Prop. 3.16 corr.]. So, (by) cutting circumference
περιφέρειανἐλάσσονατῆςΑΔ.λελείφθω,καὶἔστωἡΛΔ, BAD in half, and the half of it in half, and doing this conκαὶἀπὸτοῦΛἐπὶτὴνΒΔκάθετοςἤχθωἡΛΜκαὶδιήχθω
tinually, we will (eventually) leave a circumference less
498

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
ἐπὶτὸΝ,καὶἐπεζεύχθωσαναἱΛΔ,ΔΝ·ἴσηἄραἐστὶνἡ than AD [Prop. 10.1]. Let it have been left, and let it be
ΛΔτῇΔΝ.καὶἐπεὶπαράλληλόςἐστινἡΛΝτῇΑΓ,ἡδὲΑΓ LD. And let LM have been drawn, from L, perpendicu-
ἐφάπτεταιτοῦΕΖΗΘκύκλου,ἡΛΝἄραοὐκἐφάπτεταιτοῦ lar to BD, and let it have been drawn through to N. And
ΕΖΗΘκύκλου·πολλῷἄρααἱΛΔ,ΔΝοὐκἐφάπτονταιτοῦ let LD and DN have been joined. Thus, LD is equal to
ΕΖΗΘκύκλου.ἐὰνδὴτῇΛΔεὐθείᾳἴσαςκατὰτὸσυνεχὲς DN [Props. 3.3, 1.4]. And since LN is parallel to AC
ἐναρμόσωμενεἰςτὸνΑΒΓΔκύκλον,ἐγγραφήσεταιεἰςτὸν [Prop. 1.28], and AC touches circle EFGH, LN thus
ΑΒΓΔκύκλονπολύγωνονἰσόπλευρόντεκαὶἀρτιόπλευρον does not touch circle EFGH. Thus, even more so, LD
μὴ ψαῦον τοῦ ἐλάσσονοςκύκλουτοῦΕΖΗΘ· ὅπερ ἔδει and DN do not touch circle EFGH. And if we continuποιῆσαι.
ously insert (straight-lines) equal to straight-line LD into
circle ABCD [Prop. 4.1] then an equilateral and evensided
polygon, not touching the lesser circle EFGH, will
have been inscribed in circle ABCD. † (Which is) the very
thing it was required to do.
† Note that the chord of the polygon, LN, does not touch the inner circle either.
Þ.
Proposition 17
Δύο σφαιρῶν περὶ τὸ αὐτὸ κέντρον οὐσῶν εἰς τὴν There being two spheres about the same center, to inμείζονασφαῖρανστερεὸνπολύεδρονἐγγράψαιμὴψαῦοντῆς
scribe a polyhedral solid in the greater sphere, not touch-
Ä Å
ἐλάσσονοςσφαίραςκατὰτὴνἐπιφάνειαν.
Í
ing the lesser sphere on its surface.
E
M
Ã
Ï É Ê
L
T
S
U
À
K
R O
Q
P
 Æ
F
X W
B
Y V G A
D
H
N
C
ΝενοήσθωσανδύοσφαῖραιπερὶτὸαὐτὸκέντροντὸΑ· Let two spheres have been conceived about the same
δεῖδὴεἰςτὴνμείζονασφαῖρανστερεὸνπολύεδρονἐγγράψαι center, A. So, it is necessary to inscribe a polyhedral solid
μὴψαῦοντῆςἐλάσσονοςσφαίραςκατὰτὴνἐπιφάνειαν. in the greater sphere, not touching the lesser sphere on
Τετμήσθωσαναἱσφαῖραιἐπιπέδῳτινὶδιὰτοῦκέντρου· its surface.
ἔσονται δὴ αἱ τομαὶ κύκλοι, ἐπειδήπερ μενούσης τῆς Let the spheres have been cut by some plane through
διαμέτρου καὶ περιφερομένου τοῦ ἡμικυκλίουἐγιγνετοἡ the center. So, the sections will be circles, inasmuch
σφαῖρα· ὥστε καὶ καθ᾿ οἵας ἂν θέσεως ἐπινοήσωμεν τὸ as a sphere is generated by the diameter remaining be-
ἡμικύκλιον, τὸ δι᾿ αὐτοῦἐκβαλλόμενονἐπίπεδονποιήσει hind, and a semi-circle being carried around [Def. 11.14].
ἐπὶ τῆς ἐπιφανείας τῆς σφαίρας κύκλον. καὶ φανερόν, And, hence, whatever position we conceive (of for) the
ὅτικαὶμέγιστον,ἐπειδήπερἡδιάμετροςτῆςσφαίρας,ἥτις semi-circle, the plane produced through it will make a
Ç Ë Ì È
499

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
ἐστὶκαὶτοῦἡμικυκλίουδιάμετροςδηλαδὴκαὶτοῦκύκλου, circle on the surface of the sphere. And (it is) clear
μείζωνἐστὶπασῶντῶνεἰςτὸνκύκλονἢτὴνσφαῖρανδια- that (it is) also a great (circle), inasmuch as the diamγομένων
[εὐθειῶν]. ἔστωοὖνἐνμὲντῇμείζονισφαίρᾳ eter of the sphere, which is also manifestly the diameκύκλοςὁΒΓΔΕ,ἐνδὲτῇἐλάσσονισφαίρᾳκύκλοςὁΖΗΘ,
ter of the semi-circle and the circle, is greater than all
καὶἤχθωσαναὐτῶνδύοδιάμετροιπρὸςὀρθὰςἀλλήλαιςαἱ of the (other) [straight-lines] drawn across in the cir-
ΒΔ,ΓΕ,καὶδύοκύκλωνπερὶτὸαὐτὸκέντρονὄντωντῶν cle or the sphere [Prop. 3.15]. Therefore, let BCDE be
ΒΓΔΕ,ΖΗΘεἰςτὸνμείζονακύκλοντὸνΒΓΔΕπολύγωνον the circle in the greater sphere, and FGH the circle in
ἰσόπλευρονκαὶἀρτιόπλευρονἐγγεγράφθωμὴψαῦοντοῦ the lesser sphere. And let two diameters of them have
ἐλάσσονοςκύκλουτοῦΖΗΘ,οὗπλευραὶἔστωσανἐντῷΒΕ been drawn at right-angles to one another, (namely),
τεταρτημορίῳαἱΒΚ,ΚΛ,ΛΜ,ΜΕ,καὶἐπιζευχθεῖσαἡΚΑ BD and CE. And there being two circles about the
διήχθωἐπὶτὸΝ,καὶἀνεστάτωἀπὸτοῦΑσημείουτῷτοῦ same center—(namely), BCDE and FGH—let an equi-
ΒΓΔΕκύκλουἐπιπέδῳπρὸςὀρθὰςἡΑΞκαὶσυμβαλλέτω lateral and even-sided polygon have been inscribed in
τῇἐπιφανείᾳτῆςσφαίραςκατὰτὸΞ,καὶδιὰτῆςΑΞκαὶ the greater circle, BCDE, not touching the lesser circle,
ἑκατέραςτῶνΒΔ,ΚΝἐπίπεδαἐκβεβλήσθω·ποιήσουσιδὴ FGH [Prop. 12.16], of which let the sides in the quadδιὰτὰεἰρημέναἐπὶτῆςἐπιφανείαςτῆςσφαίραςμεγίστους
rant BE be BK, KL, LM, and ME. And, KA being
κύκλους.ποιείτωσαν,ὧνἡμικύκλιαἔστωἐπὶτῶνΒΔ,ΚΝ joined, let it have been drawn across to N. And let AO
διαμέτρωντὰΒΞΔ,ΚΞΝ.καὶἐπεὶἡΞΑὀρθήἐστιπρὸςτὸ have been set up at point A, at right-angles to the plane of
τοῦΒΓΔΕκύκλουἐπίπεδον,καὶπάνταἄρατὰδιὰτῆςΞΑ circle BCDE. And let it meet the surface of the (greater)
ἐπίπεδάἐστινὀρθὰπρὸςτὸτοῦΒΓΔΕκύκλουἐπίπεδον· sphere at O. And let planes have been produced through
ὥστεκαὶτὰΒΞΔ,ΚΞΝἡμικύκλιαὀρθάἐστιπρὸςτὸτοῦ AO and each of BD and KN. So, according to the afore-
ΒΓΔΕκύκλουἐπίπεδον.καὶἐπεὶἴσαἐστὶτὰΒΕΔ,ΒΞΔ, mentioned (discussion), they will make great circles on
ΚΞΝἡμικύκλια·ἐπὶγὰρἴσωνεἰσὶδιαμέτρωντῶνΒΔ,ΚΝ· the surface of the (greater) sphere. Let them make (great
ἴσαἐστὶκαὶτὰΒΕ,ΒΞ,ΚΞτεταρτημόριαἀλλήλοις. ὄσαι circles), of which let BOD and KON be semi-circles on
ἄραεἰσὶνἐντῷΒΕτεταρτημορίῳπλευραὶτοῦπολυγώνου, the diameters BD and KN (respectively). And since OA
τοσαῦταίεἰσικαὶἐντοῖςΒΞ,ΚΞτεταρτημορίοιςἴσαιταῖς is at right-angles to the plane of circle BCDE, all of the
ΒΚ,ΚΛ,ΛΜ,ΜΕεὐθείαις.ἐγγεγράφθωσανκαὶἔστωσαν planes through OA are thus also at right-angles to the
αἱΒΟ,ΟΠ,ΠΡ,ΡΞ,ΚΣ,ΣΤ,ΤΥ,ΥΞ,καὶἐπεζεύχθωσαν plane of circle BCDE [Prop. 11.18]. And, hence, the
αἱ ΣΟ, ΤΠ, ΥΡ, καὶ ἀπὸ τῶν Ο, Σ ἐπὶ τὸ τοῦ ΒΓΔΕ semi-circles BOD and KON are also at right-angles to
κύκλουἐπίπεδονκάθετοιἤχθωσαν·πεσοῦνταιδὴἐπὶτὰς the plane of circle BCDE. And since semi-circles BED,
κοινὰς τομὰςτῶν ἐπιπέδων τὰςΒΔ, ΚΝ, ἐπειδήπερκαὶ BOD, and KON are equal—for (they are) on the equal
τὰτῶνΒΞΔ,ΚΞΝἐπίπεδαὀρθάἐστιπρὸςτὸτοῦΒΓΔΕ diameters BD and KN [Def. 3.1]—the quadrants BE,
κύκλουἐπίπεδον. πιπτέτωσαν,καὶἔστωσαναἱΟΦ,ΣΧ, BO, and KO are also equal to one another. Thus, as
καὶἐπεζεύχθω ἡΧΦ. καὶἐπεὶἐν ἴσοιςἡμικυκλίοιςτοῖς many sides of the polygon as are in quadrant BE, so
ΒΞΔ,ΚΞΝἴσαιἀπειλημμέναιεἰσὶναἱΒΟ,ΚΣ,καὶκάθετοι many are also in quadrants BO and KO equal to the
ἠγμέναιεἰσὶναἱΟΦ,ΣΧ,ἴση[ἄρα]ἐστὶνἡμὲνΟΦτῇΣΧ, straight-lines BK, KL, LM, and ME. Let them have
ἡδὲΒΦτῇΚΧ.ἔστιδὲκαὶὅληἡΒΑὅλῃτῇΚΑἴση·καὶ been inscribed, and let them be BP , PQ, QR, RO, KS,
λοιπὴἄραἡΦΑλοιπῇτῇΧΑἐστινἴση·ἔστινἄραὡςἡΒΦ ST , TU, and UO. And let SP , TQ, and UR have been
πρὸςτὴνΦΑ,οὕτωςἡΚΧπρὸςτὴνΧΑ·παράλληλοςἄρα joined. And let perpendiculars have been drawn from P
ἐστὶνἡΧΦτῇΚΒ.καὶἐπεὶἑκατέρατῶνΟΦ,ΣΧὀρθή and S to the plane of circle BCDE [Prop. 11.11]. So,
ἐστιπρὸςτὸτοῦΒΓΔΕκύκλουἐπίπεδον,παράλληλοςἄρα they will fall on the common sections of the planes BD
ἐστὶνἡΟΦτῇΣΧ.ἐδείχθηδὲαὐτῇκαὶἴση·καὶαἱΧΦ,ΣΟ and KN (with BCDE), inasmuch as the planes of BOD
ἄραἴσαιεἰσὶκαὶπαράλληλοι. καὶἐπεὶπαράλληλόςἐστιν and KON are also at right-angles to the plane of circle
ἡΧΦτῇΣΟ,ἀλλὰἡΧΦτῇΚΒἐστιπαράλληλος,καὶ BCDE [Def. 11.4]. Let them have fallen, and let them be
ἡΣΟἄρατῇΚΒἐστιπαράλληλος. καὶἐπιζευγνύουσιν PV and SW . And let WV have been joined. And since
αὐτὰς αἱ ΒΟ, ΚΣ· τὸ ΚΒΟΣ ἄρα τετράπλευρον ἐν ἑνί BP and KS are equal (circumferences) having been cut
ἐστινἐπιπέδῳ,ἐπειδήπερ,ἐὰνὦσιδύοεὐθεῖαιπαράλληλοι, off in the equal semi-circles BOD and KON [Def. 3.28],
καὶἐφ᾿ἑκατέραςαὐτῶνληφθῇτυχόντασημεῖα,ἡἐπὶτὰ and PV and SW are perpendiculars having been drawn
σημεῖαἐπιζευγνυμένηεὐθεῖαἐντῷαὐτῷἐπιπέδῳἐστὶταῖς (from them), PV is [thus] equal to SW , and BV to KW
παραλλήλοις. διὰτὰαὐτὰδὴκαὶἑκάτεροντῶνΣΟΠΤ, [Props. 3.27, 1.26]. And the whole of BA is also equal
ΤΠΡΥ τετραπλεύρωνἐνἑνίἐστιν ἐπιπέδῳ. ἔστιδὲκαὶ to the whole of KA. And, thus, as BV is to V A, so KW
τὸΥΡΞτρίγωνονἐνἑνὶἑπιπέδῳ. ἐὰνδὴνοήσωμενἀπὸ (is) to WA. WV is thus parallel to KB [Prop. 6.2]. And
500

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
τῶνΟ,Σ,Π,Τ,Ρ,ΥσημείωνἐπὶτὸΑἐπιζευγνυμένας since PV and SW are each at right-angles to the plane
εὐθείας,συσταθήσεταίτισχῆμαστερεὸνπολύεδρονματαξὺ of circle BCDE, PV is thus parallel to SW [Prop. 11.6].
τῶνΒΞ,ΚΞπεριφερειῶνἐκπυραμίδωνσυγκείμενον,ὧν And it was also shown (to be) equal to it. And, thus,
βάσειςμὲντὰΚΒΟΣ,ΣΟΠΤ,ΤΠΡΥτετράπλευρακαὶτὸ WV and SP are equal and parallel [Prop. 1.33]. And
ΥΡΞτρίγωνον,κορυφὴδὲτὸΑσημεῖον. ἐὰνδὲκαὶἐπὶ since WV is parallel to SP , but WV is parallel to KB,
ἑκάστηςτῶνΚΛ,ΛΜ,ΜΕπλευρῶνκαθάπερἐπὶτῆςΒΚτὰ SP is thus also parallel to KB [Prop. 11.1]. And BP
αὐτὰκατασκευάσωμενκαὶἔτιτῶνλοιπῶντριῶντεταρτη- and KS join them. Thus, the quadrilateral KBPS is in
μορίων,συσταθήσεταίτισχῆμαπολύεδρονἐγγεγραμμένον one plane, inasmuch as if there are two parallel straightεἰςτὴνσφαῖρανπυραμίσιπεριεχόμενον,ὧνβάσιες[μὲν]τὰ
lines, and a random point is taken on each of them, then
εἰρημένατετράπλευρακαὶτὸΥΡΞτρίγωνονκαὶτὰὁμοταγῆ the straight-line joining the points is in the same plane
αὐτοῖς,κορυφὴδὲτὸΑσημεῖον.
as the parallel (straight-lines) [Prop. 11.7]. So, for the
Λέγω ὅτι τὸ εἰρημένον πολύεδρον οὐκ ἐφάψεται τῆς same (reasons), each of the quadrilaterals SPQT and
ἐλάσσονοςσφαίραςκατὰτὴνἐπιφάνειαν, ἐφ᾿ ἧςἐστιν ὁ TQRU is also in one plane. And triangle URO is also
ΖΗΘκύκλος.
in one plane [Prop. 11.2]. So, if we conceive straight-
῎Ηχθω ἀπὸ τοῦ Α σημείου ἐπὶ τὸ τοῦ ΚΒΟΣ τε- lines joining points P , S, Q, T , R, and U to A then
τραπλεύρουἐπίπεδονκάθετοςἡΑΨκαὶσυμβαλλέτωτῷ some solid polyhedral figure will have been constructed
ἐπιπέδῳκατὰτὸΨσημεῖον,καὶἐπεζεύχθωσαναἱΨΒ,ΨΚ. between the circumferences BO and KO, being comκαὶἐπεὶἡΑΨὀρθήἐστιπρὸςτὸτοῦΚΒΟΣτετραπλεύρου
posed of pyramids whose bases (are) the quadrilaterals
ἐπίπεδον,καὶπρὸςπάσαςἄρατὰςἁπτομέναςαὐτῆςεὐθείας KBPS, SPQT , TQRU, and the triangle URO, and apex
καὶοὔσαςἐντῷτοῦτετραπλεύρουἐπιπέδῳὀρθήἐστιν. ἡ the point A. And if we also make the same construction
ΑΨἄραὀρθήἐστιπρὸςἑκατέραντῶνΒΨ,ΨΚ.καὶἐπεὶ on each of the sides KL, LM, and ME, just as on BK,
ἴσηἐστὶνἡΑΒτῇΑΚ,ἵσονἐστὶκαὶτὸἀπὸτῆςΑΒτῷ and, further, (repeat the construction) in the remaining
ἀπὸτῆςΑΚ.καίἐστιτῷμὲνἀπὸτῆςΑΒἴσατὰἀπὸτῶν three quadrants, then some polyhedral figure which has
ΑΨ,ΨΒ·ὀρθὴγὰρἡπρὸςτῷΨ·τῷδὲἀπὸτῆςΑΚἴσατὰ been inscribed in the sphere will have been constructed,
ἀπὸτῶνΑΨ,ΨΚ.τὰἄραἀπὸτῶνΑΨ,ΨΒἴσαἐστὶτοῖς being contained by pyramids whose bases (are) the afore-
ἀπὸτῶνΑΨ,ΨΚ.κοινὸνἀφῃρήσθωτὸἀπὸτῆςΑΨ·λοιπὸν mentioned quadrilaterals, and triangle URO, and the
ἄρατὸἀπὸτῆςΒΨλοιπῷτῷἀπὸτῆςΨΚἴσονἐστίν·ἴση (quadrilaterals and triangles) similarly arranged to them,
ἄραἡΒΨτῇΨΚ.ὁμοίωςδὴδείξομεν,ὅτικαὶαἱἀπὸτοῦΨ and apex the point A.
ἐπὶτὰΟ,Σἐπιζευγνύμεναιεὐθεῖαιἴσαιεἰσὶνἑκατέρᾳτῶν So, I say that the aforementioned polyhedron will not
ΒΨ,ΨΚ.ὁἄρακέντρῳτῷΨκαὶδιαστήματιἑνὶτῶνΨΒ, touch the lesser sphere on the surface on which the circle
ΨΚγραφόμενοςκύκλοςἥξεικαὶδιὰτῶνΟ,Σ,καὶἔσται FGH is (situated).
ἐνκύκλῳτὸΚΒΟΣτετράπλευρον.
Let the perpendicular (straight-line) AX have been
ΚαὶἐπεὶμείζωνἐστὶνἡΚΒτῆςΧΦ,ἴσηδὲἡΧΦτῇ drawn from point A to the plane KBPS, and let it meet
ΣΟ,μείζωνἄραἡΚΒτῆςΣΟ.ἴσηδὲἡΚΒἑκατέρᾳτῶν the plane at point X [Prop. 11.11]. And let XB and
ΚΣ,ΒΟ·καὶἑκατέραἄρατῶνΚΣ,ΒΟτῆςΣΟμείζων XK have been joined. And since AX is at right-angles to
ἐστίν.καὶἐπεὶἐνκύκλῳτετράπλευρόνἐστιτὸΚΒΟΣ,καὶ the plane of quadrilateral KBPS, it is thus also at right-
ἴσαιαἱΚΒ,ΒΟ,ΚΣ,καὶἐλάττωνἡΟΣ,καὶἐκτοῦκέντρου angles to all of the straight-lines joined to it which are
τοῦκύκλουἐστὶνἡΒΨ,τὸἄραἀπὸτῆςΚΒτοῦἀπὸτῆς also in the plane of the quadrilateral [Def. 11.3]. Thus,
ΒΨμεῖζόνἐστινἢδιπλάσιον.ἤχθωἀπὸτοῦΚἐπὶτὴνΒΦ AX is at right-angles to each of BX and XK. And since
κάθετοςἡΚΩ.καὶἐπεὶἡΒΔτῆςΔΩἐλάττωνἐστὶνἢ AB is equal to AK, the (square) on AB is also equal to
διπλῆ,καίἐστινὡςἡΒΔπρὸςτὴνΔΩ,οὕτωςτὸὑπὸτῶν the (square) on AK. And the (sum of the squares) on AX
ΔΒ,ΒΩπρὸςτὸὑπὸ[τῶν]ΔΩ,ΩΒ,ἀναγραφομένουἀπὸ and XB is equal to the (square) on AB. For the angle
τῆςΒΩτετραγώνουκαὶσυμπληρουμένουτοῦἐπὶτῆςΩΔ at X (is) a right-angle [Prop. 1.47]. And the (sum of
παραλληλογράμμουκαὶτὸὑπὸΔΒ,ΒΩἄρατοῦὑπὸΔΩ, the squares) on AX and XK is equal to the (square) on
ΩΒἔλαττόνἐστινἢδιπλάσιον. καίἐστιτῆςΚΔἐπιζευ- AK [Prop. 1.47]. Thus, the (sum of the squares) on AX
γνυμένηςτὸμὲνὑπὸΔΒ,ΒΩἴσοντῷἀπὸτῆςΒΚ,τὸδὲ and XB is equal to the (sum of the squares) on AX and
ὑπὸτῶνΔΩ,ΩΒἴσοντῷἀπὸτῆςΚΩ·τὸἄραἀπὸτῆςΚΒ XK. Let the (square) on AX have been subtracted from
τοῦἀπὸτῆςΚΩἔλασσόνἐστινἢδιπλάσιον. ἀλλὰτὸἀπὸ both. Thus, the remaining (square) on BX is equal to the
τῆςΚΒτοῦἀπὸτῆςΒΨμεῖζόνἐστινἢδιπλάσιον·μεῖζον remaining (square) on XK. Thus, BX (is) equal to XK.
ἄρατὸἀπὸτῆςΚΩτοῦἀπὸτῆςΒΨ.καὶἐπεὶἴσηἐστὶνἡ So, similarly, we can show that the straight-lines joined
ΒΑτῇΚΑ,ἴσονἐστὶτὸἀπὸτῆςΒΑτῷἀπὸτῆςΑΚ.καί from X to P and S are equal to each of BX and XK.
501

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
ἐστιτῷμὲνἀπὸτῆςΒΑἴσατὰἀπὸτῶνΒΨ,ΨΑ,τῷδὲἀπὸ Thus, a circle drawn (in the plane of the quadrilateral)
τῆςΚΑἴσατὰἀπὸτῶνΚΩ,ΩΑ·τὰἄραἀπὸτῶνΒΨ,ΨΑ with center X, and radius one of XB or XK, will also
ἴσαἐστὶτοῖςἀπὸτῶνΚΩ,ΩΑ,ὧντὸἀπὸτῆςΚΩμεῖζον pass through P and S, and the quadrilateral KBPS will
τοῦἀπὸτῆςΒΨ·λοιπὸνἄρατὸἀπὸτῆςΩΑἔλασσόνἐστι be inside the circle.
τοῦἀπὸτῆςΨΑ.μείζωνἄραἡΑΨτῆςΑΩ·πολλῷἄραἡ And since KB is greater than WV , and WV (is) equal
ΑΨμείζωνἐστὶτῆςΑΗ.καίἐστινἡμὲνΑΨἐπὶμίαντοῦ to SP , KB (is) thus greater than SP . And KB (is)
πολυέδρουβάσιν,ἡδὲΑΗἐπὶτὴντῆςἐλάσσονοςσφαίρας equal to each of KS and BP . Thus, KS and BP are
ἐπιφάνειαν·ὥστετὸπολύεδρονοὐψαύσειτῆςἐλάσσονος each greater than SP . And since quadrilateral KBPS
σφαίραςκατὰτὴνἐπιφάνειαν.
is in a circle, and KB, BP , and KS are equal (to one
Δύοἄρασφαιρῶνπερὶτὸαὐτὸκέντρονοὐσῶνεἰςτὴν another), and PS (is) less (than them), and BX is the
μείζονασφαῖρανστερεὸνπολύεδρονἐγγέγραπταιμὴψαῦον radius of the circle, the (square) on KB is thus greater
τῆς ἐλάσσονος σφαίρας κατὰ τὴν ἐπιφάνειαν· ὅπερ ἔδει than double the (square) on BX. † Let the perpendicular
ποιῆσαι.
KY have been drawn from K to BV . ‡ And since BD is
less than double DY , and as BD is to DY , so the (rectangle
contained) by DB and BY (is) to the (rectangle
contained) by DY and Y B—a square being described
on BY , and a (rectangular) parallelogram (with short
side equal to BY ) completed on Y D—the (rectangle contained)
by DB and BY is thus also less than double the
(rectangle contained) by DY and Y B. And, KD being
joined, the (rectangle contained) by DB and BY is equal
to the (square) on BK, and the (rectangle contained) by
DY and Y B equal to the (square) on KY [Props. 3.31,
6.8 corr.]. Thus, the (square) on KB is less than double
the (square) on KY . But, the (square) on KB is greater
than double the (square) on BX. Thus, the (square) on
KY (is) greater than the (square) on BX. And since
BA is equal to KA, the (square) on BA is equal to the
(square) on AK. And the (sum of the squares) on BX
and XA is equal to the (square) on BA, and the (sum of
the squares) on KY and Y A (is) equal to the (square) on
KA [Prop. 1.47]. Thus, the (sum of the squares) on BX
and XA is equal to the (sum of the squares) on KY and
Y A, of which the (square) on KY (is) greater than the
(square) on BX. Thus, the remaining (square) on Y A
is less than the (square) on XA. Thus, AX (is) greater
than AY . Thus, AX is much greater than AG. § And AX
is (a perpendicular) on one of the bases of the polyhedron,
and AG (is a perpendicular) on the surface of the
lesser sphere. Hence, the polyhedron will not touch the
lesser sphere on its surface.
Thus, there being two spheres about the same center,
a polyhedral solid has been inscribed in the greater
sphere which does not touch the lesser sphere on its surface.
(Which is) the very thing it was required to do.
† Since KB, BP , and KS are greater than the sides of an inscribed square, which are each of length √ 2 BX.
‡ Note that points Y and V are actually identical.
§ This conclusion depends on the fact that the chord of the polygon in proposition 12.16 does not touch the inner circle.
502

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
ÈÖ×Ñ.
᾿ΕὰνδὲκαὶεἰςἑτάρανσφαῖραντῷἐντῇΒΓΔΕσφαίρᾳ And, also, if a similar polyhedral solid to that in
Corollary
στερεῷπολυέδρῳὅμοιονστερεὸνπολύεδρονἐγγραφῇ,τὸ sphere BCDE is inscribed in another sphere then the
ἐν τῇ ΒΓΔΕ σφαίρᾳ στερεὸν πολύεδρον πρὸς τὸ ἐν τῇ polyhedral solid in sphere BCDE has to the polyhedral
ἑτέρᾳσφαίρᾳστερεὸνπολύεδροντριπλασίοναλόγονἔχει, solid in the other sphere the cubed ratio that the diameter
ἤπερἡτῆςΒΓΔΕσφαίραςδιάμετροςπρὸςτὴντῆςἑτέρας of sphere BCDE has to the diameter of the other sphere.
σφαίρας διάμετρον. διαιρεθέντων γὰρ τῶν στερεῶν εἰς For if the solids are divided into similarly numbered, and
τὰςὁμοιοπληθεῖςκαὶὁμοιοταγεῖςπυραμίδαςἔσονταιαἱπυ- similarly situated, pyramids, then the pyramids will be
ραμίδεςὅμοιαι. αἱδὲὅμοιαιπυραμίδεςπρὸςἀλλήλαςἐν similar. And similar pyramids are in the cubed ratio of
τριπλασίονιλόγῳεἰσὶτῶνὁμολόγωνπλευρῶν·ἡἄραπυ- corresponding sides [Prop. 12.8 corr.]. Thus, the pyraραμίς,ἧςβάσιςμένἐστιτὸΚΒΟΣτετράπλευρον,κορυφὴ
mid whose base is quadrilateral KBPS, and apex the
δὲτὸΑσημεῖον,πρὸςτὴνἐντῇἑτέρᾳσφαίρᾳὁμοιοταγῆ point A, will have to the similarly situated pyramid in the
πυραμίδατριπλασίοναλόγονἔχει,ἤπερἡὁμόλογοςπλευρὰ other sphere the cubed ratio that a corresponding side
πρὸςτὴνὁμόλογονπλευράν,τουτέστινἤπερἡΑΒἐκτοῦ (has) to a corresponding side. That is to say, that of raκέντρουτῆςσφαίραςτῆςπερὶκέντροντὸΑπρὸςτὴνἐκτοῦ
dius AB of the sphere about center A to the radius of the
κέντρουτῆςἑτέραςσφαίρας. ὁμοίωςκαὶἑκάστηπυραμὶς other sphere. And, similarly, each pyramid in the sphere
τῶνἐντῇπερὶκέντροντὸΑσφαίρᾳπρὸςἑκάστηνὁμοταγῆ about center A will have to each similarly situated pyraπυραμίδατῶνἐντῇἑτέρᾳσφαίρᾳτριπλασίοναλόγονἕξει,
mid in the other sphere the cubed ratio that AB (has) to
ἤπερἡΑΒπρὸςτὴνἐκτοῦκέντρουτῆςἑτέραςσφαίρας. the radius of the other sphere. And as one of the leading
καὶὡςἓντῶνἡγουμένωνπρὸςἓντῶνἑπομένων,οὕτως (magnitudes is) to one of the following (in two sets of
ἅπαντατὰἡγούμεναπρὸςἅπαντατὰἑπόμενα·ὥστεὅλον proportional magnitudes), so (the sum of) all the leadτὸ
ἐν τῇ περὶ κέντρον τὸ Α σφαίρᾳ στερεὸν πολύεδρον ing (magnitudes is) to (the sum of) all of the following
πρὸςὅλοντὸἐντῇἑτέρᾳ[σφαίρᾳ]στερεὸνπολύεδροντρι- (magnitudes) [Prop. 5.12]. Hence, the whole polyhedral
πλασίοναλόγονἕξει,ἤπερἡΑΒπρὸςτὴνἐκτοῦκέντρου solid in the sphere about center A will have to the whole
τῆςἑτέραςσφαίρας,τουτέστινἤπερἡΒΔδιάμετροςπρὸς polyhedral solid in the other [sphere] the cubed ratio that
τὴντῆςἑτέραςσφαίραςδιάμετρον·ὅπερἔδειδεῖξαι.
(radius) AB (has) to the radius of the other sphere. That
is to say, that diameter BD (has) to the diameter of the
other sphere. (Which is) the very thing it was required to
show.
. Proposition 18
Αἱσφαῖραιπρὸςἀλλήλαςἐντριπλασίονιλόγῳεἰσὶτῶν Spheres are to one another in the cubed ratio of their
ἰδίωνδιαμέτρων.
Ä
respective diameters.
A
L
À Å Æ
B
C
M
N
D
 Ã
G
H K
E
F
503

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 12
Νενοήσθωσαν σφαῖραι αἱ ΑΒΓ, ΔΕΖ, διάμετροι δὲ Let the spheres ABC and DEF have been conceived,
αὐτῶναἱΒΓ,ΕΖ·λέγω,ὅτιἡΑΒΓσφαῖραπρὸςτὴνΔΕΖ and (let) their diameters (be) BC and EF (respectively).
σφαῖραντριπλασίοναλόγονἔχειἤπερἡΒΓπρὸςτὴνΕΖ. I say that sphere ABC has to sphere DEF the cubed ratio
ΕἰγὰρμὴἡΑΒΓσφαῖραπρὸςτὴνΔΕΖσφαῖραντρι- that BC (has) to EF .
πλασίοναλόγονἔχειἤπερἡΒΓπρὸςτὴνΕΖ,ἕξειἄραἡ For if sphere ABC does not have to sphere DEF the
ΑΒΓσφαῖραπρὸςἐλάσσονάτινατῆςΔΕΖσφαίραςτρι- cubed ratio that BC (has) to EF then sphere ABC will
πλασίοναλόγονἢπρὸςμείζοναἤπερἡΒΓπρὸςτὴνΕΖ. have to some (sphere) either less than, or greater than,
ἐχέτωπρότερονπρὸςἐλάσσονατὴνΗΘΚ,καὶνενοήσθωἡ sphere DEF the cubed ratio that BC (has) to EF . Let
ΔΕΖτῇΗΘΚπερὶτὸαὐτὸκέντρον,καὶἐγγεγράφθωεἰς it, first of all, have (such a ratio) to a lesser (sphere),
τὴνμείζονασφαῖραντὴνΔΕΖστερεὸνπολύεδρονμὴψαῦον GHK. And let DEF have been conceived about the
τῆς ἐλάσσονος σφαίρας τῆς ΗΘΚ κατὰ τὴν ἐπιφάνειαν, same center as GHK. And let a polyhedral solid have
ἐγγεγράφθωδὲκαὶεἰςτὴνΑΒΓσφαῖραντῷἐντῇΔΕΖ been inscribed in the greater sphere DEF , not touching
σφαίρᾳ στερεῷ πολυέδρῳὅμοιονστερεὸν πολύεδρον· τὸ the lesser sphere GHK on its surface [Prop. 12.17]. And
ἄραἐντῇΑΒΓστερεὸνπολύεδρονπρὸςτὸἐντῇΔΕΖ let a polyhedral solid, similar to the polyhedral solid in
στερεὸνπολύεδροντριπλασίοναλόγονἔχειἤπερἡΒΓπρὸς sphere DEF , have also been inscribed in sphere ABC.
τὴνΕΖ.ἔχειδὲκαὶἡΑΒΓσφαῖραπρὸςτὴνΗΘΚσφαῖραν Thus, the polyhedral solid in sphere ABC has to the
τριπλασίοναλόγονἤπερἡΒΓπρὸςτὴνΕΖ·ἔστινἄραὡς polyhedral solid in sphere DEF the cubed ratio that BC
ἡΑΒΓσφαῖραπρὸςτὴνΗΘΚσφαῖραν,οὕτωςτὸἐντῇ (has) to EF [Prop. 12.17 corr.]. And sphere ABC also
ΑΒΓσφαίρᾳστερεὸνπολύεδρονπρὸςτὸἐντῇΔΕΖσφαίρᾳ has to sphere GHK the cubed ratio that BC (has) to EF .
στερεὸνπολύεδρον·ἐναλλὰξ[ἄρα]ὡςἡΑΒΓσφαῖραπρὸς Thus, as sphere ABC is to sphere GHK, so the polyheτὸἐναὐτῇπολύεδρον,οὕτωςἡΗΘΚσφαῖραπρὸςτὸἐντῇ
dral solid in sphere ABC (is) to the polyhedral solid is
ΔΕΖσφαίρᾳστερεὸνπολύεδρον.μείζωνδὲἡΑΒΓσφαῖρα sphere DEF . [Thus], alternately, as sphere ABC (is) to
τοῦἐναὐτῇπολυέδρου· μείζων ἄρακαὶἡΗΘΚσφαῖρα the polygon within it, so sphere GHK (is) to the polyheτοῦ
ἐν τῇ ΔΕΖ σφαίρᾳ πολυέδρου. ἀλλὰ καὶ ἐλάττων· dral solid within sphere DEF [Prop. 5.16]. And sphere
ἐμπεριέχεταιγὰρὑπ᾿αὐτοῦ. οὐκἄραἡΑΒΓσφαῖραπρὸς ABC (is) greater than the polyhedron within it. Thus,
ἐλάσσονατῆςΔΕΖσφαίραςτριπλασίοναλόγονἔχειἤπερἡ sphere GHK (is) also greater than the polyhedron within
ΒΓδιάμετροςπρὸςτὴνΕΖ.ὁμοίωςδὴδείξομεν,ὅτιοὐδὲἡ sphere DEF [Prop. 5.14]. But, (it is) also less. For it is
ΔΕΖσφαῖραπρὸςἐλάσσονατῆςΑΒΓσφαίραςτριπλασίονα encompassed by it. Thus, sphere ABC does not have to
λόγονἔχειἤπερἡΕΖπρὸςτὴνΒΓ.
(a sphere) less than sphere DEF the cubed ratio that di-
Λέγωδή,ὅτιοὐδὲἡΑΒΓσφαῖραπρὸςμείζονάτινατῆς ameter BC (has) to EF . So, similarly, we can show that
ΔΕΖσφαίραςτριπλασίοναλόγονἔχειἤπερἡΒΓπρὸςτὴν sphere DEF does not have to (a sphere) less than sphere
ΕΖ.
ABC the cubed ratio that EF (has) to BC either.
Εἰγὰρδυνατόν,ἐχέτωπρὸςμείζονατὴνΛΜΝ·ἀνάπαλιν So, I say that sphere ABC does not have to some
ἄραἡΛΜΝσφαῖραπρὸςτὴνΑΒΓσφαῖραντριπλασίονα (sphere) greater than sphere DEF the cubed ratio that
λόγονἔχειἤπερἡΕΖδιάμετροςπρὸςτὴνΒΓδιάμετρον. BC (has) to EF either.
ὡςδὲἡΛΜΝσφαῖραπρὸςτὴνΑΒΓσφαῖραν,οὕτωςἡΔΕΖ For, if possible, let it have (the cubed ratio) to a
σφαῖραπρὸςἐλάσσονάτινατῆςΑΒΓσφαίρας,ἐπειδήπερ greater (sphere), LMN. Thus, inversely, sphere LMN
μείζωνἐστὶνἡΛΜΝτῆςΔΕΖ,ὡςἔμπροσθενἐδείχθη.καὶ (has) to sphere ABC the cubed ratio that diameter
ἡΔΕΖἄρασφαῖραπρὸςἐλάσσονάτινατῆςΑΒΓσφαίρας EF (has) to diameter BC [Prop. 5.7 corr.]. And as
τριπλασίονα λόγον ἔχει ἤπερ ἡ ΕΖ πρὸς τὴν ΒΓ· ὅπερ sphere LMN (is) to sphere ABC, so sphere DEF
ἀδύνατονἐδείχθη. οὐκἄραἡΑΒΓσφαῖραπρὸςμείζονά (is) to some (sphere) less than sphere ABC, inasmuch
τινατῆςΔΕΖσφαίραςτριπλασίοναλόγονἔχειἤπερἡΒΓ as LMN is greater than DEF , as was shown before
πρὸςτὴνΕΖ.ἐδείχθηδέ,ὅτιοὐδὲπρὸςἐλάσσονα. ἡἄρα [Prop. 12.2 lem.]. And, thus, sphere DEF has to some
ΑΒΓσφαῖραπρὸςτὴνΔΕΖσφαῖραντριπλασίοναλόγονἔχει (sphere) less than sphere ABC the cubed ratio that EF
ἤπερἡΒΓπρὸςτὴνΕΖ·ὅπερἔδειδεῖξαι.
(has) to BC. The very thing was shown (to be) impossible.
Thus, sphere ABC does not have to some (sphere)
greater than sphere DEF the cubed ratio that BC (has)
to EF . And it was shown that neither (does it have
such a ratio) to a lesser (sphere). Thus, sphere ABC has
to sphere DEF the cubed ratio that BC (has) to EF .
(Which is) the very thing it was required to show.
504

ELEMENTS BOOK 13
The Platonic Solids †
† The five regular solids—the cube, tetrahedron (i.e., pyramid), octahedron, icosahedron, and dodecahedron—were problably discovered by
the school of Pythagoras. They are generally termed “Platonic” solids because they feature prominently in Plato’s famous dialogue Timaeus. Many
of the theorems contained in this book—particularly those which pertain to the last two solids—are ascribed to Theaetetus of Athens.
505

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
. Proposition 1
᾿Εὰν εὐθεῖα Äγραμμὴ ἄκρον καὶ μέσον
λόγον τμηθῇ, If a straight-line is cut in extreme and mean ratio
τὸ μεῖζον τμῆμα προσλαβὸντὴν ἡμίσειαν τῆς ὅληςπεν- then the square on the greater piece, added to half of
ταπλάσιονδύναταιτοῦἀπὸτῆςἡμισείαςτετραγώνου. the whole, is five times the square on the half.
Æ
L
F
Å
Â
N
P
M H
O
C
D
A
B
Ã
Ç
À
K G E
Εὐθεῖα γὰρ γραμμὴ ἡ ΑΒ ἄκρον καὶ μέσον λόγον For let the straight-line AB have been cut in extreme
τετμήσθωκατὰτὸΓσημεῖον,καὶἔστωμεῖζοντμῆματὸ and mean ratio at point C, and let AC be the greater
ΑΓ,καὶἐκβεβλήσθωἐπ᾿εὐθείαςτῇΓΑεὐθεῖαἡΑΔ,καὶ piece. And let the straight-line AD have been produced
κείσθωτῆςΑΒἡμίσειαἡΑΔ·λέγω,ὅτιπενταπλάσιόνἐστι in a straight-line with CA. And let AD be made (equal
τὸἀπὸτῆςΓΔτοῦἀπὸτῆςΔΑ.
to) half of AB. I say that the (square) on CD is five times
Ἀναγεγράφθωσαν γὰρ ἀπὸ τῶν ΑΒ, ΔΓ τετράγωνα the (square) on DA.
τὰΑΕ,ΔΖ,καὶκαταγεγράφθωἐντῷΔΖτὸσχῆμα,καὶ For let the squares AE and DF have been described
διήχθωἡΖΓἐπὶτὸΗ.καὶἐπεὶἡΑΒἄκρονκαὶμέσον on AB and DC (respectively). And let the figure in DF
λόγοντέτμηταικατὰτὸΓ,τὸἄραὑπὸτῶνΑΒΓἴσονἐστὶ have been drawn. And let FC have been drawn across to
τῷἀπὸτῆςΑΓ.καίἐστιτὸμὲνὑπὸτῶνΑΒΓτὸΓΕ,τὸ G. And since AB has been cut in extreme and mean ratio
δὲἀπὸτῆςΑΓτὸΖΘ·ἴσονἄρατὸΓΕτῷΖΘ.καὶἐπεὶ at C, the (rectangle contained) by ABC is thus equal to
διπλῆἐστινἡΒΑτῆςΑΔ,ἴσηδὲἡμὲνΒΑτῇΚΑ,ἡδὲ the (square) on AC [Def. 6.3, Prop. 6.17]. And CE is
ΑΔτῇΑΘ,διπλῆἄρακαὶἡΚΑτῆςΑΘ.ὡςδὲἡΚΑπρὸς the (rectangle contained) by ABC, and FH the (square)
τὴνΑΘ,οὕτωςτὸΓΚπρὸςτὸΓΘ·διπλάσιονἄρατὸΓΚ on AC. Thus, CE (is) equal to FH. And since BA is
τοῦΓΘ.εἰσὶδὲκαὶτὰΛΘ,ΘΓδιπλάσιατοῦΓΘ.ἴσονἄρα double AD, and BA (is) equal to KA, and AD to AH,
τὸΚΓτοῖςΛΘ,ΘΓ.ἐδείχθηδὲκαὶτὸΓΕτῷΘΖἴσον· KA (is) thus also double AH. And as KA (is) to AH, so
ὅλονἄρατὸΑΕτετράγωνονἴσονἐστὶτῷΜΝΞγνώμονι. CK (is) to CH [Prop. 6.1]. Thus, CK (is) double CH.
καὶἐπεὶδιπλῆἐστινἡΒΑτῆςΑΔ,τετραπλάσιόνἐστιτὸ And LH plus HC is also double CH [Prop. 1.43]. Thus,
ἀπὸτῆςΒΑτοῦἀπὸτῆςΑΔ,τουτέστιτὸΑΕτοῦΔΘ. KC (is) equal to LH plus HC. And CE was also shown
ἴσονδὲτὸΑΕτῷΜΝΞγνώμονι·καὶὁΜΝΞἄραγνώμων (to be) equal to HF . Thus, the whole square AE is equal
τετραπλάσιόςἐστιτοῦΑΟ·ὅλονἄρατὸΔΖπενταπλάσιόν to the gnomon MNO. And since BA is double AD, the
ἐστιτοῦΑΟ.καίἐστιτὸμὲνΔΖτὸἀπὸτῆςΔΓ,τὸδὲΑΟ (square) on BA is four times the (square) on AD—that
τὸἀπὸτῆςΔΑ·τὸἄραἀπὸτῆςΓΔπενταπλάσιόνἐστιτοῦ is to say, AE (is four times) DH. And AE (is) equal to
ἀπὸτῆςΔΑ.
gnomon MNO. And, thus, gnomon MNO is also four
᾿Εὰνἄραεὐθεῖαἄκρονκαὶμέσονλόγοντμηθῇ,τὸμεῖζον times AP . Thus, the whole of DF is five times AP . And
τμῆμα προσλαβὸν τὴν ἡμίσειαν τῆς ὅλης πενταπλάσιον DF is the (square) on DC, and AP the (square) on DA.
δύναταιτοῦἀπὸτῆςἡμισείαςτετραγώνου·ὅπερἔδειδεῖξαι. Thus, the (square) on CD is five times the (square) on
DA.
Thus, if a straight-line is cut in extreme and mean ratio
then the square on the greater piece, added to half of
506

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
the whole, is five times the square on the half. (Which is)
the very thing it was required to show.
. Proposition 2
᾿Εὰν εὐθεῖα γραμμὴ τμήματος ἑαυτῆς πενταπλάσιον If the square on a straight-line is five times the
δύνηται,τῆςδιπλασίαςτοῦεἰρημένουτμήματοςἄκρονκαὶ
Ä
(square) on a piece of it, and double the aforementioned
μέσονλόγοντεμνομένηςτὸμεῖζοντμῆματὸλοιπὸνμέρος piece is cut in extreme and mean ratio, then the greater
ἐστὶτῆςἐξἀρχῆςεὐθείας.
piece is the remaining part of the original straight-line.
Æ
L
F
ÅÂ
N
M H
O
B
A
C
D
à ΕὐθεῖαγὰργραμμὴἡΑΒτμήματοςἑαυτῆςτοῦΑΓπεν- For let the square on the straight-line AB be five times
ταπλάσιονδυνάσθω,τῆςδὲΑΓδιπλῆἔστωἡΓΔ.λέγω, the (square) on the piece of it, AC. And let CD be double
ὅτιτῆςΓΔἄκρονκαὶμέσονλόγοντεμνομένοςτὸμεῖζον AC. I say that if CD is cut in extreme and mean ratio
τμῆμάἐστινἡΓΒ.
then the greater piece is CB.
Ἀναγεγράφθωγὰρἀφ᾿ἑκατέραςτῶνΑΒ,ΓΔτετράγωνα For let the squares AF and CG have been described
τὰΑΖ,ΓΗ,καὶκαταγεγράφθωἐντῷΑΖτὸσχῆμα,καὶ on each of AB and CD (respectively). And let the figure
διήχθωἡΒΕ.καὶἐπεὶπενταπλάσιόνἐστιτὸἀπότῆςΒΑ in AF have been drawn. And let BE have been drawn
τοῦἀπὸτῆςΑΓ,πενταπλάσιόνἐστιτὸΑΖτοῦΑΘ.τε- across. And since the (square) on BA is five times the
τραπλάσιοςἄραὁΜΝΞγνώμωντοῦΑΘ.καὶἐπεὶδιπλῆ (square) on AC, AF is five times AH. Thus, gnomon
ἐστινἡΔΓτῆςΓΑ,τετραπλάσιονἄραἐστὶτὸἀπὸΔΓτοῦ MNO (is) four times AH. And since DC is double CA,
ἀπὸΓΑ,τουτέστιτὸΓΗτοῦΑΘ.ἐδείχθηδὲκαὶὁΜΝΞ the (square) on DC is thus four times the (square) on
γνώμωντετραπλάσιοςτοῦΑΘ·ἴσοςἄραὁΜΝΞγνώμων CA—that is to say, CG (is four times) AH. And the
τῷΓΗ.καὶἐπεὶδιπλῆἐστινἡΔΓτῆςΓΑ,ἴσηδὲἡμὲν gnomon MNO was also shown (to be) four times AH.
ΔΓτῇΓΚ,ἡδὲΑΓτῇΓΘ,[διπλῆἄρακαὶἡΚΓτῆςΓΘ], Thus, gnomon MNO (is) equal to CG. And since DC is
διπλάσιονἄρακαὶτὸΚΒτοῦΒΘ.εἰσὶδὲκαὶτὰΛΘ,ΘΒ double CA, and DC (is) equal to CK, and AC to CH,
τοῦΘΒδιπλάσια·ἴσονἄρατὸΚΒτοῖςΛΘ,ΘΒ.ἐδείχθη [KC (is) thus also double CH], (and) KB (is) also douδὲκαὶὅλοςὁΜΝΞγνώμωνὅλῳτῷΓΗἴσος·καὶλοιπὸν
ble BH [Prop. 6.1]. And LH plus HB is also double HB
ἄρατὸΘΖτῷΒΗἐστινἴσον. καίἐστιτὸμὲνΒΗτὸὑπὸ [Prop. 1.43]. Thus, KB (is) equal to LH plus HB. And
τῶνΓΔΒ·ἴσηγὰρἡΓΔτῇΔΗ·τὸδὲΘΖτὸἀπὸτῆςΓΒ· the whole gnomon MNO was also shown (to be) equal
τὸἄραὑπὸτῶνΓΔΒἴσονἐστὶτῷἀπὸτῆςΓΒ.ἔστινἄρα to the whole of CG. Thus, the remainder HF is also
ὡςἡΔΓπρὸςτὴνΓΒ,οὕτωςἡΓΒπρὸςτὴνΒΔ.μείζων equal to (the remainder) BG. And BG is the (rectangle
δὲἡΔΓτῆςΓΒ·μείζωνἄρακαὶἡΓΒτῆςΒΔ.τῆςΓΔ contained) by CDB. For CD (is) equal to DG. And HF
ἄραεὐθείαςἄκρονκαὶμέσονλόγοντεμνομένηςτὸμεῖζον (is) the square on CB. Thus, the (rectangle contained)
τμῆμάἐστινἡΓΒ.
by CDB is equal to the (square) on CB. Thus, as DC
᾿Εὰνἄραεὐθεῖαγραμμὴτμήματοςἑαυτῆςπενταπλάσιον is to CB, so CB (is) to BD [Prop. 6.17]. And DC (is)
δύνηται,τῆςδιπλασίαςτοῦεἰρημένουτμήματοςἄκρονκαὶ greater than CB (see lemma). Thus, CB (is) also greater
μέσονλόγοντεμνομένηςτὸμεῖζοντμῆματὸλοιπὸνμέρος than BD [Prop. 5.14]. Thus, if the straight-line CD is cut
K
E
G
507

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
ἐστὶτῆςἐξἀρχῆςεὐθείας·ὅπερἔδειδεῖξαι.
in extreme and mean ratio then the greater piece is CB.
Thus, if the square on a straight-line is five times
the (square) on a piece of itself, and double the aforementioned
piece is cut in extreme and mean ratio, then
the greater piece is the remaining part of the original
straight-line. (Which is) the very thing it was required
to show.
ÄÑÑ.
῞Οτιδὲ ἡδιπλῆτῆςΑΓ μείζων ἐστὶ τῆςΒΓ, οὕτως And it can be shown that double AC (i.e., DC) is
Lemma
δεικτέον.
greater than BC, as follows.
Εἰγὰρμή,ἔστω,εἰδυνατόν,ἡΒΓδιπλῆτῆςΓΑ.τε- For if (double AC is) not (greater than BC), if possiτραπλάσιονἄρατὸἀπὸτῆςΒΓτοῦἀπὸτῆςΓΑ·πενταπλάσια
ble, let BC be double CA. Thus, the (square) on BC (is)
ἄρατὰἀπὸτῶνΒΓ,ΓΑτοῦἀπὸτῆςΓΑ.ὑπόκειταιδὲκαὶ four times the (square) on CA. Thus, the (sum of) the
τὸἀπὸτῆςΒΑπενταπλάσιοντοῦἀπὸτῆςΓΑ·τὸἄραἀπὸ (squares) on BC and CA (is) five times the (square) on
τῆςΒΑἴσονἐστὶτοῖςἀπὸτῶνΒΓ,ΓΑ·ὅπερἀδύνατον. CA. And the (square) on BA was assumed (to be) five
οὐκἄραἡΓΒδιπλασίαἐστὶτῆςΑΓ.ὁμοίωςδὴδείξομεν, times the (square) on CA. Thus, the (square) on BA is
ὅτιοὐδὲἡἐλάττωντῆςΓΒδιπλασίωνἐστὶτῆςΓΑ·πολλῷ equal to the (sum of) the (squares) on BC and CA. The
γὰρ[μεῖζον]τὸἄτοπον.
very thing (is) impossible [Prop. 2.4]. Thus, CB is not
῾ΗἄρατῆςΑΓδιπλῆμείζωνἐστὶτῆςΓΒ·ὅπερἔδει double AC. So, similarly, we can show that a (straightδεῖξαι.
line) less than CB is not double AC either. For (in this
case) the absurdity is much [greater].
Thus, double AC is greater than CB. (Which is) the
very thing it was required to show.
. Proposition 3
᾿Εὰν εὐθεῖα γραμμὴ ἄκρον καὶ μέσον λόγον τμηθῇ, If a straight-line is cut in extreme and mean ratio then
τὸἔλασσοντμῆμαπροσλαβὸντὴνἡμίσειαντοῦμείζονος
the square on the lesser piece added to half of the greater
τμήματοςπενταπλάσιονδύναταιτοῦἀπὸτῆςἡμισείαςτοῦ piece is five times the square on half of the greater piece.
μείζονοςτμήματοςτετραγώνου.
Ê Ç
A D C B
À
P
G
R
M
Â
O
Ã
Q
H
N
K F
Ä Ë
L S E
ΕὐθεῖαγάρτιςἡΑΒἄκρονκαὶμέσονλόγοντετμήσθω For let some straight-line AB have been cut in exκατὰ
τὸ Γ σημεῖον, καὶ ἔστω μεῖζον τμῆμα τὸ ΑΓ, καὶ treme and mean ratio at point C. And let AC be the
τετμήσθωἡΑΓδίχακατὰτὸΔ·λέγω,ὅτιπενταπλάσιόν greater piece. And let AC have been cut in half at D. I
È
Å
Æ
ἐστιτὸἀπὸτῆςΒΔτοῦἀπὸτῆςΔΓ.
say that the (square) on BD is five times the (square) on
ἈναγεγράφθωγὰρἀπὸτῆςΑΒτετράγωνοντὸΑΕ,καὶ DC.
508

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
καταγεγράφθωδιπλοῦντὸσχῆμα. ἐπεὶδιπλῆἐστινἡΑΓ For let the square AE have been described on AB.
τῆςΔΓ,τετραπλάσιονἄρατὸἀπὸτῆςΑΓτοῦἀπὸτῆςΔΓ, And let the figure have been drawn double. Since AC is
τουτέστιτὸΡΣτοῦΖΗ.καὶἐπεὶτὸὑπὸτῶνΑΒΓἴσον double DC, the (square) on AC (is) thus four times the
ἐστὶτῷἀπὸτῆςΑΓ,καίἐστιτὸὑπὸτῶνΑΒΓτὸΓΕ,τὸ (square) on DC—that is to say, RS (is four times) FG.
ἄραΓΕἴσονἐστὶτῷΡΣ.τετραπλάσιονδὲτὸΡΣτοῦΖΗ· And since the (rectangle contained) by ABC is equal to
τετραπλάσιονἄρακαὶτὸΓΕτοῦΖΗ.πάλινἐπεὶἴσηἐστὶν the (square) on AC [Def. 6.3, Prop. 6.17], and CE is the
ἡΑΔτῇΔΓ,ἴσηἐστὶκαὶἡΘΚτῇΚΖ.ὥστεκαὶτὸΗΖ (rectangle contained) by ABC, CE is thus equal to RS.
τετράγωνονἴσονἐστὶτῷΘΛτετραγώνῳ. ἴσηἄραἡΗΚ And RS (is) four times FG. Thus, CE (is) also four times
τῇΚΛ,τουτέστινἡΜΝτῇΝΕ·ὥστεκαὶτὸΜΖτῷΖΕ FG. Again, since AD is equal to DC, HK is also equal to
ἐστινἴσον.ἀλλὰτὸΜΖτῷΓΗἐστινἴσον·καὶτὸΓΗἄρα KF . Hence, square GF is also equal to square HL. Thus,
τῷΖΕἐστινἴσον.κοινὸνπροσκείσθωτὸΓΝ·ὁἄραΞΟΠ GK (is) equal to KL—that is to say, MN to NE. Hence,
γνώμωνἴσοςἐστὶτῷΓΕ.ἀλλὰτὸΓΕτετραπλάσιονἐδείχθῃ MF is also equal to FE. But, MF is equal to CG. Thus,
τοῦΗΖ·καὶὁΞΟΠἄραγνώμωντετραπλάσιόςἐστιτοῦΖΗ CG is also equal to FE. Let CN have been added to
τετραγώνου. ὁΞΟΠἄραγνώμωνκαὶτὸΖΗτετράγωνον both. Thus, gnomon OPQ is equal to CE. But, CE was
πενταπλάσιόςἐστιτοῦΖΗ.ἀλλὰὁΞΟΠγνώμωνκαὶτὸ shown (to be) equal to four times GF . Thus, gnomon
ΖΗτετράγωνόνἐστιτὸΔΝ.καίἐστιτὸμὲνΔΝτὸἀπὸ OPQ is also four times square FG. Thus, gnomon OPQ
τῆςΔΒ,τὸδὲΗΖτὸἀπὸτῆςΔΓ.τὸἄραἀπὸτῆςΔΒ plus square FG is five times FG. But, gnomon OPQ plus
πενταπλάσιόνἐστιτοῦἀπὸτῆςΔΓ·ὅπερἔδειδεῖξαι. square FG is (square) DN. And DN is the (square) on
DB, and GF the (square) on DC. Thus, the (square) on
DB is five times the (square) on DC. (Which is) the very
thing it was required to show.
. Proposition 4
᾿Εὰνεὐθεῖαγραμμὴἄκρονκαὶμέσονλόγοντμηθῇ,τὸ
If a straight-line is cut in extreme and mean ratio then
ἀπὸτῆςὅληςκαὶτοῦἐλάσσονοςτμήματος,τὰσυναμφότερα the sum of the squares on the whole and the lesser piece
τετράγωνα,τριπλάσιάἐστιτοῦἀπὸτοῦμείζονοςτμήματος is three times the square on the greater piece.
τετραγώνου.
ÄÆ Å Ã
A
C B
M
H
L F
K
Â
N
À
῎Εστω εὐθεῖαἡΑΒ, καὶτετμήσθω ἄκρον καὶμέσον Let AB be a straight-line, and let it have been cut in
λόγονκατὰτὸΓ,καὶἔστωμεῖζοντμῆματὸΑΓ·λέγω, extreme and mean ratio at C, and let AC be the greater
ὅτιτὰἀπὸτῶνΑΒ,ΒΓτριπλάσιάἐστιτοῦἀπὸτῆςΓΑ. piece. I say that the (sum of the squares) on AB and BC
ἈναγεγράφθωγὰρἀπὸτῆςΑΒτετράγωνοντὸΑΔΕΒ, is three times the (square) on CA.
καὶκαταγεγράφθωτὸσχῆμα. ἐπεὶοὖνἡΑΒἄκρονκαὶ For let the square ADEB have been described on AB,
μέσονλόγοντέτμηταικατὰτὸΓ,καὶτὸμεῖζοντμῆμάἐστιν and let the (remainder of the) figure have been drawn.
ἡΑΓ,τὸἄραὑπὸτῶνΑΒΓἴσονἐστὶτῷἀπὸτῆςΑΓ.καί Therefore, since AB has been cut in extreme and mean
ἐστιτὸμὲνὑπὸτῶνΑΒΓτὸΑΚ,τὸδὲἀπὸτῆςΑΓτὸΘΗ· ratio at C, and AC is the greater piece, the (rectangle
D
G
E
509

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
ἴσονἄραἐστὶτὸΑΚτῷΘΗ.καὶἐπεὶἴσονἐστὶτὸΑΖτῷ contained) by ABC is thus equal to the (square) on AC
ΖΕ,κοινὸνπροσκείσθωτὸΓΚ·ὅλονἄρατὸΑΚὅλῳτῷ [Def. 6.3, Prop. 6.17]. And AK is the (rectangle con-
ΓΕἐστινἴσον·τὰἄραΑΚ,ΓΕτοῦΑΚἐστιδιπλάσια.ἀλλὰ tained) by ABC, and HG the (square) on AC. Thus,
τὰΑΚ,ΓΕὁΛΜΝγνώμωνἐστὶκαὶτὸΓΚτετράγωνον·ὁ AK is equal to HG. And since AF is equal to FE
ἄραΛΜΝγνώμωνκαὶτὸΓΚτετράγωνονδιπλάσιάἐστιτοῦ [Prop. 1.43], let CK have been added to both. Thus,
ΑΚ.ἀλλὰμὴνκαὶτὸΑΚτῷΘΗἐδείχθηἴσον·ὁἄραΛΜΝ the whole of AK is equal to the whole of CE. Thus, AK
γνώμων καὶ[τὸ ΓΚτετράγωνον διπλάσιάἐστιτοῦΘΗ· plus CE is double AK. But, AK plus CE is the gnomon
ὥστεὁΛΜΝγνώμωνκαὶ]τὰΓΚ,ΘΗτετράγωνατριπλάσιά LMN plus the square CK. Thus, gnomon LMN plus
ἐστιτοῦΘΗτετραγώνου.καίἐστινὁ[μὲν]ΛΜΝγνώμων square CK is double AK. But, indeed, AK was also
καὶτὰΓΚ,ΘΗτετράγωναὅλοντὸΑΕκαὶτὸΓΚ,ἅπερἐστὶ shown (to be) equal to HG. Thus, gnomon LMN plus
τὰἀπὸτῶνΑΒ,ΒΓτετράγωνα,τὸδὲΗΘτὸἀπὸτῆςΑΓ [square CK is double HG. Hence, gnomon LMN plus]
τετράγωνον.τὰἄραἀπὸτῶνΑΒ,ΒΓτετράγωνατριπλάσιά the squares CK and HG is three times the square HG.
ἐστιτοῦἀπὸτῆςΑΓτετραγώνου·ὅπερἔδειδεῖξαι. And gnomon LMN plus the squares CK and HG is the
whole of AE plus CK—which are the squares on AB
and BC (respectively)—and GH (is) the square on AC.
Thus, the (sum of the) squares on AB and BC is three
times the square on AC. (Which is) the very thing it was
required to show.
. Proposition 5
᾿Εὰνεὐθεῖαγραμμὴἄκρονκαὶμέσονλόγοντμηθῇ,καὶ If a straight-line is cut in extreme and mean ratio, and
προστεθῇαὐτῇἴσητῷμείζονιτμήματι,ἡὅληεὐθεῖαἄκρον a (straight-line) equal to the greater piece is added to it,
καὶμέσονλόγοντέτμηται,καὶτὸμεῖζοντμῆμάἐστινἡἐξ then the whole straight-line has been cut in extreme and
ἀρχῆςεὐθεῖα.
mean ratio, and the original straight-line is the greater
piece.
D A C B
Ä
Â
Ã
Εὐθεῖα γὰρ γραμμὴ ἡ ΑΒ ἄκρον καὶ μέσον λόγον For let the straight-line AB have been cut in extreme
τετμήσθω κατὰτὸΓσημεῖον, καὶἔστω μεῖζοντμῆμαἡ and mean ratio at point C. And let AC be the greater
ΑΓ,καὶτῇΑΓἴση[κείσθω]ἡΑΔ.λέγω,ὅτιἡΔΒεὐθεῖα piece. And let AD be [made] equal to AC. I say that the
ἄκρονκαὶμέσονλόγοντέτμηταικατὰτὸΑ,καὶτὸμεῖζον straight-line DB has been cut in extreme and mean ratio
τμῆμάἐστινἡἐξἀρχῆςεὐθεῖαἡΑΒ.
at A, and that the original straight-line AB is the greater
ἈναγεγράφθωγὰρἀπὸτῆςΑΒτετράγωνοντὸΑΕ,καὶ piece.
καταγεγράφθωτὸσχῆμα.ἐπεὶἡΑΒἄκρονκαὶμέσονλόγον For let the square AE have been described on AB,
τέτμηταικατὰτὸΓ,τὸἄραὑπὸΑΒΓἴσονἐστὶτῷἀπὸΑΓ. and let the (remainder of the) figure have been drawn.
καίἐστιτὸμὲνὑπὸΑΒΓτὸΓΕ,τὸδὲἀπὸτῆςΑΓτὸΓΘ· And since AB has been cut in extreme and mean ratio at
ἴσονἄρατὸΓΕτῷΘΓ.ἀλλὰτῷμὲνΓΕἴσονἐστὶτὸΘΕ, C, the (rectangle contained) by ABC is thus equal to the
τῷδὲΘΓἴσοντὸΔΘ·καὶτὸΔΘἄραἴσονἐστὶτῷΘΕ (square) on AC [Def. 6.3, Prop. 6.17]. And CE is the
[κοινὸνπροσκείσθωτὸΘΒ].ὅλονἄρατὸΔΚὅλῳτῷΑΕ (rectangle contained) by ABC, and CH the (square) on
ἐστινἴσον. καίἐστιτὸμὲνΔΚτὸὑπὸτῶνΒΔ,ΔΑ·ἴση AC. But, HE is equal to CE [Prop. 1.43], and DH equal
L
H
K
E
510

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
γὰρἡΑΔτῇΔΛ·τὸδὲΑΕτὸἀπὸτῆςΑΒ·τὸἄραὑπὸ to HC. Thus, DH is also equal to HE. [Let HB have
τῶνΒΔΑἴσονἐστὶτῷἀπὸτῆςΑΒ.ἔστινἄραὡςἡΔΒ been added to both.] Thus, the whole of DK is equal to
πρὸςτὴνΒΑ,οὕτωςἡΒΑπρὸςτὴνΑΔ.μείζωνδὲἡΔΒ the whole of AE. And DK is the (rectangle contained)
τῆςΒΑ·μείζωνἄρακαὶἡΒΑτῆςΑΔ.
by BD and DA. For AD (is) equal to DL. And AE (is)
῾ΗἄραΔΒἄκρονκαὶμέσονλόγοντέτμηταικατὰτὸΑ, the (square) on AB. Thus, the (rectangle contained) by
καὶτὸμεῖζοντμῆμάἐστινἡΑΒ·ὅπερἔδειδεῖξαι. BDA is equal to the (square) on AB. Thus, as DB (is)
to BA, so BA (is) to AD [Prop. 6.17]. And DB (is)
greater than BA. Thus, BA (is) also greater than AD
[Prop. 5.14].
Thus, DB has been cut in extreme and mean ratio at
A, and the greater piece is AB. (Which is) the very thing
it was required to show.
¦. Proposition 6
᾿Εὰν εὐθεῖα ῥητη ἄκρον καὶ μέσον λόγον τμηθῇ, If a rational straight-line is cut in extreme and mean
ἑκάτεροντῶντμημάτωνἄλογόςἐστινἡκαλουμένηἀπο- ratio then each of the pieces is that irrational (straightτομή.
line) called an apotome.
C
D A B
῎ΕστωεὐθεῖαῥητὴἡΑΒκαὶτετμήσθωἄκρονκαὶμέσον Let AB be a rational straight-line cut in extreme and
λόγονκατὰτὸΓ,καὶἔστωμεῖζοντμῆμαἡΑΓ·λέγω,ὅτι mean ratio at C, and let AC be the greater piece. I say
ἑκατέρατῶνΑΓ,ΓΒἄλογόςἐστινἡκαλουμένηἀποτομή. that AC and CB is each that irrational (straight-line)
᾿ΕκβεβλήσθωγὰρἡΒΑ,καὶκείσθωτῆςΒΑἡμίσεια called an apotome.
ἡΑΔ.ἐπεὶοὖνεὐθεῖαἡΑΒτέτμηταιἄκρονκαὶμέσον For let BA have been produced, and let AD be made
λόγονκατὰτὸΓ,καὶτῷμείζονιτμήματιτῷΑΓπρόσκειται (equal) to half of BA. Therefore, since the straight-
ἡΑΔἡμίσειαοὖσατῆςΑΒ,τὸἄραἀπὸΓΔτοῦἀπὸΔΑ line AB has been cut in extreme and mean ratio at C,
πενταπλάσιόνἐστιν.τὸἄραἀπὸΓΔπρὸςτὸἀπὸΔΑλόγον and AD, which is half of AB, has been added to the
ἔχει,ὃνἀριθμὸςπρὸςἀριθμόν·σύμμετρονἄρατὸἀπὸΓΔ greater piece AC, the (square) on CD is thus five times
τῷἀπὸΔΑ.ῥητὸνδὲτὸἀπὸΔΑ·ῥητὴγάρ[ἐστιν]ἡΔΑ the (square) on DA [Prop. 13.1]. Thus, the (square) on
ἡμίσειαοὖσατῆςΑΒῥητῆςοὔσης·ῥητὸνἄρακαὶτὸἀπὸ CD has to the (square) on DA the ratio which a number
ΓΔ·ῥητὴἄραἐστὶκαὶἡΓΔ.καὶἐπεὶτὸἀπὸΓΔπρὸς (has) to a number. The (square) on CD (is) thus comτὸἀπὸΔΑλόγονοὐκἔχει,ὃντετράγωνοςἀριθμὸςπρὸς
mensurable with the (square) on DA [Prop. 10.6]. And
τετράγωνονἀριθμόν,ἀσύμμετροςἄραμήκειἡΓΔτῇΔΑ·αἱ the (square) on DA (is) rational. For DA [is] rational,
ΓΔ,ΔΑἄραῥηταίεἰσιδυνάμειμόνονσύμμετροι·ἀποτομὴ being half of AB, which is rational. Thus, the (square)
ἄραἐστὶνἡΑΓ.πάλιν,ἐπεὶἡΑΒἄκρονκαὶμέσονλόγον on CD (is) also rational [Def. 10.4]. Thus, CD is also
τέτμηται,καὶτὸμεῖζοντμῆμάἐστινἡΑΓ,τὸἄραὑπὸΑΒ, rational. And since the (square) on CD does not have
ΒΓτῷἀπὸΑΓἴσονἐστίν. τὸἄραἀπὸτῆςΑΓἀποτομῆς to the (square) on DA the ratio which a square numπαρὰτὴνΑΒῥητὴνπαραβληθὲνπλάτοςποιεῖτὴνΒΓ.τὸ
ber (has) to a square number, CD (is) thus incommensuδὲἀπὸἀποτομῆςπαρὰῥητὴνπαραβαλλόμενονπλάτοςποιεῖ
rable in length with DA [Prop. 10.9]. Thus, CD and DA
ἀποτομὴνπρώτην·ἀποτομὴἄραπρώτηἐστὶνἡΓΒ.ἐδείχθη are rational (straight-lines which are) commensurable in
δὲκαὶἡΓΑἀποτομή. square only. Thus, AC is an apotome [Prop. 10.73].
᾿Εὰνἄραεὐθεῖαῥητὴἄκρονκαὶμέσονλόγοντμηθῇ, Again, since AB has been cut in extreme and mean ratio,
ἑκάτεροντῶντμημάτωνἄλογόςἐστινἡκαλουμένηἀπο- and AC is the greater piece, the (rectangle contained) by
τομή·ὅπερἔδειδεῖξαι. AB and BC is thus equal to the (square) on AC [Def. 6.3,
Prop. 6.17]. Thus, the (square) on the apotome AC, applied
to the rational (straight-line) AB, makes BC as
width. And the (square) on an apotome, applied to a
rational (straight-line), makes a first apotome as width
[Prop. 10.97]. Thus, CB is a first apotome. And CA was
also shown (to be) an apotome.
511

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
Thus, if a rational straight-line is cut in extreme and
mean ratio then each of the pieces is that irrational
(straight-line) called an apotome.
Þ. Proposition 7
᾿Εὰν πενταγώνου ἰσοπλεύρουαἱ
τρεῖς γωνίαι ἤτοι αἱ If three angles, either consecutive or not consecutive,
κατὰτὸἑξῆςἢαἱμὴκατὰτὸἑξῆςἴσαιὦσιν,ἰσογώνιον of an equilateral pentagon are equal then the pentagon
ἔσταιτὸπεντάγωνον.
will be equiangular.
A
F
B
E
Πενταγώνου γὰρ ἰσοπλεύρον τοῦ ΑΒΓΔΕ αἱ τρεῖς For let three angles of the equilateral pentagon
γωνίαιπρότεροναἱκατὰτὸἑξῆςαἱπρὸςτοῖςΑ,Β,Γἴσαι ABCDE—first of all, the consecutive (angles) at A, B,
ἀλλήλαιςἔστωσαν·λέγω,ὅτιἰσογώνιόνἐστιτὸΑΒΓΔΕ and C—-be equal to one another. I say that pentagon
πεντάγωνον.
ABCDE is equiangular.
᾿ΕπεζεύχθωσανγὰραἱΑΓ,ΒΕ,ΖΔ.καὶἐπεὶδύοαἱ For let AC, BE, and FD have been joined. And since
ΓΒ,ΒΑδυσὶταῖςΒΑ,ΑΕἴσαιἐισὶνἑκατέραἑκατέρᾳ,καὶ the two (straight-lines) CB and BA are equal to the two
γωνίαἡὑπὸΓΒΑγωνίᾳτῇὑπὸΒΑΕἐστινἴση,βάσιςἄραἡ (straight-lines) BA and AE, respectively, and angle CBA
ΑΓβάσειτῇΒΕἐστινἴση,καὶτὸΑΒΓτρίγωνοντῷΑΒΕ is equal to angle BAE, base AC is thus equal to base
τριγώνῳἴσον,καὶαἱλοιπαὶγωνίαιταῖςλοιπαῖςγωνίαιςἴσαι BE, and triangle ABC equal to triangle ABE, and the
ἔσονται,ὑφ᾿ἃςαἱἴσαιπλευραὶὑποτείνουσιν,ἡμὲνὑπὸΒΓΑ remaining angles will be equal to the remaining angles
τῇὑπὸΒΕΑ,ἡδὲὑπὸΑΒΕτῇὑπὸΓΑΒ·ὥστεκαὶπλευρὰ which the equal sides subtend [Prop. 1.4], (that is), BCA
ἡΑΖπλευρᾷτῇΒΖἐστινἴση. ἐδείχθηδὲκαὶὅληἡΑΓ (equal) to BEA, and ABE to CAB. And hence side AF
ὅλῃτῇΒΕἴση·καὶλοιπὴἄραἡΖΓλοιπῇτῇΖΕἐστινἴση. is also equal to side BF [Prop. 1.6]. And the whole of AC
ἔστιδὲκαὶἡΓΔτῇΔΕἴση. δύοδὴαἱΖΓ,ΓΔδυσὶταῖς was also shown (to be) equal to the whole of BE. Thus,
ΖΕ,ΕΔἴσαιεἰσίν·καὶβάσιςαὐτῶνκοινὴἡΖΔ·γωνίαἄρα the remainder FC is also equal to the remainder FE.
ἡὑπὸΖΓΔγωνίᾳτῇὑπὸΖΕΔἐστινἴση. ἐδείχθηδὲκαὶ And CD is also equal to DE. So, the two (straight-lines)
ἡὑπὸΒΓΑτῇὑπὸΑΕΒἴση·καὶὅληἄραἡὑπὸΒΓΔὅλῃ FC and CD are equal to the two (straight-lines) FE and
τῇὑπὸΑΕΔἴση.ἀλλ᾿ἡὑπὸΒΓΔἴσηὑπόκειταιταῖςπρὸς ED (respectively). And FD is their common base. Thus,
τοῖςΑ,Βγωνίαις·καὶἡὑπὸΑΕΔἄραταῖςπρὸςτοῖςΑ,Β angle FCD is equal to angle FED [Prop. 1.8]. And BCA
γωνίαιςἴσηἐστίν. ὁμοίωςδὴδείξομεν,ὅτικαὶἡὑπὸΓΔΕ was also shown (to be) equal to AEB. And thus the
γωνίαἴσηἐστὶταῖςπρὸςτοῖςΑ,Β,Γγωνίαις·ἰσογώνιον whole of BCD (is) equal to the whole of AED. But,
ἄραἐστὶτὸΑΒΓΔΕπεντάγωνον.
(angle) BCD was assumed (to be) equal to the angles at
Ἀλλὰδὴμὴἔστωσανἴσαιαἱκατὰτὸἑξῆςγωνίαι,ἀλλ᾿ A and B. Thus, (angle) AED is also equal to the angles
ἔστωσανἴσαιαἱπρὸςτοῖςΑ,Γ,Δσημείοις·λέγω,ὅτικαὶ at A and B. So, similarly, we can show that angle CDE
οὕτωςἰσογώνιόνἐστιτὸΑΒΓΔΕπεντάγωνον.
is also equal to the angles at A, B, C. Thus, pentagon
᾿ΕπεζεύχθωγὰρἡΒΔ.καὶἐπεὶδύοαἱΒΑ,ΑΕδυσὶ ABCDE is equiangular.
ταῖςΒΓ,ΓΔἴσαιεἰσὶκαὶγωνίαςἴσαςπεριέχουσιν,βάσις And so let consecutive angles not be equal, but let
ἄραἡΒΕβάσειτῇΒΔἴσηἐστίν,καὶτὸΑΒΕτρίγωνοντῷ the (angles) at points A, C, and D be equal. I say that
ΒΓΔτριγώνῳἴσονἐστίν,καὶαἱλοιπαὶγωνίαιταῖςλοιπαῖς pentagon ABCDE is also equiangular in this case.
γωνίαιςἴσαιἔσονται,ὑφ᾿ἃςαἱἴσαιπλευραὶὑποτείνουσιν· For let BD have been joined. And since the two
C
D
512

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
ἴσηἄραἐστὶνἡὑπὸΑΕΒγωνίατῇὑπὸΓΔΒ.ἔστιδὲκαὶ (straight-lines) BA and AE are equal to the (straight-
ἡὑπὸΒΕΔγωνίατῇὑπὸΒΔΕἴση,ἐπεὶκαὶπλευρὰἡΒΕ lines) BC and CD, and they contain equal angles, base
πλευρᾷτῇΒΔἐστινἴση. καὶὅληἄραἡὑπὸΑΕΔγωνία BE is thus equal to base BD, and triangle ABE is equal
ὅλῃτῇὑπὸΓΔΕἐστινἴση.ἀλλὰἡὑπὸΓΔΕταῖςπρὸςτοῖς to triangle BCD, and the remaining angles will be equal
Α,Γγωνίαιςὑπόκειταιἴση·καὶἡὑπὸΑΕΔἄραγωνίαταῖς to the remaining angles which the equal sides subtend
πρὸςτοῖςΑ,Γἴσηἐστίν.διὰτὰαὐτὰδὴκαὶἡὑπὸΑΒΓἴση [Prop. 1.4]. Thus, angle AEB is equal to (angle) CDB.
ἐστὶταῖςπρὸςτοῖςΑ,Γ,Δγωνίαις.ἰσογώνιονἄραἐστὶτὸ And angle BED is also equal to (angle) BDE, since side
ΑΒΓΔΕπεντάγωνον·ὅπερἔδειδεῖξαι.
BE is also equal to side BD [Prop. 1.5]. Thus, the whole
angle AED is also equal to the whole (angle) CDE. But,
(angle) CDE was assumed (to be) equal to the angles at
A and C. Thus, angle AED is also equal to the (angles)
at A and C. So, for the same (reasons), (angle) ABC is
also equal to the angles at A, C, and D. Thus, pentagon
ABCDE is equiangular. (Which is) the very thing it was
required to show.
. Proposition 8
᾿Εὰνπενταγώνουἰσοπλεύρουκαὶἰσογωνίουτὰςκατὰ If straight-lines subtend two consecutive angles of an
τὸἑξῆςδύογωνίαςὑποτείνωσινεὐθεῖαι,ἄκρονκαὶμέσον
equilateral and equiangular pentagon then they cut one
λόγοντέμνουσινἀλλήλας,καὶτὰμείζονααὐτῶντμήματα another in extreme and mean ratio, and their greater
ἴσαἐστὶτῇτοῦπενταγώνουπλευρᾷ.
pieces are equal to the sides of the pentagon.
A
Â
E
B
H
ΠενταγώνουγὰρἰσοπλεύρονκαὶἰσογωνίουτοῦΑΒΓΔΕ For let the two straight-lines, AC and BE, cutting one
δύο γωνίας τὰς κατὰτὸ ἑξῆςτὰς πρὸς τοῖςΑ, Β ὑπο- another at point H, have subtended two consecutive anτεινέτωσανεὐθεῖαιαἱΑΓ,ΒΕτέμνουσαιἀλλήλαςκατὰτὸ
gles, at A and B (respectively), of the equilateral and
Θσημεὶον·λέγω,ὅτιἑκατέρααὐτῶνἄκρονκαὶμέσονλόγον equiangular pentagon ABCDE. I say that each of them
τέτμηταικατὰτὸΘσημεῖον,καὶτὰμείζονααὐτῶντμήματα has been cut in extreme and mean ratio at point H, and
ἴσαἐστὶτῇτοῦπενταγώνουπλευρᾷ.
that their greater pieces are equal to the sides of the pen-
ΠεριγεγράφθωγὰρπερὶτὸΑΒΓΔΕπεντάγωνονκύκλος tagon.
ὁ ΑΒΓΔΕ. καὶ ἐπεὶ δύο εὐθεῖαι αἱ ΕΑ, ΑΒ δυσὶ ταῖς For let the circle ABCDE have been circumscribed
ΑΒ,ΒΓἴσαιεἰσὶκαὶγωνίαςἴσαςπεριέχουσιν,βάσιςἄρα about pentagon ABCDE [Prop. 4.14]. And since the two
ἡΒΕβάσειτῇΑΓἴσηἐστίν, καὶτὸΑΒΕτρίγωνοντῷ straight-lines EA and AB are equal to the two (straight-
ΑΒΓτριγώνῳἴσονἐστίν,καὶαἱλοιπαὶγωνίαιταῖςλοιπαῖς lines) AB and BC (respectively), and they contain equal
γωνίαιςἴσαιἔσονταιἑκατέραἑκατέρᾳ,ὑφ᾿ἃςαἱἴσαιπλευραὶ angles, the base BE is thus equal to the base AC, and tri-
ὑποτείνουσιν.ἴσηἄραἐστὶνἡὑπὸΒΑΓγωνίατῇὑπὸΑΒΕ· angle ABE is equal to triangle ABC, and the remaining
διπλῆἄραἡὑπὸΑΘΕτῆςὑπὸΒΑΘ.ἔστιδὲκαὶἡὑπὸΕΑΓ angles will be equal to the remaining angles, respectively,
τῆςὑπὸΒΑΓδιπλῆ,ἐπειδήπερκαὶπεριφέρειαἡΕΔΓπερι- which the equal sides subtend [Prop. 1.4]. Thus, angle
φερείαςτῆςΓΒἐστιδιπλῆ·ἴσηἄραἡὑπὸΘΑΕγωνίατῇ BAC is equal to (angle) ABE. Thus, (angle) AHE (is)
ὑπὸΑΘΕ·ὥστεκαὶἡΘΕεὐθεῖατῇΕΑ,τουτέστιτῇΑΒ double (angle) BAH [Prop. 1.32]. And EAC is also dou-
D
C
513

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
ἐστινἴση. καὶἐπεὶἴσηἐστὶνἡΒΑεὐθεῖατῇΑΕ,ἴσηἐστὶ ble BAC, inasmuch as circumference EDC is also douκαὶγωνίαἡὑπὸΑΒΕτῇὑπὸΑΕΒ.ἀλλὰἡὑπὸΑΒΕτῇ
ble circumference CB [Props. 3.28, 6.33]. Thus, angle
ὑπὸΒΑΘἐδείχθηἴση·καὶἡὑπὸΒΕΑἄρατῇὑπὸΒΑΘ HAE (is) equal to (angle) AHE. Hence, straight-line
ἐστινἴση. καὶκοινὴτῶνδύοτριγώνωντοῦτεΑΒΕκαὶ HE is also equal to (straight-line) EA—that is to say,
τοῦΑΒΘἐστινἡὑπὸΑΒΕ·λοιπὴἄραἡὑπὸΒΑΕγωνία to (straight-line) AB [Prop. 1.6]. And since straight-line
λοιπῇτῇὑπὸΑΘΒἐστινἴση·ἰσογώνιονἄραἐστὶτὸΑΒΕ BA is equal to AE, angle ABE is also equal to AEB
τρίγωνοντῷΑΒΘτριγώνῳ·ἀνάλογονἄραἐστὶνὡςἡΕΒ [Prop. 1.5]. But, ABE was shown (to be) equal to BAH.
πρὸςτὴνΒΑ,οὕτωςἡΑΒπρὸςτὴνΒΘ.ἴσηδὲἡΒΑτῇ Thus, BEA is also equal to BAH. And (angle) ABE is
ΕΘ·ὡςἄραἡΒΕπρὸςτὴνΕΘ,οὕτωςἡΕΘπρὸςτὴνΘΒ. common to the two triangles ABE and ABH. Thus, the
μείζωνδὲἡΒΕτῆςΕΘ·μείζωνἄρακαὶἡΕΘτῆςΘΒ.ἡ remaining angle BAE is equal to the remaining (angle)
ΒΕἅραἄκρονκαὶμέσονλόγοντέτμηταικατὰτὸΘ,καὶτὸ AHB [Prop. 1.32]. Thus, triangle ABE is equiangular to
μεῖζοντμῆματὸΘΕἴσονἐστὶτῇτοῦπενταγώνουπλευρᾷ. triangle ABH. Thus, proportionally, as EB is to BA, so
ὁμοίωςδὴδείξομεν,ὅτικαὶἡΑΓἄκρονκαὶμέσονλόγον AB (is) to BH [Prop. 6.4]. And BA (is) equal to EH.
τέτμηταικατὰτὸΘ,καὶτὸμεῖζοναὐτῆςτμῆμαἡΓΘἴσον Thus, as BE (is) to EH, so EH (is) to HB. And BE
ἐστὶτῇτοῦπενταγώνουπλευρᾷ·ὅπερἔδειδεῖξαι. (is) greater than EH. EH (is) thus also greater than
HB [Prop. 5.14]. Thus, BE has been cut in extreme and
mean ratio at H, and the greater piece HE is equal to
the side of the pentagon. So, similarly, we can show that
AC has also been cut in extreme and mean ratio at H,
and that its greater piece CH is equal to the side of the
pentagon. (Which is) the very thing it was required to
show.
. Proposition 9
᾿Εὰνἡτοῦἑξαγώνουπλευρὰκαὶἡτοῦδεκαγώνουτῶν If the side of a hexagon and of a decagon inscribed
εἰςτὸναὐτὸνκύκλονἐγγραφομένωνσυντεθῶσιν, ἡὅλη in the same circle are added together then the whole
εὐθεῖαἄκρονκαὶμέσονλόγοντέτμηται,καὶτὸμεῖξοναὐτῆς straight-line has been cut in extreme and mean ratio (at
τμῆμάἐστινἡτοῦἑξαγώνουπλευρά.
the junction point), and its greater piece is the side of the
hexagon. †
B
E
A
῎ΕστωκύκλοςὁΑΒΓ,καὶτῶνεἰςτὸνΑΒΓκύκλον Let ABC be a circle. And of the figures inscribed in
ἐγγραφομένωνσχημάτων,δεκαγώνουμὲνἔστωπλευρὰἡ circle ABC, let BC be the side of a decagon, and CD (the
ΒΓ,ἑξαγώνουδὲἡΓΔ,καὶἔστωσανἐπ᾿εὐθείας·λέγω,ὅτι side) of a hexagon. And let them be (laid down) straight-
ἡὅληεὐθεῖαἡΒΔἄκρονκαὶμέσονλόγοντέτμηται,καὶτὸ on (to one another). I say that the whole straight-line
μεῖζοναὐτῆςτμῆμάἐστινἡΓΔ.
BD has been cut in extreme and mean ratio (at C), and
ΕἰλήφθωγὰρτὸκέντροντοῦκύκλουτὸΕσημεῖον,καὶ that CD is its greater piece.
ἐπεζεύχθωσαναἱΕΒ,ΕΓ,ΕΔ,καὶδιήχθωἡΒΕἐπὶτὸ For let the center of the circle, point E, have been
C
D
F
514

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
Α.ἐπεὶδεκαγώνουἰσοπλεύρονπλευράἐστινἡΒΓ,πεντα- found [Prop. 3.1], and let EB, EC, and ED have been
πλασίωνἄραἡΑΓΒπεριφέρειατῆςΒΓπεριφερείας· τε- joined, and let BE have been drawn across to A. Since
τραπλασίωνἄραἡΑΓπεριφέρειατῆςΓΒ.ὡςδὲἡΑΓπε- BC is a side on an equilateral decagon, circumference
ριφέρειαπρὸςτὴνΓΒ,οὕτωςἡὑπὸΑΕΓγωνίαπρὸςτὴν ACB (is) thus five times circumference BC. Thus, cir-
ὑπὸΓΕΒ·τετραπλασίωνἄραἡὑπὸΑΕΓτῆςὑπὸΓΕΒ.καὶ cumference AC (is) four times CB. And as circumference
ἐπεὶἴσηἡὑπὸΕΒΓγωνίατῇὑπὸΕΓΒ,ἡἄραὑπὸΑΕΓ AC (is) to CB, so angle AEC (is) to CEB [Prop. 6.33].
γωνίαδιπλασίαἐστὶτῆςὑπὸΕΓΒ.καὶἐπεὶἴσηἐστὶνἡΕΓ Thus, (angle) AEC (is) four times CEB. And since angle
εὐθεῖατῇΓΔ·ἑκατέραγὰραὐτῶνἴσηἐστὶτῇτοῦἑξαγώνου EBC (is) equal to ECB [Prop. 1.5], angle AEC is thus
πλευρᾷτοῦεἰςτὸνΑΒΓκύκλον[ἐγγραφομένου]·ἴσηἐστὶ double ECB [Prop. 1.32]. And since straight-line EC is
καὶἡὑπὸΓΕΔγωνίατῇὑπὸΓΔΕγωνίᾳ· διπλασίαἄρα equal to CD—for each of them is equal to the side of the
ἡὑπὸΕΓΒγωνίατῆςὑπὸΕΔΓ.ἀλλὰτῆςὑπὸΕΓΒδι- hexagon [inscribed] in circle ABC [Prop. 4.15 corr.]—
πλασίαἐδείχθηἡὑπὸΑΕΓ·τετραπλασίαἄραἡὑπὸΑΕΓ angle CED is also equal to angle CDE [Prop. 1.5]. Thus,
τῆςὑπὸΕΔΓ.ἐδείχθηδὲκαὶτῆςὑπὸΒΕΓτετραπλασία angle ECB (is) double EDC [Prop. 1.32]. But, AEC
ἡὑπὸΑΕΓ·ἴσηἄραἡὑπὸΕΔΓτῇὑπὸΒΕΓ.κοινὴδὲ was shown (to be) double ECB. Thus, AEC (is) four
τῶνδύοτριγώνων,τοῦτεΒΕΓκαὶτοῦΒΕΔ,ἡὑπὸΕΒΔ times EDC. And AEC was also shown (to be) four times
γωνία·καὶλοιπὴἄραἡὑπὸΒΕΔτῇὑπὸΕΓΒἐστινἴση· BEC. Thus, EDC (is) equal to BEC. And angle EBD
ἰσογώνιονἄραἐστὶτὸΕΒΔτρίγωνοντῷΕΒΓτριγώνῳ. (is) common to the two triangles BEC and BED. Thus,
ἀνάλογονἄραἐστὶνὡςἡΔΒπρὸςτὴνΒΕ,οὕτωςἡΕΒ the remaining (angle) BED is equal to the (remaining
πρὸςτὴνΒΓ.ἴσηδὲἡΕΒτῇΓΔ.ἔστινἄραὡςἡΒΔπρὸς angle) ECB [Prop. 1.32]. Thus, triangle EBD is equianτὴνΔΓ,οὕτωςἡΔΓπρὸςτὴνΓΒ.μείζωνδὲἡΒΔτῆς
gular to triangle EBC. Thus, proportionally, as DB is to
ΔΓ·μείζωνἄρακαὶἡΔΓτῆςΓΒ.ἡΒΔἄραεὐθεῖαἄκρον BE, so EB (is) to BC [Prop. 6.4]. And EB (is) equal
καὶμέσονλόγοντέτμηται[κατὰτὸΓ],καὶτὸμεῖζοντμῆμα to CD. Thus, as BD is to DC, so DC (is) to CB. And
αὐτῆςἐστινἡΔΓ·ὅπερἔδειδεῖξαι.
BD (is) greater than DC. Thus, DC (is) also greater
than CB [Prop. 5.14]. Thus, the straight-line BD has
been cut in extreme and mean ratio [at C], and DC is its
greater piece. (Which is), the very thing it was required
to show.
† If the circle is of unit radius then the side of the hexagon is 1, whereas the side of the decagon is (1/2) ( √ 5 − 1).
. Proposition 10
᾿Εὰνεἰςκύκλονπεντάγωνονἰσόπλευρονἐγγραφῇ,ἡτοῦ If an equilateral pentagon is inscribed in a circle then
πενταγώνουπλευρὰδύναταιτήντετοῦἑξαγώνουκαὶτὴν the square on the side of the pentagon is (equal to) the
τοῦδεκαγώνουτῶνεἰςτὸναὐτὸνκύκλονἐγγραφομένων. (sum of the squares) on the (sides) of the hexagon and of
à Ä
the decagon inscribed in the same circle. †
M A
K
L
N
H
B
E
F
Å
 Æ
À
C
G
D
515

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
῎ΕστωκύκλοςὁΑΒΓΔΕ,καὶεἰςτὸΑΒΓΔΕκύκλον Let ABCDE be a circle. And let the equilateral penπεντάγωνον
ἰσόπλευρονἐγγεγράφθω τὸ ΑΒΓΔΕ. λέγω, tagon ABCDE have been inscribed in circle ABCDE. I
ὅτιἡτοῦΑΒΓΔΕπενταγώνουπλευρὰδύναταιτήντετοῦ say that the square on the side of pentagon ABCDE is
ἑξαγώνου καὶ τὴν τοῦ δεκαγώνου πλευρὰν τῶν εἰς τὸν the (sum of the squares) on the sides of the hexagon and
ΑΒΓΔΕκύκλονἐγγραφομένων.
of the decagon inscribed in circle ABCDE.
ΕἰλήφθωγὰρτὸκέντροντοῦκύκλουτὸΖσημεὶον,καὶ For let the center of the circle, point F , have been
ἐπιζευχθεῖσαἡΑΖδιήχθωἐπὶτὸΗσημεῖον,καὶἐπεζεύχθω found [Prop. 3.1]. And, AF being joined, let it have been
ἡΖΒ,καὶἀπὸτοῦΖἐπὶτὴνΑΒκάθετοςἤχθωἡΖΘ,καὶ drawn across to point G. And let FB have been joined.
διήχθωἐπὶτὸΚ,καὶἐπεζεύχθωσαναἱΑΚ,ΚΒ,καὶπάλιν And let FH have been drawn from F perpendicular to
ἀπὸτοῦΖἐπὶτὴνΑΚκάθετοςἤχθωἡΖΛ,καὶδιήχθωἐπὶ AB. And let it have been drawn across to K. And let AK
τὸΜ,καὶἐπεζεύχθωἡΚΝ.
and KB have been joined. And, again, let FL have been
᾿Επεὶ ἴση ἐστὶν ἡ ΑΒΓΗ περιφέρεια τῇ ΑΕΔΗ περι- drawn from F perpendicular to AK. And let it have been
φερείᾳ, ὧν ἡΑΒΓτῇΑΕΔ ἐστιν ἴση, λοιπὴἄραἡΓΗ drawn across to M. And let KN have been joined.
περιφέρειαλοιπῇτῇΗΔἐστινἴση. πενταγώνουδὲἡΓΔ· Since circumference ABCG is equal to circumference
δεκαγώνουἄραἡΓΗ.καὶἐπεὶἴσηἐστὶνἡΖΑτῇΖΒ,καὶ AEDG, of which ABC is equal to AED, the remainκάθετοςἡΖΘ,ἴσηἄρακαὶἡὑπὸΑΖΚγωνίατῇὑπὸΚΖΒ.
ing circumference CG is thus equal to the remaining
ὥστεκαὶπεριφέρειαἡΑΚτῇΚΒἐστινἴση·διπλῆἄραἡ (circumference) GD. And CD (is the side) of the pen-
ΑΒπεριφέρειατῆςΒΚπεριφερείας·δεκαγώνουἄραπλευρά tagon. CG (is) thus (the side) of the decagon. And since
ἐστινἡΑΚεὐθεῖα.διὰτὰαὐτὰδὴκαὶἡΑΚτῆςΚΜἐστι FA is equal to FB, and FH is perpendicular (to AB),
διπλῆ. καὶἐπεὶδιπλῆἐστινἡΑΒπεριφέρειατῆςΒΚπερι- angle AFK (is) thus also equal to KFB [Props. 1.5,
φερείας,ἴσηδὲἡΓΔπεριφέρειατῇΑΒπεριφερείᾳ,διπλῆ 1.26]. Hence, circumference AK is also equal to KB
ἄρακαὶἡΓΔπεριφέρειατῆςΒΚπεριφερείας.ἔστιδὲἡΓΔ [Prop. 3.26]. Thus, circumference AB (is) double cirπεριφέρειακαὶτῆςΓΗδιπλῆ·ἴσηἄραἡΓΗπεριφέρειατῇ
cumference BK. Thus, straight-line AK is the side of
ΒΚπεριφερείᾳ. ἀλλὰἡΒΚτῆςΚΜἐστιδιπλῆ,ἐπεὶκαὶ the decagon. So, for the same (reasons, circumference)
ἡΚΑ·καὶἡΓΗἄρατῆςΚΜἐστιδιπλῆ. ἀλλὰμὴνκαὶἡ AK is also double KM. And since circumference AB
ΓΒπεριφέρειατῆςΒΚπεριφερείαςἐστὶδιπλῆ·ἴσηγὰρἡ is double circumference BK, and circumference CD (is)
ΓΒπεριφέρειατῇΒΑ.καὶὅληἄραἡΗΒπεριφέρειατῆς equal to circumference AB, circumference CD (is) thus
ΒΜἐστιδιπλῆ·ὥστεκαὶγωνίαἡὑπὸΗΖΒγωνίαςτῆςὑπὸ also double circumference BK. And circumference CD
ΒΖΜ[ἐστι]διπλῆ. ἔστιδὲἡὑπὸΗΖΒκαὶτῆςὑπὸΖΑΒ is also double CG. Thus, circumference CG (is) equal
διπλῆ·ἴσηγὰρἡὑπὸΖΑΒτῇὑπὸΑΒΖ.καὶἡὑπὸΒΖΝἄρα to circumference BK. But, BK is double KM, since
τῇὑπὸΖΑΒἐστινἴση. κοινὴδὲτῶνδύοτριγώνων,τοῦ KA (is) also (double KM). Thus, (circumference) CG
τεΑΒΖκαὶτοῦΒΖΝ,ἡὑπὸΑΒΖγωνία·λοιπὴἄραἡὑπὸ is also double KM. But, indeed, circumference CB is
ΑΖΒλοιπῇτῇὑπὸΒΝΖἐστινἴση·ἴσογώνιονἄραἐστὶτὸ also double circumference BK. For circumference CB
ΑΒΖτρίγωνοντῷΒΖΝτριγώνῳ.ἀνάλογονἄραἐστὶνὡςἡ (is) equal to BA. Thus, the whole circumference GB
ΑΒεὐθεῖαπρὸςτὴνΒΖ,οὕτωςἡΖΒπρὸςτὴνΒΝ·τὸἄρα is also double BM. Hence, angle GFB [is] also dou-
ὑπὸτῶνΑΒΝἴσονἐστὶτῷἀπὸΒΖ.πάλινἐπεὶἴσηἐστὶνἡ ble angle BFM [Prop. 6.33]. And GFB (is) also dou-
ΑΛτῇΛΚ,κοινὴδὲκαὶπρὸςὀρθὰςἡΛΝ,βάσιςἄραἡΚΝ ble FAB. For FAB (is) equal to ABF . Thus, BFN
βάσειτῇΑΝἐστινἴση·καὶγωνίαἄραἡὑπὸΛΚΝγωνίᾳτῇ is also equal to FAB. And angle ABF (is) common to
ὑπὸΛΑΝἐστινἴση. ἀλλὰἡὑπὸΛΑΝτῇὑπὸΚΒΝἐστιν the two triangles ABF and BFN. Thus, the remain-
ἴση·καὶἡὑπὸΛΚΝἄρατῇὑπὸΚΒΝἐστινἴση.καὶκοινὴ ing (angle) AFB is equal to the remaining (angle) BNF
τῶνδύοτριγώνωντοῦτεΑΚΒκαὶτοῦΑΚΝἡπρὸςτῷ [Prop. 1.32]. Thus, triangle ABF is equiangular to trian-
Α.λοιπὴἄραἡὑπὸΑΚΒλοιπῇτῇὑπὸΚΝΑἐστινἴση· gle BFN. Thus, proportionally, as straight-line AB (is)
ἰσογώνιονἄραἐστὶτὸΚΒΑτρίγωνοντῷΚΝΑτριγώνῳ. to BF , so FB (is) to BN [Prop. 6.4]. Thus, the (rectan-
ἀνάλογονἄραἐστὶνὡςἡΒΑεὐθεῖαπρὸςτὴνΑΚ,οὕτως gle contained) by ABN is equal to the (square) on BF
ἡΚΑπρὸςτὴνΑΝ·τὸἄραὑπὸτῶνΒΑΝἴσονἐστὶτῷ [Prop. 6.17]. Again, since AL is equal to LK, and LN
ἀπὸτῆςΑΚ.ἐδείχθηδὲκαὶτὸὑπὸτῶνΑΒΝἴσοντῷἀπὸ is common and at right-angles (to KA), base KN is thus
τῆςΒΖ·τὸἄραὑπὸτῶνΑΒΝμετὰτοῦὑπὸΒΑΝ,ὅπερ equal to base AN [Prop. 1.4]. And, thus, angle LKN
ἐστὶτὸἀπὸτῆςΒΑ,ἴσονἐστὶτῷἀπὸτῆςΒΖμετὰτοῦἀπὸ is equal to angle LAN. But, LAN is equal to KBN
τῆςΑΚ.καίἐστινἡμὲνΒΑπενταγώνουπλευρά,ἡδὲΒΖ [Props. 3.29, 1.5]. Thus, LKN is also equal to KBN.
ἑξαγώνου,ἡδὲΑΚδεκαγώνου.
And the (angle) at A (is) common to the two triangles
῾Η ἄρα τοῦ πενταγώνου πλευρὰ δύναται τήν τε τοῦ AKB and AKN. Thus, the remaining (angle) AKB is
516

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
ἑξαγώνουκαὶτὴντοῦδεκαγώνουτῶνεἰςτὸναὐτὸνκύκλον equal to the remaining (angle) KNA [Prop. 1.32]. Thus,
ἐγγραφομένων·ὅπερἔδειδεῖξαι. triangle KBA is equiangular to triangle KNA. Thus,
proportionally, as straight-line BA is to AK, so KA (is) to
AN [Prop. 6.4]. Thus, the (rectangle contained) by BAN
is equal to the (square) on AK [Prop. 6.17]. And the
(rectangle contained) by ABN was also shown (to be)
equal to the (square) on BF . Thus, the (rectangle contained)
by ABN plus the (rectangle contained) by BAN,
which is the (square) on BA [Prop. 2.2], is equal to the
(square) on BF plus the (square) on AK. And BA is the
side of the pentagon, and BF (the side) of the hexagon
[Prop. 4.15 corr.], and AK (the side) of the decagon.
Thus, the square on the side of the pentagon (inscribed
in a circle) is (equal to) the (sum of the squares)
on the (sides) of the hexagon and of the decagon inscribed
in the same circle.
† If the circle is of unit radius then the side of the pentagon is (1/2) p 10 − 2 √ 5.
. Proposition 11
᾿Εὰνεἰςκύκλονῥητὴνἔχοντατὴνδιάμετρονπεντάγω-
If an equilateral pentagon is inscribed in a circle which
νονἰσόπλευρονἐγγραφῇ,ἡτοῦπενταγώνουπλευρὰἄλογός has a rational diameter then the side of the pentagon is
ἐστινἡκαλουμένηἐλάσσων.
that irrational (straight-line) called minor.
A
Å
B
E
M
Æ Â
F K
H
À
N
L
C
D
G
ΕἰςγὰρκύκλοντὸνΑΒΓΔΕῥητὴνἔχοντατὴνδίαμετρον For let the equilateral pentagon ABCDE have been
πεντάγωνον ἰσόπλευρονἐγγεγράφθω τὸ ΑΒΓΔΕ· λέγω, inscribed in the circle ABCDE which has a rational di-
ὅτιἡτοῦ[ΑΒΓΔΕ]πενταγώνουπλευρὰἄλογόςἐστινἡ ameter. I say that the side of pentagon [ABCDE] is that
Ã
Ä
καλουμένηἐλάσσων.
irrational (straight-line) called minor.
Εἰλήφθω γὰρ τὸ κέντρον τοῦκύκλουτὸ Ζ σημεῖον, For let the center of the circle, point F , have been
καὶἐπεζεύχθωσαναἱΑΖ,ΖΒκαὶδιήχθωσανἐπὶτὰΗ,Θ found [Prop. 3.1]. And let AF and FB have been joined.
σημεῖα,καὶἐπεζεύχθωἡΑΓ,καὶκείσθωτῆςΑΖτέταρτον And let them have been drawn across to points G and H
μέροςἡΖΚ.ῥητὴδὲἡΑΖ·ῥητὴἄρακαὶἡΖΚ.ἔστιδὲ (respectively). And let AC have been joined. And let FK
καὶἡΒΖῥητή· ὅληἄραἡΒΚῥητήἐστιν. καὶἐπεὶἴση made (equal) to the fourth part of AF . And AF (is) ratio-
ἐστὶνἡΑΓΗπεριφέρειατῇΑΔΗπεριφερείᾳ,ὧνἡΑΒΓ nal. FK (is) thus also rational. And BF is also rational.
τῇΑΕΔἐστινἴση, λοιπὴἄραἡΓΗλοιπῇτῇΗΔἐστιν Thus, the whole of BK is rational. And since circum-
ἴση. καὶἐὰνἐπιζεύξωμεντὴνΑΔ,συνάγονταιὀρθαὶαἱ ference ACG is equal to circumference ADG, of which
517

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
πρὸςτῷΛγωνίαι, καὶδιπλῆἡΓΔτῆςΓΛ.διὰτὰαὐτὰ ABC is equal to AED, the remainder CG is thus equal
δὴκαὶαἱπρὸςτῷΜὀρθαίεἰσιν,καὶδιπλῆἡΑΓτῆςΓΜ. to the remainder GD. And if we join AD then the angles
ἐπεὶοὖνἴσηἐστὶνἡὑπὸΑΛΓγωνίατῇὑπὸΑΜΖ,κοινὴ at L are inferred (to be) right-angles, and CD (is inferred
δὲ τῶνδύοτριγώνων τοῦτεΑΓΛκαὶτοῦΑΜΖἡὑπὸ to be) double CL [Prop. 1.4]. So, for the same (reasons),
ΛΑΓ,λοιπὴἄραἡὑπὸΑΓΛλοιπῇτῇὑπὸΜΖΑἐστινἴση· the (angles) at M are also right-angles, and AC (is) dou-
ἰσογώνιονἄραἐστὶτὸΑΓΛτρίγωνοντῷΑΜΖτριγώνῳ· ble CM. Therefore, since angle ALC (is) equal to AMF ,
ἀνάλογονἄραἐστὶνὡςἡΛΓπρὸςΓΑ,οὕτωςἡΜΖπρὸς and (angle) LAC (is) common to the two triangles ACL
ΖΑ· καὶτῶν ἡγουμένωντὰδιπλάσια· ὡς ἄραἡτῆςΛΓ and AMF , the remaining (angle) ACL is thus equal to
διπλῆπρὸςτὴνΓΑ,οὕτωςἡτῆςΜΖδιπλῆπρὸςτὴνΖΑ. the remaining (angle) MFA [Prop. 1.32]. Thus, triangle
ὡςδὲἡτῆςΜΖδιπλῆπρὸςτὴνΖΑ,οὕτωςἡΜΖπρὸςτὴν ACL is equiangular to triangle AMF . Thus, proportion-
ἡμίσειαντῆςΖΑ·καὶὡςἄραἡτῆςΛΓδιπλῆπρὸςτὴνΓΑ, ally, as LC (is) to CA, so MF (is) to FA [Prop. 6.4]. And
οὕτωςἡΜΖπρὸςτὴνἡμίσειαντῆςΖΑ·καὶτῶνἑπομένων (we can take) the doubles of the leading (magnitudes).
τὰἡμίσεα·ὡςἄραἡτῆςΛΓδιπλῆπρὸςτὴνἡμίσειαντῆς Thus, as double LC (is) to CA, so double MF (is) to
ΓΑ,οὕτωςἡΜΖπρὸςτὸτέτατροντῆςΖΑ.καίἐστιτῆς FA. And as double MF (is) to FA, so MF (is) to half of
μὲνΛΓδιπλῆἡΔΓ,τῆςδὲΓΑἡμίσειαἡΓΜ,τῆςδὲΖΑ FA. And, thus, as double LC (is) to CA, so MF (is) to
τέτατρονμέροςἡΖΚ·ἔστινἄραὡςἡΔΓπρὸςτὴνΓΜ, half of FA. And (we can take) the halves of the following
οὕτωςἡΜΖπρὸςτὴνΖΚ.συνθέντικαὶὡςσυναμφότερος (magnitudes). Thus, as double LC (is) to half of CA, so
ἡΔΓΜπρὸςτὴνΓΜ,οὕτωςἡΜΚπρὸςΚΖ·καὶὡςἄρατὸ MF (is) to the fourth of FA. And DC is double LC, and
ἀπὸσυναμφοτέρουτῆςΔΓΜπρὸςτὸἀπὸΓΜ,οὕτωςτὸ CM half of CA, and FK the fourth part of FA. Thus,
ἀπὸΜΚπρὸςτὸἀπὸΚΖ.καὶἐπεὶτῆςὑπὸδύοπλευρὰςτοῦ as DC is to CM, so MF (is) to FK. Via composition, as
πενταγώνουὑποτεινούσης,οἷοντῆςΑΓ,ἄκρονκαὶμέσον the sum of DCM (i.e., DC and CM) (is) to CM, so MK
λόγοντεμνομένηςτὸμεῖζοντμῆμαἴσονἐστὶτῇτοῦπεν- (is) to KF [Prop. 5.18]. And, thus, as the (square) on the
ταγώνουπλευρᾷ,τουτέστιτῇΔΓ,τὸδὲμεῖζοντμῆμαπρο- sum of DCM (is) to the (square) on CM, so the (square)
σλαβὸντὴνἡμίσειαντῆςὅλῆςπενταπλάσιονδύναταιτοῦ on MK (is) to the (square) on KF . And since the greater
ἀπὸτῆςἡμισείαςτῆςὅλης,καίἐστινὅληςτῆςΑΓἡμίσεια piece of a (straight-line) subtending two sides of a pen-
ἡΓΜ,τὸἄραἀπὸτῆςΔΓΜὡςμιᾶςπενταπλάσιόνἐστι tagon, such as AC, (which is) cut in extreme and mean
τοῦἀπὸτῆςΓΜ.ὡςδὲτὸἀπὸτῆςΔΓΜὡςμιᾶςπρὸςτὸ ratio is equal to the side of the pentagon [Prop. 13.8]—
ἀπὸτῆςΓΜ,οὕτωςἐδείχθητὸἀπὸτῆςΜΚπρὸςτὸἀπὸ that is to say, to DC—-and the square on the greater piece
τῆςΚΖ·πενταπλάσιονἄρατὸἀπὸτῆςΜΚτοῦἀπὸτῆς added to half of the whole is five times the (square) on
ΚΖ.ῥητὸνδὲτὸἀπὸτῆςΚΖ·ῥητὴγὰρἡδιάμετρος·ῥητὸν half of the whole [Prop. 13.1], and CM (is) half of the
ἄρακαὶτὸἀπὸτῆςΜΚ·ῥητὴἄραἐστὶνἡΜΚ[δυνάμει whole, AC, thus the (square) on DCM, (taken) as one,
μόνον]. καὶἐπεὶτετραπλασίαἐστὶνἡΒΖτῆςΖΚ,πεντα- is five times the (square) on CM. And the (square) on
πλασίαἄραἐστὶνἡΒΚτῆςΚΖ·εἰκοσιπενταπλάσιονἄρατὸ DCM, (taken) as one, (is) to the (square) on CM, so
ἀπὸτῆςΒΚτοῦἀπὸτῆςΚΖ.πενταπλάσιονδὲτὸἀπὸτῆς the (square) on MK was shown (to be) to the (square)
ΜΚτοῦἀπὸτῆςΚΖ·πενταπλάσιονἄρατὸἀπὸτῆςΒΚ on KF . Thus, the (square) on MK (is) five times the
τοῦἀπὸτῆςΚΜ·τὸἄραἀπὸτῆςΒΚπρὸςτὸἀπὸΚΜ (square) on KF . And the square on KF (is) rational.
λόγονοὐκἔχει,ὃντετράγωνοςἀριθμὸςπρὸςτετράγωνον For the diameter (is) rational. Thus, the (square) on
ἀριθμόν·ἀσύμμετροςἄραἐστὶνἡΒΚτῇΚΜμήκει.καίἐστι MK (is) also rational. Thus, MK is rational [in square
ῥητὴἑκατέρααὐτῶν. αἱΒΚ,ΚΜἄραῥηταίεἰσιδυνάμει only]. And since BF is four times FK, BK is thus five
μόνονσύμμετροι.ἐὰνδὲἀπὸῥητῆςῥητὴἀφαιρεθῇδυνάμει times KF . Thus, the (square) on BK (is) twenty-five
μόνονσύμμετροςοὖσατῇὅλῃ,ἡλοιπὴἄλογόςἐστινἀπο- times the (square) on KF . And the (square) on MK
τομή·ἀποτομὴἄραἐστὶνἡΜΒ,προσαρμόζουσαδὲαὐτῇἡ (is) five times the square on KF . Thus, the (square)
ΜΚ.λέγωδή,ὅτικαὶτετάρτη. ᾧδὴμεῖζόνἐστιτὸἀπὸ on BK (is) five times the (square) on KM. Thus, the
τῆςΒΚτοῦἀπὸτῆςΚΜ,ἐκείνῳἴσονἔστωτὸἀπὸτῆςΝ· (square) on BK does not have to the (square) on KM
ἡΒΚἄρατῆςΚΜμεῖζονδύναταιτῇΝ.καὶἐπεὶσύμμετρός the ratio which a square number (has) to a square num-
ἐστινἡΚΖτῇΖΒ,καὶσυνθέντισύμμετρόςἐστιἡΚΒτῇ ber. Thus, BK is incommensurable in length with KM
ΖΒ.ἀλλὰἡΒΖτῇΒΘσύμμετρόςἐστιν·καὶἡΒΚἄρατῇ [Prop. 10.9]. And each of them is a rational (straight-
ΒΘσύμμετρόςἐστιν. καὶἐπεὶπενταπλάσιόνἐστιτὸἀπὸ line). Thus, BK and KM are rational (straight-lines
τῆςΒΚτοῦἀπὸτῆςΚΜ,τὸἄραἀπὸτῆςΒΚπρὸςτὸ which are) commensurable in square only. And if from
ἀπὸτῆςΚΜλόγονἔχει,ὃνεπρὸςἕν. ἀναστρέψαντιἄρα a rational (straight-line) a rational (straight-line) is subτὸἀπὸτῆςΒΚπρὸςτὸἀπὸτῆςΝλόγονἔχει,ὃνεπρὸς
tracted, which is commensurable in square only with the
518

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
δ,οὐχὃντετράγωνοςπρὸςτετράγωνον·ἀσύμμετροςἄρα whole, then the remainder is that irrational (straight-line
ἐστὶνἡΒΚτῇΝ·ἡΒΚἄρατῆςΚΜμεῖζονδύναταιτῷἀπὸ called) an apotome [Prop. 10.73]. Thus, MB is an apo-
ἀσυμμέτρουἑαυτῇ.ἐπεὶοὖνὅληἡΒΚτῆςπροσαρμοζούσης tome, and MK its attachment. So, I say that (it is) also
τῆςΚΜμεῖζονδύναταιτῷἀπὸἀσυμμέτρουἑαυτῇ,καὶὅλη a fourth (apotome). So, let the (square) on N be (made)
ἡΒΚσύμμετρόςἐστιτῇἐκκειμένῃῥητῇτῇΒΘ,ἀποτομὴ equal to that (magnitude) by which the (square) on BK
ἄρατετάρτηἐστὶνἡΜΒ.τὸδὲὑπὸῥητῆςκαὶἀποτομῆς is greater than the (square) on KM. Thus, the square on
τετάρτηςπεριεχόμενονὀρθογώνιονἄλογόνἐστιν,καὶἡδυ- BK is greater than the (square) on KM by the (square)
ναμένηαὐτὸἄλογόςἐστιν,καλεῖταιδὲἐλάττων. δύναται on N. And since KF is commensurable (in length) with
δὲτὸὑπὸτῶνΘΒΜἡΑΒδιὰτὸἐπιζευγνυμένηςτῆςΑΘ FB then, via composition, KB is also commensurable (in
ἰσογώνιονγίνεσθαιτὸΑΒΘτρίγωνοντῷΑΒΜτριγώνῳ length) with FB [Prop. 10.15]. But, BF is commensuκαὶεἶναιὡςτὴνΘΒπρὸςτὴνΒΑ,οὕτωςτὴνΑΒπρὸςτὴν
rable (in length) with BH. Thus, BK is also commen-
ΒΜ.
surable (in length) with BH [Prop. 10.12]. And since
῾ΗἄραΑΒτοῦπενταγώνουπλευρὰἄλογόςἐστινἡκα- the (square) on BK is five times the (square) on KM,
λουμένηἐλάττων·ὅπερἔδειδεῖξαι.
the (square) on BK thus has to the (square) on KM the
ratio which 5 (has) to one. Thus, via conversion, the
(square) on BK has to the (square) on N the ratio which
5 (has) to 4 [Prop. 5.19 corr.], which is not (that) of a
square (number) to a square (number). BK is thus incommensurable
(in length) with N [Prop. 10.9]. Thus,
the square on BK is greater than the (square) on KM
by the (square) on (some straight-line which is) incommensurable
(in length) with (BK). Therefore, since the
square on the whole, BK, is greater than the (square) on
the attachment, KM, by the (square) on (some straightline
which is) incommensurable (in length) with (BK),
and the whole, BK, is commensurable (in length) with
the (previously) laid down rational (straight-line) BH,
MB is thus a fourth apotome [Def. 10.14]. And the
rectangle contained by a rational (straight-line) and a
fourth apotome is irrational, and its square-root is that
irrational (straight-line) called minor [Prop. 10.94]. And
the square on AB is the rectangle contained by HBM,
on account of joining AH, (so that) triangle ABH becomes
equiangular with triangle ABM [Prop. 6.8], and
(proportionally) as HB is to BA, so AB (is) to BM.
Thus, the side AB of the pentagon is that irrational
(straight-line) called minor. † (Which is) the very thing it
was required to show.
† If the circle has unit radius then the side of the pentagon is (1/2) p 10 − 2 √ 5. However, this length can be written in the “minor” form (see
Prop. 10.94) (ρ/ √ q
2) 1 + k/ √ 1 + k 2 − (ρ/ √ q
2) 1 − k/ √ 1 + k 2 , with ρ = p 5/2 and k = 2.
. Proposition 12
᾿Εὰνεἰςκύκλοντρίγωνονἰσόπλευρονἐγγραφῇ,ἡτοῦ If an equilateral triangle is inscribed in a circle then
τριγώνου πλευρὰ δυνάμει τριπλασίων ἐστὶ τῆς ἐκ τοῦ the square on the side of the triangle is three times the
κέντρουτοῦκύκλου.
(square) on the radius of the circle.
᾿ΕστωκύκλοςὁΑΒΓ,καὶεἰςαὐτὸντρίγωνονἰσόπλευρ- Let there be a circle ABC, and let the equilateral triονἐγγεγράφθωτὸΑΒΓ·λέγω,ὅτιτοῦΑΒΓτριγώνουμία
angle ABC have been inscribed in it [Prop. 4.2]. I say
πλευρὰδυνάμειτριπλασίωνἐστὶτῆςἐκτοῦκέντρουτοῦ that the square on one side of triangle ABC is three times
ΑΒΓκύκλου.
the (square) on the radius of circle ABC.
519

ËÌÇÁÉÁÏƺ ELEMENTS
A
BOOK 13
ΕἰλήφθωγὰρτὸκέντροντοῦΑΒΓκύκλουτὸΔ,καὶ For let the center, D, of circle ABC have been found
ἐπιζευχθεῖσαἡΑΔδιήχθωἐπὶτὸΕ,καὶἐπεζεύχθωἡΒΕ. [Prop. 3.1]. And AD (being) joined, let it have been
ΚαὶἐπεὶἰσόπλευρόνἐστιτὸΑΒΓτρίγωνον,ἡΒΕΓἄρα drawn across to E. And let BE have been joined.
περιφέρειατρίτονμέροςἐστὶτῆςτοῦΑΒΓκύκλουπερι- And since triangle ABC is equilateral, circumference
φερείας. ἡἄραΒΕπεριφέρειαἕκτονἐστὶμέροςτῆςτοῦ BEC is thus the third part of the circumference of cirκύκλουπεριφερείας·ἑξαγώνουἄραἐστὶνἡΒΕεὐθεῖα·ἴση
cle ABC. Thus, circumference BE is the sixth part of
ἄραἐστὶτῇἐκτοῦκέντρουτῇΔΕ.καὶἐπεὶδιπλῆἐστινἡ the circumference of the circle. Thus, straight-line BE is
ΑΕτῆςΔΕ,τετραπλάσιονἐστιτὸἀπὸτῆςΑΕτοῦἀπὸτῆς (the side) of a hexagon. Thus, it is equal to the radius
ΕΔ,τουτέστιτοῦἀπὸτῆςΒΕ.ἴσονδὲτὸἀπὸτῆςΑΕτοῖς DE [Prop. 4.15 corr.]. And since AE is double DE, the
ἀπὸτῶνΑΒ,ΒΕ·τὰἄραἀπὸτῶνΑΒ,ΒΕτετραπλάσιάἐστι (square) on AE is four times the (square) on ED—that
τοῦἀπὸτῆςΒΕ.διελόντιἄρατὸἀπὸτῆςΑΒτριπλάσιόν is to say, of the (square) on BE. And the (square) on
ἐστιτοῦἀπὸΒΕ.ἴσηδὲἡΒΕτῇΔΕ·τὸἄραἀπὸτῆςΑΒ AE (is) equal to the (sum of the squares) on AB and BE
τριπλάσιόνἐστιτοῦἀπὸτῆςΔΕ.
[Props. 3.31, 1.47]. Thus, the (sum of the squares) on
῾Ηἄρατοῦτριγώνουπλευρὰδυνάμειτριπλασίαἐστὶτῆς AB and BE is four times the (square) on BE. Thus,
ἐκτοῦκέντρου[τοῦκύκλου]·ὅπερἔδειδεῖξαι.
via separation, the (square) on AB is three times the
(square) on BE. And BE (is) equal to DE. Thus, the
(square) on AB is three times the (square) on DE.
Thus, the square on the side of the triangle is three
times the (square) on the radius [of the circle]. (Which
is) the very thing it was required to show.
. Proposition 13
Πυραμίδασυστήσασθαικαὶσφαίρᾳπεριλαβεῖντῇδοθείσῃ To construct a (regular) pyramid (i.e., a tetrahedron),
καὶδεῖξαι,ὅτιἡτῆςσφαίραςδιάμετροςδυνάμειἡμιολίαἐστὶ and to enclose (it) in a given sphere, and to show that
τῆςπλευρᾶςτῆςπυραμίδος.
the square on the diameter of the sphere is one and a
half times the (square) on the side of the pyramid.
B
E
D
C
520

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
D
Ã
A
K
C
E
B
Â
Ä À
H
F
L
G
᾿ΕκκείσθωἡτῆςδοθείσηςσφαίραςδίαμετροςἡΑΒ,καὶ Let the diameter AB of the given sphere be laid out,
τετμήσθωκατὰτὸΓσημεῖον,ὥστεδιπλασίανεἶναιτὴνΑΓ and let it have been cut at point C such that AC is double
τῆςΓΒ·καὶγεγράφθωἐπὶτῆςΑΒἡμικύκλιοντὸΑΔΒ, CB [Prop. 6.10]. And let the semi-circle ADB have been
καὶἤχθωἀπὸτοῦΓσημείουτῇΑΒπρὸςὀρθὰςἡΓΔ, drawn on AB. And let CD have been drawn from point C
καὶἐπεζεύχθωἡΔΑ·καὶἐκκείσθωκύκλοςὁΕΖΗἴσην at right-angles to AB. And let DA have been joined. And
ἔχωντὴνἐκτοῦκέντρουτῇΔΓ,καὶἐγγεγράφθωεἰςτὸν let the circle EFG be laid down having a radius equal
ΕΖΗκύκλοντρίγωνονἰσόπλευροντὸΕΖΗ·καὶεἰλήφθω to DC, and let the equilateral triangle EFG have been
τὸκέντροντοῦκύκλουτὸΘσημεῖον,καὶἐπεζεύχθωσαν inscribed in circle EFG [Prop. 4.2]. And let the center
αἱΕΘ,ΘΖ,ΘΗ·καὶἀνεστάτωἀπὸτοῦΘσημείουτῷτοῦ of the circle, point H, have been found [Prop. 3.1]. And
ΕΖΗκύκλουἐπιπέδῳπρὸςὀρθὰςἡΘΚ,καὶἀφῃρήσθωἀπὸ let EH, HF , and HG have been joined. And let HK
τῆςΘΚτῇΑΓεὐθείᾳἴσηἡΘΚ,καὶἐπεζεύχθωσαναἱΚΕ, have been set up, at point H, at right-angles to the plane
ΚΖ,ΚΗ.καὶἐπεὶἡΚΘὀρθήἐστιπρὸςτὸτοῦΕΖΗκύκλου of circle EFG [Prop. 11.12]. And let HK, equal to the
ἐπίπεδον,καὶπρὸςπάσαςἄρατὰςἁπτομέναςαὐτῆςεὐθείας straight-line AC, have been cut off from HK. And let
καὶοὔσαςἐντῷτοῦΕΖΗκύκλουἐπιπέδῳὀρθὰςποιήσει KE, KF , and KG have been joined. And since KH is at
γωνίας.ἅπτεταιδὲαὐτῆςἑκάστητῶνΘΕ,ΘΖ,ΘΗ·ἡΘΚ right-angles to the plane of circle EFG, it will thus also
ἄραπρὸςἑκάστητῶνΘΕ,ΘΖ,ΘΗὀρθήἐστιν.καὶἐπεὶἴση make right-angles with all of the straight-lines joining it
ἐστὶνἡμὲνΑΓτῇΘΚ,ἡδὲΓΔτῇΘΕ,καὶὀρθὰςγωνίας (which are) also in the plane of circle EFG [Def. 11.3].
περιέχουσιν,βάσιςἄραἡΔΑβάσειτῇΚΕἐστινἴση. διὰ And HE, HF , and HG each join it. Thus, HK is at
τὰαὐτὰδὴκαὶἑκατέρατῶνΚΖ,ΚΗτῇΔΑἐστινἴση·αἱ right-angles to each of HE, HF , and HG. And since
τρεῖςἄρααἱΚΕ,ΚΖ,ΚΗἴσαιἀλλήλαιςεἰσίν.καὶἐπεὶδιπλῆ AC is equal to HK, and CD to HE, and they contain
ἐστινἡΑΓτῆςΓΒ,τριπλῆἄραἡΑΒτῆςΒΓ.ὡςδὲἡΑΒ right-angles, the base DA is thus equal to the base KE
πρὸςτὴνΒΓ,οὕτωςτὸἀπὸτῆςΑΔπρὸςτὸἀπὸτῆςΔΓ, [Prop. 1.4]. So, for the same (reasons), KF and KG is
ὡςἑξῆςδειχθήσεται. τριπλάσιονἄρατὸἀπὸτῆςΑΔτοῦ each equal to DA. Thus, the three (straight-lines) KE,
ἀπὸτῆςΔΓ.ἔστιδὲκαὶτὸἀπὸτῆςΖΕτοῦἀπὸτῆςΕΘ KF , and KG are equal to one another. And since AC is
τριπλάσιον,καίἐστινἴσηἡΔΓτῇΕΘ·ἴσηἄρακαὶἡΔΑ double CB, AB (is) thus triple BC. And as AB (is) to
τῇΕΖ.ἀλλὰἡΔΑἑκάστῃτῶνΚΕ,ΚΖ,ΚΗἐδείχθηἴση· BC, so the (square) on AD (is) to the (square) on DC,
καὶἑκάστηἄρατῶνΕΖ,ΖΗ,ΗΕἑκάστῃτῶνΚΕ,ΚΖ,ΚΗ as will be shown later [see lemma]. Thus, the (square)
ἐστινἴση·ἰσόπλευραἄραἐστὶτὰτέσσαρατρίγωνατὰΕΖΗ, on AD (is) three times the (square) on DC. And the
ΚΕΖ,ΚΖΗ,ΚΕΗ.πυραμὶςἄρασυνέσταταιἐκτεσσάρων (square) on FE is also three times the (square) on EH
τριγώνωνἰσοπλέυρων,ἧςβάσιςμένἐστιτὸΕΖΗτρίγωνον, [Prop. 13.12], and DC is equal to EH. Thus, DA (is)
521

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
κορυφὴδὲτὸΚσημεῖον.
also equal to EF . But, DA was shown (to be) equal to
Δεῖ δὴ αὐτὴν καὶ σφαίρᾳ περιλαβεῖν τῇ δοθείσῃ καὶ each of KE, KF , and KG. Thus, EF , FG, and GE are
δεῖξαι,ὅτιἡτῆςσφαίραςδιάμετροςἡμιολίαἐστὶδυνάμει equal to KE, KF , and KG, respectively. Thus, the four
τῆςπλευρᾶςτῆςπυραμίδος.
triangles EFG, KEF , KFG, and KEG are equilateral.
᾿Εκβεβλήσθωγὰρἐπ᾿εὐθείαςτῇΚΘεὐθεῖαἡΘΛ,καὶ Thus, a pyramid, whose base is triangle EFG, and apex
κείσθωτῇΓΒἴσηἡΘΛ.καὶἐπείἐστινὡςἡΑΓπρὸςτὴν the point K, has been constructed from four equilateral
ΓΔ,οὕτωςἡΓΔπρὸςτὴνΓΒ,ἴσηδὲἡμὲνΑΓτῇΚΘ,ἡδὲ triangles.
ΓΔτῇΘΕ,ἡδὲΓΒτῇΘΛ,ἔστινἄραὡςἡΚΘπρὸςτὴνΘΕ, So, it is also necessary to enclose it in the given
οὕτωςἡΕΘπρὸςτὴνΘΛ·τὸἄραὑπὸτῶνΚΘ,ΘΛἴσον sphere, and to show that the square on the diameter of
ἐστὶτῷἀπὸτῆςΕΘ.καίἐστινὀρθὴἑκατέρατῶνὑπὸΚΘΕ, the sphere is one and a half times the (square) on the side
ΕΘΛγωνιῶν·τὸἄραἐπὶτῆςΚΛγραφόμενονἡμικύκλιον of the pyramid.
ἥξεικαὶδιὰτοῦΕ[ἐπειδήπερἐὰνἐπιζεύξωμεντὴνΕΛ,ὀρθὴ For let the straight-line HL have been produced in
γίνεταιἡὑπὸΛΕΚγωνίαδιὰτὸἰσογώνιονγίνεσθαιτὸ a straight-line with KH, and let HL be made equal to
ΕΛΚτρίγωνονἑκατέρῳτῶνΕΛΘ,ΕΘΚτριγώνων]. ἐὰν CB. And since as AC (is) to CD, so CD (is) to CB
δὴμενούσηςτῆςΚΛπεριενεχθὲντὸἡμικύκλιονεἰςτὸαὐτὸ [Prop. 6.8 corr.], and AC (is) equal to KH, and CD to
πάλινἀποκατασταθῇ,ὅθενἤρξατοφέρεσθαι,ἥξεικαὶδιὰ HE, and CB to HL, thus as KH is to HE, so EH (is)
τῶνΖ,ΗσημείωνἐπιζευγνυμένωντῶνΖΛ,ΛΗκαὶὀρθῶν to HL. Thus, the (rectangle contained) by KH and HL
ὁμοίωςγινομένωντῶνπρὸςτοῖςΖ,Ηγωνιῶν·καὶἔσται is equal to the (square) on EH [Prop. 6.17]. And each
ἡπυραμὶςσφαίρᾳπεριειλημμένητῇδοθείσῇ. ἡγὰρΚΛ of the angles KHE and EHL is a right-angle. Thus,
τῆςσφαίραςδιάμετροςἴσηἐστὶτῇτῆςδοθείσηςσφαίρας the semi-circle drawn on KL will also pass through E
διαμετρῳτῇΑΒ,ἐπειδήπερτῇμὲνΑΓἴσηκεῖταιἡΚΘ,τῇ [inasmuch as if we join EL then the angle LEK beδὲΓΒἡΘΛ.
comes a right-angle, on account of triangle ELK becom-
Λέγω δή, ὅτι ἡ τῆς σφαίρας διάμετρος ἡμιολία ἐστὶ ing equiangular to each of the triangles ELH and EHK
δυνάμειτῆςπλευρᾶςτῆςπυραμίδος.
[Props. 6.8, 3.31] ]. So, if KL remains (fixed), and the
᾿ΕπεὶγὰρδιπλῆἐστινἡΑΓτῆςΓΒ,τριπλῆἄραἐστὶν semi-circle is carried around, and again established at the
ἡΑΒτῆςΒΓ·ἀναστρέψαντιἡμιολίαἄραἐστὶνἡΒΑτῆς same (position) from which it began to be moved, it will
ΑΓ.ὡςδὲἡΒΑπρὸςτὴνΑΓ,οὕτωςτὸἀπὸτῆςΒΑπρὸς also pass through points F and G, (because) if FL and
τὸἀπὸτῆςΑΔ[ἐπειδήπερἐπιζευγνμένηςτῆςΔΒἐστιν LG are joined, the angles at F and G will similarly be-
ὡςἡΒΑπρὸςτὴνΑΔ,οὕτωςἡΔΑπρὸςτὴνΑΓδιὰ come right-angles. And the pyramid will have been enτὴνὁμοιότητατῶνΔΑΒ,ΔΑΓτριγώνων,καὶεἶναιὡςτὴν
closed by the given sphere. For the diameter, KL, of the
πρώτηνπρὸςτὴντρίτην,οὕτωςτὸἀπὸτῆςτρώτηςπρὸςτὸ sphere is equal to the diameter, AB, of the given sphere—
ἀπὸτῆςδευτέρας]. ἡμιόλιονἄρακαὶτὸἀπὸτῆςΒΑτοῦ inasmuch as KH was made equal to AC, and HL to CB.
ἀπὸτῆςΑΔ.καίἐστινἡμὲνΒΑἡτῆςδοθείσηςσφαίρας So, I say that the square on the diameter of the sphere
διάμετρος,ἡδὲΑΔἴσητῇπλευρᾷτῆςπυραμίδος. is one and a half times the (square) on the side of the
῾Ηἄρατῆςσφαίραςδιάμετροςἡμιολίαἐστὶτῆςπλευρᾶς pyramid.
τῆςπυραμίδος·ὅπερἔδειδεῖξαι.
For since AC is double CB, AB is thus triple BC.
Thus, via conversion, BA is one and a half times AC.
And as BA (is) to AC, so the (square) on BA (is) to the
(square) on AD [inasmuch as if DB is joined then as BA
is to AD, so DA (is) to AC, on account of the similarity
of triangles DAB and DAC. And as the first is to the
third (of four proportional magnitudes), so the (square)
on the first (is) to the (square) on the second.] Thus,
the (square) on BA (is) also one and a half times the
(square) on AD. And BA is the diameter of the given
sphere, and AD (is) equal to the side of the pyramid.
Thus, the square on the diameter of the sphere is one
and a half times the (square) on the side of the pyramid. †
(Which is) the very thing it was required to show.
† If the radius of the sphere is unity then the side of the pyramid (i.e., tetrahedron) is p 8/3.
522

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
D
A
C
B
ÄÑÑ.
Δεικτέον,ὅτιἐστὶνὡςἡΑΒπρὸςτὴνΒΓ,οὕτωςτὸ It must be shown that as AB is to BC, so the (square)
E
F
Lemma
ἀπὸτῆςΑΔπρὸςτὸἀπὸτῆςΔΓ.
on AD (is) to the (square) on DC.
᾿Εκκείσθω γὰρ ἡ τοῦ ἡμικυκλίου καταγραφή, καὶ For, let the figure of the semi-circle have been set
ἐπεζεύχθωἡΔΒ,καὶἀναγεγράφθωἀπὸτῆςΑΓτετράγωνον out, and let DB have been joined. And let the square
τὸ ΕΓ, καὶ συμπεπληρώσθω τὸ ΖΒ παραλληλόγραμμον. EC have been described on AC. And let the parallel-
ἐπεὶοὖνδιὰτὸἰσογώνιονεἶναιτὸΔΑΒτρίγωνοντῷΔΑΓ ogram FB have been completed. Therefore, since, on
τριγώνῳἐστὶνὡςἡΒΑπρὸςτὴνΑΔ,οὕτωςἡΔΑπρὸς account of triangle DAB being equiangular to triangle
τὴνΑΓ,τὸἄραὑπὸτῶνΒΑ,ΑΓἴσονἐστὶτῷἀπὸτῆς DAC [Props. 6.8, 6.4], (proportionally) as BA is to AD,
ΑΔ.καὶἐπείἐστινὡςἡΑΒπρὸςτὴνΒΓ,οὕτωςτὸΕΒ so DA (is) to AC, the (rectangle contained) by BA and
πρὸςτὸΒΖ,καίἐστιτὸμὲνΕΒτὸὑπὸτῶνΒΑ,ΑΓ·ἴση AC is thus equal to the (square) on AD [Prop. 6.17].
γὰρἡΕΑτῇΑΓ·τὸδὲΒΖτὸὑπὸτῶνΑΓ,ΓΒ,ὡςἄραἡ And since as AB is to BC, so EB (is) to BF [Prop. 6.1].
ΑΒπρὸςτὴνΒΓ,οὕτωςτὸὑπὸτῶνΒΑ,ΑΓπρὸτὸὑπὸ And EB is the (rectangle contained) by BA and AC—for
τῶνΑΓ,ΓΒ.καίἐστιτὸμὲνὑπὸτῶνΒΑ,ΑΓἴσοντῷἀπὸ EA (is) equal to AC. And BF the (rectangle contained)
τῆςΑΔ,τὸδὲὑπὸτῶνΑΓΒἴσοντῷἀπὸτῆςΔΓ·ἡγὰρ by AC and CB. Thus, as AB (is) to BC, so the (rectan-
ΔΓκάθετοςτῶντῆςβάσεωςτμημάτωντῶνΑΓ,ΓΒμέση gle contained) by BA and AC (is) to the (rectangle con-
ἀνάλογόνἐστιδιὰτὸὀρθὴνεἶναιτὴνὑπὸΑΔΒ.ὡςἄραἡ tained) by AC and CB. And the (rectangle contained)
ΑΒπρὸςτὴνΒΓ,οὕτωςτὸἀπὸτῆςΑΔπρὸςτὸἀπὸτῆς by BA and AC is equal to the (square) on AD, and the
ΔΓ·ὅπερἔδειδεῖξαι.
(rectangle contained) by ACB (is) equal to the (square)
on DC. For the perpendicular DC is the mean proportional
to the pieces of the base, AC and CB, on account
of ADB being a right-angle [Prop. 6.8 corr.]. Thus, as
AB (is) to BC, so the (square) on AD (is) to the (square)
on DC. (Which is) the very thing it was required to show.
. Proposition 14
᾿Οκτάεδρονσυστήσασθαικαὶσφαίρᾳπεριλαβεῖν,ᾗκαὶ To construct an octahedron, and to enclose (it) in a
τὰπρότερα,καὶδεῖξαι,ὅτιἡτῆςσφαίραςδιάμετροςδυνάμει (given) sphere, like in the preceding (proposition), and
διπλασίαἐστὶτῆςπλευρᾶςτοῦὀκταέδρου.
to show that the square on the diameter of the sphere is
᾿Εκκείσθω ἡ τῆς δοθείσης σφαίρας διάμετρος ἡ ΑΒ, double the (square) on the side of the octahedron.
καὶτετμήσθωδίχακατὰτὸΓ,καὶγεγράφθωἐπὶτῆςΑΒ Let the diameter AB of the given sphere be laid out,
ἡμικύκλιοντὸΑΔΒ,καὶἤχθωἀπὸτοῦΓτῇΑΒπρὸςὀρθὰς and let it have been cut in half at C. And let the semi-
ἡ ΓΔ, καὶ ἐπεζεύχθω ἡ ΔΒ, καὶ ἐκκείσθω τετράγωνον circle ADB have been drawn on AB. And let CD be
τὸ ΕΖΗΘ ἴσην ἔχον ἑκάστην τῶν πλευρῶντῇ ΔΒ, καὶ drawn from C at right-angles to AB. And let DB have
523

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
ἐπεζεύχθωσαναἱΘΖ,ΕΗ,καὶἀνεστάτωἀπὸτοῦΚσημείου been joined. And let the square EFGH, having each of
τῷτοῦΕΖΗΘτετραγώνουἐπιπέδῳπρὸςὀρθὰςεὐθεῖαἡ its sides equal to DB, be laid out. And let HF and EG
ΚΛκαὶδιήχθωἐπὶτὰἕτεραμέρητοῦἐπιπέδουὡςἡΚΜ, have been joined. And let the straight-line KL have been
καὶἀφῃρήσθωἀφ᾿ἑκατέραςτῶνΚΛ,ΚΜμιᾷτῶνΕΚ,ΖΚ, set up, at point K, at right-angles to the plane of square
ΗΚ,ΘΚἴσηἑκατέρατῶνΚΛ,ΚΜ,καὶἐπεζεύχθωσαναἱ EFGH [Prop. 11.12]. And let it have been drawn across
ΛΕ,ΛΖ,ΛΗ,ΛΘ,ΜΕ,ΜΖ,ΜΗ,ΜΘ.
on the other side of the plane, like KM. And let KL and
KM, equal to one of EK, FK, GK, and HK, have been
cut off from KL and KM, respectively. And let LE, LF ,
LG, LH, ME, MF , MG, and MH have been joined.
D
Ä
Ã
À
A
E
L
C
H
B
Å
M
ΚαὶἐπεὶἴσηἐστὶνἡΚΕτῇΚΘ,καίἐστινὀρθὴἡὑπὸ And since KE is equal to KH, and angle EKH is a
ΕΚΘγωνία,τὸἄραἀπὸτῆςΘΕδιπλάσιόνἐστιτοῦἀπὸτῆς right-angle, the (square) on the HE is thus double the
ΕΚ.πάλιν,ἐπεὶἴσηἐστὶνἡΛΚτῇΚΕ,καίἐστινὀρθὴἡ (square) on EK [Prop. 1.47]. Again, since LK is equal
ὑπὸΛΚΕγωνία,τὸἄραἀπὸτῆςΕΛδιπλάσιόνἐστιτοῦἀπὸ to KE, and angle LKE is a right-angle, the (square) on
ΕΚ.ἐδείχθηδὲκαὶτὸἀπὸτῆςΘΕδιπλάσιοντοῦἀπὸτῆς EL is thus double the (square) on EK [Prop. 1.47]. And
ΕΚ·τὸἄραἀπὸτῆςΛΕἴσονἐστὶτῷἀπὸτῆςΕΘ·ἴσηἄρα the (square) on HE was also shown (to be) double the
ἐστὶνἡΛΕτῇΕΘ.διὰτὰαὐτὰδὴκαὶἡΛΘτῇΘΕἐστιν (square) on EK. Thus, the (square) on LE is equal to
ἴση· ἰσόπλευρονἄραἐστὶτὸΛΕΘτρίγωνον. ὁμοίωςδὴ the (square) on EH. Thus, LE is equal to EH. So, for
δείξομεν,ὅτικαὶἕκαστοντῶνλοιπῶντριγώνων,ὧνβάσεις the same (reasons), LH is also equal to HE. Triangle
μένεἰσιναἱτοῦΕΖΗΘτετραγώνουπλευραί,κορυφαὶδὲτὰ LEH is thus equilateral. So, similarly, we can show that
Λ,Μσημεῖα,ἰσόπλευρόνἐστιν·ὀκτάεδρονἄρασυνέσταται each of the remaining triangles, whose bases are the sides
ὑπὸὀκτὼτριγώνωνἰσοπλεύρωνπεριεχόμενον. of the square EFGH, and apexes the points L and M,
Δεῖδὴαὐτὸκαὶσφαίρᾳπεριλαβεῖντῇδοθείσῃκαὶδεῖξαι, are equilateral. Thus, an octahedron contained by eight
ὅτιἡτῆςσφαίραςδιάμετροςδυνάμειδιπλασίωνἐστὶτῆςτοῦ equilateral triangles has been constructed.
ὀκταέδρουπλευρᾶς.
So, it is also necessary to enclose it by the given
᾿ΕπεὶγὰραἱτρεῖςαἱΛΚ,ΚΜ,ΚΕἴσαιἀλλήλαιςεἰσίν, sphere, and to show that the square on the diameter of
τὸἄραἐπὶτῆςΛΜγραφόμενονἡμικύκλιονἥξεικαὶδιὰ the sphere is double the (square) on the side of the octaτοῦΕ.καὶδιὰτὰαὐτά,ἐὰνμενούσηςτῆςΛΜπεριενεχθὲν
hedron.
τὸ ἡμικύκλιον εἰς τὸ αὐτὸ ἀποκατασταθῇ, ὅθεν ἤρξατο For since the three (straight-lines) LK, KM, and KE
φέρεσθαι, ἥξεικαὶδιὰτῶν Ζ, Η, Θ σημείων, καὶἔσται are equal to one another, the semi-circle drawn on LM
σφαίρᾳπεριειλημμένοντὸὀκτάεδρον. λέγωδή,ὅτικαὶτῇ will thus also pass through E. And, for the same (reaδοθείσῃ.
ἐπεὶγὰρἴσηἐστὶνἡΛΚτῇΚΜ,κοινὴδὲἡΚΕ, sons), if LM remains (fixed), and the semi-circle is car-
F
K
G
524

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
καὶγωνίαςὀρθὰςπεριέχουσιν,βάσιςἄραἡΛΕβάσειτῇ ried around, and again established at the same (position)
ΕΜἐστινἴση. καὶἐπεὶὀρθήἐστινἡὑπὸΛΕΜγωνία·ἐν from which it began to be moved, then it will also pass
ἡμικυκλίῳγάρ·τὸἄραἀπὸτῆςΛΜδιπλάσιόνἐστιτοῦἀπὸ through points F , G, and H, and the octahedron will
τῆςΛΕ.πάλιν,ἐπεὶἴσηἐστὶνἡΑΓτῇΓΒ,διπλασίαἐστὶν have been enclosed by a sphere. So, I say that (it is)
ἡΑΒτῆςΒΓ.ὡςδὲἡΑΒπρὸςτὴνΒΓ,οὕτωςτὸἀπὸ also (enclosed) by the given (sphere). For since LK is
τῆςΑΒπρὸςτὸἀπὸτῆςΒΔ·διπλάσιονἄραἐστὶτὸἀπὸ equal to KM, and KE (is) common, and they contain
τῆςΑΒτοῦἀπὸτῆςΒΔ.ἐδείχθηδὲκαὶτὸἀπὸτῆςΛΜ right-angles, the base LE is thus equal to the base EM
διπλάσιοντοῦἀπὸτῆςΛΕ.καίἐστινἴσοντὸἀπὸτῆςΔΒ [Prop. 1.4]. And since angle LEM is a right-angle—for
τῷἀπὸτῆςΛΕ·ἴσηγὰρκεῖταιἡΕΘτῇΔΒ.ἴσονἄρακαὶ (it is) in a semi-circle [Prop. 3.31]—the (square) on LM
τὸἀπὸτῆςΑΒτῷἀπὸτῆςΛΜ·ἴσηἄραἡΑΒτῇΛΜ.καί is thus double the (square) on LE [Prop. 1.47]. Again,
ἐστινἡΑΒἡτῆςδοθείσηςσφαίραςδιάμετρος·ἡΛΜἄρα since AC is equal to CB, AB is double BC. And as AB
ἴσηἐστὶτῇτῆςδοθείσηςσφαίραςδιαμέτρῳ.
(is) to BC, so the (square) on AB (is) to the (square)
Περιείληπταιἄρατὸὀκτάεδροντῇδοθείσῃσφαίρᾳ.καὶ on BD [Prop. 6.8, Def. 5.9]. Thus, the (square) on AB is
συναποδέδεικται,ὅτιἡτῆςσφαίραςδιάμετροςδυνάμειδι- double the (square) on BD. And the (square) on LM was
πλασίωνἐστὶτῆςτοῦὀκταέδρουπλευρᾶς·ὅπερἔδειδεῖξαι. also shown (to be) double the (square) on LE. And the
(square) on DB is equal to the (square) on LE. For EH
was made equal to DB. Thus, the (square) on AB (is)
also equal to the (square) on LM. Thus, AB (is) equal to
LM. And AB is the diameter of the given sphere. Thus,
LM is equal to the diameter of the given sphere.
Thus, the octahedron has been enclosed by the given
sphere, and it has been simultaneously proved that the
square on the diameter of the sphere is double the
(square) on the side of the octahedron. † (Which is) the
very thing it was required to show.
† If the radius of the sphere is unity then the side of octahedron is √ 2.
. Proposition 15
Κύβονσυστήσασθαικαὶσφαίρᾳπεριλαβεῖν,ᾗκαὶτὴν To construct a cube, and to enclose (it) in a sphere,
πυραμίδα,καὶδεῖξαι,ὅτιἡτῆςσφαίραςδιάμετροςδυνάμει like in the (case of the) pyramid, and to show that the
τριπλασίωνἐστὶτῆςτοῦκύβουπλευρᾶς.
square on the diameter of the sphere is three times the
᾿ΕκκείσθωἡτῆςδοθείσηςσφαίραςδιάμετροςἡΑΒκαὶ (square) on the side of the cube.
τετμήσθωκατὰτὸΓὥστεδιπλῆνεἶναιτὴνΑΓτῆςΓΒ,καὶ Let the diameter AB of the given sphere be laid out,
γεγράφθωἐπὶτῆςΑΒἡμικύκλιοντὸΑΔΒ,καὶἀπὸτοῦΓ and let it have been cut at C such that AC is double
τῇΑΒπρὸςὀρθὰςἤχθωἡΓΔ,καὶἐπεζεύχθωἡΔΒ,καὶ CB. And let the semi-circle ADB have been drawn on
ἐκκείσθωτετράγωνοντὸΕΖΗΘἴσηνἔχοντὴνπλευρὰντῇ AB. And let CD have been drawn from C at right-
ΔΒ,καὶἀπὸτῶνΕ,Ζ,Η,ΘτῷτοῦΕΖΗΘτετραγώνου angles to AB. And let DB have been joined. And let the
ἐπιπέδῳπρὸςὀρθὰςἤχθωσαναἱΕΚ,ΖΛ,ΗΜ,ΘΝ,καὶ square EFGH, having (its) side equal to DB, be laid out.
ἀφῃρήσθωἀπὸἑκάστηςτῶνΕΚ,ΖΛ,ΗΜ,ΘΝμιᾷτῶν And let EK, FL, GM, and HN have been drawn from
ΕΖ,ΖΗ,ΗΘ,ΘΕἴσηἑκάστητῶνΕΚ,ΖΛ,ΗΜ,ΘΝ,καὶ (points) E, F , G, and H, (respectively), at right-angles to
ἐπεζεύχθωσαναἱΚΛ,ΛΜ,ΜΝ,ΝΚ·κύβοςἄρασυνέσταται the plane of square EFGH. And let EK, FL, GM, and
ὁΖΝὑπὸἓξτετραγώνωνἴσωνπεριεχόμενος. HN, equal to one of EF , FG, GH, and HE, have been
Δεῖ δὴ αὐτὸν καὶ σφαίρᾳ περιλαβεῖν τῇ δοθείσῃ καὶ cut off from EK, FL, GM, and HN, respectively. And let
δεῖξαι,ὅτιἡτῆςσφαίραςδιάμετροςδυνάμειτριπλασίαἐστὶ KL, LM, MN, and NK have been joined. Thus, a cube
τῆςπλευρᾶςτοῦκύβου.
contained by six equal squares has been constructed.
So, it is also necessary to enclose it by the given
sphere, and to show that the square on the diameter of
the sphere is three times the (square) on the side of the
cube.
525

ËÌÇÁÉÁÏƺ ELEMENTS
Ã
Æ Â
K
E
BOOK 13
N
H
Ä
À Å
L
F
M
D
G
᾿ΕπεζεύχθωσανγὰραἱΚΗ,ΕΗ.καὶἐπεὶὀρθήἐστιν For let KG and EG have been joined. And since an-
ἡὑπὸΚΕΗγωνίαδιὰτὸκαὶτὴνΚΕὀρθὴνεἶναιπρὸς gle KEG is a right-angle—on account of KE also being
τὸΕΗἐπίπεδονδηλαδὴκαὶπρὸςτὴνΕΗεὐθεῖαν,τὸἄρα at right-angles to the plane EG, and manifestly also to
ἐπὶ τῆς ΚΗ γραφόμενον ἡμικύκλιον ἥξει καὶ διὰ τοῦ Ε the straight-line EG [Def. 11.3]—the semi-circle drawn
σημείου.πάλιν,ἐπεὶἡΗΖὀρθήἐστιπρὸςἑκατέραντῶνΖΛ, on KG will thus also pass through point E. Again, since
ΖΕ,καὶπρὸςτὸΖΚἄραἐπίπεδονὀρθήἐστινἡΗΖ·ὥστε GF is at right-angles to each of FL and FE, GF is thus
καὶἐὰνἐπιζεύξωμεντὴνΖΚ,ἡΗΖὀρθὴἔσταικαὶπρὸς also at right-angles to the plane FK [Prop. 11.4]. Hence,
τὴνΖΚ·καὶδὶατοῦτοπάλιντὸἐπὶτῆςΗΚγραφόμενον if we also join FK then GF will also be at right-angles
ἡμικύκλιονἥξεικαὶδιὰτοῦΖ.ὁμοίωςκαὶδὶατῶνλοιπῶν to FK. And, again, on account of this, the semi-circle
τοῦκύβουσημείωνἥξει. ἐὰνδὴμενούσηςτῆςΚΗπε- drawn on GK will also pass through point F . Similarly,
ριενεχθὲντὸἡμικύκλιονεἰςτὸαὐτὸἀποκατασταθῇ,ὅθεν it will also pass through the remaining (angular) points of
ἤρξατο φέρεσθαι, ἔσται σφαίρᾳ περιειλημμένοςὁκύβος. the cube. So, if KG remains (fixed), and the semi-circle is
λέγωδή,ὅτικαὶτῇδοθείσῃ. ἐπεὶγὰρἴσηἐστὶνἡΗΖτῇ carried around, and again established at the same (posi-
ΖΕ,καίἐστινὀρθὴἡπρὸςτῷΖγωνία,τὸἄραἀπὸτῆςΕΗ tion) from which it began to be moved, then the cube will
διπλάσιόνἐστιτοῦἀπὸτῆςΕΖ.ἴσηδὲἡΕΖτῇΕΚ·τὸἄρα have been enclosed by a sphere. So, I say that (it is) also
ἀπὸτῆςΕΗδιπλάσιόνἐστιτοῦἀπὸτῆςΕΚ·ὥστετὰἀπὸ (enclosed) by the given (sphere). For since GF is equal
τῶνΗΕ,ΕΚ,τουτέστιτὸἀπὸτῆςΗΚ,τριπλάσιόνἐστιτοῦ to FE, and the angle at F is a right-angle, the (square)
ἀπὸτῆςΕΚ.καὶἐπεὶτριπλασίωνἐστὶνἡΑΒτῆςΒΓ,ὡς on EG is thus double the (square) on EF [Prop. 1.47].
δὲἡΑΒπρὸςτὴνΒΓ,οὕτωςτὸἀπὸτῆςΑΒπρὸςτὸἀπὸ And EF (is) equal to EK. Thus, the (square) on EG
τῆςΒΔ,τριπλάσιονἄρατὸἀπὸτῆςΑΒτοῦἀπὸτῆςΒΔ. is double the (square) on EK. Hence, the (sum of the
ἐδείχθηδὲκαὶτὸἀπὸτῆςΗΚτοῦἀπὸτῆςΚΕτριπλάσιον. squares) on GE and EK—that is to say, the (square) on
καὶκεῖταιἴσηἡΚΕτῇΔΒ·ἴσηἄρακαὶἡΚΗτῇΑΒ.καί GK [Prop. 1.47]—is three times the (square) on EK.
ἐστινἡΑΒτῆςδοθείσηςσφαίραςδιάμετρος·καὶἡΚΗἄρα And since AB is three times BC, and as AB (is) to
ἴσηἐστὶτῇτῆςδοθείσηςσφαίραςδιαμέτρῳ.
BC, so the (square) on AB (is) to the (square) on BD
Τῇδοθείσῃἄρασφαίραπεριείληπταιὁκύβος·καὶσυ- [Prop. 6.8, Def. 5.9], the (square) on AB (is) thus three
ναποδέδεικται, ὅτιἡτῆςσφαίραςδιάμετροςδυνάμειτρι- times the (square) on BD. And the (square) on GK was
πλασίωνἐστὶτῆςτοῦκύβουπλευρᾶς·ὅπερἔδειδεῖξαι. also shown (to be) three times the (square) on KE. And
KE was made equal to DB. Thus, KG (is) also equal to
AB. And AB is the radius of the given sphere. Thus, KG
is also equal to the diameter of the given sphere.
Thus, the cube has been enclosed by the given sphere.
And it has simultaneously been shown that the square on
the diameter of the sphere is three times the (square) on
A
C
B
526

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
† If the radius of the sphere is unity then the side of the cube is p 4/3.
the side of the cube. † (Which is) the very thing it was
required to show.
¦. Proposition 16
Εἰκοσάεδρονσυστήσασθαικαὶσφαίρᾳπεριλαβεῖν,ᾗκαὶ To construct an icosahedron, and to enclose (it) in a
τὰπροειρημένασχήματα,καὶδεῖξαι,ὅτιἡτοῦεἰκοσαέδρου
sphere, like the aforementioned figures, and to show that
πλευρὰἄλογόςἐστινἡκαλουμένηἐλάττων.
the side of the icosahedron is that irrational (straightline)
called minor.
D
C
᾿ΕκκείσθωἡτῆςδοθείσηςσφαίραςδιάμετροςἡΑΒκαὶ Let the diameter AB of the given sphere be laid out,
τετμήσθωκατὰτὸΓὥστετετραπλῆνεἶναιτὴνΑΓτῆςΓΒ, and let it have been cut at C such that AC is four times
καὶγεγράφθωἐπὶτῆςΑΒἡμικύκλιοντὸΑΔΒ,καὶἤχθω CB [Prop. 6.10]. And let the semi-circle ADB have been
ἀπὸτοῦΓτῇΑΒπρὸςορθὰςγωνίαςεὐθεῖαγραμμὴἡΓΔ, drawn on AB. And let the straight-line CD have been
καίἐπεζεύχθωἡΔΒ, καὶἐκκείσθωκύκλοςὁΕΖΗΘΚ, drawn from C at right-angles to AB. And let DB have
οὗἡἐντοῦκέντρουἴσηἔστωτῇΔΒ,καὶἐγγεγράφθω been joined. And let the circle EFGHK be set down,
εἰς τὸν ΕΖΗΘΚ κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ and let its radius be equal to DB. And let the equilat-
ἰσογώνιοντὸΕΖΗΘΚ,καὶτετμήσθωσαναἱΕΖ,ΖΗ,ΗΘ, eral and equiangular pentagon EFGHK have been in-
ΘΚ,ΚΕπεριφέρειαιδίχακατὰτὸΛ,Μ,Ν,Ξ,Οσημεῖα,καὶ scribed in circle EFGHK [Prop. 4.11]. And let the cir-
ἐπεζεύχθωσαναἱΛΜ,ΜΝ,ΝΞ,ΞΟ,ΟΛ,ΕΟ.ἴσόπλευρον cumferences EF , FG, GH, HK, and KE have been cut
ἄραἐστὶκαὶτὸΛΜΝΞΟπεντάγωνον, καὶδεκαγώνουἡ in half at points L, M, N, O, and P (respectively). And
ΕΟ εὐθεῖα. καὶ ἀνεστάτωσαν ἄπὸ τῶν Ε, Ζ, Η, Θ, Κ let LM, MN, NO, OP , PL, and EP have been joined.
σημείωντῷτοῦκύκλουἐπιπέδῳπρὸςὀρθὰςγωνίαςεὐθεῖαι Thus, pentagon LMNOP is also equilateral, and EP (is)
αἱΕΠ,ΖΡ,ΗΣ,ΘΤ,ΚΥἴσαιοὖσαιτῇἐκτοῦκέντρουτοῦ the side of the decagon (inscribed in the circle). And let
ΕΖΗΘΚκύκλου,καὶἐπεζεύχθωσαναἱΠΡ,ΡΣ,ΣΤ,ΤΥ, the straight-lines EQ, FR, GS, HT , and KU, which are
ΥΠ,ΠΛ,ΛΡ,ΡΜ,ΜΣ,ΣΝ,ΝΤ,ΤΞ,ΞΥ,ΥΟ,ΟΠ. equal to the radius of circle EFGHK, have been set up
ΚαὶἐπεὶἑκατέρατῶνΕΠ,ΚΥτῷαὐτῷἐπιπέδῳπρὸς at right-angles to the plane of the circle, at points E, F ,
ὀρθάς ἐστιν, παράλληλοςἄρα ἐστὶν ἡ ΕΠ τῇ ΚΥ. ἔστι G, H, and K (respectively). And let QR, RS, ST , TU,
δὲαὐτῇκαὶἴση· αἱδὲτὰςἴσαςτεκαὶπαραλλήλουςἐπι- UQ, QL, LR, RM, MS, SN, NT , TO, OU, UP , and PQ
ζευγνύουσαιἐπὶτὰαὐτὰμέρηεὐθεῖαιἴσαιτεκαὶπαράλληλοί have been joined.
εἰσιν.ἡΠΥἄρατῇΕΚἴσητεκαὶπαράλληλόςἐστιν. πεν- And since EQ and KU are each at right-angles to the
ταγώνουδὲἰσοπλεύρουἡΕΚ·πενταγώνουἄραἰσοπλεύρου same plane, EQ is thus parallel to KU [Prop. 11.6]. And
καὶ ἡ ΠΥ τοῦ εἰς τὸν ΕΖΗΘΚ κύκλον ἐγγραφομένου. it is also equal to it. And straight-lines joining equal and
διὰτὰαὐτὰδὴκαὶἑκάστητῶν ΠΡ, ΡΣ, ΣΤ,ΤΥπεν- parallel (straight-lines) on the same side are (themselves)
ταγώνου ἐστὶν ἰσοπλεύρουτοῦ εἰς τὸν ΕΖΗΘΚ κύκλον equal and parallel [Prop. 1.33]. Thus, QU is equal and
ἐγγραφομένου· ἰσόπλευρονἄρατὸΠΡΣΤΥπεντάγωνον. parallel to EK. And EK (is the side) of an equilateral
καὶἐπεὶἑξαγώνουμένἐστινἡΠΕ,δεκαγώνουδὲἡΕΟ, pentagon (inscribed in circle EFGHK). Thus, QU (is)
καίἐστινὀρθὴἡὑπὸΠΕΟ,πενταγώνουἄραἐστὶνἡΠΟ·ἡ also the side of an equilateral pentagon inscribed in circle
γὰρτοῦπενταγώνουπλευρὰδύναταιτήντετοῦἑξαγώνου EFGHK. So, for the same (reasons), QR, RS, ST , and
καὶτὴντοῦδεκαγώνουτῶνεἰςτὸναὐτὸνκύκλονἐγγρα- TU are also the sides of an equilateral pentagon inscribed
φομένων. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΟΥ πενταγώνου ἐστὶ in circle EFGHK. Pentagon QRSTU (is) thus equilat-
A
B
527

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
πλευρά. ἔστιδὲκαὶἡΠΥπενταγώνου· ἰσόπλευρονἄρα eral. And side QE is (the side) of a hexagon (inscribed
ἐστὶτὸΠΟΥτρίγωνον. διὰτὰαὐτὰδὴκαὶἕκαστοντῶν in circle EFGHK), and EP (the side) of a decagon, and
ΠΛΡ,ΡΜΣ,ΣΝΤ,ΤΞΥἰσόπλευρόνἐστιν. καὶἐπεὶπεν- (angle) QEP is a right-angle, thus QP is (the side) of a
ταγώνουἐδείχθηἑκατέρατῶνΠΛ,ΠΟ,ἔστιδὲκαὶἡΛΟ pentagon (inscribed in the same circle). For the square
πενταγώνου,ἰσόπλευρονἄραἐστὶτὸΠΛΟτρίγωνον. διὰ on the side of a pentagon is (equal to the sum of) the
τὰ αὐτὰ δὴ καὶ ἕκαστον τῶν ΛΡΜ, ΜΣΝ, ΝΤΞ, ΞΥΟ (squares) on (the sides of) a hexagon and a decagon inτριγώνωνἰσόπλευρόνἐστιν.
scribed in the same circle [Prop. 13.10]. So, for the same
(reasons), PU is also the side of a pentagon. And QU
is also (the side) of a pentagon. Thus, triangle QPU is
equilateral. So, for the same (reasons), (triangles) QLR,
RMS, SNT , and TOU are each also equilateral. And
since QL and QP were each shown (to be the sides) of a
pentagon, and LP is also (the side) of a pentagon, triangle
QLP is thus equilateral. So, for the same (reasons),
Ê È
triangles LRM, MSN, NTO, and OUP are each also
equilateral.
Ä
R
Q
L
Å É Ï
E
Ç
F
Z
M
P
Ë À Ã
W
a
Í
V
Æ Â
G X
K
S
U
N
O
H
Ì
T
ΕἰλήφθωτὸκέντροντοῦΕΖΗΘΚκύκλουτὸΦσημεῖον· Let the center, point V , of circle EFGHK have been
καὶἀπὸτοῦΦτῷτοῦκύκλουἐπιπέδῳπρὸςὀρθὰςἀνεστάτω found [Prop. 3.1]. And let V Z have been set up, at
ἡΦΩ,καὶἐκβεβλήσθωἐπὶτὰἕτεραμέρηὡςἡΦΨ,καὶ (point) V , at right-angles to the plane of the circle. And
ἀφῃρήσθωἑξαγώνουμὲνἡΦΧ,δεκαγώνουδὲἑκατέρατῶν let it have been produced on the other side (of the cir-
ΦΨ,ΧΩ,καὶἐπεζεύχθωσαναἱΠΩ,ΠΧ,ΥΩ,ΕΦ,ΛΦ,ΛΨ, cle), like V X. And let V W have been cut off (from XZ
ΨΜ.
so as to be equal to the side) of a hexagon, and each of
ΚαὶἐπεὶἑκατέρατῶνΦΧ,ΠΕτῷτοῦκύκλουἐπιπέδῳ V X and WZ (so as to be equal to the side) of a decagon.
πρὸςὀρθάςἐστιν,παράλληλοςἄραἐστὶνἡΦΧτῇΠΕ.εἰσὶ And let QZ, QW , UZ, EV , LV , LX, and XM have been
δὲ καὶἴσαι· καὶαἱΕΦ, ΠΧ ἄραἴσαιτε καὶπαράλληλοί joined.
εἰσιν. ἑξαγώνουδὲἡΕΦ·ἑξαγώνουἄρακαὶἡΠΧ.καὶ And since V W and QE are each at right-angles
ἐπεὶἑξαγώνουμένἐστινἡΠΧ,δεκαγώνουδὲἡΧΩ,καὶ to the plane of the circle, V W is thus parallel to QE
ὀρθήἐστινἡὑπὸΠΧΩγωνία, πενταγώνουἄραἐστὶνἡ [Prop. 11.6]. And they are also equal. EV and QW are
ΠΩ.διὰτὰαὐτὰδὴκαὶἡΥΩπενταγώνουἐστίν,ἐπειδήπερ, thus equal and parallel (to one another) [Prop. 1.33].
528

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
ἐὰνἐπιζεύξωμεντὰςΦΚ,ΧΥ,ἴσαικαὶἀπεναντίονἔσον- And EV (is the side) of a hexagon. Thus, QW (is) also
ται, καί ἐστιν ἡ ΦΚ ἐκ τοῦ κέντρου οὖσα ἑξαγώνου. (the side) of a hexagon. And since QW is (the side) of
ἑξαγώνουἄρακαὶἡΧΥ.δεκαγώνουδὲἡΧΩ,καὶὀρθὴ a hexagon, and WZ (the side) of a decagon, and angle
ἡὑπὸΥΧΩ·πενταγώνουἄραἡΥΩ.ἔστιδὲ καὶἡΠΥ QWZ is a right-angle [Def. 11.3, Prop. 1.29], QZ is thus
πενταγώνου·ἰσόπλευρονἄραἐστὶτὸΠΥΩτρίγωνον. διὰ (the side) of a pentagon [Prop. 13.10]. So, for the same
τὰαὐτὰδὴκαὶἕκαστοντῶνλοιπῶντριγώνων,ὧνβάσεις (reasons), UZ is also (the side) of a pentagon—inasmuch
μένεἰσιναἱΠΡ,ΡΣ,ΣΤ,ΤΥεὐθεῖαι,κορυφὴδὲτὸΩ as, if we join V K and WU then they will be equal and
σημεῖον, ἰσόπλευρόνἐστιν. πάλιν, ἐπεὶἑξαγώνουμὲνἡ opposite. And V K, being (equal) to the radius (of the cir-
ΦΛ, δεκαγώνου δὲ ἡ ΦΨ, καὶ ὀρθή ἐστιν ἡ ὑπὸ ΛΦΨ cle), is (the side) of a hexagon [Prop. 4.15 corr.]. Thus,
γωνία,πενταγώνουἄραἐστὶνἡΛΨ.διὰτὰαὐτὰδὴἐὰν WU (is) also the side of a hexagon. And WZ (is the side)
ἐπιζεύξωμεντὴνΜΦοὖσανἑξαγώνου,συνάγεταικαὶἡΜΨ of a decagon, and (angle) UWZ (is) a right-angle. Thus,
πενταγώνου. ἔστιδὲκαὶἡΛΜπενταγώνου·ἰσόπλευρον UZ (is the side) of a pentagon [Prop. 13.10]. And QU
ἄραἐστὶτὸΛΜΨτρίγωνον. ὁμοίωςδὴδειχθήσεται,ὅτι is also (the side) of a pentagon. Triangle QUZ is thus
καὶἕκαστοντῶνλοιπῶντριγώνων,ὧνβάσειςμένεἰσιναἱ equilateral. So, for the same (reasons), each of the re-
ΜΝ,ΝΞ,ΞΟ,ΟΛ,κορυφὴδὲτὸΨσημεὶον,ἰσόπλευρόν maining triangles, whose bases are the straight-lines QR,
ἐστιν. συνέσταταιἄραεἰκοσάεδρονὑπὸεἴκοσιτριγώνων RS, ST , and TU, and apexes the point Z, are also equi-
ἰσοπλεύρωνπεριεχόμενον.
lateral. Again, since V L (is the side) of a hexagon, and
Δεῖδὴαὐτὸκαὶσφαίρᾳπεριλαβεῖντῇδοθείσῃκαὶδεῖξαι, V X (the side) of a decagon, and angle LV X is a right-
ὅτιἡτοῦεἰκοσαέδρουπλευρὰἄλογόςἐστινἡκαλουμένη angle, LX is thus (the side) of a pentagon [Prop. 13.10].
ἐλάσσων.
So, for the same (reasons), if we join MV , which is (the
᾿ΕπεὶγὰρἑξαγώνουἐστὶνἡΦΧ,δεκαγώνουδὲἡΧΩ,ἡ side) of a hexagon, MX is also inferred (to be the side)
ΦΩἄραἄκρονκαὶμέσονλόγοντέτμηταικατὰτὸΧ,καὶτὸ of a pentagon. And LM is also (the side) of a pentagon.
μεῖζοναὐτῆςτμῆμάἐστινἡΦΧ·ἔστινἄραὡςἡΩΦπρὸς Thus, triangle LMX is equilateral. So, similarly, it can
τὴνΦΧ,οὕτωςἡΦΧπρὸςτὴνΧΩ.ἴσηδὲἡμὲνΦΧτῇ be shown that each of the remaining triangles, whose
ΦΕ,ἡδὲΧΩτῇΦΨ·ἔστινἄραὡςἡΩΦπρὸςτὴνΦΕ, bases are the (straight-lines) MN, NO, OP , and PL,
οὕτωςἡΕΦπρὸςτὴνΦΨ.καίεἰσινὀρθαὶαἱὑπὸΩΦΕ, and apexes the point X, are also equilateral. Thus, an
ΕΦΨγωνίαι·ἐὰνἄραἐπιζεύξωμεντὴνΕΩεὐθεὶαν,ὀρθὴ icosahedron contained by twenty equilateral triangles has
ἔσταιἡὑπὸΨΕΩγωνίαδιὰτὴνὁμοιότητατῶνΨΕΩ,ΦΕΩ been constructed.
τριγώνων. διὰτὰαὐτὰδὴἐπείἐστινὡςἡΩΦπρὸςτὴν So, it is also necessary to enclose it in the given
ΦΧ,οὕτωςἡΦΧπρὸςτὴνΧΩ,ἴσηδὲἡμὲνΩΦτῇΨΧ, sphere, and to show that the side of the icosahedron is
ἡδὲΦΧτῇΧΠ,ἔστινἄραὡςἡΨΧπρὸςτὴνΧΠ,οὕτως that irrational (straight-line) called minor.
ἡΠΧπρὸςτὴνΧΩ.καὶδιὰτοῦτοπάλινἐὰνἐπιζεύξωμεν For, since V W is (the side) of a hexagon, and WZ
τὴνΠΨ,ὀρθὴἔσταιἡπρὸςτῷΠγωνία·τὸἄραἐπὶτῆς (the side) of a decagon, V Z has thus been cut in ex-
ΨΩγραφόμενονἡμικύκλιονἥξεικαὶδὶατοῦΠ.καὶἐὰν treme and mean ratio at W , and V W is its greater piece
μενούσηςτῆςΨΩπεριενεχθὲντὸἡμικύκλιονεἰςτὸαὐτὸ [Prop. 13.9]. Thus, as ZV is to V W , so V W (is) to WZ.
πάλινἀποκατασταθῇ,ὅθενἤρξατοφέρεσθαι,ἥξεικαὶδιὰ And V W (is) equal to V E, and WZ to V X. Thus, as
τοῦΠκαὶτῶνλοιπῶνσημείωντοῦεἰκοσαέδρου,καὶἔσται ZV is to V E, so EV (is) to V X. And angles ZV E and
σφαίρᾳπεριειλημμένοντὸεἰκοσάεδρον. λέγωδή,ὅτικαὶ EV X are right-angles. Thus, if we join straight-line EZ
τῇδοθείσῃ. τετμήσθωγὰρἡΦΧδίχακατὰτὸα. καὶἐπεὶ then angle XEZ will be a right-angle, on account of the
εὐθεῖαγραμμὴἡΦΩἄκρονκαὶμέσονλόγοντέτμηταικατὰ similarity of triangles XEZ and V EZ. [Prop. 6.8]. So,
τὸΧ,καὶτὸἔλασσοναὐτῆςτμῆμάἐστινἡΩΧ,ἡἄραΩΧ for the same (reasons), since as ZV is to V W , so V W
προσλαβοῦσατὴνἡμίσειαντοῦμείζονοςτμήματοςτὴνΧα (is) to WZ, and ZV (is) equal to XW , and V W to WQ,
πενταπλάσιονδύναταιτοῦἀπὸτῆςἡμισείαςτοῦμείζονος thus as XW is to WQ, so QW (is) to WZ. And, again,
τμήματος·πενταπλάσιονἄραἐστὶτὸἀπὸτῆςΩατοῦἀπὸ on account of this, if we join QX then the angle at Q will
τῆςαΧ.καίἐστιτῆςμὲνΩαδιπλῆἡΩΨ,τὴςδὲαΧδιπλῆ be a right-angle [Prop. 6.8]. Thus, the semi-circle drawn
ἡΦΧ·πενταπλάσιονἄραἐστὶτὸἀπὸτῆςΩΨτοῦἀπὸτῆς on XZ will also pass through Q [Prop. 3.31]. And if XZ
ΧΦ.καὶἐπεὶτετραπλῆἐστινἡΑΓτῆςΓΒ,πενταπλῆἄρα remains fixed, and the semi-circle is carried around, and
ἐστὶνἡΑΒτῆςΒΓ.ὡςδὲἡΑΒπρὸςτὴνΒΓ,οὕτωςτὸ again established at the same (position) from which it
ἀπὸτῆςΑΒπρὸςτὸἀπὸτῆςΒΔ·πενταπλάσιονἄραἐστὶ began to be moved, then it will also pass through (point)
τὸἀπὸτῆςΑΒτοῦἀπὸτῆςΒΔ.ἐδείχθηδὲκαὶτὸἀπὸτῆς Q, and (through) the remaining (angular) points of the
ΩΨπενταπλάσιοντοῦἀπὸτῆςΦΧ.καίἐστινἴσηἡΔΒτῇ icosahedron. And the icosahedron will have been en-
529

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
ΦΧ·ἑκατέραγὰραὐτῶνἴσηἐστὶτῇἐκτοῦκέντρουτοῦ closed by a sphere. So, I say that (it is) also (enclosed)
ΕΖΗΘΚκύκλου·ἴσηἄρακαὶἡΑΒτῇΨΩ.καίἐστινἡΑΒ by the given (sphere). For let V W have been cut in half
ἡτῆςδοθείσηςσφαίραςδιάμετρος·καὶἡΨΩἄραἴσηἐστὶ at a. And since the straight-line V Z has been cut in exτῇτῆςδοθείσηςσφαίραςδιαμέτρῳ·τῇἄραδοθείσῃσφαίρᾳ
treme and mean ratio at W , and ZW is its lesser piece,
περιείληπταιτὸεἰκοσάεδρον.
then the square on ZW added to half of the greater piece,
Λέγωδή,ὅτιἡτοῦεἰκοσαέδρουπλευρὰἄλογόςἐστινἡ Wa, is five times the (square) on half of the greater piece
καλουμένηἐλάττων. ἐπεὶγὰρῥητήἐστινἡτῆςσφαίρας [Prop. 13.3]. Thus, the (square) on Za is five times the
διάμετρος, καί ἐστι δυνάμει πενταπλασίων τῆς ἐκ τοῦ (square) on aW . And ZX is double Za, and V W double
κέντρου τοῦ ΕΖΗΘΚ κύκλου, ῥητὴ ἄρα ἐστὶ καὶ ἡ ἑκ aW . Thus, the (square) on ZX is five times the (square)
τοῦκέντρουτοῦΕΖΗΘΚκύκλου·ὥστεκαὶἡδιάμετρος on WV . And since AC is four times CB, AB is thus
αὐτοῦῥητήἐστιν. ἐὰνδὲεἰςκύκλονῥητὴνἔχοντατὴν five times BC. And as AB (is) to BC, so the (square)
διάμετρον πεντάγωνον ἰσόπλευρον ἐγγραφῃ, ἡ τοῦ πεν- on AB (is) to the (square) on BD [Prop. 6.8, Def. 5.9].
ταγώνουπλευρὰἄλογόςἐστινἡκαλουμένηἐλάττων.ἡδὲ Thus, the (square) on AB is five times the (square) on
τοῦΕΖΗΘΚπενταγώνουπλευρὰἡτοῦεἰκοσαέδρουἐστίν. BD. And the (square) on ZX was also shown (to be)
ἡἄρατοῦείκοσαέδρουπλευρὰἄλογόςἐστινἡκαλουμένη five times the (square) on V W . And DB is equal to V W .
ἐλάττων.
For each of them is equal to the radius of circle EFGHK.
Thus, AB (is) also equal to XZ. And AB is the diameter
of the given sphere. Thus, XZ is equal to the diameter
of the given sphere. Thus, the icosahedron has been enclosed
by the given sphere.
So, I say that the side of the icosahedron is that irrational
(straight-line) called minor. For since the diameter
of the sphere is rational, and the square on it is five times
the (square) on the radius of circle EFGHK, the radius
of circle EFGHK is thus also rational. Hence, its diameter
is also rational. And if an equilateral pentagon
is inscribed in a circle having a rational diameter then
the side of the pentagon is that irrational (straight-line)
called minor [Prop. 13.11]. And the side of pentagon
EFGHK is (the side) of the icosahedron. Thus, the side
of the icosahedron is that irrational (straight-line) called
minor.
Corollary
ÈÖ×Ñ.
᾿Εκδὴτούτουφανερόν, ὅτιἡτῆςσφαίραςδιάμετρος So, (it is) clear, from this, that the square on the diδυνάμειπενταπλασίωνἐστὶτῆςἐκτοῦκέντρουτοῦκύκλου,
ameter of the sphere is five times the square on the ra-
ἀφ᾿οὗτὸεἰκοσάεδρονἀναγέγραπται,καὶὅτιἡτῆςσφαίρας dius of the circle from which the icosahedron has been
διάμετροςσύγκειταιἔκτετῆςτοῦἑξαγώνουκαὶδύοτῶν described, and that the the diameter of the sphere is the
τοῦδεκαγώνουτῶνεἰςτὸναὐτὸνκύκλονἐγγραφομένων. sum of (the side) of the hexagon, and two of (the sides)
ὅπερἔδειδεῖξαι. of the decagon, inscribed in the same circle. †
† If the radius of the sphere is unity then the radius of the circle is 2/ √ 5, and the sides of the hexagon, decagon, and pentagon/icosahedron are
2/ √ 5, 1 − 1/ √ 5, and (1/ √ p
5) 10 − 2 √ 5, respectively.
Þ. Proposition 17
Δωδεκάεδρονσυστήσασθαικαὶσφαίρᾳπεριλαβεῖν,ᾗκαὶ To construct a dodecahedron, and to enclose (it) in a
τὰπροειρημένασχήματα,καὶδεῖξαι,ὅτιἡτοῦδωδεκαέδρου sphere, like the aforementioned figures, and to show that
πλευρὰἄλογόςἐστινἡκαλουμένηἀποτομή.
the side of the dodecahedron is that irrational (straightline)
called an apotome.
530

ËÌÇÁÉÁÏƺ ELEMENTS
À
Æ ÊÍ
Å Ë Ç
Ì
 Ï
É
È Ã
Ä
G
A
N
E
L
T
BOOK 13
᾿Εκκείσθωσαν τοῦ προειρημένου κύβου δύο ἐπίπεδα Let two planes of the aforementioned cube [Prop.
πρὸς ὀρθὰς ἀλλήλοις τὰ ΑΒΓΔ, ΓΒΕΖ, καὶ τετμήσθω 13.15], ABCD and CBEF , (which are) at right-angles
ἑκάστητῶνΑΒ,ΒΓ,ΓΔ,ΔΑ,ΕΖ,ΕΒ,ΖΓπλευρῶνδίχα to one another, be laid out. And let the sides AB, BC,
κατὰτὰΗ,Θ,Κ,Λ,Μ,Ν,Ξ,καὶἐπεζεύχθωσαναἱΗΚ, CD, DA, EF , EB, and FC have each been cut in half at
ΘΛ,ΜΘ,ΝΞ,καὶτετηήσθωἑκάστητῶνΝΟ,ΟΞ,ΘΠ points G, H, K, L, M, N, and O (respectively). And let
ἄκρονκαὶμέσονλόγονκατὰτὰΡ,Σ,Τσημεῖα,καὶἔστω GK, HL, MH, and NO have been joined. And let NP ,
αὐτῶνμείζονατμήματατὰΡΟ,ΟΣ,ΤΠ,καὶἀνεστάτωσαν PO, and HQ have each been cut in extreme and mean
ἀπὸτῶνΡ,Σ,Τσημείωντοῖςτοῦκύβουἐπιπέδοιςπρὸς ratio at points R, S, and T (respectively). And let their
ὀρθὰςἐπὶτὰἐκτὸςμέρητοῦκύβουαἱΡΥ,ΣΦ,ΤΧ,καὶ greater pieces be RP , PS, and TQ (respectively). And
κείσθωσανἴσαιταῖςΡΟ,ΟΣ,ΤΠ,καὶἐπεζεύχθωσαναἱ let RU, SV , and TW have been set up on the exterior
ΥΒ,ΒΧ,ΧΓ,ΓΦ,ΦΥ.
side of the cube, at points R, S, and T (respectively), at
Λέγω,ὅτιτὸΥΒΧΓΦπεντάγωνονἰσόπλευρόντεκαὶἐν right-angles to the planes of the cube. And let them be
ἑνὶἐπιπέδῳκαὶἔτιἰσογώνιόνἐστιν. ἐπεζεύχθωσανγὰραἱ made equal to RP , PS, and TQ. And let UB, BW , WC,
ΡΒ,ΣΒ,ΦΒ.καὶἐπεὶεὐθεῖαἡΝΟἄκρονκαὶμέσονλόγον CV , and V U have been joined.
τέτμηταικατὰτὸΡ,καὶτὸμεῖζοντμῆμάἐστινἡΡΟ,τὰἄρα I say that the pentagon UBWCV is equilateral, and
ἀπὸτῶνΟΝ,ΝΡτριπλάσιάἐστιτοῦἀπὸτῆςΡΟ.ἴσηδὲἡ in one plane, and, further, equiangular. For let RB, SB,
μὲνΟΝτῇΝΒ,ἡδὲΟΡτῇΡΥ·τὰἄραἀπὸτῶνΒΝ,ΝΡ and V B have been joined. And since the straight-line NP
τριπλάσιάἐστιτοῦἀπὸτῆςΡΥ.τοῖςδὲἀπὸτῶνΒΝ,ΝΡτὸ has been cut in extreme and mean ratio at R, and RP is
ἀπὸτῆςΒΡἐστινἴσον·τὸἄραἀπὸτῆςΒΡτριπλάσιόνἐστι the greater piece, the (sum of the squares) on PN and
τοῦἀπὸτῆςΡΥ·ὥστετὰἀπὸτῶνΒΡ,ΡΥτετραπλάσιά NR is thus three times the (square) on RP [Prop. 13.4].
ἐστιτοῦἀπὸτῆςΡΥ.τοῖςδὲἀπὸτῶνΒΡ,ΡΥἴσονἐστιτὸ And PN (is) equal to NB, and PR to RU. Thus, the
ἀπὸτῆςΒΥ·τὸἄραἄπὸτῆςΒΥτετραπλάσιόνἐστιτοῦἀπὸ (sum of the squares) on BN and NR is three times the
τῆςΥΡ·διπλῆἄραἐστὶνἡΒΥτῆςΡΥ.ἔστιδὲκαὶἡΦΥτῆς (square) on RU. And the (square) on BR is equal to
ΥΡδιπλῆ,ἐπειδήπερκαὶἡΣΡτῆςΟΡ,τουτέστιτῆςΡΥ, the (sum of the squares) on BN and NR [Prop. 1.47].
ἐστιδιπλῆ·ἴσηἄραἡΒΥτῇΥΦ.ὁμοίωςδὴδειχθήσεται, Thus, the (square) on BR is three times the (square) on
ὅτικαὶἑκάστητῶνΒΧ,ΧΓ,ΓΦἑκατέρᾳτῶνΒΥ,ΥΦ RU. Hence, the (sum of the squares) on BR and RU
ἐστινἴση. ἰσόπλευρονἄραἐστὶτὸΒΥΦΓΧπεντάγωνον. is four times the (square) on RU. And the (square) on
λέγωδή,ὅτικαὶἐνἑνίἐστινἐπιπέδῳ.ἤχθωγὰρἀπὸτοῦΟ BU is equal to the (sum of the squares) on BR and RU
ἑκατέρᾳτῶνΡΥ,ΣΦπαράλληλοςἐπὶτὰἐκτὸςτοῦκύβου [Prop. 1.47]. Thus, the (square) on BU is four times the
μέρηἡΟΨ,καὶἐπεζεύχθωσαναἱΨΘ,ΘΧ·λέγω,ὅτιἡ (square) on UR. Thus, BU is double RU. And V U is also
ΨΘΧεὐθεῖάἐστιν.ἐπεὶγὰρἡΘΠἄκρονκαὶμέσονλόγον double UR, inasmuch as SR is also double PR—that is
τέτμηταικατὰτὸΤ,καὶτὸμεῖζοναὐτῆςτμῆμάἐστινἡΠΤ, to say, RU. Thus, BU (is) equal to UV . So, similarly, it
ἔστινἄραὡςἡΘΠπρὸςτὴνΠΤ,οὕτωςἡΠΤπρὸςτὴν can be shown that each of BW , WC, CV is equal to each
B
R
U
H
W
Q
X
P
Z
M
V
S
C
K
D
O
F
531

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
ΤΘ.ἴσηδὲἡμὲνΘΠτῇΘΟ,ἡδὲΠΤἑκατέρᾳτῶνΤΧ, of BU and UV . Thus, pentagon BUV CW is equilateral.
ΟΨ·ἔστινἄραὡςἡΘΟπρὸςτὴνΟΨ,οὕτωςἡΧΤπρὸς So, I say that it is also in one plane. For let PX have
τὴνΤΘ.καίἐστιπαράλληλοςἡμὲνΘΟτῇΤΧ·ἑκατέρα been drawn from P , parallel to each of RU and SV , on
γὰραὐτῶντῷΒΔἐπιπέδῳπρὸςὀρθάςἐστιν·ἡδὲΤΘτῇ the exterior side of the cube. And let XH and HW have
ΟΨ·ἑκατέραγὰραὐτῶντῷΒΖἐπιπέδῳπρὸςὀρθάςἐστιν. been joined. I say that XHW is a straight-line. For since
ἐὰνδὲδύοτρίγωνασυντεθῇκατὰμίανγωνίαν,ὡςτὰΨΟΘ, HQ has been cut in extreme and mean ratio at T , and
ΘΤΧ,τὰςδύοπλευρὰςταῖςδυνὶνἀνάλογονἔχοντα,ὥστε QT is its greater piece, thus as HQ is to QT , so QT (is)
τὰςὁμολόγουςαὐτῶνπλευρὰςκαὶπαραλλήλουςεἶναι,αἱ to TH. And HQ (is) equal to HP , and QT to each of
λοιπαὶεὐθεῖαιἐπ᾿εὐθείαςἔσονται·ἐπ᾿εὐθείαςἄραἐστὶνἡ TW and PX. Thus, as HP is to PX, so WT (is) to
ΨΘτῇΘΧ.πᾶσαδὲεὐθεῖαἐνἑνίἐστινἐπιπέδῳ·ἐνἑνὶἄρα TH. And HP is parallel to TW. For of each of them is
ἐπιπέδῳἐστὶτὸΥΒΧΓΦπεντάγωνον.
at right-angles to the plane BD [Prop. 11.6]. And TH
Λέγωδή,ὅτικαὶἰσογώνιόνἐστιν.
(is parallel) to PX. For each of them is at right-angles
᾿ΕπεὶγὰρεὐθεῖαγραμμὴἡΝΟἄκρονκαὶμέσονλόγον to the plane BF [Prop. 11.6]. And if two triangles, like
τέτμηταικατὰτὸΡ,καὶτὸμεῖζοντμῆμάἐστινἡΟΡ[ἔστιν XPH and HTW, having two sides proportional to two
ἄραὡςσυναμφότεροςἡΝΟ,ΟΡπρὸςτὴνΟΝ,οὕτωςἡ sides, are placed together at a single angle such that their
ΝΟπρὸςτὴνΟΡ],ἴσηδὲἡΟΡτῇΟΣ[ἔστινἄραὡςἡΣΝ corresponding sides are also parallel then the remaining
πρὸςτὴνΝΟ,οὕτωςἡΝΟπρὸςτὴνΟΣ],ἡΝΣἄραἄκρον sides will be straight-on (to one another) [Prop. 6.32].
καὶμέσονλόγοντέτμηταικατὰτὸΟ,καὶτὸμεῖζοντμῆμά Thus, XH is straight-on to HW . And every straight-line
ἐστινἡΝΟ·τὰἄραἀπὸτῶνΝΣ,ΣΟτριπλάσιάἐστιτοῦ is in one plane [Prop. 11.1]. Thus, pentagon UBWCV is
ἀπὸτῆςΝΟ.ἴσηδὲἡμὲνΝΟτῇΝΒ,ἡδὲΟΣτῇΣΦ·τὰ in one plane.
ἄραἀπὸτῶνΝΣ,ΣΦτετράγωνατριπλάσιάἐστιτοῦἀπὸ So, I say that it is also equiangular.
τῆςΝΒ·ὥστετὰἀπὸτῶνΦΣ,ΣΝ,ΝΒτετραπλάσιάἐστι For since the straight-line NP has been cut in extreme
τοῦἀπὸτῆςΝΒ.τοῖςδὲἀπὸτῶνΣΝ,ΝΒἴσονἐστὶτὸἀπὸ and mean ratio at R, and PR is the greater piece [thus as
τῆςΣΒ·τὰἄραἀπὸτῶνΒΣ,ΣΦ,τουτέστιτὸἀπὸτῆςΒΦ the sum of NP and PR is to PN, so NP (is) to PR], and
[ὀρθὴγὰρἡὑπὸΦΣΒγωνία],τετραπλάσιόνἐστιτοῦἀπὸ PR (is) equal to PS [thus as SN is to NP , so NP (is) to
τῆςΝΒ·διπλῆἄραἐστὶνἡΦΒτῆςΒΝ.ἔστιδὲκαὶἡΒΓ PS], NS has thus also been cut in extreme and mean
τῆςΒΝδιπλῆ·ἴσηἄραἐστὶνἡΒΦτῇΒΓ.καὶἐπεὶδύοαἱ ratio at P , and NP is the greater piece [Prop. 13.5].
ΒΥ,ΥΦδυσὶταῖςΒΧ,ΧΓἴσαιεἰσίν,καὶβάσιςἡΒΦβάσει Thus, the (sum of the squares) on NS and SP is three
τῇΒΓἴση,γωνίαἄραἡὑπὸΒΥΦγωνίᾳτῇὑπὸΒΧΓἐστιν times the (square) on NP [Prop. 13.4]. And NP (is)
ἴση.ὁμοίωςδὴδείξομεν,ὅτικαὶἡὑπὸΥΦΓγωνίαἴσηἐστὶ equal to NB, and PS to SV . Thus, the (sum of the)
τῇὑπὸΒΧΓ·αἱἄραὑπὸΒΧΓ,ΒΥΦ,ΥΦΓτρεῖςγωνίαι squares on NS and SV is three times the (square) on
ἴσαιἀλλήλαιςεἰσίν.ἐὰνδὲπενταγώνουἰσοπλεύρουαἱτρεῖς NB. Hence, the (sum of the squares) on V S, SN, and
γωνίαιἴσαιἀλλήλαιςὦσιν,ἰσογώνιονἔσταιτὸπεντάγωνον· NB is four times the (square) on NB. And the (square)
ἰσογώνιονἄραἐστὶτὸΒΥΦΓΧπεντάγωνον.ἐδείχθηδὲκαὶ on SB is equal to the (sum of the squares) on SN and
ἰσόπλευρον·τὸἄραΒΥΦΓΧπεντάγωνονἰσόπλευρόνἐστι NB [Prop. 1.47]. Thus, the (sum of the squares) on BS
καὶἰσογώνιον,καίἐστινἐπὶμιᾶςτοῦκύβουπλευρᾶςτῆς and SV —that is to say, the (square) on BV [for angle
ΒΓ.ἐὰνἄραἐφ᾿ἑκάστηςτῶντοῦκύβουδώδεκαπλευρῶν V SB (is) a right-angle]—is four times the (square) on
τὰαὐτὰκατασκευάσωμεν,συσταθήσεταίτισχῆμαστερεὸν NB [Def. 11.3, Prop. 1.47]. Thus, V B is double BN.
ὑπὸδώδεκαπενταγώνωνἰσοπλεύρωντεκαὶἰσογωνίωνπε- And BC (is) also double BN. Thus, BV is equal to BC.
ριεχόμενον,ὃκαλεῖταιδωδεκάεδρον.
And since the two (straight-lines) BU and UV are equal
Δεῖδὴαὐτὸκαὶσφαίρᾳπεριλαβεῖντῇδοθείσῃκαὶδεῖξαι, to the two (straight-lines) BW and WC (respectively),
ὅτιἡτοῦδωδεκαέδρουπλευρὰἄλογόςἐστινἡκαλουμένη and the base BV (is) equal to the base BC, angle BUV
ἀποτομή.
is thus equal to angle BWC [Prop. 1.8]. So, similarly, we
᾿ΕκβεβλήσθωγὰρἡΨΟ,καὶἔστωἡΨΩ·συμβάλλειἄρα can show that angle UV C is equal to angle BWC. Thus,
ἡΟΩτῇτοῦκύβουδιαμέτρῳ,καὶδίχατέμνουσινἀλλήλας· the three angles BWC, BUV , and UV C are equal to one
τοῦτο γὰρ δέδεικται ἐν τῷ παρατελεύτῳ θεωρήματι τοῦ another. And if three angles of an equilateral pentagon
ἑνδεκάτου βιβλίου. τεμνέτωσαν κατὰ τὸ Ω· τὸ Ω ἄρα are equal to one another then the pentagon is equianguκέντρονἐστὶτῆςσφαίραςτῆςπεριλαμβανούσηςτὸνκύβον,
lar [Prop. 13.7]. Thus, pentagon BUV CW is equianguκαὶἡΩΟἡμίσειατῆςπλευρᾶςτοῦκύβου.ἐπεζεύχθωδὴἡ
lar. And it was also shown (to be) equilateral. Thus, pen-
ΥΩ.καὶἐπεὶεὐθεῖαγραμμὴἡΝΣἄκρονκαὶμέσονλόγον tagon BUV CW is equilateral and equiangular, and it is
τέτμηταικατὰτὸΟ,καὶτὸμεῖζοναὐτῆςτμῆμάἐστινἡΝΟ, on one of the sides, BC, of the cube. Thus, if we make the
532

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
τὰἄραἀπὸτῶνΝΣ,ΣΟτριπλάσιάἐστιτοῦἀπὸτῆςΝΟ. same construction on each of the twelve sides of the cube
ἴσηδὲἡμὲνΝΣτῇΨΩ,ἐπειδήπερκαὶἡμὲνΝΟτῇΟΩ then some solid figure contained by twelve equilateral
ἐστινἴση,ἡδὲΨΟτῇΟΣ.ἀλλὰμὴνκαὶἡΟΣτῇΨΥ,ἐπεὶ and equiangular pentagons will have been constructed,
καὶτῇΡΟ·τὰἄραἀπὸτῶνΩΨ,ΨΥτριπλάσιάἐστιτοῦἀπὸ which is called a dodecahedron.
τῆςΝΟ.τοῖςδὲἀπὸτῶνΩΨ,ΨΥἴσονἐστὶτὸἀπὸτῆςΥΩ· So, it is necessary to enclose it in the given sphere,
τὸἄραἀπὸτῆςΥΩτριπλάσιόνἐστιτοῦἀπὸτῆςΝΟ.ἔστι and to show that the side of the dodecahedron is that
δὲκαὶἡἐκτοῦκέντρουτῆςσφαίραςτῆςπεριλαμβανούσης irrational (straight-line) called an apotome.
τὸνκύβονδυνάμειτριπλασίωντῆςἡμισείαςτῆςτοῦκύβου For let XP have been produced, and let (the produced
πλευρᾶς·προδέδεικταιγὰρκύβονσυστήσασθαικαὶσφαίρᾳ straight-line) be XZ. Thus, PZ meets the diameter of the
περιλαβεῖνκαὶδεῖξαι,ὅτιἡτῆςσφαίραςδιάμετροςδυνάμει cube, and they cut one another in half. For, this has been
τριπλασίωνἐστὶτῆςπλευρᾶςτοῦκύβου.εἰδὲὅλητῆςὅλης, proved in the penultimate theorem of the eleventh book
καὶ[ἡ]ἡμίσειατῆςἡμισείας·καίἐστινἡΝΟἡμίσειατῆςτοῦ [Prop. 11.38]. Let them cut (one another) at Z. Thus,
κύβουπλευρᾶς·ἡἄραΥΩἴσηἐστὶτῇἐκτοῦκέντρουτῆς Z is the center of the sphere enclosing the cube, and ZP
σφαίραςτῆςπεριλαμβανούσηςτὸνκύβον. καίἐστιτὸΩ (is) half the side of the cube. So, let UZ have been joined.
κέντροντῆςσφαίραςτῆςπεριλαμβανούσηςτὸνκύβον·τὸ And since the straight-line NS has been cut in extreme
Υἄρασημεῖονπρὸςτῇἐπιφανείᾳἐστιτῆςσφαίρας.ὁμοίως and mean ratio at P , and its greater piece is NP , the
δὴδείξομεν,ὅτικαὶἑκάστητῶνλοιπῶνγωνιῶντοῦδω- (sum of the squares) on NS and SP is thus three times
δεκαέδρουπρὸςτῇἐπιφανείᾳἐστὶτῆςσφαίρας·περιείληπται the (square) on NP [Prop. 13.4]. And NS (is) equal to
ἄρατὸδωδεκαέδροντῇδοθείσῃσφαίρᾳ.
XZ, inasmuch as NP is also equal to PZ, and XP to
Λέγωδή,ὅτιἡτοῦδωδεκαέδρουπλευρὰἄλογόςἐστιν PS. But, indeed, PS (is) also (equal) to XU, since (it
ἡκαλουμένηἀποτομή.
is) also (equal) to RP . Thus, the (sum of the squares)
᾿ΕπεὶγὰρτῆςΝΟἄκρονκαὶμέσονλόγοντετμημένηςτὸ on ZX and XU is three times the (square) on NP . And
μεῖζοντμῆμάἐστινὁΡΟ,τῆςδὲΟΞἄκρονκαὶμέσονλόγον the (square) on UZ is equal to the (sum of the squares)
τετμημένηςτὸμεῖζοντμῆμάἐστινἡΟΣ,ὅληςἄρατῆςΝΞ on ZX and XU [Prop. 1.47]. Thus, the (square) on UZ
ἄκρονκαὶμέσονλόγοντεμνομένηςτὸμεῖζοντμῆμάἐστινἡ is three times the (square) on NP . And the square on
ΡΣ.[οἷονἐπείἐστινὡςἡΝΟπρὸςτὴνΟΡ,ἡΟΡπρὸςτὴν the radius of the sphere enclosing the cube is also three
ΡΝ,καὶτὰδιπλάσια·τὰγὰρμέρητοῖςἰσάκιςπολλαπλασίοις times the (square) on half the side of the cube. For it
τὸναὐτὸνἔχειλόγον·ὡςἄραἡΝΞπρὸςτὴνΡΣ,οὕτωςἡ has previously been demonstrated (how to) construct the
ΡΣπρὸςσυναμφότεροντὴνΝΡ,ΣΞ.μείζωνδὲἡΝΞτῆς cube, and to enclose (it) in a sphere, and to show that
ΡΣ·μείζωνἄρακαὶἡΡΣσυναμφοτέρουτῆςΝΡ,ΣΞ·ἡΝΞ the square on the diameter of the sphere is three times
ἄραἄκρονκαὶμέσονλόγοντέτμηται,καὶτὸμεῖζοναὐτῆς the (square) on the side of the cube [Prop. 13.15]. And
τμῆμάἐστινἡΡΣ.]ἴσηδὲἡΡΣτῇΥΦ·τῆςἄραΝΞἄκρον if the (square on the) whole (is three times) the (square
καὶμέσονλόγοντεμνομένηςτὸμεῖζοντμῆμάἐστινἡΥΦ. on the) whole, then the (square on the) half (is) also
καὶἐπεὶῥητήἐστιντῆςσφαίραςδιάμετροςκαίἐστιδυνάμει (three times) the (square on the) half. And NP is half
τριπλασίωντῆςτοῦκύβουπλευρᾶς,ῥητὴἄραἐστὶνἡΝΞ of the side of the cube. Thus, UZ is equal to the radius
πλευρὰοὖσατοῦκύβου. ἐὰνδὲῥητὴγραμμὴἄκρονκαὶ of the sphere enclosing the cube. And Z is the center of
μέσονλόγοντμηθῇ,ἑκάτεροντῶντμημάτωνἄλογόςἐστιν the sphere enclosing the cube. Thus, point U is on the
ἀποτομή.
surface of the sphere. So, similarly, we can show that
῾ΗΥΦἄραπλευρὰοὖσατοῦδωδεκαέδρουἄλογόςἐστιν each of the remaining angles of the dodecahedron is also
ἀποτομή.
on the surface of the sphere. Thus, the dodecahedron has
been enclosed by the given sphere.
So, I say that the side of the dodecahedron is that
irrational straight-line called an apotome.
For since RP is the greater piece of NP , which has
been cut in extreme and mean ratio, and PS is the
greater piece of PO, which has been cut in extreme and
mean ratio, RS is thus the greater piece of the whole
of NO, which has been cut in extreme and mean ratio.
[Thus, since as NP is to PR, (so) PR (is) to RN, and
(the same is also true) of the doubles. For parts have the
same ratio as similar multiples (taken in corresponding
533

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
order) [Prop. 5.15]. Thus, as NO (is) to RS, so RS (is)
to the sum of NR and SO. And NO (is) greater than
RS. Thus, RS (is) also greater than the sum of NR and
SO [Prop. 5.14]. Thus, NO has been cut in extreme and
mean ratio, and RS is its greater piece.] And RS (is)
equal to UV . Thus, UV is the greater piece of NO, which
has been cut in extreme and mean ratio. And since the
diameter of the sphere is rational, and the square on it
is three times the (square) on the side of the cube, NO,
which is the side of the cube, is thus rational. And if
a rational (straight)-line is cut in extreme and mean ratio
then each of the pieces is the irrational (straight-line
called) an apotome.
Thus, UV , which is the side of the dodecahedron,
is the irrational (straight-line called) an apotome [Prop.
13.6].
ÈÖ×Ñ.
᾿Εκ δὴ τούτου φανερόν, ὅτι τῆς τοῦ κύβου πλευρᾶς So, (it is) clear, from this, that the side of the dodeca-
Corollary
ἄκρονκαὶμέσονλόγοντεμνομένηςτὸμεῖζοντμῆμάἐστιν hedron is the greater piece of the side of the cube, when
ἡτοῦδωδεκαέδρουπλευρά.ὅπερἔδειδεῖξαι.
it is cut in extreme and mean ratio. † (Which is) the very
thing it was required to show.
† If the radius of the circumscribed sphere is unity then the side of the cube is p 4/3, and the side of the dodecahedron is (1/3) ( √ 15 − √ 3).
. Proposition 18
À
Τὰςπλευρὰςτῶνπέντεσχημάτωνἐκθέσθαικαὶσυγκρῖν- To set out the sides of the five (aforementioned) figαιπρὸςἀλλήλας.
ures, and to compare (them) with one another. †
G
Â
Å
Æ
H
E
F
M
N
à Ä
A K C D L B
᾿ΕκκείσθωἡτῆςδοθείσηςσφαίραςδιάμετροςἡΑΒ,καὶ Let the diameter, AB, of the given sphere be laid out.
τετμήσθωκατὰτὸΓὥστεἴσηνεἶναιτὴνΑΓτῇΓΒ,κατὰδὲ And let it have been cut at C, such that AC is equal to
τὸΔὥστεδιπλασίοναεἶναιτὴνΑΔτῆςΔΒ,καὶγεγράφθω CB, and at D, such that AD is double DB. And let the
ἐπὶτῆςΑΒἡμικύκλιοντὸΑΕΒ,καὶἀπὸτῶνΓ,ΔτῇΑΒ semi-circle AEB have been drawn on AB. And let CE
πρὸςὀρθὰςἤχθωσαναἱΓΕ,ΔΖ,καὶἐπεζεύχθωσαναἱΑΖ, and DF have been drawn from C and D (respectively),
ΖΒ,ΕΒ.καὶἐπεὶδιπλῆἐστινἡΑΔτῆςΔΒ,τριπλῆἄρα at right-angles to AB. And let AF , FB, and EB have
ἐστὶνἡΑΒτῆςΒΔ.ἀναστρέψαντιἡμιολίαἄραἐστὶνἡΒΑ been joined. And since AD is double DB, AB is thus
τῆςΑΔ.ὡςδὲἡΒΑπρὸςτὴνΑΔ,οὕτωςτὸἀπὸτῆςΒΑ triple BD. Thus, via conversion, BA is one and a half
534

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
πρὸςτὸἀπὸτῆςΑΖ·ἰσογώνιονγάρἐστιτὸΑΖΒτρίγωνον times AD. And as BA (is) to AD, so the (square) on
τῷΑΖΔτριγώνῳ·ἡμιόλιονἄραἐστὶτὸἀπὸτῆςΒΑτοῦἀπὸ BA (is) to the (square) on AF [Def. 5.9]. For triangle
τῆςΑΖ.ἔστιδὲκαὶἡτῆςσφαίραςδιάμετροςδυνάμειἡμιολία AFB is equiangular to triangle AFD [Prop. 6.8]. Thus,
τῆςπλευρᾶςτῆςπυραμίδος.καίἐστινἡΑΒἡτῆςσφαίρας the (square) on BA is one and a half times the (square)
διάμετρος·ἡΑΖἄραἴσηἐστὶτῇπλευρᾷτῆςπυραμίδος. on AF . And the square on the diameter of the sphere is
Πάλιν,ἐπεὶδιπλασίωνἐστὶνἡΑΔτῆςΔΒ,τριπλῆἄρα also one and a half times the (square) on the side of the
ἐστὶνἡΑΒτῆςΒΔ.ὡςδὲἡΑΒπρὸςτὴνΒΔ,οὕτωςτὸ pyramid [Prop. 13.13]. And AB is the diameter of the
ἀπὸτῆςΑΒπρὸςτὸἀπὸτῆςΒΖ·τριπλάσιονἄραἐστὶτὸ sphere. Thus, AF is equal to the side of the pyramid.
ἀπὸτῆςΑΒτοῦἀπὸτῆςΒΖ.ἔστιδὲκαὶἡτῆςσφαίρας Again, since AD is double DB, AB is thus triple BD.
διάμετροςδυνάμειτριπλασίωντῆςτοῦκύβουπλευρᾶς.καί And as AB (is) to BD, so the (square) on AB (is) to the
ἐστινἡΑΒἡτῆςσφαίραςδιάμετρος·ἡΒΖἄρατοῦκύβου (square) on BF [Prop. 6.8, Def. 5.9]. Thus, the (square)
ἐστὶπλευρά.
on AB is three times the (square) on BF . And the square
ΚαὶἐπεὶἴσηἐστὶνἡΑΓτῇΓΒ,διπλῆἄραἐστὶνἡΑΒ on the diameter of the sphere is also three times the
τῆςΒΓ.ὡςδὲἡΑΒπρὸςτὴνΒΓ,οὕτωςτὸἀπὸτῆςΑΒ (square) on the side of the cube [Prop. 13.15]. And AB
πρὸςτὸἀπὸτῆςΒΕ·διπλάσιονἄραἐστὶτὸἀπὸτῆςΑΒτοῦ is the diameter of the sphere. Thus, BF is the side of the
ἀπὸτῆςΒΕ.ἔστιδὲκαὶἡτῆςσφαίραςδιάμετροςδυνάμει cube.
διπλασίωντῆςτοῦὀκταέδρουπλευρᾶς. καὶἐστινἡΑΒἡ And since AC is equal to CB, AB is thus double BC.
τῆςδοθείσηςσφαίραςδιάμετρος·ἡΒΕἄρατοῦὀκταέδρου And as AB (is) to BC, so the (square) on AB (is) to the
ἐστὶπλευρά.
(square) on BE [Prop. 6.8, Def. 5.9]. Thus, the (square)
῎ΗχθωδὴἀπὸτοῦΑσημείουτῇΑΒεὐθείᾳπρὸςὀρθὰς on AB is double the (square) on BE. And the square
ἡΑΗ,καὶκείσθωἡΑΗἴσητῇΑΒ,καὶἐπεζεύχθωἡΗΓ, on the diameter of the sphere is also double the (square)
καὶἀπὸτοῦΘἐπὶτὴνΑΒκάθετοςἤχθωἡΘΚ.καὶἐπεὶ on the side of the octagon [Prop. 13.14]. And AB is the
διπλῆἐστινἡΗΑτῆςΑΓ·ἴσηγὰρἡΗΑτῇΑΒ·ὡςδὲἡ diameter of the given sphere. Thus, BE is the side of the
ΗΑπρὸςτὴνΑΓ,οὕτωςἡΘΚπρὸςτὴνΚΓ,διπλῆἄρα octagon.
καὶἡΘΚτῆςΚΓ.τετραπλάσιονἄραἐστὶτὸἀπὸτῆςΘΚ So let AG have been drawn from point A at rightτοῦἀπὸτῆςΚΓ·τὰἄραἀπὸτῶνΘΚ,ΚΓ,ὅπερἐστὶτὸ
angles to the straight-line AB. And let AG be made equal
ἀπὸτῆςΘΓ,πενταπλάσιόνἐστιτοῦἀπὸτῆςΚΓ.ἴσηδὲ to AB. And let GC have been joined. And let HK have
ἡΘΓτῇΓΒ·πενταπλάσιονἄραἐστὶτὸἀπὸτῆςΒΓτοῦ been drawn from H, perpendicular to AB. And since GA
ἀπὸ τῆς ΓΚ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ΑΒ τῆς ΓΒ, ὧν ἡ is double AC. For GA (is) equal to AB. And as GA (is)
ΑΔτῆςΔΒἐστιδιπλῆ,λοιπὴἄραἡΒΔλοιπῆςτῆςΔΓ to AC, so HK (is) to KC [Prop. 6.4]. HK (is) thus also
ἐστι διπλῆ. τριπλῆἄραἡΒΓτῆςΓΔ·ἐνναπλάσιονἄρα double KC. Thus, the (square) on HK is four times the
τὸἀπὸτῆςΒΓτοῦἀπὸτῆςΓΔ.πενταπλάσιονδὲτὸἀπὸ (square) on KC. Thus, the (sum of the squares) on HK
τῆςΒΓτοῦἀπὸτῆςΓΚ·μεῖζονἄρατὸἀπὸτῆςΓΚτοῦ and KC, which is the (square) on HC [Prop. 1.47], is
ἀπὸτῆςΓΔ.μείζωνἄραἐστὶνἡΓΚτῆςΓΔ.κείσθωτῇ five times the (square) on KC. And HC (is) equal to CB.
ΓΚἴσηἡΓΛ,καὶἀπὸτοῦΛτῇΑΒπρὸςὀρθὰςἤχθωἡ Thus, the (square) on BC (is) five times the (square) on
ΛΜ,καὶἐπεζεύχθωἡΜΒ.καὶἐπεὶπενταπλάσιόνἐστιτὸ CK. And since AB is double CB, of which AD is double
ἀπὸτῆςΒΓτοῦἀπὸτῆςΓΚ,καίἐστιτῆςμὲνΒΓδιπλῆ DB, the remainder BD is thus double the remainder DC.
ἡΑΒ,τῆςδὲΓΚδιπλῆἡΚΛ,πενταπλάσιονἄραἐστὶτὸ BC (is) thus triple CD. The (square) on BC (is) thus
ἀπὸτῆςΑΒτοῦἀπὸτῆςΚΛ.ἔστιδὲκαὶἡτῆςσφαίρας nine times the (square) on CD. And the (square) on BC
διάμετροςδυνάμειπενταπλασίωντῆςἐκτοῦκέντρουτοῦ (is) five times the (square) on CK. Thus, the (square)
κύκλου,ἀφ᾿οὗτὸεἰκοσάεδρονἀναγέγραπται. καίἐστινἡ on CK (is) greater than the (square) on CD. CK is thus
ΑΒἡτῆςσφαίραςδιάμετρος·ἡΚΛἄραἐκτοῦκέντρου greater than CD. Let CL be made equal to CK. And
ἐστὶ τοῦ κύκλου, ἀφ᾿ οὗ τὸ εἰκοσάεδρον ἀναγέγραπται· let LM have been drawn from L at right-angles to AB.
ἡΚΛἄραἑξαγώνουἐστὶπλευρὰτοῦεἰρημένουκύκλου. And let MB have been joined. And since the (square) on
καὶἐπεὶἡτῆςσφαίραςδιάμετροςσύγκειταιἔκτετῆςτοῦ BC is five times the (square) on CK, and AB is double
ἑξαγώνουκαὶδύοτῶντοῦδεκαγώνουτῶνεἰςτὸνεἰρημένον BC, and KL double CK, the (square) on AB is thus five
κύκλονἐγγραφομένων,καίἐστινἡμὲνΑΒἡτῆςσφαίρας times the (square) on KL. And the square on the diamδιάμετρος,ἡδὲΚΛἑξαγώνουπλευρά,καὶἴσηἡΑΚτῇ
eter of the sphere is also five times the (square) on the
ΛΒ, ἑκατέρα ἄρα τῶν ΑΚ, ΛΒ δεκαγώνου ἐστὶ πλευρὰ radius of the circle from which the icosahedron has been
τοῦἐγγραφομένουεἰςτὸνκύκλον,ἀφ᾿οὗτὸεἰκοσάεδρον described [Prop. 13.16 corr.]. And AB is the diameter
ἀναγέγραπται. καὶἐπεὶδεκαγώνουμὲνἡΛΒ,ἑξαγώνου of the sphere. Thus, KL is the radius of the circle from
535

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
δὲἡΜΛ·ἴσηγάρἐστιτῇΚΛ,ἐπεὶκαὶτῇΘΚ·ἴσονγὰρ which the icosahedron has been described. Thus, KL is
ἀπέχουσινἀπὸτοῦκέντρου· καίἐστινἑκατέρατῶνΘΚ, (the side) of the hexagon (inscribed) in the aforemen-
ΚΛδιπλασίωντῆςΚΓ·πενταγώνουἄραἐστὶνἡΜΒ.ἡδὲ tioned circle [Prop. 4.15 corr.]. And since the diameter of
τοῦπενταγώνουἐστὶνἡτοῦεἰκοσαέδρου·εἰκοσαέδρουἄρα the sphere is composed of (the side) of the hexagon, and
ἐστὶνἡΜΒ.
two of (the sides) of the decagon, inscribed in the afore-
ΚαὶἐπεὶἡΖΒκύβουἐστὶπλευρά,τετμήσθωἄκρονκαὶ mentioned circle, and AB is the diameter of the sphere,
μέσονλόγονκατὰτὸΝ,καὶἔστωμεῖζοντμῆματὸΝΒ·ἡ and KL the side of the hexagon, and AK (is) equal to
ΝΒἄραδωδεκαέδρουἐστὶπλευρά.
LB, thus AK and LB are each sides of the decagon in-
Καὶ ἐπεὶ ἡ τῆς σφαίρας διάμετρος ἐδείχθη τῆς μὲν scribed in the circle from which the icosahedron has been
ΑΖ πλευρᾶς τῆς πυραμίδος δυνάμει ἡμιολία, τῆς δὲ τοῦ described. And since LB is (the side) of the decagon.
ὀκταέδρουτῆςΒΕδυνάμειδιπλασίων,τῆςδὲτοῦκύβουτῆς And ML (is the side) of the hexagon—for (it is) equal to
ΖΒδυνάμειτριπλασίων,οἵωνἄραἡτῆςσφαίραςδιάμετρος KL, since (it is) also (equal) to HK, for they are equally
δυνάμειἕξ,τοιούτωνἡμὲντῆςπυραμίδοςτεσσάρων,ἡδὲ far from the center. And HK and KL are each double
τοῦὀκταέδρουτριῶν,ἡδὲτοῦκύβουδύο. ἡμὲνἄρατῆς KC. MB is thus (the side) of the pentagon (inscribed
πυραμίδοςπλευρὰτῆςμὲντοῦὀκταέδρουπλευρᾶςδυνάμει in the circle) [Props. 13.10, 1.47]. And (the side) of the
ἐστὶνἐπίτριτος,τῆςδὲτοῦκύβουδυνάμειδιπλῆ,ἡδὲτοῦ pentagon is (the side) of the icosahedron [Prop. 13.16].
ὀκταέδρου τῆς τοῦ κύβου δυνάμει ἡμιολία. αἱ μὲν οὖν Thus, MB is (the side) of the icosahedron.
εἰρημέναιτῶντριῶνσχημάτωνπλευραί,λέγωδὴπυραμίδος And since FB is the side of the cube, let it have been
καὶὀκταέδρουκαὶκύβου, πρὸςἀλλήλαςεἰσὶν ἐν λόγοις cut in extreme and mean ratio at N, and let NB be the
ῥητοῖς. αἱδὲλοιπαὶδύο,λέγωδὴἥτετοῦεἰκοσαέδρου greater piece. Thus, NB is the side of the dodecahedron
καὶἡτοῦδωδεκαέδρου,οὔτεπρὸςἀλλήλαςοὔτεπρὸςτὰς [Prop. 13.17 corr.].
προειρημέναςεἰσὶνἐνλόγοιςῥητοῖς·ἄλογοιγάρεἰσιν,ἡμὲν And since the (square) on the diameter of the sphere
ἐλάττων,ἡδὲἀποτομή.
was shown (to be) one and a half times the square on the
῞ΟτιμείζωνἐστὶνἡτοῦεἰκοσαέδρουπλευρὰἡΜΒτῆς side, AF , of the pyramid, and twice the square on (the
τοῦδωδεκαέδρουτῆςΝΒ,δείξομενοὕτως.
side), BE, of the octagon, and three times the square
᾿ΕπεὶγὰρἰσογώνιόνἐστιτὸΖΔΒτρίγωνοντῷΖΑΒ on (the side), FB, of the cube, thus, of whatever (parts)
τριγώνῳ, ἀνάλογόνἐστινὡςἡΔΒπρὸςτὴνΒΖ,οὕτως the (square) on the diameter of the sphere (makes) six,
ἡΒΖπρὸςτὴνΒΑ.καὶἐπεὶτρεῖςεὐθεῖαιἀνάλογόνεἰσιν, of such (parts) the (square) on (the side) of the pyramid
ἔστινὡςἡπρώτηπρὸςτὴντρίτην,οὕτωςτὸἀπὸτῆςπρώτης (makes) four, and (the square) on (the side) of the ocπρὸςτὸἀπὸτῆςδευτέρας·ἔστινἄραὡςἡΔΒπρὸςτὴνΒΑ,
tagon three, and (the square) on (the side) of the cube
οὕτωςτὸἀπὸτῆςΔΒπρὸςτὸἀπὸτῆςΒΖ·ἀνάπαλινἄρα two. Thus, the (square) on the side of the pyramid is one
ὡςἡΑΒπρὸςτὴνΒΔ,οὕτωςτὸἀπὸτῆςΖΒπρὸςτὸἀπὸ and a third times the square on the side of the octagon,
τῆςΒΔ.τριπλῆδὲἡΑΒτῆςΒΔ·τριπλάσιονἄρατὸἀπὸ and double the square on (the side) of the cube. And the
τῆςΖΒτοῦἀπὸτῆςΒΔ.ἔστιδὲκαὶτὸἀπὸτῆςΑΔτοῦ (square) on (the side) of the octahedron is one and a half
ἀπὸτῆςΔΒτετραπλάσιον·διπλῆγὰρἡΑΔτῆςΔΒ·μεῖζον times the square on (the side) of the cube. Therefore,
ἄρατὸἀπὸτῆςΑΔτοῦἀπὸτῆςΖΒ·μείζωνἄραἡΑΔτῆς the aforementioned sides of the three figures—I mean, of
ΖΒ·πολλῷἄραἡΑΛτῆςΖΒμείζωνἐστίν. καὶτῆςμὲν the pyramid, and of the octahedron, and of the cube—
ΑΛἄκρονκαὶμέσονλόγοντεμνομένηςτὸμεῖζοντμῆμά are in rational ratios to one another. And (the sides
ἐστινἡΚΛ,ἐπειδήπερἡμὲνΛΚἑξαγώνουἐστίν,ἡδὲΚΑ of) the remaining two (figures)—I mean, of the icosaheδεκαγώνου·τῆςδὲΖΒἄκρονκαὶμέσονλόγοντεμνομένης
dron, and of the dodecahedron—are neither in rational
τὸμεῖζοντμῆμάἐστινἡΝΒ·μείζωνἄραἡΚΛτῆςΝΒ. ratios to one another, nor to the (sides) of the aforemen-
ἴσηδὲἡΚΛτῇΛΜ·μείζωνἄραἡΛΜτῆςΝΒ[τῆςδὲ tioned (three figures). For they are irrational (straight-
ΛΜμείζωνἐστὶνἡΜΒ].πολλῷἄραἡΜΒπλευρὰοὖσα lines): (namely), a minor [Prop. 13.16], and an apotome
τοῦεἰκοσαέδρουμείζωνἐστὶτῆςΝΒπλευρᾶςοὔσηςτοῦ [Prop. 13.17].
δωδεκαέδρου·ὅπερἔδειδεῖξαι.
(And), we can show that the side, MB, of the icosahedron
is greater that the (side), NB, or the dodecahedron,
as follows.
For, since triangle FDB is equiangular to triangle
FAB [Prop. 6.8], proportionally, as DB is to BF , so BF
(is) to BA [Prop. 6.4]. And since three straight-lines are
(continually) proportional, as the first (is) to the third,
536

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
so the (square) on the first (is) to the (square) on the
second [Def. 5.9, Prop. 6.20 corr.]. Thus, as DB is to
BA, so the (square) on DB (is) to the (square) on BF .
Thus, inversely, as AB (is) to BD, so the (square) on
FB (is) to the (square) on BD. And AB (is) triple BD.
Thus, the (square) on FB (is) three times the (square)
on BD. And the (square) on AD is also four times the
(square) on DB. For AD (is) double DB. Thus, the
(square) on AD (is) greater than the (square) on FB.
Thus, AD (is) greater than FB. Thus, AL is much greater
than FB. And KL is the greater piece of AL, which is
cut in extreme and mean ratio—inasmuch as LK is (the
side) of the hexagon, and KA (the side) of the decagon
[Prop. 13.9]. And NB is the greater piece of FB, which
is cut in extreme and mean ratio. Thus, KL (is) greater
than NB. And KL (is) equal to LM. Thus, LM (is)
greater than NB [and MB is greater than LM]. Thus,
MB, which is (the side) of the icosahedron, is much
greater than NB, which is (the side) of the dodecahedron.
(Which is) the very thing it was required to show.
† If the radius of the given sphere is unity then the sides of the pyramid (i.e., tetrahedron), octahedron, cube, icosahedron, and dodecahedron,
respectively, satisfy the following inequality: p 8/3 > √ 2 > p 4/3 > (1/ √ 5) p 10 − 2 √ 5 > (1/3) ( √ 15 − √ 3).
Λέγωδή,ὅτιπαρὰτὰεἰρημέναπέντεσχήματαοὐσυ- So, I say that, beside the five aforementioned figures,
σταθήσεταιἕτερονσχῆμαπεριεχόμενονὑπὸἰσοπλεύρωντε no other (solid) figure can be constructed (which is) conκαὶἰσογωνίωνἴσωνἀλλήλοις.
tained by equilateral and equiangular (planes), equal to
῾Υπὸμὲνγὰρδύοτριγώνωνἢὅλωςἐπιπέδωνστερεὰ one another.
γωνίαοὐσυνίσταται. ὑπὸδὲτριῶντριγώνωνἡτῆςπυ- For a solid angle cannot be constructed from two triραμίδος,ὑπὸδὲτεσσάρωνἡτοῦὀκταέδρου,ὑπὸδὲπέντε
angles, or indeed (two) planes (of any sort) [Def. 11.11].
ἡ τοῦ εἰκοσαέδρου· ὑπὸ δὲ ἓξ τριγώνων ἰσοπλεύρων τε And (the solid angle) of the pyramid (is constructed)
καὶἰσογωνίων πρὸςἑνὶ σημείῳσυνισταμένων οὐκἔσται from three (equiangular) triangles, and (that) of the ocστερεὰ
γωνία· οὔσηςγὰρ τῆςτοῦ ἰσοπλεύρουτριγώνου tahedron from four (triangles), and (that) of the icosaheγωνίαςδιμοίρουὀρθῆςἔσονταιαἱἓξτέσσαρσινὀρθαῖςἴσαι·
dron from (five) triangles. And a solid angle cannot be
ὅπερἀδύνατον·ἅπασαγὰρστερεὰγωνίαὑπὸἐλασσόνων (made) from six equilateral and equiangular triangles set
ἢτεσσάρωνὀρθῶνπερέχεται. διὰτὰαὐτὰδὴοὐδὲὑπὸ up together at one point. For, since the angles of a equiπλειόνωνἢἓξγωνιῶνἐπιπέδωνστερεὰγωνίασυνίσταται.
lateral triangle are (each) two-thirds of a right-angle, the
ὑπὸδὲτετραγώνωντριῶνἡτοῦκύβουγωνίαπεριέχεται· (sum of the) six (plane) angles (containing the solid an-
ὑπὸδὲτεσσάρωνἀδύνατον· ἔσονταιγὰρπάλιντέσσαρες gle) will be four right-angles. The very thing (is) impos-
ὀρθαί.ὑπὸδὲπενταγώνωνἰσοπλεύρωνκαὶἰσογωνίων,ὑπὸ sible. For every solid angle is contained by (plane angles
μὲντριῶνἡτοῦδωδεκαέδρου·ὑπὸδὲτεσσάρωνἀδύνατον· whose sum is) less than four right-angles [Prop. 11.21].
οὔσηςγὰρτῆςτοῦπενταγώνουἰσοπλεύρουγωνίαςὀρθῆς So, for the same (reasons), a solid angle cannot be conκαὶπέμπτου,ἔσονταιαἱτέσσαρεςγωνίαιτεσσάρωνὀρθῶν
structed from more than six plane angles (equal to twoμείζους·ὅπερἀδύνατον.οὐδὲμὴνὑπὸπολυγώνωνἑτέρων
thirds of a right-angle) either. And the (solid) angle of
σχημάτωνπερισχεθήσεταιστερεὰγωνίαδιὰτὸαὐτὸἄτο- a cube is contained by three squares. And (a solid angle
πον.
contained) by four (squares is) impossible. For, again, the
Οὐκἄραπαρὰτὰεἰρημέναπέντεσχήματαἕτερονσχῆμα (sum of the plane angles containing the solid angle) will
στερεὸνσυσταθήσεταιὑπὸἰσοπλεύρωντεκαὶἰσογωνίων be four right-angles. And (the solid angle) of a dodecπεριεχόμενον·ὅπερἔδειδεῖξαι.
ahedron (is contained) by three equilateral and equiangular
pentagons. And (a solid angle contained) by four
537

ËÌÇÁÉÁÏƺ ELEMENTS
BOOK 13
(equiangular pentagons is) impossible. For, the angle of
an equilateral pentagon being one and one-fifth of rightangle,
four (such) angles will be greater (in sum) than
four right-angles. The very thing (is) impossible. And,
on account of the same absurdity, a solid angle cannot
be constructed from any other (equiangular) polygonal
figures either.
Thus, beside the five aforementioned figures, no other
solid figure can be constructed (which is) contained by
equilateral and equiangular (planes). (Which is) the very
thing it was required to show.
A
E
F
B
D
C
ÄÑÑ.
Lemma
῞Οτιδὲἡτοῦἰσοπλεύρουκαὶἰσογωνίουπενταγώνου It can be shown that the angle of an equilateral and
γωνίαὀρθήἐστικαὶπέμπτου,οὕτωδεικτέον.
equiangular pentagon is one and one-fifth of a right-
῎Εστωγὰρπεντάγωνονἰσόπλευρονκαὶἰσογώνιοντὸ angle, as follows.
ΑΒΓΔΕ,καὶπεριγεγράφθωπερὶαὐτὸκύκλοςὁΑΒΓΔΕ, For let ABCDE be an equilateral and equiangular
καὶεἰλήφθωαὐτοῦτὸκέντροντὸΖ,καὶἐπεζεύχθωσαναἱ pentagon, and let the circle ABCDE have been circum-
ΖΑ,ΖΒ,ΖΓ,ΖΔ,ΖΕ.δίχαἄρατέμνουσιτὰςπρὸςτοῖςΑ, scribed about it [Prop. 4.14]. And let its center, F , have
Β,Γ,Δ,Ετοῦπενταγώνουγωνίας.καὶἐπεὶαἱπρὸςτῷΖ been found [Prop. 3.1]. And let FA, FB, FC, FD,
πέντεγωνίαιτέσσαρσινὀρθαῖςἴσαιεἰσὶκαίεἰσινἴσαι,μία and FE have been joined. Thus, they cut the angles
ἄρααὐτῶν,ὡςἡὑπὸΑΖΒ,μιᾶςὀρθῆςἐστιπαρὰπέμπτον· of the pentagon in half at (points) A, B, C, D, and E
λοιπαὶἄρααἱὑπὸΖΑΒ,ΑΒΖμιᾶςεἰσινὀρθῆςκαὶπέμπτου. [Prop. 1.4]. And since the five angles at F are equal (in
ἴσηδὲἡὑπὸΖΑΒτῇὑπὸΖΒΓ·καὶὅληἄραἡὑπὸΑΒΓτοῦ sum) to four right-angles, and are also equal (to one anπενταγώνουγωνίαμιᾶςἐστινὀρθῆςκαὶπέμπτου·ὅπερἔδει
other), (any) one of them, like AFB, is thus one less a
δεῖξαι.
fifth of a right-angle. Thus, the (sum of the) remaining
(angles in triangle ABF ), FAB and ABF , is one plus a
fifth of a right-angle [Prop. 1.32]. And FAB (is) equal
to FBC. Thus, the whole angle, ABC, of the pentagon
is also one and one-fifth of a right-angle. (Which is) the
very thing it was required to show.
538

GREEK-ENGLISH LEXICON
539

ËÌÇÁÉÁÏÆ
GREEK–ENGLISH LEXICON
ABBREVIATIONS: act - active; adj - adjective; adv - adverb; conj
- conjunction; fut - future; gen - genitive; imperat - imperative;
impf - imperfect; ind - indeclinable; indic - indicative; intr - intransitive;
mid - middle; neut - neuter; no - noun; par - particle;
part - participle; pass - passive; perf - perfect; pre - preposition;
pres - present; pro - pronoun; sg - singular; tr - transitive; vb -
verb.
Û,ÜÛ¸ÓÒ¸¹Õ¸Ñ¸ÕÒ: vb, lead, draw (a line).
ÒØÓ¹ÓÒ: adj, impossible.
: adv, always, for ever.
ÖÛ¸Ö×Û¸℄ÐÓÒ¸Ö¸ÖѸíÖÒ: vb, grasp.
ØÛ¸Ø×Û¸Ø׸ظØѸîØ: vb, postulate.
ØѹØÓ¸Ø: no, postulate.
ÐÓÙÓ¹ÓÒ: adj, analogous, consequent on, in conformity
with.
ÖÓ¹¹ÓÒ: adj, outermost, end, extreme.
ÐÐ: conj, but, otherwise.
ÐÓÓ¹ÓÒ: adj, irrational.
Ñ: adv, at once, at the same time, together.
ÑÐÙôÒÓ¹ÓÒ: adj, obtuse-angled;ØÑÐÙôÒÓÒ, no, obtuse
angle.
Ñй¹: adj, obtuse.
ÑØÖÓ¹¹ÓÒ: pro, both.
ÒÖÛ: vb, describe (a figure); seeÖÛ. ÒÐÓ¸é: no, proportion, (geometric) progression.
ÒÐÓÓ¹ÓÒ: adj, proportional.
ÒÔÐÒ: adv, inverse(ly).
ÒÔÐÖÛ: vb, fill up.
Ò×ØÖÛ: vb, turn upside down, convert (ratio); see×ØÖÛ. Ò×ØÖÓ¸é: no, turning upside down, conversion (of ratio).
ÒÙÖÛ: vb, take away in turn; seeÖÛ. Ò×ØÑ: vb, set up; see×ØÑ. Ò×Ó¹ÓÒ: adj, unequal, uneven.
ÒØÔ×ÕÛ: vb, be reciprocally proportional; seeÔ×ÕÛ. ÜÛÒ¹ÓÒÓ¸: vb, axis.
ÔÜ: adv, once.
Ô¸Ô׸ÔÒ: adj, quite all, the whole.
ÔÖÓ¹ÓÒ: adj, infinite.
ÔÒÒØÓÒ: ind, opposite.
ÔÕÛ: vb, be far from, be away from; seeÕÛ. ÔÐع: adj, without breadth.
Ôܹ۸é: no, proof.
ÔÓ×ØÑ: vb, re-establish, restore; see×ØÑ. ÔÓÐÑÒÛ:vb, take from, subtract from, cut off from; see
ÐÑÒÛ. ÔÓØÑÒÛ: vb, cut off, subtend.
ÔØÑѹØÓ¸Ø: no, piece cut off, segment.
ÔÓØÓѸé: vb, piece cut off, apotome.
ÔØÛ¸Ý۸ݸ—,ÑÑ, — : vb, touch, join, meet.
ÔôØÖÓ¹¹ÓÒ: adj,
ÖѸ: no,
ÖØ: adv,
ÖØÔÐÙÖÓ¹ÓÒ: adj,
ÖÕÛ¸ÖÜÛ¸ÖܸÖÕ¸ÖѸÖÕÒ: vb,
×ÑÑØÖÓ¹ÓÒ: adj,
×ÑÔØÛØÓ¹ÓÒ: adj,
ÖØÓ¹¹ÓÒ: adj,
ØÑØÓ¹ÓÒ: adj,
ØÔÓ¹ÓÒ: adj,
ØÒ: adv,
ÖÛ: vb,
¸é: no,
Ó¹Ó¸Ø: no,
ÒÛ¸¹×ÓѸ¹Ò¸¸—, —
ÐÐÛ¸Ðü¸ÐÓҸиÐѸÐÒ: vb,
×¹Û¸é: no,
℄ÒÓѸÒ×ÓѸÒÑÒ¸ÓÒ¸ÒѸ— :
ÒôÑÛÒ¹ÓÒÓ¸é: no,
ÖÑѸé: no,
ÖÛ¸ÖÝÛ¸ÖÝ/℄¸Ö¸ÖÑѸÖÝÑÒ: vb,
ÛÒ¸é: no,
ÒÙѸÜ۸ܸոѸÕÒ: vb,
ØÓÒ: ind,
ܹ۸é: no,
ôÒÓ¹ÓÒ: adj, ten-sided;ØôÒÓÒ, ÕÓѸÜÓѸÜÑÒ¸—,ѸÕÒ: vb,
Ð: ind,
ÐÓ¹¹ÓÒ: adj,
ÐÓÒØ: adv,
Û: vb,
ôÒÓ¹ÓÒ: adj,
ÐÔÛ: vb,
ÑØÖÓ¹ÓÒ: adj,
Ö×¹Û¸é: no,
further off.
Ö: par, thus, as it seems (inferential).
number.
an even number of times.
rule; mid., begin.
having a even number of sides.
incommensurable.
not touching, not meeting.
even, perfect.
uncut.
absurd, paradoxical.
immediately, obviously.
take from, subtract from, cut off from; seeÖÛ.
point of contact.
depth, height.
(of angle).
base (of a triangle).
Ö: conj, for (explanatory).
happen, become.
line.
gnomon.
: vb, walk; perf, stand
throw.
vb,
draw (a figure).
angle.
: vb, be necessary;,it is necessary;, it was necassary;
ÓÒ, being necessary.
show,
demonstrate.
one must show.
proof.
no, decagon.
receive,
accept.
: conj, so (explanatory).
quite clear, manifest.
clear.
manifestly.
carry over, draw through, draw across; seeÛ.
diagonal.
leave an interval between.
diametrical;éÑØÖÓ, no, diameter, diagonal.
division, separation.
540

ËÌÇÁÉÁÏÆ ÖÛ: vb, divide (in two);ÖÒØÓ¹¹ÓÒ, (ratio); seeÖÛ. ×ØѹØÓ¸Ø: no,
ÖÛ: vb, differ; seeÖÛ. ÛѸô×Û¸Û¸Û¸ÓѸÒ: vb,
ÑÓÖÓ¹ÓÒ: adj,
ÔÐ×ÞÛ: vb,
ÔÐ×Ó¹¹ÓÒ: adj,
ÔÐ×ÛÒ¹ÓÒ: adj,
ÔÐÓ¹¹ÓÒ: adj,
Õ: adv,
ÕÓÖÓѸé: no,
Ù¹Ó¸é: no,
ÒÑ:vb, be
squared;ÙÒÑÒ¸é, no,
Òѹ۸é: no,
ÙÒع¹Ò: adj,
ÛÖÓ¹ÓÒ: adj,
ÙØÓ¹¹Ó: adj,
ÛÒ¹ÓÒ: adj,
ÖÛ: vb,
Ó¹Ó¸Ø: no,
Ó×ÖÓ¹ÓÒ: adj,
ÖÛ/ÐÛ¸Öü/ÖÛ¸ÔÓÒ¸Ö¸ÖѸÖÖÒ: vb,
say, speak; per pass part,ÖÑÒÓ¹¹ÓÒ, Ø. . .Ø: ×ØÓ¹¹ÓÒ: pro,
ØÖÓ¹¹ÓÒ: pro,
ÐÐÛ: vb,
Û: vb,
Ñ: vb,
ØÑ: vb, set out; seeØÑ. Ø: pre
Ð××ÛÒ»ÐØØÛÒ¹ÓÒ: adj,
ÐÕ×ØÓ¹¹ÓÒ: adj,
ÐÐÔÛ: vb,
ÑÔÔØÛ: vb,
ÑÔÖÓ×Ò: adv,
ÒÐÐÜ: adv,
ÒÖÑÞÛ: vb,
ÒÕÓÑ: vb,
ÒÒ: ind,
ÒÒÔÐ×Ó¹¹ÓÒ: adj,
ÒÒÓ¸é: no,
adj, separated
radius.
two-thirds.
double.
double, twofold.
double, twofold.
double.
: adv, twice.
in two, in half.
point of bisection.
the number two, dyad.
give.
able, be capable, generate, square, be when
square-root (of area)—i.e., straight-line
whose square is equal to a given area.
power (usually 2nd power when used in
mathematical sence, hence), square.
possible.
twelve-sided.
of him/her/it/self, his/her/its/own.
nearer, nearest.
inscribe; seeÖÛ.
figure, form, shape.
twenty-sided.
adj, said, aforementioned.
ind, either . . . or.
each, every one.
each (of two).
produce (a line); seeÐÐÛ.
set out.
be set out, be taken; seeÑ.
+ gen, outside, external.
less, lesser.
least.
be less than, fall short of.
meet (of lines), fall on; seeÔÔØÛ.
in front.
alternate(ly).
insert; perf indic pass 3rd sg,ÒÖÑÓ×Ø.
admit, allow.
on account of, for the sake of.
nine-fold, nine-times.
notion.
GREEK–ENGLISH LEXICON
ÒÔÖÕÛ: vb, encompass; seeÕÛ. ÒÔÔØÛ: seeÑÔÔØÛ. ÒØ: pre
ÜÛÒÓ¹ÓÒ: adj,
ÜÔÐ×Ó¹¹ÓÒ: adj,
ÜÛÒ: adv,
ÔÒÛ: adv,
Ô¸é: no,
ÔÔÖ: ind,
ÔÞÒѸÔÞÜÛ¸ÔÞÙÜ, —,ÔÞÙѸÔÞÕÒ: ÔÐÓÞÓÑ: vb,
ÔÒÓÛ: vb,
ÔÔÓ¹ÓÒ: adj, level, flat, plane;ØÔÔÓÒ, Ô×ÔØÓÑ: vb,
Ô×ݹ۸é: no,
ÔØ××Û: vb, put upon, enjoin;ØÔØÕÒ, ÔØÖØÓ¹ÓÒ: adj,
ÔÒ¸é: no,
ÔÓÑ: vb,
ÖÕÓѸÐ×ÓѸÐÓÒ¸ÐÐÙ, —,
×ÕØÓ¹¹ÓÒ: adj,
ØÖѹ: adj, oblong;ØØÖÑ, ØÖÓ¹¹ÓÒ: adj,
ÖÑÞÛ¸ÖÑ×Û¸ÖÑÓ׸ÑÓ¸ÑÓ×ѸÑ×Ò
ÖÑÑÓ¹ÓÒ: adj, rectilinear;ØÖÑÑÓÒ, ¹¹: adj,
, in
Ö×Û¸Ö×Û¸ÖÓÒ¸Ö¸ÖѸÖÒ: vb,
ÔØÛ: vb,
seeÔØÛ. Ü: adv,
×ØÑ: vb,
ÕÛ¸ÜÛ¸×ÕÓÒ¸×Õ¸¹×ÕÑ, —
éÓѸé×ÓѸé×ÑÒ¸éÑ, —,éÒ: vb,
Û¸ÜÛ, —,
éÑÐÓÒ¸Ø: no,
éÑÐÓ¹¹ÓÒ: adj,
Ñ×Ù¹¹Ù: adj,
ÔÖ=+ÔÖ: conj,
no, hexagon.
+ gen, inside, interior, within, internal.
hexagonal;Ø ÜÛÒÓÒ, sixfold.
Ü: adv, in order, successively, consecutively.
outside, extrinsic.
above.
point of contact.
Ô: conj, since (causal).
inasmuch as, seeing that.
vb, join (by a line).
conclude.
think of, contrive.
no, plane.
investigate.
inspection, investigation.
no, the (thing)
prescribed; seeØ××Û.
one and a third times.
surface.
follow.
— : vb, come, go.
outermost, uttermost, last.
no, rectangle.
other (of two).
Ø: par, yet, still, besides.
no, rectilinear
figure.
straight;é, no, straight-line;Ô
a straight-line, straight-on.
find.
bind to; mid, touch;é ÔØÓÑÒ, no, tangent;
: vb, coincide; pass, be applied.
in order, adjacent.
set, stand, place upon; see×ØÑ.
: vb, have.
lead.
: ind, already, now.
—, —, — : vb, have come, be present.
semi-circle.
times.
containing one and a half, one and a half
half.
than, than indeed.
541

ËÌÇÁÉÁÏÆ
GREEK–ENGLISH LEXICON
ØÓ. . .: par, surely, either . . . or; in fact, either . . . or.
×¹Û¸é: no,
ÛÖѹØÓ¸Ø: no,
Ó¹¹ÓÒ: adj,
×:adv, the same number of times;×ÔÓÐÐÔÐ×, ×ÓôÒÓ¹ÓÒ: adj,
×ÔÐÙÖÓ¹ÓÒ: adj,
×ÓÔй: adj,
×Ó¹¹ÓÒ: adj, equal;Ü×ÓÙ, ×Ó×й: adj,
×ØѸ×Ø×Û¸×Ø×, —, —,×ØÒ: ×ØѸ×Ø×Û¸×ØÒ¸×ظ×ØѸ×ØÒ: vb
×Óݹ: adj,
ÔÖ: ind,
ØÓ¹ÓÒ: adj,
ÐÓÙ: adv,
ÐÛ: vb,
ÒÓ=ÒÓ. Ò=Ò: ind,
ØÖ¸é: no,
ØÖÛ: vb,
ØÓÐÓÙÛ: vb,
ØÐÔÛ: vb, leave behind; seeÐÔÛ;ØØÐÔÑÒ, ØÐÐÐÓ¹ÓÒ: adj,
ØÑØÖÛ: vb,
ØÒØÛ: vb,
Ø×ÙÞÛ: vb,
Ѹ×ÓÑ, —,
made; seeØÑ. ÒØÖÓÒ¸Ø: no,
ÐÛ: vb,
ÐÒÛ¸ÐÒÛ¸ÐҸиÐѸÐÒ: vb,
Ð×¹Û¸é: no,
ÓÐÓ¹¹ÓÒ: adj,
ÓÖÙ¸é: no, top, summit, apex;ØÓÖÙÒ, ÖÒÛ¸ÖÒü¸ÖÒ¸Ö¸ÖѸÖÒ: vb,
Ó¸: no,
ÐÓ¸: no,
ÐÒÖÓ¸: no,
ÙÖع¹Ò: adj,
üÒÓ¸: no,
placing, setting, position.
theorem.
one’s own.
the same multiples, equal multiples.
equiangular.
equilateral.
equal in number.
equally, evenly.
isosceles.
vb tr, stand (something).
intr, stand
up (oneself); Note: perfect I have stood up can be taken
to mean present I am standing.
of equal height.
according as, just as.
perpendicular.
on the whole, in general.
call.
even if, and if.
diagram, figure.
describe/draw, inscribe (a figure); seeÖÛ.
follow after.
no,
remainder.
in succession, in corresponding order.
measure (exactly).
come to, arrive at.
furnish, construct.
—, —, — : vb, have been placed, lie, be
center.
break off, inflect.
lean, incline.
inclination, bending.
hollow, concave.
vertically
opposite (of angles).
judge.
cube.
circle.
cylinder.
cone.
convex.
542
ÐÑÒÛ¸ÐÝÓѸÐÓÒ¸ÐÐÑѸÐÒ: vb,
take.
ÐÛ: vb, say; pres pass part,ÐÑÒÓ¹¹ÓÒ, seeÖÛ. ÐÔÛ¸ÐÝÛ¸ÐÔÓÒ¸ÐÐÓÔ¸ÐÐÑѸÐÒ: vb,
ÐÑÑØÓÒ¸Ø: no,
ÐÑѹØÓ¸Ø: no,
Ðݹ۸é: no,
ÐÓ¸: no,
ÐÓÔ¹¹Ò: adj,
ÑÒÒÛ¸Ñ×ÓѸÑÓÒ¸ÑÑ, —,
ÑÓ¹Ó¸Ø: no,
ÑÞÛÒ¹ÓÒ: adj,
ÑÒÛ¸ÑÒü¸ÑÒ¸ÑÑÒ, —,
ÑÖÓ¹ÓÙ¸Ø: no,
Ñ×Ó¹¹ÓÒ: adj, middle, mean, medial;ÓÑ×ÛÒ, ÑØÐÑÒÛ: vb,
ÑØÜ: adv,
ÑØÛÖÓ¹ÓÒ: adj,
ÑØÖÛ: vb,
ÑØÖÓÒ¸Ø: no,
ѸÑѸÑÒ: adj,
ÑÔÓØ: adv,
ÑØÖÓ¹¹ÓÒ: pro,
ÑÓ¹Ó¸Ø: no,
ÑÓÒ¹Ó¸é: no,
ÑÓÒÕ¹¹Ò: adj,
ÑÓÒÕü: adv,
ÑÒÓ¹¹ÓÒ: adj,
ÒÓÛ, —,Ò׸ÒÒ¸ÒÒѸÒÓÒ: vb,
ÓÓ¹¹ÓÒ: pre,
ØÖÓ¹ÓÒ: adj,
ÐÓ¹¹ÓÒ: adj,
ÑÓÒ¹: adj,
ÑÓÓ¹¹ÓÒ: adj,
ÑÓÓÔй: adj,
ÑÓÓع: adj,
ÑÓعØÓ¸é: no
ÑÓÛ: adv,
ÑÐÓÓ¹ÓÒ: adj,
ÑÓع: adj,
ÑôÒÙÑÓ¹ÓÒ: adj,
ÒÓѹØÓ¸Ø: no, name;ÓÒÓÑØÛÒ,
adj, so-called;
leave,
leave behind.
diminutive ofÐÑÑ.
lemma.
taking, catching.
— : vb, stay, remain.
part, direction, side.
bime-
take up.
dial.
ratio, proportion, argument.
remaining.
greater.
magnitude, size.
— : vb, learn.
between.
raised off the ground.
measure.
measure.
never.
neither (of two).
length.
ÑÒ: par, truely, indeed.
unit, unity.
unique.
uniquely.
alone.
conceive.
such as, of what sort.
eight-sided.
whole.
not even one, (neut.) nothing.
of the same kind.
similar.
similarly.
similar in number.
similarly arranged.
similarity.
corresponding, homologous.
apprehend,
ranged in the same row or line.
having the same name.
binomial.

ËÌÇÁÉÁÏÆ ÜÙôÒÓ¹ÓÒ: adj,
ܹ¹: adj,
ÔÓÓ×ÓÒ=ÔÓÓ¹¹ÓÒ+ÓÒ: adj,
Ô×Ó¹¹ÓÒ: pro,
ÔÓ×Ó×ÔÓØÓÒ=Ô×Ó¹¹ÓÒ++ÔÓØ+ÓÒ: adj,
ÔÓ×Ó×ÓÒ=Ô×Ó¹¹ÓÒ+ÓÒ: adj,
ÔØÖÓ¹¹ÓÒ: pro,
ÖÓôÒÓÒ¸Ø: no,
Ö¹¹Ò: adj,
ÖÛÒ, at
ÖÓ¸: no,
×ÔÓØÓÒ=×++ÔÓØ+ÓÒ: ind,
×: ind,
×ÔÐ×Ó¹ÓÒ: pro,
×Ó¹¹ÓÒ: pro,
×ÔÖ¸ÔÖ¸ÔÖ: pro,
×ظظØ: pro,
ØÒ: adv,
ØÓÒ: ind,
Ó¸ÓѸÓÒ: pro,
ÓØÖÓ¹¹ÓÒ: pro,
ÓØÖÓ: seeÓØÖÓ. ÓÒ: ind,
ÓØÛ: adv,
ÔÐÒ: adv,
ÔÒØÛ: adv,
ÔÖ: prep
ÔÖÐÐÛ: vb,
ÔÖÓиé: no,
ÔÖÑ: vb,
ÔÖÐÐ××Û¸ÔÖÐÐÜÛ¸—,ÔÖÐÐÕ, —,
ÔÖÐÐÐÔÔÓ¸¹ÓÒ: adj,
ÐÐÔÔÓÒ, no,
ÔÖÐÐÐÖÑÑÓ¹ÓÒ: adj,
ÖÐÐÐÖÑÑÓÒ, no,
ÔÖÐÐÐÓ¹ÓÒ: adj, parallel;ØÔÖÐÐÐÓÒ, ÔÖÔÐÖÛѹØÓ¸Ø: no,
ÔÖØÐÙØÓ¹ÓÒ: adj,
ÔÖ: prep
ÔÖÑÔÔØÛ: vb, insert; seeÔÔØÛ.
acute-angled;ØÜÙôÒÓÒ, no, acute angle.
acute.
of whatever kind, any
kind whatsoever.
as many, as many as.
of
whatever number, any number whatsoever.
of whatever number,
any number whatsoever.
either (of two), which (of two).
rectangle, right-angle.
straight, right-angled, perpendicular;ÔÖ
right-angles.
boundary, definition, term (of a ratio).
any number
whatsoever.
as many times as, as often as.
which.
as many times as.
as many as.
when, whenever.
whatsoever.
the very man who, the very thing
anyone who, anything which.
not one, nothing.
not either.
nothing.
ÓÒ: adv, therefore, in fact.
thusly, in this case.
back, again.
in all ways.
+ acc, parallel to.
apply (a figure); seeÐÐÛ.
application.
lie beside, apply (a figure); seeÑ.
— : vb, miss, fall
awry.
with parallel surfaces;ØÔÖй
parallelepiped.
bounded by parallel lines;ØÔ¹
parallelogram.
no, parallel,
parallel-line.
complement (of a parallelogram).
+ gen, except.
penultimate.
GREEK–ENGLISH LEXICON
Ô×ÕÛ¸Ô×ÓѸÔÓÒ¸ÔÔÓÒ, —, — : vb, suffer.
ÔÒØÛÒÓ¹ÓÒ: adj, pentagonal;ØÔÒØÛÒÓÒ, ÔÒØÔÐ×Ó¹¹ÓÒ: adj,
ÔÒØÛÒÓÒ¸Ø: no,
ÔÔÖ×ÑÒÓ¹¹ÓÒ: adj,
ÔÖÒÛ¸ÔÖÒü¸ÔÖÒ, —,ÔÔÖ×ѸÔÖÒÒÒ: vb,
ÔÖ¹ØÓ¸Ø: no,
ÔÖØÛ, —,
ÔÖÖÛ: vb,
ÔÖÕÛ: vb,
ÔÖÐÑÒÛ: vb, enclose; seeÐÑÒÛ. ÔÖ××: adv,
ÔÖ××¹¹Ò: adj,
ÔÖÖ¸é: no,
ÔÖÖÛ: vb,
ÔÐعØÓ¸é: no,
ÔÔØÛ¸Ô×ÓѸÔ×ÓÒ¸ÔÔØÛ, —,
ÔÐØÓ¹Ó¸Ø: no,
ÔÐÛÒ¹ÓÒ: adj,
ÔÐÙÖ¸é: no,
ÔÐÓ¹Ó¸Ø: no,
ÔÐÒ: adv
ÔÓ¹¹Ò: adj,
ÔÓÐÐÔÐ×ÞÛ: vb,
ÔÓÐÐÔÐ××Ѹ: no,
ÔÓÐÐÔÐ×ÓÒ¸Ø: no,
ÔÓÐÖÓ¹ÓÒ: adj, polyhedral;ØÔÓÐÖÓÒ, ÔÓÐÛÒÓ¹ÓÒ: adj, polygonal;ØÔÓÐÛÒÓÒ, ÔÓÐÔÐÙÖÓ¹ÓÒ: adj,
ÔÖ×ѹØÓ¸Ø: no,
ÔÓØ: ind,
ÔÖ×ѹØÓ¸Ø: no,
ÔÖÓÒÛ: vb,
ÔÖÓÒÙÑ: vb,
ÔÖÓØÑ: vb,
ÔÖÓÖÛ: vb,
ÔÖÓ×ÒÔÐÖÛ: vb,
ÔÖÓ×ÒÖÛ: vb,
ÔÖÓ×ÖÑÞÛ: vb,
ÔÖÓ×ÐÐÛ: vb,
ÔÖÓ×ÙÖ×Û: vb,
ÔÖÓ×ÐÑÑÛ: vb,
ÔÖÑ: vb,
ÔÖ×Ñ: vb,
no, pentagon.
five-fold, five-times.
fifteen-sided figure.
finite, limited; seeÔÖÒÛ.
bring
to end, finish, complete; pass, be finite.
end, extremity.
—, —, —, — : vb, bring to an end.
circumscribe; seeÖÛ.
encompass, surround, contain, comprise; seeÕÛ.
an odd number of times.
odd.
circumference.
carry round; seeÖÛ.
magnitude, size.
— : vb, fall.
breadth, width.
more, several.
side.
great number, multitude, number.
& prep + gen, more than.
of a certain nature, kind, quality, type.
multiply.
multiplication.
multiple.
no, polyhedron.
no, polygon.
multilateral.
part,ÔÖÓÖÑÒÓ¹
corollary.
at some time.
prism.
step forward, advance.
show previously; seeÒÙÑ.
set forth beforehand; seeØÑ.
say beforehand; perf pass
¹ÓÒ, adj, aforementioned; seeÖÛ.
fill up, complete.
complete (tracing of); seeÖÛ.
fit to, attach to.
produce (a line); seeÐÐÛ.
find besides, find; seeÖ×Û.
add.
set before, prescribe; seeÑ.
be laid on, have been added to; seeÑ.
543

construct (a figure), set up together; perf im-
ËÌÇÁÉÁÏÆ ÔÖÓ×ÔÔØÛ: vb,
ÔÖÓØ×¹Û¸é: no,
ÔÖÓ×Ø××Û: vb, prescribe, enjoin;ØØÖÓ×ØÕÒ, ÔÖÓ×ØÑ: vb, add; seeØÑ. ÔÖØÖÓ¹¹ÓÒ: adj,
ÔÖÓØÑ: vb, assign; seeØÑ. ÔÖÓÕÛÖÛ: vb,
ÔÖüØÓ¹¹ÓÒ: adj,
ÔÙÖѹӸé: no,
ع¹Ò: adj,
ÓÑÓ¹: adj, rhomboidal;ØÓÑÓ, ÑÓ¸no, rhombus.
×ÑÓÒ¸Ø: no,
×ÐÒ¹¹Ò: adj,
×ØÖ¹¹Ò: adj, solid;Ø×ØÖÒ, ×ØÓÕÓÒ¸Ø: no,
×ØÖÛ¸¹×ØÖÝÛ¸×ØÖݸ—,×ØÑѸ×ØÒ: vb,
×Ñ: vb,
×ÙÑÒÓ¹¹ÓÒ, adj,
seeÑ. ×ÖÒÛ: vb, compare; seeÖÒÛ. ×ÙÑÒÛ: vb,
×ÙÑÐÐÛ: vb,
×ÑÑØÖÓ¹ÓÒ: adj,
×ÑÔ¹ÒØÓ¸: no,
×ÙÑÔÔØÛ: vb,
×ÙÑÔÐÖÛ: vb,
×ÙÒÛ: vb,
×ÙÒÑØÖÓ¹¹: adj, both together;×ÙÒÑØÖÓ, ×ÙÒÔÓÒÙÑ: no,
×ÙÒ¸é: no,
×ÒÙÓ¸Ó¸¸Ø: no,
×ÙÒÕ¹: adj, continuous;ØØ×ÙÒÕ, ×Ò×¹Û¸é: no,
×ÒØÓ¹ÓÒ: adj,
×ÙÒ℄×ØÑ: vb,
perat pass 3rd sg,×ÙÒ×ØØÛ; see×ØÑ. ×ÙÒØÑ:vb, put
seeØÑ. ×Õ×¹Û¸é: no,
×ÕѹØÓ¸Ø: no,
×Ö¹¸é: no,
Øܹ۸é: no,
ØÖ××Û¸ØÖÜÛ¸—, —,ØØÖѸØÖÕÒ: vb,
ble, disturbe;ØØÖÑÒÓ¹¹ÓÒ,
fall on, fall toward, meet; seeÔÔØÛ.
proposition.
no, the
thing prescribed; seeØ××Û.
first (comparative), before, former.
go/come forward, advance.
first, prime.
pyramid.
expressible, rational.
no, romboid.
point.
scalene.
no, solid, solid body.
element.
turn.
lie together, be the sum of, be composed;
composed (ratio), compounded;
come to pass, happen, follow; seeÒÛ.
throw together, meet; seeÐÐÛ.
commensurable.
sum, whole.
meet together (of lines); seeÔÔØÛ.
complete (a figure), fill in.
conclude, infer; seeÛ.
no,
sum (of two things).
demonstrate together; seeÒÙÑ.
point of junction.
two together, in pairs.
continuously.
putting together, composition.
composite.
together, add together, compound (ratio);
state, condition.
figure.
sphere.
arrangement, order.
stir, trou-
adj, disturbed, perturbed.
544
GREEK–ENGLISH LEXICON
Ø××Û¸ØÜÛ¸ØܸØØÕ¸ØØѸØÕÒ: vb, arrange,
ØÐÓ¹¹ÓÒ: adj,
ØÑÒÛ¸ØÑÒü¸ØÑÓÒ¸¹ØØѸØØÑѸØÑÒ: vb,
ØØÖØÑÓÖÓÒ¸Ø: no,
ØØÖÛÒÓ¹ÓÒ: adj, square;ØØØÖÛÒÓÒ, ØØÖ: adv,
ØØÖÔÐ×Ó¹¹ÓÒ: adj,
ØØÖÔÐÙÖÓ¹ÓÒ: adj,
ØØÖÔÐÓ¹¹ÓÒ: adj,
ØѸ×Û¸¸Ø¸Ñ¸ØÒ: vb,
ØÑѹØÓ¸Ø: no,
ØÓÒÙÒ: par,
ØÓÓØӹعÓØÓ: pro,
ØÓѹ۸: no,
ØÓѸé: no,
ØÔÓ¸: no,
ØÓ×ÙØ: adv,
ØÓ×ÙØÔÐ×Ó¹¹ÓÒ: pro,
ØÓ×ÓØӹعÓØÓ: pro,
ØÓÙØ×Ø=ØÓØ×Ø: par,
ØÖÔÞÓÒ¸Ø: no,
ØÖÛÒÓ¹ÓÒ: adj,
ØÖÔÐ×Ó¹¹ÓÒ: adj,
ØÖÔÐÙÖÓ¹ÓÒ: adj,
ØÖÔйӹ¹ÓÒ: adj,
ØÖÔÓ¸: no,
ØÙÕÒÛ¸ØÜÓѸØÙÕÓÒ¸ØØÕ¸ØØÙѸØÕÒ: ÔÖÕÛ: vb,
ÔÜÖ×¹Û¸é: ÔÓØÒÛ¸ÔÓØÒü¸ÔØÒ¸ÔÓØظÔÓØØѸÔØÒ
no,
ÔÖÐÐÛ: vb,
ÔÖÓÕ¸é: no,
ÔÖÕÛ: vb,
Ô×¹Û¸é: no,
ÔÑ: vb,
ÔÓÐÔÛ: vb,
ÝÓ¹Ó¸Ø: no,
ÒÖ¹¹Ò: adj,
Ѹ×Û¸Ò, —, —, — : vb, say;ÑÒ, ÖÛ¸Ó×Û¸ÒÓÒ¸ÒÒÓÕ¸ÒÒѸêÒÕÒ: vb,
ÕôÖÓÒ¸Ø: no,
ÕÛÖ: pre
ÝÛ: vb,
ØÙÕÒ: par,
×ØÛ: adv,
draw up.
perfect.
cut;
pres/fut indic act 3rd sg,ØÑ.
quadrant.
no, square.
four times.
quadruple.
quadrilateral.
fourfold.
place, put.
part cut off, piece, segment.
accordingly.
such as this.
sector (of circle).
cutting, stump, piece.
place, space.
so many times.
so many times.
so many.
that is to say.
trapezium.
triangular;ØØÖÛÒÓÒ, no, triangle.
triple, threefold.
trilateral.
triple.
way.
vb, hit, happen to be at (a place).
begin, be, exist; seeÖÕÛ.
removal.
overshoot, exceed; seeÐÐÛ.
excess, difference.
exceed; seeÕÛ.
hypothesis.
underlie, be assumed (as hypothesis); seeÑ.
leave remaining.
: vb, subtend.
height.
visible, manifest.
we said.
carry.
place, spot, area, figure.
+ gen, apart from.
touch.
: par, as, like, for instance.
at random.
in the same manner, just so.
õ×Ø: conj, so that (causal), hence.