A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

THE COLLECTION. BOOKS III, IV 369
this way. If d be the diameter of the sphere, set out two
straight lines x, y such that d, x, y are in the ratio of the sides
of the regular pentagon, hexagon and decagon respectively
described in one and the same circle. The smaller pair of
circles have r as radius where v 2 = ^y
2
,
and the larger pair
have r' as radius where r 2 — \x 2 .
(e) In the case of the dodecahedron the saw e four parallel
circular sections are drawn as in
the case of the icosaheclrori.
Inscribed pentagons set the opposite way are inscribed in the
two smaller circles ; these pentagons form opposite faces.
Regular pentagons inscribed in the larger circles with vertices
at the proper points (and again set the opposite way) determine
ten more vertices of the inscribed dodecahedron.
The constructions
are quite different from those in Euclid
XIII. 13, 15, 14, 16, 17 respectively, where the problem is first
to construct the particular regular solid and then to 'comprehend
it in a sphere ', i. e. to determine the circumscribing
sphere in each case. I have set out Pappus's propositions in
detail elsewhere. 1
Book IV.
At the beginning of Book IV the title and preface are
missing, and the first section of the Book begins immediately
with an enunciation. The first section (pp. 176-208) contains
Propositions 1-12 which, with the exception of Props. 8-10,
seem to be isolated propositions given for their own sakes and
not connected by any general plan.
Section (1). Extension of the theorem of Pythagoras.
The first proposition is of great interest, being the generalization
of Eucl. I. 47, as Pappus himself calls it, which is by this
time pretty widely known to mathematicians. The enunciation
is as follows.
'If ABC be a triangle and on AB, AC any parallelograms
whatever be described, as ABLE, ACFG, and if DE, FG
produced meet in H and HA be joined, then the parallelograms
ACFG are together equal to the parallelogram
ABDEy
1
Vide notes to Euclid's propositions in The Thirteen Books of Euclid's
Elements, pp. 473, 480, 477, 489-91, 501-3.
1523 2 B b
370 PAPPUS OF ALEXANDRIA
contained by BC, HA in an angle which is equal to the sum of
the angles ABC, DHA!
Produce HA to meet BC in K, draw BL, CM parallel to KH
meeting BE in L and FG in M, and join LNM.
Then BLHA is a parallelogram, and HA is equal and
parallel to BL.
Similarly HA, CM are equal and parallel ;
are equal and parallel.
Therefore BLMC is a parallelogram ; and
therefore BL, CM
its angle LBK is
equal to the sum of the angles ABC, DHA.
Now a ABBE — BLHA, in the same parallels,
= O BLNK, for the same reason.
Similarly ACFG = O jiOAffl' = TOCif.
Therefore, by addition, ABDE+C3 ACFG = a 5ZM7.
It has been observed (by Professor Cook Wilson *) that the
parallelograms on A B, AC need not necessarily be erected
outwards from AB, AC. If one of them, e.g. that on AC, be
drawn inwards, as in the second figure above, and Pappus's
construction be made, we have a similar result with a negative
sign, namely,
o BLMC = BLNK - o CMTO
Again, if both ABBE and ACFG were drawn inwards, their
sum would be equal to BLMC drawn outwards. Generally, if
the areas of the parallelograms described outwards are regarded
as of opposite sign to those of parallelograms drawn inwards,
1
Mathematical Gazette, vii, p. 107 (May 1913).