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

AD)
. EC
THE COLLECTION. BOOK VII 417
The other problem (Prop. 117, pp. 848-50) is, as we have
seen, equivalent to the following : Given a circle arid three
'points D, E, F in a straight line external to it, to inscribe in
the circle a triangle ABC such that its sides pass sever ally
through the three 'points D, E F. For the solution, see
y
pp. 182-4, above.
(e) The Lemmas to the Plane Loci of Apollonius (Props.
119-26, pp. 852-64) are mostly propositions in geometrical
algebra worked out by the methods of Eucl., Books II and VI.
We may mention the following :
Prop. 122 is the well-known proposition that, if D be the
middle point of the side BC in a triangle ABC,
BA 2 + AC 2 = 2 (AD 2 + DC 2 ).
Props. 123 and 124 are two cases of the same proposition,
the enunciation being marked by an expression which is also
found in Euclid's Data. Let AB : BC be a given ratio, and
let the rectangle CA .
A DEC B
AD
be given ; then, if BE is a mean
proportional between DB, BC, the square on AE ' is greater
by the rectangle CA .
square on EC\ by which is meant that
or (AE 2 - CA .
AD than in the ratio of AB to BC to the
A R
AE = CA.AD+ ~ 2
c
:
2 ,
EC = AB 2 :
BC.
The algebraical equivalent may be expressed thus (if AB=a,
BC = b, AD = c, BE=x):
jo /, zt ,1 (aTx) 2 -(a — b)c a
It x = v(a — c)b, then ; _: —— = -*•
v ;
(x + bf b
Prop. 125 is remarkable : If C, D be two points on a straight
line AB,
Aft
/ID
^ AD 2 + . DB = 2 AC 2 + AC.CB+ == . CD\
1823,8
E e
418 PAPPUS OF ALEXANDRIA
This is equivalent to the general relation between four
points on a straight line discovered by Simson and therefore
wrongly known as Stewart's theorem
AD 2 . BC+BD 2 . CA + CD 2 . AB + BC.CA . AB
= 0.
(Simson discovered this theorem for the more general case
where D is a point outside the line ABC)
An algebraical equivalent is the identity
(d _ a f (b-c) + (d - b) 2
(c-a) + (d - c ) 2 (a - b)
+ (b — c) (c — a) (a — b) = 0.
Pappus's proof of the last-mentioned lemma is perhaps
worth giving.
A c D B
C, D being two points on the straight line AB, take the
point F on it such that
Then FB :
FD:DB = AC:CB. (1)
BD = AB :
BC,
and (AB-FB) : (BC-BD) = AB.BC,
or
and therefore
From (1) we derive
AF:CD = AB:BC,
AF.CD:CD 2 = AB:BC. [2)
AG
and from (2)
. DB
or
We have now to prove that
2 = FD. DB,
^. CD 2 = AF.CD.
AD 2 + BD.DF= AC 2 + AC.CB + AF.CD,
AD 2 + BD.DF= CA.AB + AF.CD,