A history of Greek mathematics - Wilbourhall.org

THE COLLECTION. BOOK VII 407
Two proofs are given of the first theorem.
We will give the
first (Prop. 26) because it is a case of theoretical analysis
followed by synthesis. Describe a circle about ABD :
produce
EA, CA to meet the circle again in F, G, and join BF, FG.
Substituting GC . CA for BC . CD and FE . EA for BE. ED,
we have to inquire whether GC . CA : CA 2 = FE . EA : AE 2 ,
i. e. whether
t
GC:CA = FE: EA,
i.e. whether GA :
i.e.
AC = FA :
AE,
whether the triangles GAF, CAE are similar or, in other
words, whether GF is parallel to BC.
But GF is parallel to BC, because, the angles BAC, DAE
being supplementary, Z DAE = Z GAB — Z GFB, while at the
same time Z DAE = suppt. of Z FAD = Z FBD.
The synthesis is obvious.
An alternative proof (Prop. 27) dispenses with the circle,
and only requires EKH to be drawn parallel to CA to meet
AB, AD in H, K.
Similarly (Prop. 28) for case (b)
it is only necessary to draw
FG through D parallel to AC meeting BA in F and AE
produced in G.
Then, I FAG, Z ADF (= IDAC) being both right angles,
FD . DG = DA 2 .
Therefore CA 2 : AD = 2 CA 2 :FD.DG = (CA : FD) . (CA : DG)
= (BC:BD).(CE:DE)
= BC.CE:BD.DE.
In case (c) a circle is circumscribed to ADE cutting AB in F
and AC in G. Then, since Z FAD = L GAE, the arcs DF, EG
are equal and therefore FG is parallel to DE. The proof is
like that of case (a).