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 ABB: 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 GC : CA = FE: EA, 
i.e. whether GA : AG = FA : AE, 
i.e. 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 ZDAE = suppt. of Z FAD — ¿FBD. 
The synthesis is obvious. 
An alternative proof (Prop. 27) dispenses with the circle, 
and only requires EKH to be drawn parallel to GA to meet 
AB, AD in H, K. 
Similarly (Prop. 28) for case (h) it is only necessary to draw 
FG through D parallel to AG meeting BA in F and AE 
produced in G. 
A 
Then, ¿FAG, Z ADF { = ¿DAG) being both right angles, 
FD.DG = DA 2 . 
Therefore CA 2 : AD 2 = CA 2 : FD . DG = {CA : FD). {CA : DG) 
= {BG:BD).{GE:DE) 
= BC. GE: BD. DE. • 
In case (c) a circle is circumscribed to ADE cutting AB in F 
and AC in G. Then, since Z FAD = Z GAE, the arcs DF, EG 
are equal and therefore FG is parallel to DE. The proof is 
like that of case {a).