GEOMETRY 
313 
cut into four others ADGE, DF, FGGB, CE, so that DF, CE 
are equal, the common vertex G will lie on the diagonal AB. 
Heron produces AG to meet GF \n H, and then proves that 
AHB is a straight line. 
Since DF, CE are equal, so are 
the triangles DGF, EGG. Adding 
the triangle GGF, we have the 
triangles ECF, DCF equal, and 
DE, CF are parallel. 
But (by I. 34, 29, 26) the tri 
angles AKE, GKD are congruent, 
so that EK — KD ; and by lemma (1) it follows that GH—HF. 
Now, in the triangles FHB, GHG, two sides {BF, FH and 
GO, GH) and the included angles are equal ; therefore the 
triangles are congruent, and the angles BHF, GHG are equal. 
Add to each the angle GHF, and 
Z BHF + Z FUG = Z GHG + Z GHF = two right angles. 
To prove his substantive proposition Heron draws AKL 
perpendicular to BG, and joins EG meeting AK in M. Then 
we have only to prove that BMG is a straight line. 
A D 
Complete the parallelogram FAHO, and draw the diagonals 
OA, FH meeting in Y. Through M draw PQ, SR parallel 
respectively to BA, AG.