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.