A history of Greek mathematics - Wilbourhall.org

cut into
GEOMETRY 313
four others ADGE, I)F, FGGB, GE, so that DF, GE
are equal, the common vertex G will lie on the diagonal AB.
Heron produces AG to meet GF in H, and then proves that
AHB is a straight line.
Since DF, GE are equal, so are
the triangles D GF, EGG. A dding
the triangle GGF, we have the
triangles EGF, JDGF equal, and
DE, GF are parallel.
But (by I. 34, 29, 26) the triangles
AKE, GKD are congruent,
so that EK=KD
;
and by lemma ( 1) it follows that CH=HF.
Now, in
the triangles FHB, CHG, two sides (BF, FH and
GC, 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 FHG = Z CHG + Z GHF = two right angles.
To prove his substantive proposition Heron draws AKL
Then
perpendicular to BG, and joins EG meeting AK in M.
we have only to prove that BMG is a straight line.
Complete the parallelogram FAHG, and draw the diagonals
OA, FH meeting in F. Through M draw PQ, SR parallel
respectively to BA, AG.