In triangle ABC the bisector of ∠BAC meets BC at D. The point E is chosen on AC so that DE is parallel to AB. The point F is chosen on AB so that DF is parallel to AC. Show that (AB)^2/(AC)^2 = BF/CE

Answer 1

AB / AC = BD / DC Angle bisector Theorem

Then

[ AB / AC ]² = [BD / DC ]² ----> AB² / AC² = BD² / DC² ( E q 1)

Triangles BFD and DEC are similar . Then the ratio of their areas are equal to the square of any pair of the corresponding sides of the similar triangles.

That is BD² / DC² = (Area BFD) / (Area DEC) = ( BF* DM) / ( DC*DL) = BF / DC . DM and DL being the heights of the similar triangles BFD and DEC and since D on the bisector of the angle BAC they are equal in length.

Therefore BD² / DC²= BF / DC ( E q 2)

From ( E q 1) and ( E q 2) we have AB² / AC² = BF / DC .