ABC is a triangle. Let D be a point on side BC produced beyond B such that BD=BA. Let M be the mid-point of AC. The bisector of angle ABC meets DM at P. Prove that angle BAP=angle ACB.
If you know the theorem of Menelaus, then it's easy.
Define Q the be the point where AP intersects BC. You have to prove that triangles CBA and ABQ are similar by side-angle-side. Menelaus's Theorem applied to the transversal MPD of triangle CAQ says that the product
(CM/MA) times (AP/PQ) times (QD/DC) = 1. Use the given information now to prove that AB/BQ = CB/BA.
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