Let $ABC$ be an isosceles triangle with $AC = BC$, and let $M$ be the midpoint of $AB$. Let $P$ be a point inside the triangle such that $\angle PAB =\angle PBC$. Prove that $\angle APM + \angle BPC = 180^o$.
Problem
Source: Indian Postal Coaching 2007 set 6 p1
Tags: angles, geometry, isosceles, equal angles