Let $ABC$ be an isosceles triangle with $AB = AC$. Let P be a point in the interior of $ABC$ such that $PB > PC$ and $\angle PBA = \angle PCB$. Let $M$ be the midpoint of the side $BC$. Let $O$ be the circumcenter of the triangle $APM$. Prove that $\angle OAC=2 \angle BPM$ .
Problem
Source: 2017 Baltic Way Shortlist G7 BW https://artofproblemsolving.com/community/c2659348_baltic_way_shortlist_2017_geometry
Tags: equal angles, geometry
