MEMO 2019 Problem 5

Let $ABC$ be an acute-angled triangle such that $AB<AC$ . Let $D$ be the point of intersection of the perpendicular bisector of the side $BC$ with the side $AC$ . Let $P$ be a point on the shorter arc $AC$ of the circumcircle of the triangle $ABC$ such that $DP \parallel BC$ . Finally, let $M$ be the midpoint of the side $AB$ . Prove that $\angle APD=\angle MPB$ .

Proposed by Dominik Burek, Poland