Question 17
Let $ABC$ be an acute-angled triangle such that $AB \lt AC$. 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 $\angle APD = \angle MPB$.