IR-MO-R2 2015 Problem 3

Consider a triangle $ABC$ . The points $D,E$ are on sides $AB,AC$ such that $BDEC$ is a cyclic quadrilateral. Let $P$ be the intersection of $BE$ and $CD$ . $H$ is a point on $AC$ such that $\angle PHA = 90^{\circ}$ . Let $M,N$ be the midpoints of $AP,BC$ . Prove that: $ACD \sim MNH$ .