Problem

Source: Iran Second Round 2015 - Problem 3 Day 1

Tags: geometry, 2015



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 $.