TH-TMO 2011 Problem 3

Given a $\Delta ABC$ where $\angle C = 90^{\circ}$, $D$ is a point in the interior of $\Delta ABC$ and lines $AD$ $,$ $BD$ and $CD$ intersect $BC$, $CA$ and $AB$ at points $P$, $Q$ and $R$,respectively. Let $M$ be the midpoint of $\overline{PQ}$. Prove that, if $\angle BRP$ $=$ $\angle PRC$ then $MR=MC$.