SMO 2011 Problem 1

In the acute-angled non-isosceles triangle $ABC$ , $O$ is its circumcenter, $H$ is its orthocenter and $AB>AC$ . Let $Q$ be a point on $AC$ such that the extension of $HQ$ meets the extension of $BC$ at the point $P$ . Suppose $BD=DP$ , where $D$ is the foot of the perpendicular from $A$ onto $BC$ . Prove that $\angle ODQ=90^{\circ}$ .