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SMO 2011 Open Round 2 Question 1

SMO · SMO 2011 Open Round 2

Question 1

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