Question 22
In the acute-angled non-isosceles triangle $ABC$, $O$ is its circumcentre, $H$ is 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 the point $P$. Suppose $BD = DP$, where $D$ is the foot of the perpendicular from $A$ onto $BC$. Prove that $\angle ODQ = 90^\circ$.