Extended Law of Sines

(2016 USAJMO #5) Let $\triangle ABC$ be an acute triangle, with $O$ as its circumcenter. Point $H$ is the foot of the perpendicular from $A$ to line $\overleftrightarrow{BC},$ and points $P$ and $Q$ are the feet of the perpendiculars from $H$ to the lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{AC},$ respectively.

Given that $$AH^2 = 2\cdot AO^2,$$ prove that the points $O, P,$ and $Q$ are collinear.