Question 64
Let $ABC$ be an acute triangle. The altitudes $BE$ and $CF$ intersect at the orthocenter $H$, and point $O$ denotes the circumcenter. Point $P$ is chosen so that $\angle APH = \angle OPE = 90^{\circ}$, and point $Q$ is chosen so that $\angle AQH = \angle OQF = 90^{\circ}$. Lines $EP$ and $FQ$ meet at point $T$. Prove that points $A, T, O$ are collinear.