Question 145
Let $ABC$ be a triangle inscribed in circle $(O)$, with its altitudes $BE$, $CF$ intersecting at orthocenter $H$ ($E \in AC$, $F \in AB$). Let $M$ be the midpoint of $BC$, $K$ be the orthogonal projection of $H$ on $AM$. $EF$ intersects $BC$ at $P$. Let $Q$ be the intersection of the tangent to $(O)$ which passes through $A$ with $BC$, $T$ be the reflection of $Q$ through $P$. Prove that $\angle OKT = 90^{\circ}$.