Let $ABC$ be an acute triangle with orthocentre $H$ and circumcircle $Γ$. Let $BH$ intersect $AC$ at $E$, and let $CH$ intersect $AB$ at $F$. Let AH intersect $Γ$ again at $P$ which is not equal to $A$. Let PE intersect $Γ$ again at $Q$ which is not equal to $P$. Prove that $BQ$ bisects segment $EF$.
This is an exam question which I gave yesterday. So there is no point for me trying it now... I want to see how it will be done..
Edit: Let the reflection of $H$ over $AC$ be $X$. Clearly triangles $BFE$ and $PHX$ are similar. Since $PE$ is a median in triangle $PHX$ it follows that $BQ$ is a median in triangle $BFE$. Hence $BQ$ bisects $EF$. This is what I have done... Am I correct?