Prove PQ is parallel to BC with angle chase? I've been working on this geometry problem for a long time, but I haven't been able to make any progress. I have tried angle chase but that didn't get me anywhere.
Acute triangle $ABC$ has $AB<AC.$ Let $D, E, F$ be the foots of perpendiculars from $A, B, C  \text{ and } H $ is orthocenter. $EF$ and $AD$ intersect at $P.$ The reflection of $AB$ in $AD$ intersects BE at $Q$, making $\angle DAB =  \angle DAB' $
I want to prove that line joining $P,Q$ is parallel to $BC.$ How can I prove?


Note that $AFHE$ is cyclic because $\angle HFA = 90^\circ = \angle AEH$. Hence $\angle FEH = \angle FAH = \angle HAQ$. This shows that $\angle PEQ = \angle PAQ$, hence $APQE$ is cyclic. Since $\angle AEQ = 90^\circ$, it follows that $\angle QPA = 90^\circ$. Hence $PQ \perp AH$. Since $BC \perp AH$, it follows that $PQ \parallel BC$.
