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?

 A: 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 $PQ \parallel BC$.
A: Let D be the origin and DA be the $y$-axis. Suppose $A(0,a)$, $B(b,0)$, $C(c,0)$, with $a\not=0$ and $b\not=c$.
$AB$: $x/b+y/a=1$
$CH$: $y=b/a(x-c)$
$F$: $(\frac{b(a^2+bc)}{a^2+b^2}, \frac{ab(b-c)}{a^2+b^2})$
$AC$: $x/c+y/a=1$
$BH$: $y=c/a(x-b)$
$E$: $(\frac{c(a^2+bc)}{a^2+c^2}, \frac{ac(c-b)}{a^2+c^2})$
The $y$-coordinate of $P$ is $\frac{\frac{b(a^2+bc)}{a^2+b^2} \frac{ac(c-b)}{a^2+c^2}-\frac{c(a^2+bc)}{a^2+c^2}\frac{ab(b-c)}{a^2+b^2}}{\frac{b(a^2+bc)}{a^2+b^2}-\frac{c(a^2+bc)}{a^2+c^2}}=\frac{-2abc}{a^2-bc}$
$AB'$: $x/(-b)+y/a=1$
$BH$: $y=c/a(x-b)$
The $y$-coordinate of $Q$ is $\frac{\frac{-2cb}a}{1-\frac{bc}{a^2}}=\frac{-2abc}{a^2-bc}$.
Hence $PQ$ is parallel to the $x$-axis, i.e., $BC$.

Because of the generic nature of the coordinates, we have proved the proposition is true for all triangles, including right triangles and obtuse triangles, except when $a^2=bc$, in which case $EF$ is parallel to $AD$ and $BH$ is parallel to $AB'$, both $P$ and $Q$ "at infinity".
