$BCHQ$ is cyclic. Hence, $\angle BCH = \angle AQH$. Let $AB \cap CH = F$. $CF \perp AB \Rightarrow \angle FBC = 90 - \angle FCB \Rightarrow \angle QAH = \angle FCB = \angle AQH \Rightarrow AH = HQ$. Similarly, $AH = HP$ So $HQ = HP$ and thus $H$ is the circumcenter of $\triangle AQP$.

We have\begin{align*}\angle APB & =180^\circ -\angle BPC \\ & =180^\circ -\angle BHC \\ & =\angle BAC \\ & =\angle BAP, \end{align*}so $BA=BP$, i.e. $H$ is on the perpendicular bisector of $AP$. Analogously, $H$ is on the perpendicular bisector of $AQ$. Done.

We just need to notice that$$HP=HQ=HA=\cos\angle A\cdot 2R$$.

Ez complex bash (synthetic is too hard) set $a, b, c$ on the unit circle with $h = a+b+c$. we redefine point $q$ as the reflection of a about the C-altitude. we claim that $Q$ lies on the circumcircle of $BCH$. by standard equations we have: $q = b + c - \frac{ab}{c}$ to show that the points are cyclic it suffices to show that: $ T := \frac{(h-b)(h-c)}{(q-b)(q-c)} = \frac{(a+c)(b-\frac{ab}{c})}{(a+b)(c-\frac{ab}{c})}$ is real. This is clearly true by checking it is equal to its own conjugate. This works analagously for $P$, thus we have that $P, Q$ are the original points in the problem and we are done.