Line obtained by reflection passes through orthocentre

[Tahmid Hasan]

Some time ago, I was requested by Fahim vai to find a synthetic solution to the following problem:
Let $P$ be a point on the circumcircle of $\triangle ABC$. Then the reflections of $P$ wrt the sides of $\triangle ABC$ lie on a line that passes through the orthocentre of $\triangle ABC$.
I thought I'd rather post the solution here in case someone else finds it useful.
Observe that a homothety with centre $P$ and ratio $2$ sends the simson line of $\triangle ABC$ to the desired line. So it suffices to prove that $H$, the orthocentre lies on the line i.e. the line through any two of the reflection points.
Let the reflection of $P$ wrt $BC,AC$ be $P_a,P_b$ respectively.
Let the reflection of $H,P_b$ in $BC$ be $H',R$ respectively.
It is well-known that $H' \in \odot ABC$.
So our requirement shifts into proving $H',P,R$ are collinear.
Let $PP_b \cap AC=R_b,RP_b \cap BC=R_a,PP_b \cap BC=K$.
Note that $PR \parallel R_aR_b$.
So proving $R_aR_b \parallel H'P$ will complete the proof.
$R_aR_b \parallel H'P \Leftrightarrow \angle KR_bR_a=\angle HP'K$
$\Leftrightarrow 180^{\circ}-\angle R_bKC-\angle R_bR_aC=\angle HP'K$
$\Leftrightarrow \angle HP'K+\angle H'AC+\angle R_bP_bC=180^{\circ}$
$\Leftrightarrow \angle H'AC+\angle H'PK+\angle KPC=180^{\circ}$, which is actually true.