Proving an angle to be 90 
$H$ is the orthocenter of acute triangle $A B C$. Let $\omega$ be the circumcircle of $B H C$ with center $O^{\prime}. \Omega$ is the nine-point circle of $A B C . X$ is an arbitrary point on arc $B H C$ of $\omega$ and $A X$ intersects $\Omega$ at $Y . P$ is a point on $\Omega$ such that $PX= PY$. Prove that $ \angle O^{\prime}PX=90°$.

I extended $XP$ to intersect circumcircle of $ \triangle BHC$ at $M$ and then tried proving $\angle XYM$ a right angle. Kept angle chasing but didn't get any satisfactory result.
 A: First, consider the homothety at $A$ with scale factor $2$. This maps three points on the nine-point circle (the midpoints of $AB$, $AC$, and $AH$) to three points on $(HBC)$, so it maps the nine-point circle to $(HBC)$. The image of $Y$ is thus a point on both line $AX$ and $(HBC)$; it isn't $X$ (based on the diagram; if it is $X$ the problem is in fact false). So, if $AX$ intersects $(HBC)$ again at $Z\neq X$, then $Y$ is the midpoint of $AZ$. Also, if $N$ is the center of the nine-point circle, then the homothety takes $N$ to $O'$, and so $N$ is the midpoint of $AO'$.
We can now "simplify" the problem as follows:

The points $X$ and $Z$ lie on a circle centered at $O'$, and a point $A$ lies in line $XZ$. $Y$ is the midpoint of $AZ$, and $N$ is the midpoint of $AO'$. Show that  (1) the circle centered at $N$ passing through $Y$, (2) the circle with diameter $O'X$, and (3) the perpendicular bisector of $XY$ concur, as long as (1) and (3) intersect.

(We are given that (1) and (3) intersect since $P$ exists; this is the same as the original problem since $P$ is on the circle with diameter $O'X$ if and only if $\angle XPO'=90^\circ$.) To solve this restated problem, let $R$ be the midpoint of $O'X$, so that (1) and (2) are the circles centered at $N$ passing through $Y$ and centered at $R$ passing through $X$, respectively. Note that $$\measuredangle RXZ=\measuredangle O'XZ=\measuredangle XZO'=\measuredangle AZO'=\measuredangle AYN=\measuredangle XYN,$$
that $R$ and $N$ lie on the same side of line $XY$ as each other, and that
$$XR=\frac{O'X}2=\frac{O'Z}2=NY,$$
so $XRNY$ is an isosceles trapezoid with $RN\|XY$. This means that $N$ and $R$ are reflections over (3), so circles (1) and (2) are as well; this means that any intersection of (1) and (3) also lies on (2), as desired.
