Iranian Mathematical Olympiad

Let $X^{\prime }$ be the reflection of $X$ with respect to $AH$ and $X^{\prime \prime }$ be the reflection of $H$ with respect to $X^{\prime }$. Notice that $X^{\prime }$ lies on the nine point circle (Euler circle) and $X^{\prime \prime }$ lies on the circumcircle. Lines $X^{\prime \prime }H$ and $XH$ intersect the circumcircle of triangle $ABC$ for the second time at $Y^{\prime }$ and $D$. Let $M$ be the midpoint of $HD$, so $$XH\cdot HD=X^{\prime \prime }H\cdot H{Y}^{\prime }⟹H{Y}^{\prime }=\frac{1}{2}HD=HM$$ Since the point $M$ lies on the nine point circle, and $\angle XHA=\angle X^{\prime \prime }HA$, $Y^{\prime }$ is the reflection of $M$ with respect to $AH$, so $Y^{\prime }\equiv Y$ and the result follows.