Nice problem! Claim 1 : $AXQY$ is concyclic and $QX=QY$ Just note that $\angle AXQ + \angle AYQ= 180^{\circ}$, as desired. Now, by Fact 5, we get $QX=QY$ .$\square$ Claim 2 : $BH \parallel QY$ Let $H'$, $Y'$ and $Q'$ be reflections of $H$,$Y$ and $Q$ about $M$, its well-known that $H'$ lies on $(ABC)$ and $AH'$ is the diameter of $(ABC)$, hence $\angle ACH' = 90^{\circ}$. To prove $BH \parallel QY$, it suffices to show that $BH \parallel CH' \parallel Q'Y'$. Now, just note that $\angle HBY' = H'CY=90^{\circ}=\angle QYC = \angle BY'Q'$, as desired $\square$ Since, $QX=QY$ we get $\angle QXY = QYX = \frac{A}{2}$. Hence, $\angle BHY = 180 - \frac{A}{2}$. Now, just note that $\angle BHX=\frac{A}{2}$. Hence, $\angle BHY+\angle BHX=180$. Thus, $X,H,Y$ are collinear , as desired $\blacksquare$

Let the line through $H$ perpendicular to $MH$ cut $AB, AC$ at $S, T$ respectively. Lemma. $\overline{MS}=\overline{MT}$. Proof. well-known Hence $\overline{QS}=\overline{QT}\to Q\in (AST)$ and by Simson Line thm $H\in XY$.

It's nothing but ISL 2005 G5