41st Balkan Mathematical Olympiad

Let $ABC$ be an acute-angled triangle with $AB<AC$, orthocentre $H$, circumcircle $\Gamma$ and circumcentre $O$. Let $M$ be the midpoint of $BC$ and let $D$ be a point such that $ADOH$ is a parallelogram. Suppose that there exists a point $X$ on $\Gamma$ and on the opposite side of $DH$ to $A$ such that $\angle DXH+\angle DHA=90^{\circ }$. Let $Y$ be the midpoint of $OX$. Prove that if $MY=OA$ then $OA=2OH$.