In the acute triangle $ABC$ the orthocenter $H$ and the center of the circumscribed circle $O$ were noted. The line $AO$ intersects the side $BC$ at the point $D$. A perpendicular drawn to the side $BC$ at the point $D$ intersects the heights from the vertices $B$ and $C$ of the triangle $ABC$ at the points $X$ and $Y$ respectively. Prove that the center of the circumscribed circle $\Delta HXY$ is equidistant from the points $B$ and $C$. (Danilo Hilko)