At the altitude $AH_1$ of an acute non-isosceles triangle $ABC$ chose a point $X$ , from which draw the perpendiculars $XN$ and $XM$ on the sides $AB$ and $AC$ respectively. It turned out that $H_1A$ is the angle bisector $MH_1N$. Prove that $X$ is the point of intersection of the altitudes of the triangle $ABC$.