Let $Z$ be the midpoint of segment $XY$, and $N$ and $T$ be the midpoints of segments $AB$ and $AC$ respectively (see the figure).

As, according to the condition, $AX = XC$ and $AY = YB$, $XT$ and $YN$ are bisectors to segments $AC$ and $AB$. Then triangles $XTY$ and $XNY$ are right triangles, so $XZ = ZN = ZT = ZY$, because the median of a right triangle, drawn to the hypotenuse, equals its half. So, point $Z$ is equidistant from $N$ and $T$.

On the other side, $NT \parallel BC$, because $NT$ is the midline of $\triangle ABC$. $MT$ is also a center line of $\triangle ABC$, so $MT = \frac{1}{2}AB$. $NH = \frac{1}{2}AB$ as the median, drawn to the hypotenuse in a right triangle. So, $HNTM$ is an isosceles trapezoid. Then its bases have a common bisector. We have proved that point $Z$ belongs to bisector $NT$. Then it belongs to bisector $HM$.