We prove the statement indirectly. Namely suppose that $Z$ is the intersection of $BI$ and $XY$, we will prove that the perpendicular bisector of $CH$ intersect $XY$ at $Z$. Suppose that $CI$ intersect $XY$ at $W$. Note that $\angle AXY = 90 - \frac{\angle A}{2}$ making $\angle ZXY = \angle BXY = 90 + \frac{\angle A}{2}, \angle XBY = \frac{\angle B}{2}$ and hence $\angle WZB = \frac{\angle C}{2} =\angle WCB$, and hence $WZCB$ is a cyclic quad. Now, $\angle XWI = \angle ZWC = \frac{\angle B}{2} = \angle XBI$ making $XWBI$ cyclic. Hence $\angle BZC = \angle BWC = \angle BWI = \angle BXI = 90$. We can conclude also that $BHZC$ is cyclic, with the center at the midpoint of $BC$. Since $HBZ = \angle HBC$, we can conclude that $HZ = ZC$ and hence the perpendicular bisector of HC intersect $Z$.