Let $AA'$, $DD'$ be altitudes in the triangle $\triangle APD$ and let $BB'$, $CC'$ be altitudes in the triangle $\triangle BPC$.
We have $\angle XAC = \angle XBD$ and $\angle XCA = \angle XDB$, and thus $\triangle XAC \sim \triangle XBD$. We prove that the line $H_1H_2$ passes through the point $X$ if and only if the above triangles are congruent.

We consider the circles $\omega_1, \omega_2$ of diameter $AC$ and $BD$ respectively. The points $H_1, H_2$ belong to the radical axis of these two circles, because $A, A' \in \omega_1, D, D' \in \omega_2$

şi $P_{\omega_1}(H_1) = -AH_1 \cdot A'H_1 = -DH_1 \cdot D'H_1 = P_{\omega_2}(H_1)$; and similar for $H_2$. Thus, $X \in H_1H_2$ if and only if $P_{\omega_1}(X) = P_{\omega_2}(X)$.
We have $P_{\omega_1}(X) = XM^2 - MC^2$ and $P_{\omega_2}(X) = XN^2 - ND^2$, were $M, N$ are the centers of the circles $\omega_1$ and $\omega_2$ respectively. We notice that $(XM^2 - MC^2)/(XN^2 - ND^2)$ is the square of the similarity ratio of the triangles $AXC$ and $BXD$, or $XM^2 - MC^2 = XN^2 - ND^2 = 0$. In the former case, the triangles $AXC$ and $BXD$ should be similar and right angles in $X$. But now the line $CD$ corresponds to the line $AB$ by a similarity of center $X$ and angle $90^\circ$, so $AB \perp CD$ – contradiction. Thus $(XM^2 - MC^2)/(XN^2 - ND^2)$ is the square of the similarity ratio of the triangles $AXC$ and $BXD$.
We obtain that $X \in H_1H_2$ if and only if the similarity ratio of the triangles $AXC$ and $BXD$ is 1, that is $\Delta AXC \equiv \Delta BXD$. But this is equivalent to $AC = BD$.
In the same manner, $Y \in H_1H_2$ if and only if $AC = BD$, and we get the conclusion of the problem.