Let $A^{\prime}$ be the intersection of lines $B X$ and $C Y$. By power of a point, it suffices to prove that $A^{\prime} B \cdot A^{\prime} X = A^{\prime} C \cdot A^{\prime} Y$, or, equivalently, that $A^{\prime}$ lies on the radical axis of the circles $A B D X$ and $A C E Y$. From $D A = D X$ it follows that in circle $A B D X$, point $D$ bisects one of the arcs $A X$. Therefore, depending on the order of points, the line $B C$ is either the internal or external bisector of $\angle A B X$. In both cases, line $B X$ is the reflection of $B A$ in line $B D C$. Analogously, line $C Y$ is the reflection of $C A$ in line $B C$; we can see that $A^{\prime}$ is the reflection of $A$ in line $B C$, so $A, F$ and $A^{\prime}$ are collinear. By $P D \| A C$ and $P E \| A B$ we have $\frac{F D}{F C} = \frac{F P}{F A} = \frac{F E}{F B}$, hence $F D \cdot F B = F E \cdot F C$. So, point $F$ has equal powers with respect to circles $A B D X$ and $A C E Y$. Point $A$, being a common point of the two circles, is another point with equal powers, so the radical axis of circles $A B D X$ and $A C E Y$ is the altitude $A F$ that passes through $A^{\prime}$. !