Let $(ADXY)$ intersect $BC$ again at $Z$. As $Z$ moves linearly, $ZX$ and $ZY$ are parallel to fixed lines, so $X$ and $Y$ also move linearly. The $X$ and $Y$ altitudes of $AXY$ move linearly since they are parallel to $CF$ and $BE$, which means that $H$ also moves linearly. When $Z=B$, $H=E$, and when $Z=C$, $H=F$, so $H$ lies on $EF$.

Note that the Orthocenter of $AXY$ lies on the Steiner line of $D$ wrt. $AXY$, which goes through the reflections of $D$ over $AB$ and $AC$. Now to verify that this is $EF$, we simply notice that $AC$ is an angle bisector of $\angle FED$ and thus the reflection of $D$ over it lies on $FE$. Also, this is very similar to Imo 1965 P5a

I'm so sorry By phantom points it suffices to prove that for every point $H' \in \overline{EF}$, if we let $X',Y'$ lie on $\overline{AB},\overline{AC}$ respectively such that $\overline{X'H'} \perp \overline{AC}$ and $\overline{Y'H'} \perp \overline{AB}$ then $AX'Y'D$ is cyclic. Place the problem in the coordinate plane with $D$ as the origin and $\overline{BC}$ and the $x$-axis. As we vary $H$, since the slopes of $\overline{H'X'}$ and $\overline{H'Y'}$ are fixed, $X'$ and $Y'$ move linearly, so the slopes of $\overline{DX'}$ and $\overline{DY'}$ are described by rational functions whose numerator and denominator are linear. Then by the tangent subtraction formula, the tangent of $\angle X'DY'$ is described by a rational function whose numerator and denominator are quadratic. It suffices to show that this is always equal to the negative of the tangent of $\angle X'AY'$, which is fixed, so we only have to check three cases. When $H=E$, $X'=B$ and $Y'=E$, and when $H=F$, $X'=F$ and $Y'=C$, so these cases are easy. For our third case, we pick $H'$ such that $\overline{X'Y'} \parallel \overline{BC}$. I claim that in this case $H'=\overline{EF} \cap \overline{AD}$. If $\overline{AD}$ intersects $(ABC)$ again at $K$ and $H_0$ is the orthocenter of $\triangle ABC$, then for homothety reasons it suffices to show that $\frac{AH_0}{AK}=\frac{AH}{AD} \iff \frac{AH_0}{AH}=\frac{AD}{AK}$. This is true, since if $A'$ is the point diametrically opposite $A$ in $(ABC)$ and $\overline{AA'} \cap \overline{BC}=T$ then $\frac{AH_0}{AH}=\frac{AT}{AA'}$ by triangle similarity and $\frac{AT}{AA'}=\frac{AD}{AK}$ since $\overline{A'K} \parallel \overline{BC}$. $\blacksquare$

Let $DF \cap (AXY) = K$ and $DE \cap (AXY)=L$ Claim 1: $K$ and $L$ are reflections of orthocenter of $\triangle AXY$

Proof

Claim 2: $F-H-E$

Proof