Prove that $XY$ crosses the midpoints $\triangle ABC$ has altitudes $AD$, $BE$, $CF$. The reflections of $E$, $F$ in $H$ are $E'$, $F'$. The circle $DE'F'$ intersects $BE$, $CF$ at $X$, $Y$. Prove that $XY$ goes through the midpoints of $AB$, $AC$. 

I can show that $XY$ is parallel to $BC$ by simply angle-chasing. $EYFX$ is cyclic as well as $APFY$. 

I also tried showing that $AP$=$PH+HD$
Take point $C$ as the origin of a Cartesian coordinate system. Then let $|PD|=a$. It suffices to note that since the line $XPY$ is parallel to the base of the triangle, its equation is $y=a$. Now since the midpoints of $AC$ and $AB$ have coordinates $(x_1,a)$ and $(x_2,a)$ respectively, it's clear that they lie on the line of interest. $\square$
Take point $C$ as the origin of a Cartesian coordinate system. Then let $|PD|=a$. It suffices to note that since the line $XPY$ is parallel to the base of the triangle, its equation is $y=a$. Now since the midpoints of $AC$ and $AB$ have coordinates $(x_1,a)$ and $(x_2,a)$ respectively, it's clear that they lie on the line of interest. $\square$