Japan Mathematical Olympiad

Since $\angle AXD=\angle AYD=90^{\circ }$, points $A$, $D$, $X$ and $Y$ are concyclic. Therefore, we have $\angle FAD=\angle XAD=\angle XYD=\angle FPD$, and by the converse of the inscribed angle theorem, points $A$, $D$, $F$ and $P$ are concyclic. According to the midpoint theorem in triangle $ABC$, we find that $FD$ is parallel to $AC$. Hence, we have $\angle FDP=90^{\circ }$ and consequently $\angle FAP=90^{\circ }$. This implies that lines $AB$ and $AP$ are perpendicular. Furthermore, applying the midpoint theorem in triangle $ABC$ again, we find that $DE$ is parallel to $AB$. Thus, lines $DE$ and $AP$ are perpendicular. Now, since lines $AE$ and $DP$ are perpendicular, $E$ is the orthocenter of triangle $ADP$. Therefore, we can conclude that lines $AD$ and $EP$ are perpendicular.