49th Mathematical Olympiad in Ukraine

Let $M$ be a midpoint of the segment $AP$. Let us the line $EM$ intersects the segment $AD$ at a point $Q$. $MPE$ and $BPE$ are the congruent triangles because both of them are right and both of them have the equal legs. Then $\angle QAM=\angle DAB=\angle DEB=\angle PEB=\angle PEM$ (Fig.9).

Since $\angle QMA=\angle EMP$ then triangles $QMA$ and $EMP$ have in twos equal angles. Consequently both of them have equal third angles, then $\angle AQM=\angle MPE=90^{\circ }$, then $EQ$ and $AP$ are the altitudes of the triangle $ADE$ which proves that $M$ is the orthocenter of that triangle. That completes the proof.

Fig.9