Solution:

Let $H$ be the point at which the line $EQ$ meets $AD$; we must show that the angle $A\widehat{H}E$ is right. Draw the segment $BE$. The triangle $BQE$ is isosceles because the altitude $EP$ is also a median; indeed $P$, the foot of the altitude $EP$, is the midpoint of $BQ$ since $PQ = \frac{1}{2} AP = PB$. $EP$ is therefore also the bisector of the angle $B\widehat{E}Q$, that is, the two angles $P\widehat{E}Q$, $P\widehat{E}B$ are congruent. Next, the angles $D\widehat{E}B$, $D\widehat{A}B$ are congruent because they are inscribed angles subtending the same arc; it follows that the angles $P\widehat{E}Q$, $D\widehat{A}P$ are congruent. The triangles $AHQ$, $EPQ$ therefore have congruent angles at $A$ and at $E$; further, their respective angles with vertex at $Q$ are congruent, being vertical angles. The triangles $AHQ$, $EPQ$ are therefore similar, and in particular the angles with vertices at $P$ and $H$ are congruent. Since the angle $E\widehat{P}Q$ is right by construction,

the angle $A\widehat{H}E$ is also right, as was to be proved.