A first reflex is to draw a figure where the triangle $ABC$ is isosceles right-angled, but without indicating that it is. Many symmetries then appear. Let $X$ be the intersection point of the lines $(AC)$ and $(EP)$, $\ell$ the perpendicular bisector of $[PQ]$, $O$ the center of the circumcircle of $ABC$, and $M$ the midpoint of $[BC]$: it is a matter of proving that $M=O$.
Since $PQX$ is isosceles at $X$, the symmetry with axis $\ell$ exchanges the lines $(AQ)$ and $(PE)$, so that $\widehat{AQP}=\widehat{QPE}$. Since $A, E, P$, and $Q$ are concyclic, the quadrilateral $AEPQ$ is therefore an isosceles trapezoid, which is easily verified since

$$(AE, PQ) = (AE, AQ) + (AQ, PQ) = (PE, PQ) + (QA, QP) = 0^{\circ}$$

The symmetry with axis $\ell$ therefore exchanges the points $A$ and $E$, and $\ell$ is in fact the perpendicular bisector of $[AE]$. It therefore contains, in particular, the centers $O$ and $O'$ of the circumcircles of $ABC$ and $AEPQ$.
Given the right angles at $P, Q$, and $M$, the latter circle is in fact the circle with diameter $[AD]$, and it contains $D$ and $M$. Moreover, $(AM)$ is the angle bisector of $\widehat{BAC}$, so $M$ is the South pole of $A$ in the triangle $APQ$, and $PQM$ is isosceles at $M$. Thus, $M$ lies on $\ell$.
In conclusion, $M$ and $O$ both belong to $(AM)$ and to $\ell$. Since $D \neq M$, we also know that $O'$, the midpoint of $[AD]$, is not on $(AM)$. Since $\ell$ contains $O'$, it is therefore not coincident with $(AM)$, and $M$ must coincide with $O$, which concludes.
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Comment from the graders: The exercise was very well done. Many students manage, through the development of simple ideas like angle chasing, to make significant progress in solving the problem, while some others force the use of very advanced results like Pascal's or Desargues' theorems but without obtaining significant results.

The figures provided are often very precise, and it was sometimes frustrating to see that the midpoint of the segment $[BC]$ belongs to the circle passing through the points $A, P$, and $Q$ appear on them without the student mentioning this midpoint in their attempt.
We noted a significant number of incorrect solutions. The errors are often due to the use of a property that is evident on the figure but not yet proven. Often, to avoid this pitfall, it is good to ask oneself if all the hypotheses of the statement have been used. Thus, several students believe they have solved the exercise without using either that the triangle $ABC$ is isosceles or that the point $E$ belongs to its circumcircle.