Czech-Polish-Slovak Mathematical Match competition problems

Let $\omega_B, \omega_C$ be the circumcircles of triangles $FEB$ and $DGC$ respectively. Since $PQ$ is the radical axis of $\omega_B$ and $\omega_C$, it is sufficient to prove that the powers of $A$ with respect to $\omega_B$ and $\omega_C$ are equal. If we denote by $X$ the second intersection of $\omega_B$ and $AC$, and by $Y$ the second intersection of $\omega_C$ and $AB$, then this is equivalent to
$$AX \cdot AE = AY \cdot AD.$$
Lines $DE$ and $BC$ are parallel, yielding $\frac{AB}{AD} = \frac{AC}{AE}$, so the above is equivalent to
$$AX \cdot AC = AY \cdot AB.$$
This, in turn, is equivalent to $B, Y, X, C$ being concyclic. From angles in $\omega_B$ and $\omega_C$ we have
$$\angle BYC = \angle DYC = \angle DGC \quad \text{and} \quad \angle BXC = \angle BXE = \angle BFE.$$
On the other hand, trapezoid $BFGC$ is inscribed in the circumcircle of $ABC$, so it is isosceles, hence $\angle BFE = \angle DGC$. We conclude that $\angle BYC = \angle BXC$, so $B, Y, X, C$ are indeed concyclic and we are done. □