Czech-Polish-Slovak Mathematical Match

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 $$\angle BYC=\angle DYC=\angle DGC\text{and}\angle BXC=\angle BXE=\angle BFE.$$ 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. □