62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour

Denote by $d(Z,XY)$ the distance from the point $Z$ to the line $XY$. From the equality of inscribed angles it follows that (fig. 6) $$\angle PEB=\angle ADB=\angle ACB=\angle PFA\Rightarrow PE=PF.$$ Furthermore, $$\angle EDP=\angle BAP=\angle BDC,\angle FCP=\angle PBA=\angle DCP,$$ so $P$ is the intersection of bisectors of angles $\angle EDC$ and $\angle DCF$, so $d(P,DE)=d(P,DC)=d(P,CF)$, so $PQ=PR$. Then $△FPQ\cong △EPR$ as right triangles with equal cathetus and hypotenuse. From the equality $PQ=PR$ it follows that $\angle PQR=\angle PRQ$, so $\angle FQR=\angle QRE$. Similarly, $\angle QFE=\angle REF$. This means that $EQRF$ is an isosceles trapezoid, so $QR\parallel EF$, as desired.