62nd Ukrainian National Mathematical Olympiad

Let $A^{\prime }$ be the point diametrically opposite to $A$ in $\omega$, and let $R$ be the projection of $A$ onto $EF$ (fig. 12). Since $\angle BA{A}^{\prime }=90^{\circ }-\angle ACB=90^{\circ }-\angle AEF$, the line $AR$ passes through $A^{\prime }$. Lines $PE$ and $QF$ also pass through $A^{\prime }$. Since the quadrilaterals $AREP$ and $ARFQ$ are inscribed in circles with diameters $AE$ and $AF$ respectively, we have $A^{\prime }E\cdot A^{\prime }P=A^{\prime }A\cdot A^{\prime }R=A^{\prime }F\cdot A^{\prime }Q$, so the quadrilateral $EFPQ$ is cyclic. Then the lines $BC$, $EF$ and $PQ$ intersect in the radial center of the circumscribed circles of the quadrilaterals $EFPQ$, $BCQP$ and $BCEF$