Inside the acute-angled triangle $ABC$ we take $P$ and $Q$ two isogonal conjugate points. The perpendicular lines on the interior angle-bisector of $\angle BAC$ passing through $P$ and $Q$ intersect the segments $AC$ and $AB$ at the points $B_p\in AC$, $B_q\in AC$, $C_p\in AB$ and $C_q\in AB$, respectively. Let $W$ be the midpoint of the arc $BAC$ of the circle $(ABC)$. The line $WP$ intersects the circle $(ABC)$ again at $P_1$ and the line $WQ$ intersects the circle $(ABC)$ again at $Q_1$. Prove that the points $P_1$, $Q_1$, $B_p$, $B_q$, $C_p$ and $C_q$ lie on a circle. Proposed by P. Bibikov
