Question 170
Let $L$ be the foot of the internal angle bisector of $\angle B$ in an acute-angled triangle $ABC$. The points $D$ and $E$ are the midpoints of the smaller arcs $AB$ and $BC$, respectively, in the circumcircle $\omega$ of $\triangle ABC$. Points $P$ and $Q$ are marked on the extensions of the segments $BD$ and $BE$ beyond $D$ and $E$, respectively, so that $\angle APB = \angle CQB = 90^\circ$. Prove that the midpoint of $BL$ lies on the line $PQ$. (A. Kuznetsov)