RU-ALLRUSSIAN 2019 Problem 106

Let $L$ be the foot of the internal 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 $\measuredangle APB=\measuredangle CQB=90^{\circ}.$ Prove that the midpoint of $BL$ lies on the line $PQ.$