Let $B_a, B_c$ be the reflections of $B$ over $P, Q$ respectively. Then, we just need to show that $B_a, L, B_c$ are collinear. It's easy to see by angle-chasing that $\angle BAB_a = 180 - \angle C$ and $\angle BCB_c = 180 - \angle A$, which give that $AB_a || CB_c.$ Now, since we have that $\frac{AB_a}{CB_c} = \frac{AB}{BC} = \frac{AL}{LC},$ we have that $\triangle LAB_a \sim \triangle LCB_c \Rightarrow \angle ALB_a = \angle CLB_c,$ and so we're done. $\square$

k.vasilev wrote: 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 $PQ.$ Let $M_a,M_c$ be the midpoints of segments $BC,AB$. By changsing angle we easy to see $PM_c \parallel QM_c$. $M_aM_c$ cuts $PQ$ at $G$. So by Thales we obtain $\dfrac{GM_c}{GM_a}=\dfrac{PM_c}{QM_a}=\dfrac{AB}{BC}$, so $G$ lies on $AL$. and we know $G$ is also midpoint of $AL$. Q.E.D

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Use Complex Numbers...

Remarks

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Here's the solution with complex numbers

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In the end ,we just subtract -1 from both sides which leads to a cancellation of a factor of $a-c$ and making it more suitable to expand. Thanks to mroreojuce to provide the latex. Edit:I tried to skip the calculation part in an intuitive but wrong sense but then found the correct way.

Let $X,Y$ be midpoints of $BA, BC$. Since $(PX, AB) = \angle ACB$ and $(QY, BC) = \angle BAC$, we have $(PX, QY) = (PX, AB) + (QY, AC) + (BA, BC) = 0$ so $PX || QY$. Let $Z = PQ \cap XY$. We have $\frac{XZ}{YZ} = \frac{AX}{CY} = \frac{AB}{BC} = \frac{AL}{LC}$ which implies that $Z \in BL$, as desired. $\blacksquare$