Note $P$ lie on $(A_1B_1C)$. As $BA_1 \cap AB_1 =C$ and $Q=(A_1B_1C) \cap (ABC)$ We get spiral similarity at $Q$ send $A_1B_1 \rightarrow AB$ hence if $N$ is midpoint of $AB$ $$\angle NQM = \angle BQA_1$$We prove $\overline{Q-P-N}$ Let $X =QP \cap (ABC)$ $$\angle QCB = \angle QXB = \angle QCA_1 = \angle QPA_1 = \angle XPA$$ hence $PA \parallel BX$ and similarly $PB \parallel AX$ and hence we get $N= PX \cap AB$

A great problem for JBMO training! (Writing $N$ for the midpoint of $A_1B_1$ instead of $M$, as I like it more.) Angle chasing gives $\triangle ABQ \sim \triangle B_1A_1Q$, while $\angle PQN = \angle BQA_1$ is equivalent to $\angle A_1QN = \angle BQP$. Hence if $QP \cap AB = M$, by the similarity it suffices to show that $M$ is the midpoint of $AB$. (In particular, $N$ is no longer needed in the diagram.) Note that $\angle APM = \angle QPA_1 = \angle QCA_1 = \angle QCB = \angle QAB$, so $MA^2 = MP \cdot MQ$. Analogously $MB^2 = MP \cdot MQ$ and we are done.

Note that $P$ lies on the reflection of $(ABC)$ over $\overline{AB}$. For spiral similarity reasons it suffices to show that $\overline{PQ}$ passes through the midpoint of $\overline{AB}$. If $P'$ is the second intersection of $\overline{PQ}$ with $(ABC)$ then $\measuredangle QPB_1=\measuredangle QA_1B_1=\measuredangle QBA=\measuredangle QP'A$, so $\overline{P'A} \parallel \overline{PB_1}$. Likewise, $\overline{P'B} \parallel \overline{PA_1}$. Thus $APBP'$ is a parallelogram so $\overline{PP'}$ bisects $\overline{AB}$ as desired.

VicKmath7 wrote: Let $ABC$ be scalene and acute triangle with $CA>CB$ and let $P$ be an internal point, satisfying $\angle APB=180^{\circ}-\angle ACB$; the lines $AP, BP$ meet $BC, CA$ at $A_1, B_1$. If $M$ is the midpoint of $A_1B_1$ and $(A_1B_1C)$ meets $(ABC)$ at $Q$, show that $\angle PQM=\angle BQA_1$. Let $N$ be the intersection of lines $QP$, $AB$ Cuz $\angle APB=180^{\circ}-\angle ACB$ so $Q$, $A_1$, $B_1$, $C$, $P$ are concyclic Lead to $\angle NPB = \angle B_1PQ = \angle B_1CQ = \angle ABQ$ Therefore, $\triangle NBP \sim \triangle NBQ$, $NB^2 = NP. NQ$. Similar, $NA^2 = NP. NQ$ So we can see that $N$ is the midpoint of segment $AB$ In the order hand, since $(A_1B_1C)$ meets $(ABC)$ at $Q$, we have $Q$ is the center of the spiral similarity carries $AB$ to $A_1B_1$. Which means $\triangle QMN \sim^+ \triangle QA_1B \sim^+ \triangle QB_1A$ Thus $\angle PQM = \angle NQM = \angle BQA_1$, done

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