My solution: Let $N$ is the midpoint of $BQ$ then exists be a point $X$ in $AB$ such that $AM=MX$ and $XB=XC$ since $BM=AM+BC$ $\Longrightarrow $ $\angle CXB=\angle BCX=\angle QBA$ and $\angle CQB=\angle CAB$ $\Longrightarrow $ $\triangle XAC\sim \triangle BCQ$ with $M$ and $N$ are midpoints of $AX$ and $QB$ $\Longrightarrow $ $\angle CMX=\angle CNB$ since $\triangle XAC\sim \triangle BCQ$ $\Longrightarrow $ $B,N,M,C$ are concyclic $\Longrightarrow $ the circumcircle of $\triangle BMC$ through midpoint of $BQ$...

Nice problem! Denote by $P$ the midpoint of arc $AC$ containing $B$, $S$ the midpoint of $BQ$, $O$ the center of $(ABC)$ and $E \equiv OS \cap PM$ By Archimedes lemma we have that $PM \perp AB$. Note that $90^{\circ}=\angle BME=\angle BSE$ $\implies$ $BMSE$ cyclic Also since $\angle OPC=\angle BPM=\frac{\angle ABC}{2}$, We have $\frac{CP}{CO}=2.\frac{PM}{PB}=\frac{PM}{OS}$ This implies that $POEC$ is cyclic, Since $C$ is the center of spiral sim taking $PM$ to $OS$ by observing that $\angle CPM= \angle COS= \frac{\angle ABC}{2}-\angle BAC $ Then $\angle OEC=180^{\circ}-\angle OPC=180^{\circ}-\angle SBC$ $\implies$ $BMSEC$ is a cyclic pentagon and we are done.

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