Let $ L$ be the midpoint of $AB$ and $ P$ the second intersection of $ \odot (AOC)$ with $ BQ.$ $ \angle APQ = \angle BNA = \angle ABC.$ If $ R \equiv BQ \cap AC$ and $D \equiv BN \cap AP,$ then it follows that $PDNR$ is cyclic. Notice that $ PQ$ bisects $ \angle APC,$ since $ Q$ bisects the arc $ AC$ of $\odot(AOC)$ $\Longrightarrow$ $ \angle BPC = \pi - \angle ABC = \angle BNC$ $ \Longrightarrow$ $BPNC$ is cyclic, thus $ \angle NPR = \angle ACB.$ But since $ PDNR$ is cyclic, it follows that $\angle NPR = \angle NDR = \angle ACB = \angle ABN$ $\Longrightarrow$ $ DR \parallel BA.$ Therefore, cevian $NP$ of $ \triangle BNA$ passes through the midpoint $ L$ of $ AB$ $ \Longrightarrow$ $ M$ lies on $NP$ as desired.