This is just angle chasing. And until the very last step $M$ can be anywhere on $BC$. Let $T$ be the intersection of $BD$ and $CE$. Angle chase to show angle $ETD$ equals angle $BAC$. Hence $T$ lies on circle $(ADE)$, and also points $M,D,T,C$ lie on the same circle. Let $P$ be the circumcentre of $(ADMB)$ and $Q$ be the circumcentre of $(MDTC)$. Angle chase to show that angle $POQ$ plus angle $PMQ$ sum to $180$. So $O,P,M,Q$ lie on the same circle, and hence $OM$ is perpendicular to $BC$ Now, finally using that $M$ is the midpoint of $BC$, the conclusion follows