Firstly, by Spiral Similarity, $\triangle MEF \sim \triangle MAC$ but $MA=MC$ so $ME=MF$. Angle chasing also gives $\angle EMA=\angle CMF=\angle MCF=\angle MBD$. Let these angles be $x^\circ$. Consider a clockwise rotation centered at $M$. It sends the line $MC$ to $MF$, line $MA$ to $ME$, and $OM$ to some random line $l$. Let $l$ intersect $BD$ at $B'$. Since $OM\parallel B'D$, $\angle OMB'=x^\circ=\angle MB'D=\angle MBD$, so $B'=B$ and $l$ is in fact $MB$. Combined with the fact that $\triangle MEF \sim \triangle MAC$ and $OM\perp AC$, we have $BM\perp EF$ and then we are done.