Let $d$ be a line parallel to $BC$ and goes through $K$. $d$ intersects $AB,AC$ at $X,Y$. Since $ON \perp BC \Rightarrow XY \perp OK$ $\Rightarrow \lbrace O,K,X,D \rbrace, \lbrace O,K,E,Y \rbrace$ are $2$ sets of concyclic points. $\Rightarrow \begin{cases} \widehat{ODK}=\widehat{OXK} \\ \widehat{OEK}=\widehat{OYK} \end{cases}$ $\Rightarrow \widehat{OXK}=\widehat{OYK}$ Therefore, $\Delta OXY$ is isoceles triangle at $O$. Since $OK \perp XY$, $K$ is the midpoint of $XY$. Using Thales theorem we have $\dfrac{XK}{KY}=\dfrac{BM}{MC}$, therefore $M$ is the midpoint of $BC$.

EGMO lemma 4.17