Let $D,E$ be the feet of the altitudes issuing from $A,C,$ respectively. By Butterfly theorem for the chords $AE,CD$ in the circle $(M)$ with diameter $AC,$ the perpendicular to $MB$ through $B$ cuts lines $AD,CE$ at $K,N$ such that $\overline{BN}=-\overline{BK}.$ Since $\angle BAK=\angle BCN,$ it follows that $\odot( ABK) \equiv (O_1)$ and $\odot(CBN) \equiv (O_2)$ are symmetric about $BM.$ Let $P$ be the 2nd intersection of $(O_1)$ and $(O_2).$ Thus, $P \in BM$ and $\angle BPC=\angle ABM$ $\Longrightarrow$ $PC \parallel AB$ $\Longrightarrow$ $\overline{MO_1}=-\overline{MO_2}.$

Let point $D$ be such that $ABCD$ is a parallelogram. Since $\angle AKB=\angle DBC=\angle BDA$, it follows that $DABK$ is cyclic and $O_1$ is the midpoint of $KD$. Similarly, $O_2$ is the midpoint of $DN$. Up to now, $M$ lies on the line segment $O_1O_2$. It suffices to prove that circles $O_1$ and $O_2$ are equal, which is obvious since they are the circumcircles of the triangles $ABD$ and $BCD$, where $ABCD$ is a parallelogram.