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Let $T$ be the foot of perpendicular from $A$ to $CD$. Quadrilateral $ABDT$ is cyclic meaning $\angle BAD=\angle BTD$. From $\angle BAD=\angle BCD$ follows that triangle $BCT$ is cyclic. Let $BH$ be the altitude of the triangle. $H$ is the midpoint of $CT$, so $HM$ is the midline of triangle $ACT$, meaning $MH\parallel AT$, but from definition we have $BH\parallel AT$. This implies that points $B$, $M$, $H$ are collinear.

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