Suppose $BD$ meets ($CDM$) again at $P$. Applying Pascal's theorem to the points $MMCDDP$, we know that $MM \cap DD = E$, $MC \cap DP = B$ and $CD \cap PM$ are collinear. Thus, $MP$ passes through $BE \cap CD = F$. Now, since
$$\angle CMF = \angle ADB = \angle BAF,$$
the points $A, B, M, F$ are concyclic.

We provide another proof without using Pascal's theorem as follows. Again, the main task is to prove $M, P, F$ are collinear. We define $F'$ as the intersection point of $CD$ and $MP$. It suffices to prove $B, E, F'$ are collinear. Suppose $DE$ meets ($PDF'$) again at $X$, and $ME$ meets ($MCF'$) again at $Y$.
Firstly, since $BP \times BD = BM \times BC$, the powers of $B$ with respect to ($PDF'$) and ($MCF'$) are the same. Thus, $B$ lies on the radical axis of these circles.
Secondly, as $\angle XF'P = \angle XDP = \angle DMP = \angle DMF'$, we have $XF'//MD$. Also, as $\angle MYF' = 180^\circ - \angle MCF' = 180^\circ - \angle YMD$, we have $YF'//MD$. Therefore, $Y, X, F'$ are collinear, and $YX//MD$. Since $EM = ED$, we have $EY = EX$. From this, it follows that $EX \times ED = EY \times EM$, and hence $E$ lies on the radical axis of ($PDF'$) and ($MCF'$).
Thirdly, it is clear that $F'$ lies on the radical axis of ($PDF'$) and ($MCF'$). It follows that $B, E, F'$ are collinear as desired.