We have that $\angle BFC = \angle BEC = 90$, so $BFEC$ is cyclic. Thus, by the Radical Axis Theorem, the circumcircles of $BFEC$, $\triangle BFD$, and $\triangle DEC$ concur, so $BF$, $CE$, and $DM$ concur, so $DM$ passes through $A$ and is the angle bisector of $\angle BAC$. Now $M$ is the Miquel point of $\triangle ABC$ with respect to $\triangle DEF$, so $\triangle AFME$ is cyclic. Since $\angle EAM = \angle FAM$, we have $ME=MF$, as both $ME$ and $MF$ are chords subtending the same angles in the circumcircle of $AFME$.

Notice that $M$ is a Miquel Point, so $AFEM$ is cyclic. Moreover, the radical axes of $(BFEC), (BFMD),$ and $(CEMD)$ concur at $A.$ But the radical axis of $(BFMD)$ and $(CEMD)$ is $MD$, so $A,M,D$ are collinear, thus forcing $\angle FAM = \angle EAM,$ so $EM=FM.$

By easy angle chasing we gain that $A,F,M,E$ so M is the miquel point of $ABC$. Let apply sine theorem on $AFM$ and $AEM$. Let say circumcircle of $EFBC$ radius is $R$ and so $\frac{\sin FAM}{FM}=\frac{\sin EAM}{ME}=\frac{1}{2R}$ since $\sin FAM=\sin EAM$ we get that $ME=MF$.

Attachments:

Since $\angle BFC = \angle BEC$, $BFEC$ is cyclic. Hence $\angle AEF = \angle B$. Since $M$ is a Maquel Point, $AFME$ is cyclic. $\angle FMD + \angle FMA = 180^{\circ} - \angle B + \angle AEF = 180^{\circ}$. That is, $M$ lies on $AD$. So $\angle FAM = \angle EAM$ leading to $EM = FM$.