For distinct three points $P$, $Q$, $R$, the description $\angle QPR = \theta$ means that line $PQ$ rotated around $P$ by angle $\theta$ counterclockwise coincides with line $PR$. Here $180^\circ$ difference is ignored.
By the inscribed angle theorem, we have $\angle XBD = \angle XEC$, $\angle XDB = \angle XCE$. We also have $BD = BM = CM = CE$ hence triangle $BDX$ and triangle $ECX$ are congruent, therefore $BX = EX$.
Let $N$ be the midpoint of arc $BC$ of triangle $ABC$'s circumcircle which includes $A$. We have $\angle BXE = \angle BAE = \angle BAC = \angle BNC$, and triangle $BEX$ is an isosceles triangle with apex $X$, triangle $BCN$ is an isosceles triangle with apex $N$. Also both $\angle BXE$ and $\angle BAC$ are acute angles, hence those two triangles are similar including orientation. Therefore we obtain $BX : BE = BN : BC$ and $\angle NBX = \angle EBX + \angle NBE = \angle CBN + \angle NBE = \angle CBE$, which shows triangle $BNX$ and triangle $BCE$ are similar. Also we have
$$\angle XAB = \angle XEB = \angle NCB = \angle NAB$$
thus the points $A$, $N$, $X$ are collinear.
Furthermore, $\angle BNO = \angle BNM = \angle BCM$ holds and triangle $BNO$ is an isosceles triangle with apex $O$, triangle $BCM$ is an isosceles triangle with apex $M$ hence those two are similar. Therefore by $CE = CM = BM$ we obtain
$$NX = CE \cdot \frac{BN}{BC} = CM \cdot \frac{BO}{BM} = NO.$$
Let $T$ be the point symmetric to $N$ with respect to line $OX$, then above shows $TO = NO = NX = TX$ hence quadrilateral $NOTX$ is a rhombus. Therefore we have $\angle OTX = \angle XNO = \angle OAX$, thus $T$ is on the circumcircle of triangle $AOX$. Also by $NO = OT$, $T$ is on the circumcircle of triangle $ABC$.
Since we have $\angle MBD = \angle MCE$ and $BD = BM = CM = CE$, triangle $BDM$ and triangle $CEM$ are congruent, implying $DM = EM$. Also we have $\angle ADM = \angle CEM$ thus $E$ is on the circumcircle of triangle $ADM$. Furthermore, line $NX$ and line $OT$ are parallel and line $NX$ and line $AM$ are orthogonal, hence line $OT$ and line $AM$ are also orthogonal. Thus by $AO = MO$, we obtain $AT = MT$. Therefore, $A$ and $M$ are symmetric with respect to line $OT$, hence we have $\angle OAT = \angle TMO = \angle OTM$ and thus line $MT$ is tangent to the circumcircle of triangle $AOT$.
Since we have
$$\angle ATM = \angle ABM = \angle ADM + \angle DMB = 2\angle ADM$$

and two points $D$ and $T$ are on the same side with respect to line $AM$, $T$ is the circumcenter of triangle $ADM$. Therefore, we have $DT = ET$, thus by $DM = EM$ we have proved line $MT$ is the perpendicular bisector of segment $DE$.