Let $F$ be the intersection of $CM$ and $BD$. Looking at $\triangle AEM$ and $\triangle DMF$, we have $\angle MAE = \angle FDM$ and $\angle AME = 180^\circ - \angle BMC - \angle DMF = 180^\circ - \angle FDM - \angle DMF = \angle DFM$, so they are similar. Furthermore, by similarity, $\frac{ME}{MF} = \frac{AM}{DF} = \frac{DM}{DF}$, so $\triangle MEF$ and $\triangle DMF$ are also similar. Thus, $\angle AME = \angle DFM = \angle MFE$. Now, we have $\angle EBC = \angle AME = \angle MFE = 180^\circ - \angle EFC$, so $BCFE$ is cyclic. Hence, $\angle BEC = \angle BFC = \angle DFM = \angle AME = \angle EBC$, as desired.

Let $X = CM \cap BD$. Claim 1. $\triangle EXM \sim \triangle EMA$

proof

Claim 2. $BEXC$ is cyclic.

proof

Angle chase: $$\angle EBC = \angle EMA = \angle MXD = \angle CXB$$By claim 2, $\angle CXB = \angle CEB$. Hence, $\angle EBC = \angle CEB$ and $CB = CE$.

Let $F = AE \cap BD$ and $G = CM \cap BD$. We have $90 - \angle EFG/2 = 90 - \angle AFD/2 = 90 - (180 - 2\angle FDA)/2 = \angle FDA = \angle EMG$. Also, since $M$ lies on the angle bisector of $\angle EFG$, we conclude that $M$ is the excenter of $\triangle EFG$. So we have $\angle EGM = \angle MGD$, so $\triangle MEG \sim \triangle DMG$. Also, $\triangle DMG \sim \triangle BCG \sim \triangle MCB$, since $\angle BMC = \angle BDA = \angle DBC$. So, $\triangle MEG \sim \triangle MCB \implies BCGE$ is cyclic. Thus, $\angle CBE = \angle EGM = \angle MGD = \angle CGB = \angle CEB$ , implying that $\triangle BCE$ is isosceles.

Attachments:

No need for similarities. If $AE \cap BC = F$, then $ABFD$ is an isosceles trapezoid (so $MB = MF$) and $DMECF$ is cyclic and now quick angle chasing completes the proof.

Here is a solution with no additional points (apart from the natural $BC \cap DM = O$) by a student who does not remember his AoPS username. Edit: Here the trapezoid is $ABCD$ with $AB \parallel CD$ and $M$ is the midpoint of $AB$, sorry. We want $\angle EDC = \angle DEC$, equivalent to (as $BM \parallel CD)$ $\angle DMB = \angle CEM$. Having in mind the given $\angle ABD = \angle CED$, it suffices to show that $\triangle DMB \sim \triangle CEM$, more concretely $\displaystyle \frac{BD}{BM} = \frac{CM}{ME}$. We have $\triangle BOM \sim \triangle DOC$ (from $BM \parallel CD$), so $\displaystyle \frac{OD}{OC} = \frac{OB}{OM}$ and thus $\displaystyle \frac{BD}{CM} = \frac{OB}{OM}$. Now the given condition $\angle EAM = \angle BAE = \angle ABD = \angle MBO$ and $\angle AEM = \angle BMD - \angle EAM = \angle BMO + \angle DMC - \angle EAM = \angle BMO$ yield $\triangle MAE \sim \triangle OBM$, which gives $\displaystyle \frac{OB}{OM} = \frac{AM}{ME}$. Since $AM = BM$, we conclude $\displaystyle \frac{BD}{CM} = \frac{OB}{OM} = \frac{AM}{ME} = \frac{BM}{ME}$, i.e. $\displaystyle \frac{BD}{BM} = \frac{CM}{ME}$, as desired.