SHOW THAT $EMC = DBM$ ITS VERY EASY Q.E.D

Because $OM\bot BC, BC$ passes through $F|OF$ is a diameter of the circle $\odot (AOM)$, i.e. $AF$ is tangent to the circle $\odot (ABC)$, thus $\angle DMF=\angle DAF=\angle ACB$; with $\angle CEM=\angle BDM$ and $BM=CM\implies \triangle BDM\cong\triangle MCE$ (a.s.a. criterion). Best regards, sunken rock

Let $U$ $\in$ $BC$ which satisfies $AU$ tangents $(ABC)$ at $A$ We have: $DM = \dfrac{BM . AU}{AB} = \dfrac{CM . CU}{AC} = CE$

$$\frac{DM}{\sin \angle DBM}=\frac{BM}{\sin \angle ADM}=\frac{CM}{\sin \angle MEC}=\frac{EC}{\sin \angle EMC}=\frac{EC}{\sin 180^{\circ} - B} \implies DM=CE$$

It's too easy.

Note that $\measuredangle ECM=90^\circ+\measuredangle OAB=90^\circ+\measuredangle OMD=\measuredangle BMD$ and $\measuredangle MEC=\measuredangle MEA=\measuredangle MDB$, thus $\triangle ECM\sim\triangle DMB$, however as $MB=CM$, we get that $DM=EC$, we are done.

Let $BC$ meet $AOM$ at $K$. Note that $\angle OAK + \angle OMK = \angle 180 \implies \angle OAK = \angle 90 \implies AK$ is tangent to $ABC$ so $\angle BMD = \angle KMD = \angle KAD = \angle ACB = \angle ACM$ and $\angle MEC = \angle MDB$ and $MC = MB$ so $MEC$ and $MDB$ are congruent so $CE = DM$.

Let $A-$symmedian intersects $(ABC)$ at $K$. Notice that $(AOM)$ passes through the $K$ (inversion). Therefore $\triangle BDM\sim\triangle MEC$. $MB=MC\implies DM=CE$.