Let $A'$ be the reflection of $A$ in $D$. Notice that by angle chasing $\triangle A'BA \sim \triangle EDF$. Therefore, since $BD$ and $MD$ are corresponding medians, we have $\angle MDE = \angle DBA' = \angle ACB$. Further, since $\angle EDB = 90^{\circ} - \angle ABC = \angle OAC$, this implies the result.

See my solution to this problem on my Youtube channel here: https://www.youtube.com/watch?v=CzmfyB1yz_8

Let $A' $ be the antipodal point of $A$ w.r.t $(ABC)$. Let $Z$ be the second intersection of the circles $(ABC),(AD)$. $A,E,B$ , $A,F,C$ are collinear. It is easy to show that $Z,D,A'$ are also collinear. So by the spiral similarity lemma we have that, $\triangle ZBE \sim \triangle ZDM$. Now we have that, \[\angle ZDM = \angle ZBE = \angle ZBA = \angle ZA'A\] This implies that $DM \parallel AO$.

Beautiful geometry problem for storage! Let's extend $AO$ to intersect side $BC$ in point $G$ and circumcircle in point $G'$. Let $AD$ intersect circumcircle at point $H$. Now we have lot of cyclic quadrilaterals (here we use $ABHC$ and $DEAF$), so: $\angle HBC=\angle HAC=\angle DAF=\angle DEF$ and $\angle HCB=\angle HAB=\angle DAE=\angle DFE \implies$ Triangles $HCB$ and $DEF$ are similar so: $\frac{HB}{DE}=\frac{BC}{EF}=\frac{BD}{EM}$ because $M$ and $D$ are midpoints, so triangles $HBD$ and $DEM$ are similar. Now using this and cyclic quadrilateral $ABHC$: $\angle EDM=\angle BHD=\angle BHA=\gamma$ and from rightangled triangle $BED$ we have $\angle BDE=90^\circ-\beta$, so: $\angle BDM=\angle BDE+\angle EDM=90^\circ-\beta+\gamma$ $\angle GAC=\angle G'AC=90^\circ-\angle CG'A=90^\circ-\beta$ $\angle AGB=180^\circ-\angle AGC=\angle GCA+\angle GAC=90^\circ-\beta+\gamma \implies \angle BDM=\angle AGB \implies DM \parallel AO$

Obviously, $AEDF$ is cyclic. Let $DO\cap (AEDF) = G$. It is easy to see that $AG\parallel BC$. We have $(D,G;E,F) \stackrel{A} = (D,\infty_{BC};B,C) = -1$ so $GEDF$ is harmonic and because $M$ is the midpoint, we get $DG$ is a symmedian in $\triangle EDF$. So $\angle ACB = \angle ODF = \angle MDE \implies AO\parallel DM$.