Let $AH$ intersects $(O)$ at second point: $D$ and $BC$ at $J$. Let $MH$ is the perpendicular line through $M$, WRT $AD$, intersects $AO$ at its midpoint $I$ By $Menelaus$: $\frac{IO}{IA}.\frac{HA}{HD}.\frac{MD}{MO}=1$. Hence: $\frac{HA}{HD}=\frac{MO}{MD}$ By another way: $DB^2=DJ.DA$ and: $DM^2=DJ.DH$, hence: $\frac{DB^2}{DM^2}=\frac{DA}{DH}$ so: $\frac{BM^2}{DM^2}=\frac{AH}{DH}$, Thus: $\frac{BM^2}{DM^2}=\frac{MO}{MD}$. Thus: $BM^2=OM.DM$ So: $\widehat{BOM}=\widehat{DBM}$, from this we can caculate: $\widehat{ABC}=120^\circ$

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Very easy but nice in my opinion! Let $M$ be midpoint of $BC$, let $D$ be the midpoint of arc of $BC$ not containing $A$, let $N$ be midpoint of $MD$, let $\ell \cap AD=I$, $A$-altitude intersect $\ell$ at $H$, finally let $J$ be midpoint of $AO$, which lies on $MI$ from the given condition. We will prove that $\triangle BDC$ is equilateral, which will prove $\angle BAC=120^{\circ}$ immidiately. To prove this it suffices to prove $\frac{DO}{OM}=2$, then the centroid and circumcenter of the isosceles triangle $\triangle BDC$ will coincide, thus equilateral. Observe that $AJ=JO$ gives $AHOM$ is parallelogram. So $MJ=JH=\frac{1}{2}JI$. Also since $\angle JOM=\angle HAJ=2\angle IAJ=2\angle DAO=2\angle IDM=\angle INM$ (the last equality is because $\angle MID=90^{\circ}$.), then $AO\parallel IN$. So $\frac{NO}{OM}=\frac{IJ}{JM}=\frac{1}{2}\Rightarrow DO=2OM$, as desired.