In the figure, $D$ and $M$ are the midpoints of $AB$ and $AC$, respectively. Then prove that $2\left[\triangle BOD\right] = \left[\square COME\right]$

My Attempt
- $\left[\triangle COM\right]=\left[\triangle CME\right]$
- $\left[\triangle BCD\right]=\left[\triangle CDA\right]$
I could not move forward. Please help me to complete
Extend $AO$ to meet $BC$ at $N$.
$AN$ is median since $O$ is centroid. Therefore, $\Delta ANB=\Delta ANC$ and $\Delta ONB=\Delta ONC$.
$$\Rightarrow \left[\Delta ANB\right]-\left[\Delta ONB\right] = \left[\Delta ANC\right]-\left[\Delta ONC\right]$$
$$\Rightarrow \frac{\left[\Delta AOB\right]}{2} = \frac{\left[\Delta AOC\right]}2$$
$$\Rightarrow \left[\Delta BOD\right] = \left[\Delta COM\right] $$
$$\Rightarrow 2\left[\Delta BOD\right] = \left[COME\right]$$