Area of Triangle Formed by Medians and Midpoints

In $\triangle ABC$, $D$ and $E$ are midpoints of $AC$ and $AB$ respectively. $DE$ is parallel to $BC$ and $DE = \frac{1}{2}BC$.

Therefore, $\triangle ADE \sim \triangle ABC$ with scale factor $1/2$. Area of $\triangle ADE = \frac{1}{4} \text{Area}(\triangle ABC) = \frac{108}{4} = 27$.

$O$ is the centroid of $\triangle ABC$, which divides median $BD$ in ratio $BO : OD = 2 : 1$. In $\triangle ABD$, $E$ is the midpoint of $AB$, so $DE$ is a median of $\triangle ABD$. Area of $\triangle ADE = \frac{1}{2} \text{Area}(\triangle ABD) = \frac{1}{4} \text{Area}(\triangle ABC) = 27$.

Since $O$ lies on $BD$ such that $OD = \frac{1}{3} BD$, in $\triangle ADE$ and $\triangle EOD$, they share height from $E$ to $BD$. Thus, $\text{Area}(\triangle EOD) = \frac{1}{3} \text{Area}(\triangle EBD)$. Note that $\text{Area}(\triangle EBD) = \frac{1}{4} \text{Area}(\triangle ABC) = 27$. So $\text{Area}(\triangle EOD) = \frac{1}{3} \times 27 = 9\text{ sq. cm.}$.