# D is the midpoint of side BC of ∆ABC and E is the midpoint of BD. If O is the midpoint of AE,

Question:

Dis the midpoint of side $B C$ of $\triangle A B C$ and $E$ is the midpoint of $B D$. If $O$ is the midpoint of $A E$, prove that ar $(\triangle B O E)=\frac{1}{8}$ ar( $\triangle A B C$ ).

Solution:

$D$ is the midpoint of side $B C$ of $\triangle A B C$.

$\Rightarrow \mathrm{AD}$ is the median of $\triangle \mathrm{ABC}$.

$\Rightarrow \operatorname{ar}(\triangle \mathrm{ABD})=\operatorname{ar}(\triangle \mathrm{ACD})=\frac{1}{2} \operatorname{ar}(\triangle \mathrm{ABC})$

$E$ is the midpoint of $B D$ of $\triangle A B D$,

$\Rightarrow \mathrm{AE}$ is the median of $\triangle \mathrm{ABD}$

$\Rightarrow \operatorname{ar}(\triangle \mathrm{ABE})=\operatorname{ar}(\triangle \mathrm{AED})=\frac{1}{2} \operatorname{ar}(\triangle \mathrm{ABD})=\frac{1}{4} \operatorname{ar}(\triangle \mathrm{ABC})$

Also, O is the midpoint of AE,

$\Rightarrow B O$ is the median of $\triangle A B E$

$\Rightarrow \operatorname{ar}(\triangle \mathrm{ABO})=\operatorname{ar}(\triangle \mathrm{BOE})=\frac{1}{2} \operatorname{ar}(\triangle \mathrm{ABE})=\frac{1}{4} \operatorname{ar}(\triangle \mathrm{ABD})=\frac{1}{8} \operatorname{ar}(\triangle \mathrm{ABC})$

Thus, $\operatorname{ar}(\Delta B O E)=\frac{1}{8} \operatorname{ar}(\Delta A B C)$