$D$ is the mid-point of side $BC$ of $\triangle ABC$ and $E$ is the mid-point of $BD$. If $O$ is the mid-point of $AE$, prove that $ar(\triangle BOE) = \frac{1}{8} ar(\triangle ABC)$.
Given:
$D$ is the mid-point of side $BC$ of $\triangle ABC$ and $E$ is the mid-point of $BD$.
$O$ is the mid-point of $AE$.
To do:
We have to prove that $ar(\triangle BOE) = \frac{1}{8} ar(\triangle ABC)$.
Solution:
Join $BO, AE$ and $AD$.
In $\triangle \mathrm{ABC}$,
$\mathrm{D}$ is the mid point of $\mathrm{BC}$
$\operatorname{ar}(\triangle \mathrm{ABD})=\operatorname{ar}(\Delta \mathrm{ADC})=\frac{1}{2} \operatorname{ar}(\Delta \mathrm{ABC})$
In $\triangle \mathrm{ABD}, \mathrm{E}$ is the mid point of $\mathrm{BD}$.
This implies,
$\operatorname{ar}(\Delta \mathrm{ABE})=\frac{1}{2} \operatorname{ar}(\Delta \mathrm{ABD})$
$=\frac{1}{2}\left(\frac{1}{2} \operatorname{ar}(\Delta \mathrm{ABC})\right)$
$=\frac{1}{4} \operatorname{ar}(\triangle \mathrm{ABC})$
$\mathrm{O}$ is the mid point of $\mathrm{AE}$
Therefore,
$a r(\Delta \mathrm{BOE})=\frac{1}{2} \operatorname{ar}(\Delta \mathrm{ABE})$
$=\frac{1}{2} \times(\frac{1}{4} \operatorname{ar}(\Delta \mathrm{ABC}))$
$=\frac{1}{8} a r(\Delta \mathrm{ABC})$
$\operatorname{ar}(\Delta \mathrm{BOE})=\frac{1}{8} \operatorname{ar}(\Delta \mathrm{ABC})$
Hence proved.
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