In triangle ABC,AD is a median. E is the midp | vedclass

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(N/A) $1$. Since $AD$ is the median of $	riangle ABC$,it divides the triangle into two triangles of equal area. Therefore,$ar(ABD) = \frac{1}{2} ar(A

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(N/A) $1$. Since $AD$ is the median of $\triangle ABC$,it divides the triangle into two triangles of equal area. Therefore,$ar(ABD) = \frac{1}{2} ar(ABC)$.
$2$. In $\triangle ABD$,$AE$ is a median because $E$ is the midpoint of $BD$. Thus,$ar(ABE) = \frac{1}{2} ar(ABD)$.
$3$. Substituting the value from step $1$: $ar(ABE) = \frac{1}{2} \times (\frac{1}{2} ar(ABC)) = \frac{1}{4} ar(ABC)$.
$4$. In $\triangle ABE$,$BO$ is a median because $O$ is the midpoint of $AE$. Thus,$ar(AOB) = \frac{1}{2} ar(ABE)$.
$5$. Substituting the value from step $3$: $ar(AOB) = \frac{1}{2} \times (\frac{1}{4} ar(ABC)) = \frac{1}{8} ar(ABC)$.
$6$. Hence,it is proved that $ar(AOB) = \frac{1}{8} ar(ABC)$.

In $\triangle ABC$,$AD$ is a median. $E$ is the midpoint of $BD$ and $O$ is the midpoint of $AE$. Prove that $ar(AOB) = \frac{1}{8} ar(ABC)$.

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