If the medians of a $\Delta ABC$ intersect at $G$,show that $\operatorname{ar}(AGB) = \operatorname{ar}(AGC) = \operatorname{ar}(BGC) = \frac{1}{3} \operatorname{ar}(ABC)$.

(N/A) Given: Medians $AE, BF$ and $CD$ of $\triangle ABC$ intersect at $G$.
To prove: $\operatorname{ar}(\triangle AGB) = \operatorname{ar}(\triangle AGC) = \operatorname{ar}(\triangle BGC) = \frac{1}{3} \operatorname{ar}(\triangle ABC)$.
Construction: Draw $BP \perp AE$.
Proof: $AG = \frac{2}{3} AE$ [Since the centroid divides the median in the ratio $2:1$].
Now,$\operatorname{ar}(\triangle AGB) = \frac{1}{2} \times AG \times BP$
$= \frac{1}{2} \times \frac{2}{3} AE \times BP$
$= \frac{2}{3} \times (\frac{1}{2} \times AE \times BP)$
$= \frac{2}{3} \operatorname{ar}(\triangle ABE)$
$= \frac{2}{3} \times \frac{1}{2} \operatorname{ar}(\triangle ABC)$ [Since the median divides a triangle into two triangles of equal area]
$= \frac{1}{3} \operatorname{ar}(\triangle ABC)$.
Similarly,$\operatorname{ar}(\triangle AGC) = \frac{1}{3} \operatorname{ar}(\triangle ABC)$ and $\operatorname{ar}(\triangle BGC) = \frac{1}{3} \operatorname{ar}(\triangle ABC)$.
Therefore,$\operatorname{ar}(\triangle AGB) = \operatorname{ar}(\triangle AGC) = \operatorname{ar}(\triangle BGC) = \frac{1}{3} \operatorname{ar}(\triangle ABC)$.
Hence,proved.