In a triangle ABC, the medians BE and CF intersect at G.
Question:

In a triangle ABC, the medians BE and CF intersect at G. Prove that ar(∆BCG) = ar(AFGE).

Solution:

Figure

$C F$ is median of $\triangle A B C$.

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

Similarly, $B E$ is the median of $\triangle A B C$,

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

From (1) and (2) we have

$\operatorname{ar}(\triangle \mathrm{BCF})=\operatorname{ar}(\triangle \mathrm{ABE})$

$\Rightarrow \operatorname{ar}(\triangle B C F)-\operatorname{ar}(\triangle B F G)=\operatorname{ar}(\triangle A B E)-\operatorname{ar}(\triangle B F G)$

$\Rightarrow \operatorname{ar}(\triangle \mathrm{BCG})=\operatorname{ar}(\mathrm{AFGE})$