Question

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

Answer 1

Figure
CF is median of $△$ABC.
$\Rightarrow$ar($△$BCF) = $\frac{1}{2}$($△$ABC) .....(1)
Similarly, BE is the median of $△$ABC,
$\Rightarrow$ar($△$ABE) = $\frac{1}{2}$($△$ABC) .....(2)
From (1) and (2) we have
ar($△$BCF) = ar($△$ABE)
$\Rightarrow$ar($△$BCF) $-$ ar($△$BFG) = ar($△$ABE) $-$ ar($△$BFG)
$\Rightarrow$ar(∆BCG) = ar(AFGE)