The medians $BE$ and $CF$ of a triangle $ABC$ intersect at $G$. Prove that the area of $\triangle GBC = \text{area of the quadrilateral } AFGE.$

(N/A) Let $BE$ and $CF$ be the medians of $\triangle ABC$ intersecting at $G$. We need to prove that $\operatorname{ar}(\triangle GBC) = \operatorname{ar}(AFGE)$.
Since a median divides a triangle into two triangles of equal area,for median $CF$:
$\operatorname{ar}(\triangle BCF) = \operatorname{ar}(\triangle ACF)$
$\operatorname{ar}(\triangle GBF) + \operatorname{ar}(\triangle GBC) = \operatorname{ar}(AFGE) + \operatorname{ar}(\triangle GCE) \quad \dots(1)$
Similarly,for median $BE$:
$\operatorname{ar}(\triangle ABE) = \operatorname{ar}(\triangle CBE)$
$\operatorname{ar}(\triangle AFG) + \operatorname{ar}(\triangle BFG) = \operatorname{ar}(\triangle GCE) + \operatorname{ar}(\triangle GBC) \quad \dots(2)$
Adding $(1)$ and $(2)$:
$\operatorname{ar}(\triangle GBF) + \operatorname{ar}(\triangle GBC) + \operatorname{ar}(\triangle AFG) + \operatorname{ar}(\triangle BFG) = \operatorname{ar}(AFGE) + \operatorname{ar}(\triangle GCE) + \operatorname{ar}(\triangle GCE) + \operatorname{ar}(\triangle GBC)$
Canceling common terms $\operatorname{ar}(\triangle GBC)$ and $\operatorname{ar}(\triangle GCE)$ from both sides:
$\operatorname{ar}(\triangle GBF) + \operatorname{ar}(\triangle AFG) + \operatorname{ar}(\triangle BFG) = \operatorname{ar}(AFGE) + \operatorname{ar}(\triangle GCE)$
Actually,a simpler approach is using the property that the centroid $G$ divides the triangle into three triangles of equal area: $\operatorname{ar}(\triangle GBC) = \operatorname{ar}(\triangle GCA) = \operatorname{ar}(\triangle GAB) = \frac{1}{3} \operatorname{ar}(\triangle ABC)$.
Since $F$ and $E$ are midpoints,$\operatorname{ar}(\triangle GAF) = \operatorname{ar}(\triangle GAE) = \frac{1}{6} \operatorname{ar}(\triangle ABC)$.
Thus,$\operatorname{ar}(AFGE) = \operatorname{ar}(\triangle GAF) + \operatorname{ar}(\triangle GAE) = \frac{1}{6} \operatorname{ar}(\triangle ABC) + \frac{1}{6} \operatorname{ar}(\triangle ABC) = \frac{2}{6} \operatorname{ar}(\triangle ABC) = \frac{1}{3} \operatorname{ar}(\triangle ABC)$.
Since $\operatorname{ar}(\triangle GBC) = \frac{1}{3} \operatorname{ar}(\triangle ABC)$,it follows that $\operatorname{ar}(\triangle GBC) = \operatorname{ar}(AFGE)$.