Find the area of the triangle EGB.
Solution 1:
Let $D$ be the midpoint of $AC$. Drop the perpendicular lines from $A,$ $B,$ $C$ and $D$ to $\overleftrightarrow{EG}$.
Since $DD'$ is the midline of trapezium $AA'C'C$,
$$DD'=\frac{AA'+CC'}2.$$
As $G$ is the centroid,
$$\frac{BB'}{DD'}=\frac{BG}{GD}=\frac21.$$
Therefore, $BB'=AA'+CC'.$
Thus, $[\triangle BEG]=[\triangle AEG]+[\triangle CEG].$