Find the area of ​the triangle EGB.

Question

Through the centroid $G$ of a triangle $ABC$, a secant line is drawn, cutting sides $AB$ and $BC$. On the straight line a point $E$ is considered. Calculate the area of ​​the triangle $EBG$ if the areas of the triangles $ECG$ and $AEG$ are $20$ and $30$ respectively. (Answer: $50$)

My progress:

I can't develop much..sorry.

I know $G$ is centroid : $\therefore S_{\triangle ABG} = S_{\triangle AGC} = S_{\triangle BGC}.$

Also $\triangle AEG, \triangle EGC, \triangle BEG$ have the same base.

I don't see similarity of triangles.

The heights are different...

I didn't understand the function of the line

????

Answer 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, $[\triangle BEG]=[\triangle AEG]+[\triangle CEG].$