Question

The medians $BE$ and $CF$ of a $△ABC$ intersect at $G$. Prove that area ( $△GBC)ar=\square AFGE$.

Answer 1

$BE$ is the median of $\Delta ABC$

so, $ar\left(\Delta BEC\right)=\frac{1}{2}ar\left(\Delta ABC\right)$ __(1)

Median of triangle divides into 2 triangles of equal area.

Also $CF$ is median of $\Delta ABC$

$\Rightarrow ar\left(\Delta ACF\right)=\frac{1}{2}ar\left(\Delta ABC\right)$ __(2)

from (1) and (2)

$ar\left(\Delta ACF\right)=ar\left(\Delta BEC\right)$

$ar\left(\Delta GBC\right)+ar\left(\Delta GEC\right)=ar\left(AFGE\right)+ar\left(GEC\right)$

$\therefore ar\left(\Delta GBC\right)=ar\left(AFGE\right)$