Question

Question 3
The median BE and CF of a triangle ABC intersect at G. Prove that the area of $\Delta$GBC = area of the quadrilateral AFGE.

Answer 1

BE is the median of $\Delta ABC$ and we know that a median of a triangle divides it into two parts of equal area.

$\Rightarrow ar\left(\Delta ABE\right)=ar\left(\Delta CBE\right)$
$ar\left(\Delta ABE\right)=\frac{1}{2}\times ar\left(\Delta ABC\right)$ ...(i)

Similarly, $ar\left(\Delta ACF\right)=ar\left(\Delta BCF\right)$
$\Rightarrow ar\left(\Delta BCF\right)=\frac{1}{2}\times ar\left(\Delta ABC\right)$ ...(ii)

From eqs. (i) and (ii),
$ar\left(\Delta ABE\right)=ar\left(\Delta BCF\right)$ ....(iii)
On subtracting $ar\left(\Delta GBF\right)$ from both sides of Eq. (iii), we get,
$\Rightarrow ar\left(\Delta ABE\right)-ar\left(\Delta GBF\right)=ar\left(\Delta BCF\right)-ar\left(\Delta GBF\right)$
$\Rightarrow ar\left(AFGE\right)=ar\left(GBC\right)$