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

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Solution

BE is the median of $Δ A B C$ and we know that a median of a triangle divides it into two parts of equal area.

$\Rightarrow a r (Δ A B E) = a r (Δ C B E)$
$a r (Δ A B E) = \frac{1}{2} \times a r (Δ A B C)$ ...(i)

Similarly, $a r (Δ A C F) = a r (Δ B C F)$
$\Rightarrow a r (Δ B C F) = \frac{1}{2} \times a r (Δ A B C)$ ...(ii)

From eqs. (i) and (ii),
$a r (Δ A B E) = a r (Δ B C F)$ ....(iii)
On subtracting $a r (Δ G B F)$ from both sides of Eq. (iii), we get,
$\Rightarrow a r (Δ A B E) - a r (Δ G B F) = a r (Δ B C F) - a r (Δ G B F)$
$\Rightarrow a r (A F G E) = a r (G B C)$

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