The medians BE and CF of a $△ A B C$ intersect at G. Prove that $a r (△ A B C) = a r (q u a d r i l a t e r a l A F G E) .$

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Solution

The line joining the mid-point of the two sides of a triangle is parallel to the third side.
$∴ B C | | E F$
Triangle on the same base and between the same parallel lines are equal in area.
$∴ a r (B C F) = a r (B C E)$
$\Rightarrow a r (B C G) + a r (C E G) = a r (B C G) + a r (B F G)$
$\Rightarrow a r (C E G) = a r (B F G) . . . . . . . . (i)$
Now, the median of a triangle divides the triangle into two triangles of equal area.
BE is median of $△ A B C$
$∴ a r (B C E) = a r (A B E)$
$\Rightarrow a r (B C G) + a r (C E G) = a r (B F G) + a r (A F G E)$
$\Rightarrow a r (B C G) + a r (C E G) = a r (C E G) + a r (A F G E)$ [From (i)]
$\Rightarrow a r (B C G) = a r (A F G E)$

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