$D$ , $E$ and $F$ are the mid-points of the sides $B C$ , $C A$ and $A B$ respectively of triangle $A B C$ . Prove that area of $B D E F$ is half the area of $Δ A B C$ .

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Solution

FD is diagonal of the parallelogram BDEF
$∴ A r e a (△ F B D) = a r e a (△ D E F)$
$S i m i l a r l y A r e a (△ D E F) = A r e a (△ D C E)$
$T h e r e f o r e A r e a (△ F B D) = A r e a (△ D E F) = A r e a (△ F A E) = A r e a (△ D C E)$
$∴ △ A B C i s d i v i d e d i n t o 4 n o n - o v e r l a p p i n g t r i a n g l e s △ F B D, △ D E F, △ F A E a n d △ D C E$
$T h e r e f o r e A r e a (△ A B C) = A r e a (△ F B D) + A r e a (△ D E F) + A r e a (△ F A E) + A r e a (△ D C E) = 4 A r e a (△ D E F)$
$A r e a (△ D E F) = \frac{1}{4} a r e a (△ A B C)$
So, as per given question
$A r e a o f (B D E F) = A r e a (△ F B D) + A r e a (△ D E F)$
$= A r e a (△ D E F) + A r e a (△ D E F)$
$= 2 A r e a (△ D E F)$
$= 2 \times \frac{1}{4} a r e a (△ A B C)$
$= \frac{1}{2} a r e a (△ A B C)$

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