In a triangle $A B C, E$ is the mid-point of median $A D$ . Show that $a r e a (B E D) = \frac{1}{4} a r e a (A B C)$ .

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Solution

$A D$ is the median on $B C$ of $Δ A B C$ ...Given

$∴ A (Δ A B D) = A (Δ A C D)$ .
and $A (Δ A B D) = A (Δ A C D) = \frac{1}{2} A (Δ A B C)$ ....(1)

$E B$ is the median on AD of $Δ A B D$ ...Given

$∴ A (Δ B E D) = A (Δ B E A)$ .
and $A (Δ B E D) = A (Δ B E A) = \frac{1}{2} A (Δ A B D)$ ....(2)

From (1) and (2),
$A (Δ B E D) = \frac{1}{2} \times \frac{1}{2} A (Δ A B C)$

$A (Δ B E D) = \frac{1}{4} A (Δ A B C)$ $[h e n c e p r o v e d]$

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