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

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Solution

Let A be the origin $a r (Δ A B C) = \frac{1}{2} | \to A B \times \to A C | = \frac{1}{2} | \to b \times \to c |$

p.v.s of E and D are obtained by section formula.

$a r (Δ B D E) = \frac{1}{2} | \to B D \times \to B E |$
$= \frac{1}{2} ∣ ∣ ∣ ∣ (\frac{\to b + \to c}{2} - \to b) \times (\frac{\to b + \to c}{4} - \to b) ∣ ∣ ∣ ∣ = \frac{1}{2} ∣ ∣ ∣ \frac{- \to b + \to c}{2} \times \frac{- 3 \to b + \to c}{4} ∣ ∣ ∣$
$= \frac{1}{2} \times 8 ∣ ∣ 3 (\to b \times \to c) - (\to b \times \to c) ∣ ∣ = \frac{1}{16} (2) (| \to b \times \to c |) = \frac{1}{4} \times \frac{1}{2} | \to b \times \to c |$
$∴ a r (Δ B D E) = \frac{1}{4} a r (Δ A B C)$ .

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