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

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Solution

Prove that, the area of triangle $A B C$ is four-time the area of triangle $B E D$ .
Compute the area of triangle $A B C$ as follows:
$\Rightarrow a r (A B C) = a r (A B D) + a r (A C D) \Rightarrow a r (A B C) = \frac{1}{2} (B D) (A D) + \frac{1}{2} (C D) (A D)$
Since, $A D$ is a median. So, $B D = C D$ ,
Therefore,
$\Rightarrow a r (A B C) = \frac{1}{2} (B D) (A D) + \frac{1}{2} (B D) (A D) \Rightarrow a r (A B C) = 2 (\frac{1}{2} (B D) (A D))$
Since, $A D = A E + E D$ .
Therefore, $a r (A B C) = 2 (\frac{1}{2} (B D) (A E + E D))$
It is given that $E$ is the mid-point of median $A D$ .
So,
$\Rightarrow a r (A B C) = 2 (\frac{1}{2} (B D) (D E + D E)) \Rightarrow a r (A B C) = 2 (\frac{1}{2} \cdot 2 (B D) (D E)) \Rightarrow a r (A B C) = 4 (\frac{1}{2} (B D) (D E)) \Rightarrow a r (A B C) = 4 a r (B E D)$
Hence prove, $a r (B E D) = \frac{1}{4} a r (A B C)$

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