In given figure, $E$ is any point on median $A D$ of a $△ A B C$ . Show that $a r e a (A B E) = a r e a (A C E)$ .

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Solution

Given,
$A D$ is a median of $△ A B C$ . Thus, it will divide $△ A B C$ into two triangles of equal area.

$∴ a r e a (A B D) = a e a r (A C D)$ --- (i)

Also,

$E D$ is the median of $△ A B C$ .

$∴ a r e a (E B D) = a r e a (E C D)$ --- (ii)

Subtracting (ii) from (i),

$a r e a (A B D) - a r e a (E B D) = a r e a (A C D) - a r e a (E C D)$

$\Rightarrow a r e a (A B E) = a r e a (A C E)$ $[h e n c e p r o v e d]$

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