In fig, $A B C$ and $A B D$ are two triangles on the same base $A B$ . If line-segment $C D$ is bisected by $A B$ at $O$ , show that $a r (A B C) = a r (A B D)$ .

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Solution

In $Δ A B C$ ,
$A O$ is the median. $∣$ $C D$ is bisected by $A B$ at $O$
$∴ a r (A O C) = a r (A O D)$ --- (i)

Also,

In $△ B C D$ ,
$B O$ is the median. $∣$ $C D$ is bisected by $A B$ at $O$
$∴ a r (B O C) = a r (B O D)$ --- (ii)

Adding (i) and (ii) we get,

$a r (A O C) + a r (B O C) = a r (A O D) + a r (B O D)$

$\Rightarrow a r (A B C) = a r (A B D)$ .

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