Diagonals $A C$ and $B D$ of a trapezium $A B C D$ with $A B ∥ D C$ intersect each other at $O$ . prove that $a r (A O D) = a r (B O C)$

Open in App

Solution

$A B C D$ is trapezium, with diagonals $A C$ and $B D$ which intersect at point $O$ .
Here, $△ D A C$ and $△ D B C$ lie on the same base $D C$ and between the same parallels $A B$ and $C D .$
$∴$ $a r (△ D A C) = a r (△ D B C)$
$\Rightarrow$ $a r (△ D A C) - a r (△ D O C) = a r (△ D B C) - a r (△ D O C)$ [On subtracting $a r (△ D O C)$ from both sides]
$\Rightarrow$ $a r (△ A O D) = a r (△ B O C)$

Suggest Corrections

Similar questions

Diagonals AC and BD of a trapezium ABCD with AB $∥$ DC intersect each other at O.
Prove that ar ( $△$ AOD) = ar ( $△$ BOC). [1 MARK]

Prove that a r (△ A O D) = a r (△ B O C) .

Join BYJU'S Learning Program