Diagonals $A C$ and $B D$ of a trapezium $A B C D$ with $A B ∥ D C$ intersects each other at $O$ . Prove that ar $(△ A O D) =$ ar $(△ B O C)$ .

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Solution

Given: Diagonals AC and BD of trapezium ABCD with AB $| |$ DC intersects each others at O.

Prove: $a r (A O D) = a r (B O C)$

Proof: Here, $Δ D A C$ and $Δ D B C$

lie on same base DC & between same parallel AB & CD.

$a r (Δ D A C) = a r (Δ D B C)$

$a r (Δ D A C) - a r (Δ D O C) = a r (Δ D B C) - a r (Δ D O C)$

$a r (Δ A O D) = a r (Δ B O C)$ .

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Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).

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

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