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

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Solution

Triangles $A B C$ and $A B D$ are on the same base $A B$ and between the same parallels $A B$ and $D C$ .

Therefore,
$a r (A B D) = a r (A B C)$
Subtract area of $△ A O B$ from both the sides.
$a r (A B D) - a r (A O B) = a r (A B C) - a r (A O B)$
$\Rightarrow a r (A O D) = a r (B O C)$

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

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) .

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