Diagonal AC and BD of trapezium ABCD, in which AB ||DC, intersect each other at O. The triangle which is equal in area of $△ A O D$ is

$△ B O C$

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$△ D O C$

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$△ A D C$

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$△ A O B$

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Solution

The correct option is A
$△ B O C$

In trapezium ABCD, diagonals AC and BD intersect each other at O. Also, AB || DC

Now, $△ A B C a n d △ A B D$ are on the same base and between the same parallels.

$∴ a r (△ A B C) = a r (△ A B D)$
Subtracting $a r (△ A O B) a r (△ A B C) - a r (△ A O B) = a r (△ A D B) - a r (△ A O B) \Rightarrow a r (△ B O C) = a r (△ A O D) \Rightarrow a r (△ A O D) = a r (△ B O C)$

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