In the figure, diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O.
$Prove that a r (△ A O D) = a r (△ B O C) .$

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Solution

Given that AB || CD
$Now △ A D C and △ B C D are on the same base and between the same parallels A B | | C D .$
$∴ △ A D C = △ B C D$
$\Rightarrow △ A D C - △ C O D = △ B C D - △ C O D$
$\Rightarrow △ A O D = △ B O C [from the figure]$

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Prove that a r (△ A O D) = a r (△ 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). [1 MARK]

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