In the given figure, ABCD is a trapezium, Diagonals AC and BD interset at Q Prove that the area of $Δ$ AOD is the same as $Δ$ BOC

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Solution

Given a trapezium ABCD where $A B ∥ D C$ $ξ$ diagonal $A C$ $ξ$ $B D$ intersect each other at $0.$
We have prove that $a r (A O D) = a r (B O C)$
Consider $△ A D C$ $ξ$ $△ B D C,$ ;ie on the same base $D C$ and between the same parallel $A B$ and $C D .$
$∴ A r e a (△ A D C) = A r e a (△ B D C)$
[ Triangle with same base and between the same parallel are equal in area .]
$\Rightarrow a r (A D C) - a r (D O C) = a r (B D C) - a r (D O C)$
$a r (A O D) = a r (B O C)$
Hence, the answer is proved.

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