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

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A trapezium ABCD where AB||DC & Diagonals AC & BD intersect each other at O.
$Δ A D C$ & $Δ B D C$ Lie on the same base DC & between the same parallels AB & CD
$∴ A r e a (Δ A D C) = A r e a (Δ B D C)$
Triangle with same base & between parallels are equal in Area.
subtracting ar $(Δ D O C)$ both side
$\Rightarrow A r (Δ A D C) - A r e a (Δ D O C) = A r (Δ B D C) - A r (Δ D O C)$
$\Rightarrow A r (A O D) = A r (Δ B O C)$ proved.

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Diagonals AC and BD of a trapezium ABCD with AB $∥$ DC intersect each other at O.
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