Question

Diagonals AC and BD of a trapezium ABCD with AB parallel DC intersect each other at O. Prove that ar(AOD)=ar(BOC).
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Answer 1

$\huge{ \mathtt{ \purple{SOLUTION:-}}}$

Given:

• A trapezium ABCD in which AB parallel DC and diagonals AC and BD interest each other at O.

To prove:

• ar(ADC) = ar(BOC)

Proof:

∆ ABC and ∆BDC lie on the same base and between the same parallels AB and CD

∴ ar(ADC) = ar(BDC) [Triangles with same base and between same parallels are equal in area]

Subtracting ar(DOC) from both sides

$\large{ \implies}$ar (ADC) - ar (DOC) = ar (BDC) - ar (DOC)

$\large{ \implies}$ar (AOD) = ar (BOC)

$\large \green{ \mathtt{Hence \: proved}}$