1.

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

Answer»

\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}}


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