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

prove that ar(AOD) = ar (BOC)?

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Given that ABCD is a trapezium with AB || DC and Diagonal AC and BD intersect each other at O.

To prove: Area (AOD) = Area (BOC)

Proof: ΔADC and ΔBDC are on the same base DC and between same parallel AB and DC.

∴Area (ΔADC) = Area (ΔBDC)    [triangles on the same base and between same parallel are equal in area]

Subtract Area (ΔDOC) from both side

Area (ΔADC) – Area (ΔDOC) = Area (ΔBDC) – Area (ΔDOC)

Area (ΔAOD) = Area (ΔBOC)

Hence proved.

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  • 95

Given : ABCD is trapezium, AB parallel to DC, diagonals AC & BD intersect at O

To Prove : ar( AOD ) = ( BOC)

Proof : Since triangle ADC & BCD lies on same base DC and between same parallels AB & DC

  ar ( ADC ) = ar ( BCD )

Subtracting ar ( DOC ) from both the sides

  ar ( ADC ) - ar ( DOC ) = ar ( BCD ) - ar ( DOC )

  ar ( AOD ) = ar ( BOC )

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It can be observed that ΔDAC and ΔDBC lie on the same base DC and between the same parallels AB and CD.

∴ Area (ΔDAC) = Area (ΔDBC)

⇒ Area (ΔDAC) − Area (ΔDOC) = Area (ΔDBC) − Area (ΔDOC)

⇒ Area (ΔAOD) = Area (ΔBOC)

:)

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