In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O.If BO = OD, prove thatAr(∆ABC) = ar(∆ADC).

Given : A quadrilateral ABCD with diagonals AC and BD and BO = OD

To prove: Area of (∆ABC) = area of (∆ADC)

Proof : BO = OD [given]

Here AO is the median of ∆ABD

Area of (∆AOD) = Area of (∆AOB) ---------------- 1

And OC is the median of ∆BCD

Area of (∆COD) = Area of (∆BOC) ---------------- 2

Now by adding –1 and –2 we get

Area of (∆AOD) + Area of (∆COD) = Area of (∆AOB) + Area of (∆BOC)

Area of (∆ABC) = Area of (∆ADC)

Hence proved

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