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in Parallelograms by (58.8k points)

In the adjoining figure, ABCD is a trapezium in which AB || DC and its diagonals AC and BD intersect at O. Prove that ar (△ AOD) = ar (△ BOC).

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From the figure we know that △ AOD and △ DCB lie on the same base CD and between two parallel lines DC and AB.

We know that the triangles lying on the same base and parallels have equal area.

Consider △ CDA and △ CDB

It can be written as

Area of △ CDA = Area of △ CDB

So we get

Area of △ AOD = Area of △ ADC – Area of △ OCD

In the same way

Area of △ BOC = Area of △ CDB – Area of △ OCD

So we get

Area of △ BOC = Area of △ ADC – Area of △ OCD

Area of △ AOD = Area of △ BOC

Therefore, it is proved that ar (△ AOD) = ar (△ BOC).

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