ABCD is a rhombus in which AC and BD intersect at O then prove that ar(OAB)=1/4ar(ABCD)
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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 have equal area.
Consider △CDA and △CDB
It can be written as
Area of △CDA = Area of △CDB
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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