Diagonal AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (ARC)=ar(BOC).prove that ABCD is a Trapeziums
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Step-by-step explanation:
Here, ABCD is a quadrilateral where diagonals AC and BD intersect at O.
⇒ ar(△AOD)=ar(△BOC)
Adding ar(ODC) on both sides,
⇒ ar(AOD)+ar(ODC)=ar(BOC)+ar(ODC)
⇒ ar(ADC)=ar(BDC)
Now, △ADC and △BDC lie on the same base DC and are equal in area and they lie between the lines AB and DC
⇒ AB∥DC ( Two triangles having the same base and equal areas lie between the same parallels )
In ABCD,
⇒ AB∥DC
So, one pair of opposite sides is parallel,
∴ ABCD is trapezium.
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