diagonals ab and bd of a quadilateral intersect at O in such a way that oad =boc proove that abcd is trepezium
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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. ---- Hence proved
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