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Q : 6     In Fig. \small 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O such that \small OB=OD. If \small AB=CD, then show that:

(iii) \small DA\parallel CB or ABCD is a parallelogram.

[Hint : From D and B, draw perpendiculars to AC.]

            

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Since \DeltaDCB and \DeltaACB both lie on the same base BC and having equal areas.
Therefore, They lie between the same parallels BC and AD
\Rightarrow CB || AD 
also \angle1 = \angle2 [ already proved]
So, AB ||  CD 
Hence ABCD is a || gm

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manish

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