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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:

(ii) \small ar(DCB)=ar(ACB)

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

            

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We already proved that,
ar(\Delta DOC)=ar(\Delta AOB)
Now, add ar(\Delta BOC) on both sides we get

\\ar(\Delta DOC)+ar(\Delta BOC)=ar(\Delta AOB)+ar(\Delta BOC)\\ ar(\Delta DCB) = ar (\Delta ACB)
Hence proved.

Posted by

manish

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