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In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that (ii) ar (APD) plus ar (PBC) equals ar (APB) plus ar (PCD)

Q: 4     In Fig. \small 9.16, P is a point in the interior of a parallelogram ABCD. Show that

             (ii) \small ar(APD)+ar(PBC)=ar(APB)+ar(PCD)

             [Hint : Through P, draw a line parallel to AB.]

            

Answers (1)
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M manish


We have a ||gm ABCD and AB || CD, AD || BC.  Through P, draw a line parallel to AB
 Now, \DeltaAPD and ||gm ADGHare on the same base AD and between the same parallels GH and AD.
Therefore, ar (\DeltaAPD) = 1/2 . ar(||gm ADGH).............(i)

Similarily, ar (\DeltaPBC) = 1/2 . ar(||gm BCGH)............(ii)

By adding the equation i and eq (ii), we get

\small ar(APD)+ar(PBC)=ar(APB)+ar(PCD)

Hence proved.

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