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# In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that (i) ar (APB) + ar (PCD) = 1/2 ar (ABCD)

Q: 4     In Fig. $9.16$, P is a point in the interior of a parallelogram ABCD. Show that
(i)  $\small ar(APB)+ar(PCD) = \frac{1}{2}ar(ABCD)$

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

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We have a ||gm ABCD and AB || CD, AD || BC.  Through P, draw a line parallel to AB
Now, $\Delta$APB and ||gm ABEFare on the same base AB and between the same parallels EF and AB.
Therefore, ar ($\Delta$APB) = 1/2 . ar(ABEF)...............(i)
Similarly, ar ($\Delta$PCD ) = 1/2 . ar (EFDC) ..............(ii)
Now, by adding both equations, we get
$\small ar(APB)+ar(PCD) = \frac{1}{2}ar(ABCD)$

Hence proved.

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