Q

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.]

Views

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

Similarily, ar ($\Delta$PBC) = 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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