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# P and $ Q $ are any two points lying on the sides $ D C $ and $ A D $ respectively of a parallelogram $ \mathrm{ABCD} $. Show that ar $ (\mathrm{APB})=\operatorname{ar}(\mathrm{BQC}) $.

Given:

$P$ and \( Q \) are any two points lying on the sides \( D C \) and \( A D \) respectively of a parallelogram \( \mathrm{ABCD} \).

To do:

We have to show that ar \( (\mathrm{APB})=\operatorname{ar}(\mathrm{BQC}) \).

Solution:

$\triangle APB$ and parallelogram $ABCD$ lie on the same base $AB$ and between the same parallels $AB$ and $DC$.

This implies,

$ar(\triangle APB) =\frac{1}{2}$ ar(parallelogram $ABCD$)........(i)

Similarly,

$\triangle BQC$ and parallelogram $ABCD$ lie on the same base $BC$ and between the same parallels $AD$ and $BC$.

This implies,

$ar(\triangle BQC) = \frac{1}{2}$ ar(parallelogram $ABCD$.............(ii)

From (i) and (ii), we get,

$ar(\triangle APB) = ar(\triangle BQC)$

Hence proved.

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