# 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 $(\mathrm{APB})=\operatorname{ar}(\mathrm{BQC})$.

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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.

Updated on 10-Oct-2022 13:41:37