P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.
Show that ar(APB) = ar(BQC).

It can be observed that \(\triangle{BQC}\) and paralleogram ABCD lie on the same base BC and these are between the same parallel lines AD and BC.

\(\triangle \)APB and parallelogram ABCD lie on the same base AB and in-between same parallel AB and DC.

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

Similarly,

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

From (i) and (ii), we have

ar(\(\triangle \) APB) = ar(\(\triangle \)BQC)
Hence, proved