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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).


Answer :

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.
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\(\triangle \)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

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