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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).

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