P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD show that ar(ΔAPB)=ar Δ(BQC).
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If a parallelogram and a triangle are on the same base and between the same parallels then area of the triangle is half the area of the parallelogram.
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Given: In parallelogram ABCD, P & Q any two points lying on the sides DC and AD.
To show: ar (APB) = ar (BQC).
Proof:
Here, ΔAPB and ||gm ABCD stands on the same base AB and lie between same parallel AB and DC.
Therefore,
ar(ΔAPB) = 1/2 ar(||gm ABCD) — (i)
Similarly,
Parallelogram ABCD and ∆BQC stand on the same base BC and lie between the same parallel BC and AD.
ar(ΔBQC) = 1/2 ar(||gm ABCD) — (ii)
From eq (i) and (ii),we have
ar(ΔAPB) = ar(ΔBQC)
Hope this will help you...
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