Question

ABCD is a parallelogram.If BC=CQ,prove that Ar(BCP)=Ar(DPQ)

Answer 1

Given : ABCD is a parallelogram.
BC=CQ.

To prove :
ar(ΔBCP)=ar(ΔDPQ)

Construction :
Join AC

Proof :
ar(ΔBPC) = ar (ΔAPC) [Triangles on the same base and between same parallels ]

Similarly ,
ar(ΔADC) = ar(ΔADQ) [Triangles on the same base and between same parallels ]

ar(ΔADC) = ar(ΔADP) +ar(ΔAPC)
ar(ΔADQ) = ar(ΔADP) + ar(ΔDPQ)

ar(ΔADC) = ar(ΔADQ) and ,
ar(ΔADP) is common.

∴ ar(ΔAPC) = ar(ΔDPQ)

But ,
ar(ΔAPC) = ar(ΔBPC)

∴ ar(ΔBPC) = ar(ΔDPQ)


Hence ,proved.


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Answer 2

I hope it will help you