Question

ABCD is a parallelogram.If BC=CQ,prove that Ar(BCP)=1/2 area(bcd)

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

Where is the figure?