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ABCD is a parallelogram . Prove …

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ABCD is a parallelogram . Prove that area of BCP=area ofDPQ if BC=CQ

Posted by Avy Avs 6 years, 2 months ago

Sia ? 6 years, 2 months ago

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.

{tex}\therefore{/tex} ar(ΔAPC) = ar(ΔDPQ)

But ,
ar(ΔAPC) = ar(ΔBPC)

∴ ar(ΔBPC) = ar(ΔDPQ)
Hence ,proved.

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