ABCD is a parallelogram . Prove …
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Posted by Avy Avs 5 years, 1 month ago
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Sia ? 5 years, 1 month 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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