Question

ABCD is a parallelogram. Prove that ar(BCP) = ar(DPQ) if BC=CQ

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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