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ABCD is a IIgm in which BC is produced to E such that CE=BC. AE intersects CD at F.

1) Prove that ar(ADF) = ar(ECF)

2)If the area of DFB=3cm^2 then find the area of the IIgm ABCD.


1.AD=BC...opp. Sides of a parallelogram and BC=CE... Given
Hence, AD=CE
2.in triangle ADF and triangle ECF,
AD=CE... From 1
Ang.DAF=angFEC...alternate angles
angAFD=angCFE...vertically opp.
Hence, ADF is congruent to ECF
Hence arADF=arCFE
4.triangle DBF lies on half the base DC of parallelogram ABCD and they lie between the same parallels AB and DC
Hence arDFB=half(half(arABCD))... The area of triangle is half the area of parallelogram if they lie on Same base and between same parallels
Hence, arDBF=0.25arABCD
Hence arABCD=3*4=12cm^2
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