A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that `ar(DeltaADF) = ar (ABFC)`.

Explanation

Join BM and AC.We have
`ar(triangleADC)=(1)/(2)xxDCxxh`
and `ar(triangleABM)=(1)/(2)xxABxxh = (1)/(2)xxDCxxh` ltrgt [`therefore` AB = DC in ||gm ABCD].
`therefore ar(triangleADC)=ar(triangleABM)`
`rArr ar(triangleADC)+ar(triangleAMC)=ar(triangleABM)+ar(triangleAMC)`
`rArr ar(triangleADM)=ar(triangleABMC)`.