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27. In a parallelogram ABCD, any point E is taken on the side CDC when produced meet at a point M. Prove thatar(AADM) ar(ABMC) |
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Answer» Given.ABCD is a parallelogram.A point E is taken on the side BC.AE and DC are produced to meet at M.Prove that ar(∆ADM)=ar(∆ADM).....(1) As DC||AB,soCM||AB Since triangles on the same base and between the same parallels are equal in area,so we have ar(∆ACM)=ar(∆BCM).....(2)Adding(1) and (2),we getar(∆ADC)+ar(∆ACM)=ar(∆ABC)+ar(∆BCM)=ar(∆ADM)=ar(ABMC). Hence proved. |
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