Since, the sum of lengths of any two sides in a triangle should be greater than the length
of third side.
Therefore, In ∆ABM, AB + BM > AM … (i)
In ∆AMC, AC + MC > AM … (ii)
Adding eq. (i) and (ii),
AB + BM + AC + MC > AM + AM
⇒ AB + AC + (BM + MC) > 2AM
⇒ AB + AC + BC > 2AM
Hence, it is true.

Class 7 Maths Chapter 6 Exercise 6.4

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