Question

AM is a median of AABC. Prove that (AB + BC + CA) > 2 AM.
Hint:
[AB + BM] > AM (in AABM)
(AC + MC) > AM (in AACM)
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Answer 1

Answer:

We know,

In a triangle sum of two sides is greater than the third side

Hence,In triangle ABM

AB + BM > AM ____1

Also in triangle ACM

AC + MC > AM ___2

Adding equation 1 and 2 we get,

AB + BM + AC + MC > AM + AM

AB + AC + (BM + MC) > 2AM

AB + AC + BC > 2AM