Answer:
Yes, given AM is a median of ΔABC. So, M is the mid-point of BC and AM divides ΔABC into two triangles, ΔABC and ΔAMC. We know that, the sum of the lengths of any two sides of a triangle is greater than the length of third side. In ΔABM, AB + BM > AM ...(i) In ΔABC, CA + MC > AM ...(ii) On a adding Eqs. (i) and (ii), we get AB + BM + CA + MC > 2AM AB + (BM + MC) + CA > 2AM AB + BC + CA > 2AM [from the figure, BM + MC = BC]
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