question_answer

AM is a median of a \[\Delta \]ABC. Is AB + BC + CA > 2 AM? (Consider the sides of \[\Delta \]ABM and \[\Delta \]AMC.)

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]