AM is the median of a triangle ABC is AB+BC+CA>2AM
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Yes! AB + BC + CA > 2 AM
AM is a median. So, BM = CM
CONSTRUCTION: Extend AM to D, such that AM= MD
=> ABDC is a parallelogram ( as diagonals are bisecting
each other)
Since, AB + BM > AM……………..(1)
( The sum of 2 sides of a triangle > third side)
& BD + BM > MD ………….(2)
( The sum of 2 sides of a triangle >third side)
Now, by adding (1) & (2)
We get, AB + BD + 2 BM > AM + MD ………(3)
But BD = AC ( opposite sides of parallelogram)
& 2BM = BC
& AM = MD
SO, Eq (3) becomes
AB + AC + BC = 2AM
(Hence proved)
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