in triangle ABC , AM is the median ,is AB +BC+ CA greater than 2 AM justify
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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 same reason)
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
[ proved]
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