Question

AM is a median of triangleABC. Prove that AB+BC+CA>2 AM.

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Answer 1

yes

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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