Question

4. AM is a median of A ABC. Prove that (AB + BC +CA) > 2AM.
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Answer 1

Answer:

Yes! AB + BC + CA > 2 AM

step by step explanation:

AM is a median. So, BM = 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 = MDSO, Eq (3) Becomes AB + AC + BC = 2AM

Hence proved