Vivian Allen

answers: 1

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Answer:

Given that,

AM is the median of ∆ABC.

In other words, AM divides ABC into two triangles ABM and ACM.

In any triangle, the sum of two sides is greater than the third side.

In ∆ABM,

AB + BM > AM (1)

In ∆ACM,

AC + CM > AM (2)

Adding equations (1) and (2),

AB + (BM + CM) + AC > AM + AM

AB + BC + AC > 2AM

Hence, proved.
500

Lemelin Emmanuel

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