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Vivian Allen

AM is a median of A ABC. Prove that (AB + BC + CA) > 2AM.Now, add the two inequalities.(In ∆ ABM)(In ∆ACM)Hint. (AB + BM) > AM(AC +MC) > AM​

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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.
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Lemelin Emmanuel
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