AM is a median of ABC. Prove that (AB + BC +CA) > 2AM. Hint (ABBM) AM On ABM CAC +MC) > AM in L. ACM) Now, add the two inequalities,
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Answered by
0
We know,
In a triangle sum of two sides is greater than the third side
Hence,In triangle ABM
AB + BM > AM ____1
Also in triangle ACM
AC + MC > AM ____2
Adding equation 1 and 2 we get,
AB + BM + AC + MC > AM + AM
AB + AC + (BM + MC) > 2AM
AB + AC + BC > 2AM
Answered by
4
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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