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Question -

AM is a median of a triangle ABC.
Is AB + BC + CA > 2AM ?
(Consider the sides of triangles ∆ABM and ∆AMC)



Answer -

Yes, In AABM, we have
 
AB + BM > AM …(i)
[Sum of any two sides of a triangle is greater than the third side]
In ∆AMC, we have
AC + CM > AM …(ii)
[Sum of any two sides of a triangle is greater than the third side]
Adding eq (i) and (ii), we have
AB + AC + BM + CM > 2AM
AB + AC + BC + > 2AM
AB + BC + CA > 2AM
Hence, proved.

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