AM is a median of a triangle ABC. Is AB + BC + CA > 2 AM? (Consider the sides of triangles ∆ABM and ∆AMC.)
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Step-by-step explanation:
As we know that the sum of lengths of any two sides in a triangle should be greater than the length of third side.
Therefore,In △ABMAB+BM>AM.....(i)
In △AMCAC+MC>AM.....(2)
Adding eqn (1)&(2), we have(AB+BM)+(AC+MC)>AM+AM
⇒AB+(BM+MC)+AC>2AM
⇒AB+BC+AC>2AB
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