Is $AB + BC + CA > 2 AM?$ [Consider the sides of triangles $∆ABM$ and $∆AMC$.]" ">

$AM$ is a median of a triangle $ABC$.

Is $AB + BC + CA > 2 AM?$ [Consider the sides of triangles $∆ABM$ and $∆AMC$.]"


Given: $AM$ is a median of a triangle $ABC$.

To do: To find whether $AB + BC + CA > 2 AM?$

Solution:

Let us consider $\Delta ABM$ and $\Delta AMC$

It is a known fact that the sum of the triangle of any two sides in a triangle should be greater than the length of the third side.

In $\Delta ABM$:

$AB+BM>AM$         ......$(i)$

In $\Delta AMC$:

$AC+MC>AM$       ......$(ii)$

Let us add $(i)$ and $(ii)$

$AB+BM+AC+MC>AM+AM$

$\Rightarrow AB+AC+(BM+MC)>2AM$

$\Rightarrow AB+AC+BC>2AM$

Hence proved!

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