$AM$ is a median of a triangle $ABC$.
![](/assets/questions/media/148618-63829-1656575010.png)
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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