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Q.3 AM is a median of a \(\triangle \) ABC.
Is AB + BC + CA > 2 AM?
(Consider the sides of triangles \(\triangle \)ABM and \(\triangle \)AMC.)

Answer :

We know that,
The sum of the length of any two sides is always greater than the third side.
Now In \(\triangle \)ABM,
Here, AB + BM > AM … [equation i]
Then, In \(\triangle \)ACM
Here, AC + CM > AM … [equation ii]
By adding equation [i] and [ii] we get,
AB + BM + AC + CM > AM + AM
From the figure we have, BC = BM + CM
AB + BC + AC > 2 AM
Hence, the given expression is true.