How to solve this question and give proof
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0
Answer:
b+ cAz-y
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I am not fully sure heheh
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3
Answer:
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 △ABM
AB+BM>AM.....(i)
In △AMC
AC+MC>AM.....(2)
Adding eqn(1)&(2), we have
(AB+BM)+(AC+MC)>AM+AM
⇒AB+(BM+MC)+AC>2AM
⇒AB+BC+AC>2AB
Hence AB+BC+AC>2AB
Step-by-step explanation:
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