Question

Am is a median ofAbc then AB+BC+CA is greater than 2AM consider the sides of ABM​

Answer 1

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 eq  

n

(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:

Answer 2

Answer:

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