Question

AM is a median of a ∆ ABC.
Show that AB + BC + AC > 2AM

Answer 1

Answer:

The median AM divides ∆ABC into two triangles , namely - ∆ABM and ∆AMC.

Now,

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In ∆ABM , we have,

AB + BM > AM [ Sum of any teri sides of a triangle is always greater than the third side] .....(i)

Similarly,

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In ∆ AMC , we gave,

MC + AC > AM [ Sum of any teri sides of a triangle is always greater than the third side] .....(ii)

Now, on adding equation (i) and (ii) , we get,

AB + BM + MC + AC > AM + AM

=> AB + (BM + MC) + AC > 2AM

=> AB + BC + AC > 2AM [Because , BM + MC = BC]

Answer 2

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

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