Question

hello friends...

Answer this question....

Answer 1

Given that In an acute angled triangle ABC, S is a any point on BC.

We know that sum of angles in a triangle is always greater than the third side.

Now,

(1) In ΔABS

= > AB + BS > AS  ------ (1)

(2) In ΔACS

= > AC + CS > AS  ------- (2)

Now,

Adding (1) & (2), we get

= > AB + BS + AC + CS > AS + AS

= > AB + AC + (BS + CS) > 2AS [BC + CS = BC]

= > AB + AC + BC > 2AS.

Hope it helps!

Answer 2

hope helpful....
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