Question

O is the point in exterior of triangle ABC show that 0A+OB+OC> 1/2 (AB+BC+CA).​

Answer 1

Answer:

I think that the question above has some missing part in it. So I am considering that we have to prove  

2[OA+OB+OC] > AB+BC+CA

Given that,

O is a point in the exterior of triangle ABC

To prove: 2 [OA + OB + OC] > AB + BC + CA

We know that the sum of any two sides of a triangle is greater than the third side. Therefore, from the figure attached below, let us consider :

1. In ∆ OAB,

OA + OB > AB …… (i)

2. In ∆ OCB,

OB + OC > BC ….. (ii)

3. In ∆ OAC,

OA + OC > CA …. (iii)

Now, adding equation (i), (ii) & (iii), we get

[OA + OB + OB + OC + OA + OC] > [AB + BC + CA]

Or, [(2*OA) + (2*OB) + (2*OC)] > [AB + BC + CA]

Or, 2 [OA + OB + OC] > [AB + BC + CA]

Hence proved  

Hope this helps!!!!!