If O is a point in the exterior of ∆ABC, show that 2(OA + OB + OC) > (AB + BC + CA).

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Solution

We know that the sum of any two sides of a triangle is greater than the third side.

In $△$ AOB:
OA + OB > AB...........(1)

In $△$ BOC:
OB + OC > BC........................... (2)

In $△$ AOC:
OA + OC > CA.............................(3)

Adding (1), (2) and (3):

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

2( OA + OB + OC) > AB +BC + CA

Hence, proved.

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If O is a point in the exterior of ∆ABC, show that 2(OA + OB + OC) > (AB + BC + CA). Figure

If O is a point in the exterior of ∆ABC, show that 2(OA + OB + OC) > (AB + BC + CA).

Figure