Question

If 0 is any point in the interior angle of triangle ABC,show that AB+BC+CA greater than 2(OA+OB+OC)​

Answer 1

Answer:

sorry I can't answer it now.

I am not at my home. if I would be at home than I have surely given you the answer by writing in my copy.

Answer 2

Answer:

ABC is divided into 3 triangles AOB ,BOC and

AOC

In AOB , By inequality property of triangle ,

OA +OB > AB -------(I)

In BOC , By inequality property of triangle ,

OB +OC > BC -------(II)

In AOC , By inequality property of triangle ,

OC +OA > AC -------(III)

On Adding Eq (i) (ii) and (iii)

OA+OB+OB+OC+OC+OA > AB+BC+AC

2OA+2OB +20C > AB+BC+AC2(OA+OB+OC)> AB+BC+AC

 AB+BC+AC < 2(OA+OB+OC)

Hence Proved