Question

A point O is in the interior of triangle ABC. show that AB+BC+AC>2(OA+OB+OC)​

Answer 1

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Answer 2

Solution

In ∆ OBC

1) OB + OC > BC

Sum of 2 sides of triangle is greater than the 3rd side

Similarly , In ∆ OAC and ∆ OAB

2) OA + OC > AC

3) OA+ OB > AB

Adding 1, 2 and 3

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

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

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

Hence proved