0 is the point in the interior of a triangle ABC prove that OA+ OB+ OC >1/2 of triangle ABC
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Taking ∆OBC
1) OB+OC>BC ………Sum of the two sides of a triangle is greater than the third side
In ∆OAC
2) OA+ OC> AC ……reason same as above.
In ∆OAB
3 ) OA +OB >AB ……..reason same as above
Adding 1) 2) & 3)
OB+OC +OA +OC+OA+OB > BC +AC+ AB
2OA+2OB + 2OC >AB+ BC+ AC
Dividing both sides by 2
(OA + OB +OC)>1÷2(AB +BC+AC ) proved.
1) OB+OC>BC ………Sum of the two sides of a triangle is greater than the third side
In ∆OAC
2) OA+ OC> AC ……reason same as above.
In ∆OAB
3 ) OA +OB >AB ……..reason same as above
Adding 1) 2) & 3)
OB+OC +OA +OC+OA+OB > BC +AC+ AB
2OA+2OB + 2OC >AB+ BC+ AC
Dividing both sides by 2
(OA + OB +OC)>1÷2(AB +BC+AC ) proved.
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