Question

If o is a point in the interior of triangle ABC, Prove that OA+OB+OC>1/2 (AB+BC+CA)?

Answer 1

Given: O is a point within ∆ABC
To prove:
(i) AB+AC>OB+OC
(ii) AB+BC+CA>OA+OB+OC
(iii) OA+OB+OC>(AB+BC+CA)
Proof:

In ∆ABC,
AB +AC >BC ….(1)
And in ∆OBC,
OB + OC > BC …(2)
Subtracting 1 from 2 we get,
(AB + AC) – (OB + OC ) > (BC – BC )
Ie AB + AC > OB + OC
From ׀, AB + AC > OB + OC
Similarly, AB + BC > OA + OC
And AC + BC > OA + OB
Adding both sides of these three inequalities, we get,
(AB + AC ) + (AB + BC) + (AC + BC) > (OB + OC) + (OA + OC) + (OA + OB)
Ie. 2(AB + BC + AC ) > 2(OA + OB + OC)
∴ AB + BC + OA > OA + OB + OC
In ∆OAB,
OA + OB > AB …(1)
In ∆OBC,
OB + OC > BC …(2)
In ∆OCA
OC + OA > CA …(3)
Adding 1,2 and 3,
(OA + OB) + (OB + OC) + (OC+ OA) >AB + BC +CA
Ie. 2(OA + OB + OC) > AB + BC + CA
∴ OA + OB + OC >  ( AB + BC + CA)

Answer 2

In triangle OAB,

OA+OB>AB - - - - (i) [Sum of two sides of a triangle is greater than the third side]

Similarly,
OB+OC>BC - - - - (ii)
And, OA+OC>AC - - - - (iii)

Now,
Adding equations (i), (ii) & (iii)
OA+OB+OB+OC+OA+OC>AB+BC+CA
2(OA+OB+OC)>AB+BC+CA
OA+OB+OC>1/2 (AB+BC+CA)

Hence, Proved

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