Question

. O is any point in the interior of a triangle ABC. Prove that

1. AB + AC > OB + OC

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

3.OA + OB + OC > 1/2(AB + AC + BC)

Answer 1

Given : In ΔABC,O is any point in the interior of theΔABC,OA, OB and OC are joined

To Prove :

(i)AB+AC>OB+OC

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

(iii)OA+OB+OC>12(AB+BC+CA)

Construction : Produce BO to meet AC in D.

Proof : In ΔABD,

(i) AB+AD>BD

(Sum of any two sides of a triangle is greater than thrid)

⇒AB+AD>BO+OD...(i)Similarly,inΔODC,OD+DC>OC...(ii)Adding (i) and (ii)AB+AD+OD+DC>OB+OD+OC⇒AB+AD+DC>OB+OC⇒AB+AC>OB+OC



(ii)~Similarly, we can prove thatBC+AB+OA+OCandCA+BC>OA+OB(ii)InΔOAB,ΔOBCandΔOCA,OA+OB>ABOB+OC>BCandOC+OA>CAAdding, we get2(OA+OB+OC)>AB+BC+CA∴OA+OB+OC>12(AB+BC+CA)