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Question

O is any point in the interior of ΔABC.

Prove that

(i) AB+AC>OB+OC

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

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

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Solution

Given : In ΔABC, O is any point in the interior of the ΔABC, 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, in ΔODC,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+OCand CA+BC>OA+OB(ii) In ΔOAB, ΔOBC and ΔOCA,OA+OB>ABOB+OC>BCand OC+OA>CAAdding, we get2(OA+OB+OC)>AB+BC+CA OA+OB+OC>12(AB+BC+CA)


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