Question

If 0 is an interior point of ΔABC, then prove that

AB+BC+CA < 2 (OA+OB+OC)​

Answer 1

ANSWER:-

In triangle ABC, O is a point interior of ∆ABC. As we know that “The sum of any two sides of a triangle is greater than the third side”. OA + OB > AB …(i) OA + OC > AC …(ii) and OB + OC > BC …(iii) Now, adding (i), (ii) and (iii), we get 2(OA + OB + OC) > AB + BC + CA or AB + BC + CA < 2(OA + OB + OC) Hence proved.Read more on Sarthaks.com - https://www.sarthaks.com/781228/in-figure-o-is-an-interior-point-of-abc-show-that-ab-bc-ca-2-oa-ob-oc

Answer 2

Answer:

InΔPBO

BP+PO>OB

AddingOCanbothsides

weget,

BP+PO+OC>OB+OC

BP+PC>OB+OC

BP+PC>OB+OC→(i)[PO+OC=PC]

InΔAPC

AP+AC>PC

AddingBPonbothsides,weget

AP+AC+BP>PC+BP→(ii)[AP+PB=AB]

from(i)&(ii)

AB+AC>BP+PC>OB+OC

AB+AC>OB+OC→(iii)

similarly

BC+AC>OB+OA→(iv)

AB+BC>OA+OC→(v)

addingequation(3)(4)(5)weget

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

AB+BC+CA>OA+OB+OC