Q. 2 B5.0( 2 Votes )

O is any point in

Answer :

From the above proof, we have

AB + AC > OB + OC …….. (I)


Let us modify the above figure a bit by connecting OA



Now from Δ ABC


AC + BC > AB ……….. (i)


Similarly from Δ OAB


OA + OB > AB ………. (ii)


Now let us subtract equation (ii) from equation (i)


AC + BC – (OA + OB) > AB – AB


AC + BC – (OA + OB) > 0


AC + BC > OA + OB ………….. (II)


Now let us again consider Δ ABC,


AB + BC > AC ………….. (iii)


Similarly from Δ OAC


OA + OC > AC …………. (iv)


Now let us subtract equation (iv) from equation (iii)


AB + BC – (OA + OC) > AC – AC


AB + BC – (OA + OC) > 0


AB + BC > OA + OC ……. (III)


Now let us add equations (I), (II) and (III)


AB + BC + AC + BC + AB + AC > OB + OA + OA + OC + OC + OB


2AB + 2BC + 2AC > 2OB + 2OC + 2OA


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


(AB + BC + AC) > (OB + OC + OA)


Hence Proved.


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