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

(i) AB + AC > OB + OC


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


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


Given that, O is any point in the interior of


We have to prove:


(i) AB + AC > OB + OC


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


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


We know that,


In a triangle sum of any two sides is greater than the third side.


So, we have


In


AB + BC > AC


BC + AC > AB


AC + AB > BC


In ,


OB + OC > BC (i)


In ,


OA + OC > AC (ii)


In ,


OA + OB > AB (iii)


Now, extend BO to meet AC in D.


In , we have


AB + AD > BD


AB + AD > BO + OD (iv) [Therefore, BD = BO + OD]


Similarly,


In , we have


OD + DC > OC (v)


(i) Adding (iv) and (v), we get


AB + AD + OD + DC > BO + OD + OC


AB + (AD + DC) > OB + OC


AB + AC > OB + OC (vi)


Similarly, we have


BC + BC > OA + OC (vii)


And,


CA + CB > OA + OB (viii)


(ii) Adding (vi), (vii) and (viii), we get


AB + AC + BC + BA + CA + CB > OB + OC + OA + OC + OA + OB


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


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


AB + BC + CA > OA + OB + OC


(iii) Adding (i), (ii) and (iii), we get


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


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


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


Therefore, (OA + OB + OC) > (AB+ BC + CA)


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