Question

if o is a point within a triangle ABC, show that AB +BC +CA > OA +OB +OC

Answer 1

Given: O is a point within ∆ABC
To prove:
(i) AB+AC>OB+OC
(ii) AB+BC+CA>OA+OB+OC
(iii) OA+OB+OC>(AB+BC+CA)
Proof:

In ∆ABC,
AB +AC >BC ….(1)
And in ∆OBC,
OB + OC > BC …(2)
Subtracting 1 from 2 we get,
(AB + AC) – (OB + OC ) > (BC – BC )
Ie AB + AC > OB + OC
From ׀, AB + AC > OB + OC
Similarly, AB + BC > OA + OC
And AC + BC > OA + OB
Adding both sides of these three inequalities, we get,
(AB + AC ) + (AB + BC) + (AC + BC) > (OB + OC) + (OA + OC) + (OA + OB)
Ie. 2(AB + BC + AC ) > 2(OA + OB + OC)
∴ AB + BC + OA > OA + OB + OC
In ∆OAB,
OA + OB > AB …(1)
In ∆OBC,
OB + OC > BC …(2)
In ∆OCA
OC + OA > CA …(3)
Adding 1,2 and 3,
(OA + OB) + (OB + OC) + (OC+ OA) >AB + BC +CA
Ie. 2(OA + OB + OC) > AB + BC + CA
∴ OA + OB + OC >  ( AB + BC + CA)

Answer 2

If O is a point within ∆ ABC, show that

i) AB + AC > OB +OC

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

iii) OA + OB + OC > 1/2 (AB+BC+CA).

Hope it helps you.