Given that, O is any point in the interior of ΔABC
We have to prove:
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + OB + OC
(iii) OA + OB + OC > \(\frac{1}{2}\)(AB + BC + CA)
We know that,
In a triangle sum of any two sides is greater than the third side.
So, we have
In ΔABC
AB + BC > AC
BC + AC > AB
AC + AB > BC
In ΔOBC,
OB + OC > BC (i)
In ΔOAC,
OA + OC > AC (ii)
In ΔOAB,
OA + OB > AB (iii)
Now, extend BO to meet AC in D.
In ΔABD, we have
AB + AD > BD
AB + AD > BO + OD (iv) [Therefore, BD = BO + OD]
Similarly,
In ΔODC, 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) > \(\frac{1}{2}\)(AB+ BC + CA)
