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 + QB + OC
(iii) OA + OB + OC >1/2 x (AB + BC + CA)
We know that, in a triangle the sum of any two sides is greater than the third side
So, we have
Now, extend (or) produce BO to meet AC in D.
Now, in ΔABD,we have
Similarly in ΔODC, we have
OD+DC>OC ....(5)
(i) Adding (4) and (5), we get
(ii) Adding equation (6), (7) and (8), we get
(iii) Adding equations (1), (2) and (3)