Question

O is any point in the interior of ΔABC. Prove that
(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + QB + OC 
(iii) OA + OB + OC >` 1/2`(AB + BC + CA) 

Answer 1

Given that O is any point in the interior of ABC
We have to prove  

(i) AB + AC > OB + OC
(ii) AB + BC + CA > OA + QB + OC
(iii) OA + OB + OC >`1/2`(AB + BC + CA) 

We know that, in a triangle the 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           ...........(1) 

In ΔOAC 

OA+OC>AC           ...........(2) 

In    ΔOAB 

OA+OB>AB         ............(3) 

Now, extend (or) produce BO to meet AC in D.
Now, in ΔABD,we have 

AB+AD+BD 

⇒ AB+AD>BO+OD                .............(4) [∵ BD=BO+OD]  

Similarly in , ΔODC we have  

OD+DC>OC                              ...............(5) 

(1) Adding (4) and (5), we get 

AB+AD+OD+DC>BO+OD+OC 

⇒AB+(AD+DC)>OB+OC 

⇒ AB+AC>OB+OC                 ...............(6) 

Similarly, we have 

BC+BA>OA+OC                ...............(7) 

and CA+CB>OA+OB             ..............(8) 

(2) Adding equation (6), (7) and (8), we get 

AB+AC+BC+BA+CA+CB>OB+OC+OA+OC+OA+OB 

⇒2AB+2BC+CA>2OA+2OB+2OC 

⇒2(AB+BC+CA>OA+OB+OC) 

⇒AB+BC+CA>OA+OB+OC 

(3) Adding equations (1), (2) and (3) 

OB+OC+OA+OC+OA+OB+>BC+AC+AB 

⇒2OA+2OB+2OC>AB+BC+CA 

We get⇒ 2(OA+OB+OC)>AB+BC+CA 

∴ (OA+OB+OC)>`1/2`(AB+BC+CA)