Question

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 > 1/2 (AB + BC + CA)

Answer 1

Hence it is proved that,

(i) AB + AC > OB + OC

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

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

As, we know that, "The sum of two sides of a triangle is greater than the third side."

We have three equations,

OA + OB > AB

OB + OC > BC

OA + OC > AC

From Δ ABD,

AB + AD > BD ........(1)

AB + AD > BO + OD

In ΔCOD,

CD + OD > OC ........(2)

adding equations (1) and (2), we get,

AB + AD + CD + OD > BO + OD + OC

upon cancelling the same terms from both sides, we have

AB + (AD+ CD) > BO + OC

AB + AC > BO + OC ...........(i)

using the above three equations and result of (i), we get,

AB + AC > OB + OC

AB + BC > OA + OC

AC + BC > OA + OB

adding these equations, we get,

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

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

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

adding all the above three equations, we get,

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

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

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

Answer 2

Answer:

ABC is divided into 3 triangles AOB ,BOC and

AOC

In AOB , By inequality property of triangle ,

OA +OB > AB -------(I)

In BOC , By inequality property of triangle ,

OB +OC > BC -------(II)

In AOC , By inequality property of triangle ,

OC +OA > AC -------(III)

On Adding Eq (i) (ii) and (iii)

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

2OA+2OB +20C > AB+BC+AC2(OA+OB+OC)> AB+BC+AC

 AB+BC+AC < 2(OA+OB+OC)

Hence Proved