Page 8 - ln1

We know that AB + AC > BC ….. (1)
Consider △ OBC
We know that OB + OC > BC ….. (2)
By subtracting both the equations we get
(AB + AC) – (OB + OC) > BC - BC
So we get
(AB + AC) – (OB + OC) > 0
AB + AC > OB + OC
Therefore, it is proved that AB + AC > OB + OC.

(ii) We know that AB + AC > OB + OC
In the same way we can write
AB + BC > OA + OC and AC + BC > OA + OB
By adding all the equations we get
AB + AC + AB + BC + AC + BC > OB + OC + OA + OC + OA + OB
So we get2 (AB + BC + AC) > 2 (OA + OB + OC)
Dividing by 2 both sides
AB + BC + AC > OA + OB + OC

(iii) Consider △ OAB
We know that OA + OB > AB ….. (1)
Consider △ OBC
We know that OB + OC > BC ….. (2)
Consider △ OCA
OC + OA > CA …… (3)
By adding all the equations
OA + OB + OB + OC + OC + OA > AB + BC + CA
So we get
2 (OA + OB + OC) > AB + BC + CA
Dividing by 2
OA + OB + OC > ½ (AB + BC + CA)
Therefore, it is proved that OA + OB + OC > ½ (AB + BC + CA).

Question 6.In the given figure, D is a point on side BC of a △ABC and E is a point
such that CD = DE. Prove that AB + AC > BE.

Solution:
Consider △ ABC
We know that
AB + AC > BC