22) In the given figure, O is a point in the exterior of
a triangle ABC. O is joined to the three vertices
A, B and C. Show that OA + OB + OC >1\2(AB+BC+CA)
Answers
Given -
O is a point in the exterior of ABC.
To prove -
OA + OB + OC > ½ (AB + BC + CA)
Construction -
Join OA, OB and OC
Property Used -
Sum of two sides in a triangle is greater than the third side .
Proof -
Using the above mentioned property - .
AO + OC > AC
AO + OB > AB
OB + OC > BC
Adding these three Inequalities -
2AO + 2BO + 2CO > AB + BC + AC
2 ( OA + OB + OC ) > ( AB + BC + AC )
( OA + OB + OC ) > ½ ( AB + BC + AC )
Hence proved
___________
Additional information -
- 1. Sum of two sides of a triangle is always greater than the third side .
- 2. Difference of two sides of a triangle is always less than the third side .
__________________
Answer:
Given data:
O is a point in the exterior of triangle ABC
To prove: 2 [OA + OB + OC] > AB + BC + CA
We know that the sum of any two sides of a triangle is greater than the third side. Therefore, from the figure attached below, let us consider:
1. In ∆ OAB,
(ii) OA + OB > AB ....... (i)
2. In ∆ OCB,
(i) OB + OC > BC....... (ii)
3. In ∆ OAC,
OA + OC > CA ........ (iii)
Now, adding equation (i), (ii) & (iii), we get
[OA + OB + OB + OC + OA + OC] >
[AB + BC + CA]
Or, [(2*OA) + (2*OB) + (2*OC)] > [AB + BC + CA]
Or, 2 [OA + OB + OC] > [AB + BC + CA]
Hence proved