Question

In a triangle ABC, O is an interior point. Prove that
2(OA+OB+OC) > AB+BC+CA​

Answer 1

Given :-

• A triangle ABC, in which O is the interior point. OA, OB, OC are joined.

To Prove :-

• OA + OB + OC > AB + BC + CA

Concept Used :-

• We know, In triangle, sum of any two sides of a triangle is greater than third side.

Proof :-

Consider,

$\rm :\longmapsto\:In \: \triangle \: OAB$

OA, AB and OB are three sides.

We know, that sum of any two sides of a triangle is greater than third side.

$\red{\boxed{\bf\implies \:OA + OB > AB}} - - - - (1)$

Consider

$\rm :\longmapsto\:In \: \triangle \: OBC$

OC, BC and OB are three sides.

We know, that sum of any two sides of a triangle is greater than third side.

$\red{\boxed{\bf\implies \:OC + OB > BC}} - - - - (2)$

Consider

$\rm :\longmapsto\:In \: \triangle \: OAC$

OA, AC and OC are three sides.

We know, that sum of any two sides of a triangle is greater than third side.

$\red{\boxed{\bf\implies \:OC + OA > CA}} - - - - (3)$

On adding equation (1), (2) and (3), we get

$\rm :\longmapsto\:OA + OB + OB + OC + OC + OA > AB + BC + CA$

$\sf \:\implies\boxed{ \bf \: 2(OA + OB + OC) > AB + BC + CA}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

1. Angle opposite to longest side is always greater.

2. Side opposite to greater angle is always longest.

3. Angle opposite to smaller side is always smallest.

4. Side opposite to smallest angle is always smaller.

5. Side opposite to equal angles are equal.

6. Angle opposite to equal sides are always equal.