Question

If O is the point in the exterior of AABC, show that 2 (OA + OB + OC) > AB + BC + CA [Hint : Join OA, OB, OC)​

Answer 1

Given :-

O is the point in the exterior of triangle ABC.

To Prove :-

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

Construction :-

Join OA, OB, OC

Proof :-

In triangle AOB

We know, Sum of two sides of a triangle is greater than the third side.

So,

$\rm\implies \:OA+OB > AB - - - (1)$

Now, In triangle BOC

Again, Sum of two sides of a triangle is greater than third side.

So,

$\rm\implies \:OB+OC>BC - - - - (2)$

Now, In triangle AOC

Again, Sum of two sides of a triangle is greater than third side.

So,

$\rm\implies \:OC+OA>AC - - - - (3)$

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

$\rm \: OA+OB + OB+OC + OC+OA>AB + BC + AC$

$\rm \: (OA + OA)+(OB + OB)+(OC + OC)>AB + BC + AC$

$\rm \: 2OA+2OB+2OC>AB + BC + AC$

$\rm\implies \:\rm \: 2(OA+OB+OC>AB + BC + AC$

Hence, Proved

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ADDITIONAL INFORMATION

1. The sum of all interior angles of a triangle is supplementary.

2. The sum of all exterior angles of a triangle is 360°.

3. Exterior angle of a triangle is equals to sum of interior opposite angles.

4. Angle opposite to longest side is always greater.

5. Side opposite to greatest angle is always longest.

Answer 2

${\pmb{\frak{\underline{\red{Given:-}}}}}$

• If O is the point in the exterior of AABC, show that 2 (OA + OB + OC) > AB + BC + CA [Hint : Join OA, OB, OC)

${\pmb{\frak{\underline{\red{Find:-}}}}}$

• 2 (OA + OB + OC) > AB + BC + CA

${\pmb{\frak{\underline{\red{Solution:-}}}}}$

$\rm \longmapsto{∆AOC,AO + OB = AB ------ ( 1 )}$

$\rm\longmapsto{∆BOC,BO + OC = BC ----( 2 )}$

$\rm\longmapsto{∆AOC,AO + OC = AC ----( 3 )}$

Adding the 1,2 and 3 :-

$\rm\longmapsto{AO+OB+OB+DC+AO+OC = > AB + BC+AC}$

$\rm\longmapsto{2 AO + 2 OB +2O = > AB + BC+AC}$

$\rm\longmapsto{2 ( AO +OB+OC) = > AB + BC+AC}$

$\: \: \: \: \: \: \: \: \: \: {\pmb{\frak {\dag\red{Proved}}}}$

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@Shivam

#BeBrainly