Prove that the sum of the distances of any point,within a triangle,from its vertices is less than the perimeter of the triangle.

Asked by Sushant Kumar | 3rd Oct, 2013, 05:15: PM

Expert Answer:

We would request you to cross check the question posted. The correct question should be:
Prove that the sum of the distances of any point,within a triangle from its vertices is greater than the perimeter of the triangle. 
 
The proof is as follows:
 

The sum of any two sides of a triangle is always greater than the third side.

In OXY, OX + OY > XY ... (1)

In OYZ, OY + OZ > YZ ... (2)

In OXZ, OX + OZ > XZ ... (3)

 

Adding equations (1), (2) and (3):

(OX + OY) + (OY + OZ) + (OX + OZ) > XY + YZ + ZX

2(OX + OY + OZ) > XY + YZ + ZX

OX + OY + OZ > (XY + YZ + ZX)

OX + OY + OZ > Semi-perimeter of XYZ > perimeter of XYZ

Hence, (OX + OY + OZ) cannot be less than the perimeter of XYZ.


Answered by  | 3rd Oct, 2013, 10:26: PM

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