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