



 
Hi Omkar. In short, prove it by contradiction: Hypothesis: Let P be the perimeter of a triangle ABC and let AB ≥ P/2. The shortest distance between any two points is a straight line, so to get from A to B is at least of distance AB. To return from B to C to A therefore, we know that BC + CA ≥ AB. But the perimeter of a triangle is the sum of the lengths of the sides, so if we add AB to each side of this latest inequality, we have: AB + CA + AB ≥ 2 AB, which reduces to P/2 ≥ AB. Thus P/2 ≥ AB ≥ P/2, which means that AB = P/2. So AB must be half the perimeter (it cannot be more). However, this means BC + CA = AB, so B, C and A are colinear (three points in a straight line). This is not what we call a triangle, since in a triangle 0° < each angle < 180°. This is a contraction, so the hypothesis is erroneous. Thus each side of a triangle must be less than half the triangle's perimeter. Stephen.  


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