# Prove that the pe

Let ABC be the triangle where D, E and F are the mid-points of BC, CA and AB respectively As, we know that the sum of two sides of the triangle is greater than twice the median bisecting the third side

Similarly, BC + AC > 2CF

Also, BC + AB > 2BE

Now, by adding all these we get:

(AB + BC) + (BC + AC) + (BC + AB) > 2AD + 2CD + 2BE

2 (AB + BC + AC) > 2(AD + BE + CF)

AB + BC + AC > AD + BE + CF

Hence, the perimeter of the triangle is greater than the sum of its medians

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