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Let ABC be the triangle and D. E and F are midpoints of BC, CA and AB respectively. Recall that the sum of two sides of a triangle is greater than twice the median bisecting the third side, Hence in ΔABD, AD is a median ⇒ AB + AC > 2AD Similarly, we get BC + AC > 2CF BC + AB > 2BE On adding the above inequations, we get (AB + AC) + (BC + AC) + (BC + AB )> 2AD + 2CD + 2BE 2(AB + BC + AC) > 2(AD + BE + CF) ∴ AB + BC + AC > AD + BE + CF Thus the perimeter of a triangle is greater than the sum of the medians

Hence in ΔABD, AD is a median ⇒ AB + AC > 2AD BC + AC > 2CF BC + AB > 2BE (AB + AC) + (BC + AC) + (BC + AB )> 2AD + 2CD + 2BE 2(AB + BC + AC) > 2(AD + BE + CF) AB + BC + AC > AD + BE + CF