Question

Proof that the perimeter of a triangle is greater than sum of its three medians

Answer 1

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 > 2(AD) 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 Hence, we can say that the perimeter of a triangle is greater than the sum of the medians.

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Answer 2

This question is already solved in Brainly.