Question

Prove that the sum of the lengths of the three medians in a triangle is smaller than the perimeter of the triangle.

Answer 1

Answer:

Proven

Step-by-step explanation:

Let ABC be a triangle and D, E and F are midpoints of BC, CA and AB.

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)

Hence, we get

BC + AC > 2 (CF )

BC + AB > 2 (BE)

On adding the above inequalities, we get

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

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

AB + BC + AC > AD + BE + CF - Proven

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