Let’s prove that the perimeter of a triangle is more than the sum of length of the medians.

Median is any line in a triangle which divides the given triangle segment in two equal parts.

Let us first consider a median from point A dividing the line BC.

D is any point on the segment BC

We know one basic fact about triangles that the sum of any two sides of a triangle is always greater than twice the median which is dividing the third side in two equal halves.

AB + AC > 2 × AM ……………. (I)

Now let us consider median from point B dividing the segment AC.

Similarly we can write here as follows:

AB + BC > 2 × BN …………… (II)

Now let us consider median from point C dividing the segment AB.

AC + BC > 2 × OC …………….. (III)

Now adding equations (I), (II) and (III)

AB + AC + BC + AB + AC + BC > (2AM + 2BN + 2OC)

2AB + 2BC + 2AC > 2 × (AM + BN + OC)

2 × (AB + AC + BC) > 2 × (AM + BN + OC)

AB + AC + BC > AM + BN + OC

Hence it is proved that the perimeter of a triangle is more than the sum of length of the medians.