Question

Prove that perimeter of a triangle is greater than sum of its three medians.

Answer 1

In the triangle $ABC$, $D,E$ and $F$ are the midpoints of sides $BC,CA$ and $AB$ respectively.

$\text{We know that the sum of two sides of a triangle is greater than twice the median bisecting the third side,}$

Hence in triangle ABD, AD is a median

$\Rightarrow AB+AC>2\left(AD\right)$

Similarly, we get

$\Rightarrow BC+AC>2\left(CF\right)$

$\Rightarrow BC+AB>2\left(BE\right)$

On adding the above inequations, we get

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

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

$\therefore AB+BC+AC>AD+BE+CF$

Then perimeter of a triangle is greater than sum of its three medians