Question

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

Answer 1

Let $ABC$ be the triangle and $D,E$ and $F$ are midpoint of $BC$, $CA$ and $AB$ respectively.

Have to call that the sum of two sides of a triangle is greater than twice the median bisecting the third side.

Hence in $\Delta ABD$ $AD$ is median

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

Similarly, we get

$BC+AC>2CF$

$BC+AB>2BE$

On adding the above in equations, we get

$(AB+AC)+(BC+AC)+(BC+AB)>2AD+2CD+2DE$

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

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