Question

prove that the sum of sides of traingle is greater than sum of 3 medians​

Answer 1

$\mathfrak{\huge{\blue{\underline{\underline{Answer:-}}}}}$

In the triangle ABC, D,E and F are the midpoints of

sides BC,CA and AB respectively.

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

⇒$AB+AC>2(AD)$

$\mathfrak{\huge{\blue{\underline{\underline{similarly:-}}}}}$

⇒$BC+AC>2(CF)$

⇒$BC+AB>2(BE)$

$\mathfrak{\huge{\blue{\underline{\underline{On adding:-}}}}}$

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

⇒$2(AB+BC+AC)>2(AD+BE+CF)$

⇒∴$AB+BC+AC>AD+BE+CF$

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