# Prove thatsum of any two sides of a triangle is greater than twice the medians with respect to third side ?

Sep 22, 2017

see explanation

#### Explanation:

See Fig 1, Let $D$ be the midpoint of $B C$,
$\implies A D$ is a median of $\Delta A B C$
Now we need to prove $A B + A C > 2 A D$

See Fig2, extend $A D$ to $E$ such that $A D = D E$
$\implies \textcolor{red}{A E = 2 A D}$,
Draw lines $B E \mathmr{and} E C$, as shown in the figure.
$\implies A B E C$ is a parallelogram.
$\implies \textcolor{red}{B E = A C}$.
Consider $\Delta A B E$
Recall that the sum of any two sides of a triangle is greater than the length of the third side,
$\implies A B + B E > A E$
$\implies A B + A C > 2 A D$ ....... (proved)