Prove that the sum of any two sides of a triangle is greater than twice the median with respect to the third side.

(N/A) Let $ABC$ be a triangle with sides $AB$,$BC$,and $AC$. Let $AD$ be the median to the side $BC$,such that $BD = DC = \frac{1}{2} BC$.
Extend $AD$ to a point $E$ such that $AD = DE$. Join $CE$.
In $\triangle ABD$ and $\triangle ECD$:
$1$. $AD = DE$ (by construction)
$2$. $\angle ADB = \angle EDC$ (vertically opposite angles)
$3$. $BD = DC$ (since $AD$ is the median)
Thus,$\triangle ABD \cong \triangle ECD$ by $SAS$ congruence criterion.
Therefore,$AB = EC$ (by $CPCT$).
In $\triangle ACE$,by the triangle inequality theorem,the sum of any two sides is greater than the third side:
$AC + CE > AE$
Since $AE = AD + DE = 2AD$ and $CE = AB$,we substitute these into the inequality:
$AC + AB > 2AD$
Hence,the sum of two sides is greater than twice the median to the third side.