In $\Delta ABC$,$AD$ is a median. Prove that $AB + AC > 2AD$.

(N/A) $1$. Extend $AD$ to a point $E$ such that $AD = DE$. Join $EC$.
$2$. In $\Delta ADB$ and $\Delta EDC$:
- $AD = DE$ (by construction)
- $\angle ADB = \angle EDC$ (vertically opposite angles)
- $BD = DC$ ($AD$ is the median to $BC$)
$3$. By $SAS$ congruence criterion,$\Delta ADB \cong \Delta EDC$.
$4$. Therefore,$AB = EC$ (by $CPCT$).
$5$. In $\Delta AEC$,by the triangle inequality theorem,the sum of any two sides is greater than the third side:
- $AC + EC > AE$
$6$. Since $AE = AD + DE = AD + AD = 2AD$ and $EC = AB$,we substitute these into the inequality:
- $AC + AB > 2AD$.
$7$. Thus,$AB + AC > 2AD$ is proved.