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) Given: $\triangle ABC$ with median $AD$ to side $BC$.
To prove: $AB + AC > 2AD$.
Construction: Produce $AD$ to $E$ such that $DE = AD$ and join $EC$.
Proof: In $\triangle ADB$ and $\triangle EDC$:
$AD = ED$ (By construction)
$\angle 1 = \angle 2$ (Vertically opposite angles)
$DB = DC$ ($AD$ is the median)
By $SAS$ congruence criterion,$\triangle ADB \cong \triangle EDC$.
Therefore,$AB = EC$ $(CPCT)$.
Now,in $\triangle AEC$,the sum of any two sides is greater than the third side:
$AC + CE > AE$
Since $AE = AD + DE = AD + AD = 2AD$ and $CE = AB$:
$AC + AB > 2AD$.
Hence,it is proved that the sum of two sides is greater than twice the median to the third side.