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

Step by Step Explanation:
1. Let $AD$ be the median to the third side of the triangle $ABC$.

We need to prove that $AB + AC > 2 AD.$
2. We know that the median from a vertex to the opposite side of a triangle bisects the opposite side.
Thus, we have $BD = DC$.
3. Let's extend $AD$ to $E$ such that $AD = DE$ and join the point $E$ to the point $C.$
4. In $\triangle ADB$ and $\triangle EDC,$ we have \begin{aligned} & \angle ADB = \angle EDC && [\text{Vertically opposite angles}] \\ & AD = DE && [\text{By construction}] \\ & BD = DC && [\text{AD is the median.}] \\ \therefore \space & \triangle ADB \cong \triangle EDC && [\text{By SAS criterion}] \end{aligned} As corresponding parts of congruent triangles are equal, we have$$AB = EC \space \space \ldots (1)$$
5. We know that the sum of any two sides of a triangle is greater than the third side.

So, in $\triangle AEC,$ we have \begin{aligned} & AC + EC > AE \\ \implies & AC + AB > AE && [\text{From (1)}] \\ \implies & AC + AB > 2AD && [\because \space \text{AE = 2AD}] \end{aligned}
6. Thus, the sum of any two sides of a triangle is greater than twice the median drawn to the third side. 