Turan graphs and degree 
G is a triangle-free graph with n vertices. I want to show the minimum
  degree of G is less than half the number of vertices.

 A: Let $n=2k$. Then $T_2(n)$ is just the complete bipartite graph $K_{k,k}$. 
Let $G=(V,E)$ be a triangle-free graph on $n$ vertices with $\delta(G)=k$. Let $v\in V$ have degree $\delta(G)$. Let $S$ be the neighbours of $v$, and let $T=V\setminus(S\cup v)$. So $|S|=k$ and $|T|=k-1$
Since $G$ is triangle-free, no two vertices in $S$ share an edge, and so the only way for a vertex $s\in S$ to have degree at least $k$ is if $s$ is adjacent to every vertex in $T$.
But now you have $T_2(n)$ as a spanning subgraph of $G$ (one side $S$, the other $T\cup v$), and of course we cannot add any more edges without creating a triangle. Therefore, $G$ must be $T_2(n)$.
A: Hint: You could prove this with induction to the number of points of your graph.
In the inductive step you leave one line out and all the lines connected to that one, apply the induction hypothesis to that new graph and then put all the lines back in.
Then you use the fact there are no triangles: there cannot be points which are connected to both endpoints of the line you deleted, so this gives an upper bound for the number of lines you deleted.
Now you can count the number of lines in your original graph: it is equal to the number of deleted edges + the number of edges in your new graph.
