Let there be a Graph $C$ which is a cycle with $n+1$ vertices. Choose a random vertex $v1$ and add edges until $v1$ is connected with all the other vertices of the cycle. Show that the new Graph $G$ that emerges, has a connected tree with a diameter $k$ such as, $2 \le k \le n$.
(The diameter is the biggest distance of all the distances between the graph's vertices).
I can get why this is true theoretically: Because in the worst case where my tree (which contains no cycles) consists of all the pre-existing edges of $C$ except one ($E=(n+1)-1=n$) the biggest path would be these $n$ edges, but that's more of an intuitive solution rather than a proper mathematical proof.
Label the vertices $v_1,v_2,\ldots,v_{n+1}$ clockwise starting at the chosen vertex $v_1$. To get a tree with diameter $k$, use the edges $\{v_1,v_i\}$ for $i=2,\ldots,n+2-k$, the edges $\{v_i,v_{i+1}\}$ for $i=n+3-k,\ldots,n$ (if $k>2$), and the edge $\{v_{n+1},v_1\}$. (It helps to draw a few pictures for a reasonable value of $n$, say $n=5$.) In the resulting tree vertex $v_1$ will have degree $n+2-k$, vertices $v_i$ for $2\le i\le n+3-k$ will have degree $1$, and any remaining vertices will have degree $2$.