Prove that the line graph of a Hamiltonian simple graph is Hamiltonian.
My proof
If $G$ is hamiltonian then there is a cycle that traverse all the vertices of $G$ exactly once. Any other edges of $G$ that's not part of this cycle can become chords of $G$ or placed outside the cycle.
How do I proceed from here?How do I use the definition of Hamiltonian to prove that the line graph of $G$ is also Hamiltonian?
Let $C_H = v_1v_2\ldots v_nv_1$ be Hamiltonian cycle in $G$. For each edge $\{\,v_i, v_j\,\} \in E(G)$ that doesn't belong to $C_H$ we choose one of two ends and assign this edge to the selected end. All edges from $C_H$ form a simple cycle $C$ in $L(G)$. Now we can insert all edges of $G$ assigned to $v_i$ between edges $\{\,v_{i - 1}, v_i\,\}$ and $\{\,v_i, v_{i + 1}\,\}$ in any order and get longer simple cycle $C'$, because all edges incident to $v_i$ are pairwise adjacent. Proceeding this for each vertex $v_i$ we get simple cycle $C^{(n)}$ in $L(G)$ containing all vertices of $L(G)$ (that are edges of $G$), so $C^{(n)}$ is Hamiltonian cycle in $L(G)$.