# Degeneracy of outerplanar graphs

Does anyone know an elegant proof to the fact that every outerplanar graph has a vertex of degree at most 2 (and hence is 2-degenerate, since every subgraph is also outerplanar). I have a proof by induction (on the number of vertices) in mind, but it is long and somewhat cumbersome (it splits into a few cases). Can anyone point me to a more elegant proof? Maybe one can do it using the dual graph - if the dual graph is circle-free then we are done, but I couldn't find an easy argument for that either.
Thanks in advance for any help.

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Consider the maximum number of edges an outerplanar graph can have. – user34709 Jun 28 '12 at 20:45

Sketch of proof: by contradiction. Assume that an outerplanar graph $G$ exists whose every vertex has degree $\ge3$. $G$ is not a tree, since a tree trivially has a vertex with degree $\le2$. Therefore it encloses at least one interior region, and has no "naked branches," either, (such as would terminate with degree 1.) Consider the cycle traversing the unbounded face of the graph. If it is not Hamiltonian, then it has at least one subcycle with no vertices repeated, connected to the rest of the graph at only one of its vertices, i.e. "pinched off" from the rest of the graph. This subgraph is Hamiltonian and outerplanar, but we can show that every Hamiltonian outerplanar graph has at least two vertices of degree exactly two, and at least one of these has total degree 2 in $G$, a contradiction.