Let \(G(V,E)\) be a simple, planar, and Hamiltonian graph with \(n\) vertices. In any planar drawing \(\mathcal D\) of \(G\) let \(G^\ast_{\mathcal D}\) be the corresponding dual graph. Furthermore, let \(C^n\) be a Hamiltonian cycle drawn in \(\mathcal D\) and let \(v_1,\ldots v_m\) be exactly those vertices of \(G^\ast_{\mathcal D}\), which are enclosed by \(C_n\) in \(\mathcal D\). Then the subgraph induced in \(G^\ast_{\mathcal D}\) by these vertices, i.e. the graph \(G^\ast_{\mathcal D}[v_1,\ldots v_m]\) is always a tree.
Proofs: 1
Proofs: 1