Show that if G is a planar, simple and 3-connected graph, then the dual graph of G is simple and 3-connected I've been thinking about this question for several days now and I haven't come up with a satisfactory answer yet. The part of proving that the dual graph is simple (under the assumption that the original graph is planar, 3-connected and simple) is easy, but the 3-connectedness part seems a little more difficult to prove. 
I was trying to imagine what it would mean for the dual not to be 3-connected in terms of the faces of the graph and the paths that exist between any two of them in the dual graph, but I didn't get anywhere with that idea. 
Then, using the fact that a 3-connected (and therefore connected) graph is isomorphic to its double dual, I realized that it was enough to prove that if the dual of a graph is 3-connected, then the original graph is 3-connected, so that I would stop thinking about paths between faces in the dual graph and rather think about paths between vertices in the original graph. I can also use the fact that the dual graph of a 2-connected graph is itself 2-connected and I thought that could help me in some way, but I haven't seen how. 
Any help or ideas would be greatly appreciated. 

If You can make the dual graph disconnected by removing two edges, this means that after removing one edge, there is a bridge, i.e., an edge with the same face on both sides. This means that before removing the first edge, there were two faces sharing two edges. In the original graph this means two vertices joined by two different edges - contradicting simplicity.
