$(2n)$-regular graphs don't have bridges Let $G$ be a $k$-regular graph where $k$ is even. I want to prove $G$ doesn't have a bridge. I was thinking I could prove by contradiction and assume that G has a bridge. But from there I have no idea where to go. 

A $2k$-regular graph is Eulerian so removing an edge results in a graph with an Eulerian walk. You could also use the 2-factor theorem to show that the endpoints of any edge a lie on a cycle and therefore there exist at least two paths between them.
