Show any two edges in a 2-connected graph lie on a cycle So I found some proofs on any two vertices would lie on a cycle, but stuck on dealing with edges. We can say any two edges are connected, but does that just imply they will be on a common cycle? 
 A: Although this question is already a few months old, I will post my (hopefully working) answer, as I was searching for it for a long time as well.
We can extend the graph G by adding two vertices a, b in the middle of the edges wanted in one cycle. This does not change our graph being 2-connected, but now we can reduce the question of finding a cycle containing the two edges in G to the question of a cycle containing a and b in G'.
A: Let $G = (V, E)$ graph, $|V| \ge 3$. Suppose that there exists an edge $e = (u, v) \in E$ such that $e$ does not lie on a cycle. Then every path from $u$ to $v$ must contain $e$ (1). If $u$ or $v$ are a leaf, then removing the vertex that connects the leaf from the rest of the graph makes the leaf disconnected (!). Otherwise, define the following partition of $G$ without $e$:
$$G_u = (V_u, E_u) \quad \quad G_v = (V_v, E_v)$$
where $V_u = \{w \in V |$ there is a path from $w$ to $u$ that does not contain $e\}$, $E_u = \{(w, t) \in E | w, t \in V_u\}$ and $V_v$, $E_v$ are defined analogously (2). If we remove $u$ or $v$ from the graph, then we are also removing $e$. Note, however, that if we remove $e$, then there is no path to go from any $w \in G_u$ to any $t \in G_v$. In other words, the graph becomes disconnected (!).
Notes:
(1) Suppose that there exists a path from $u$ to $v$ that does not contain $e$. Then we can join this path to $e$ to form a cycle, but by doing so we are showing that $e$ lies on a cycle (!).
(2) It is easy to show that $G_u$ and $G_v$ are disjoint.
