Path Graph from Cycle Graph

Theorem

Removing one edge from a cycle graph leaves a path graph.

Proof

Let $G_n$ be the graph obtained by removing any edge from the cycle graph $C_n$.

As each of those edges lies on a cycle, by Condition for an Edge to be a Bridge, none of them is a bridge.

So removing any edge from $C_n$ leaves the resulting subgraph of $C_n$ connected.

We have that the cycle graph $C_n$ has $n$ edges.

So $G_n$ has $n$ nodes and $n-1$ edges, and is connected.

From Number of Edges in Tree, $G_n$ is a tree.

But $G_n$ has two nodes of degree $1$, and all the others are of degree $2$.

It follows that $G_n$ is traversable, and hence by definition is a path graph.

$\blacksquare$