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.