Proving chordal graph has unique maximal clique In my studies of graph theory I recently came across the following:

Let $ G $ be a finite simple undirected graph which is chordal and for some $ e \in E(G) $ the graph $ G-e $ is also chordal then there exists a unique maximal clique $ K $ in $ G $ such that both ends of $ e $ are in $ K $.
A finite simple undirected graph is chordal if every cycle of length greater than three has a chord . Equivalently, a graph is chordal if it has no induced cycle of length greater than three.

A useful theorem I thought might help here

A perfect elimination ordering in a graph is an ordering of the vertices of the graph such that, for each vertex v, v and the neighbors of v that occur after v in the order form a clique. A graph is chordal if and only if it has a perfect elimination ordering.

I cannot seem to relate the graph being chordal and the modified graph being chordal to the existence of any clique, let alone to a unique one with both ends of $ e $ are in this clique. I thought perhaps perfect elimination order would be a clue here but I cannot proceed any further. I certainly appreciate all help.
 A: I don't think you need to use the theorem for this - you can prove the result by contradiction.  (I also don't think we need $G$ to be chordal.)
Suppose that there is an edge $e = \{u, v\}$ such that both $G$ and $G - e$ are chordal, and that there are two maximal cliques, say $M_1$ and $M_2$, both containing $\{u,v\}$.  Now try to use these cliques to construct a cycle that will not have a chord in $G - e$.  Details are behind the spoiler below.

 As $M_1$ is a maximal clique, we must have a vertex $x \in M_1 \setminus M_2$, since otherwise $M_1 \subset M_2$.  Now there must be a vertex $y \in M_2$ such that $\{ x, y \} \notin E(G)$, as otherwise $M_2 \cup \{ x \}$ would be a larger clique, contradicting the maximality of $M_2$.

Continuing the solution below.

 Since $\{u, v\} \subset M_1 \cap M_2$, it follows that each of $x$ and $y$ is adjacent to both of $u$ and $v$.  Hence we have the $4$-cycle $(x,u,y,v)$.  Since $x$ is not adjacent to $y$, this cycle does not have a chord in $G - e$, which is a contradiction.  Hence there must be a unique maximal clique containing both $u$ and $v$, as required.

Note that we actually proved a stronger statement.  If in a simple graph $G$ there is an edge $e$ such that $G - e$ does not contain an induced $C_4$, then there is a unique maximal clique containing both endpoints of $e$.
