Class of chordal graphs is closed under edge-contraction As the title suggests, I want to prove that the class of chordal graphs is closed under edge contraction.
A chordal graph is one in which all cycles of four or more vertices have a chord (that is, an edge that is not part of the cycle but connects two non-adjacent vertices of the cycle). Equivalently, a chordal graph is one in which every induced cycle have exactly three vertices.
An edge contraction is an operation which removes an edge from a graph while simultaneously merging the two vertices that it previously joined.
Intuitively I can get a sense of get why the above statement is true, but I am having trouble in formulating a coherent, formal proof (as is very often the case for beginners in graph theory).
Any help is much appreciated. 
 A: Ok, I actually found the proof of this one. I will post it just in case anyone ever needs it.
We will need a definition and a lemma:

Definition: Let $G=(V,E)$ be a graph and let $(T, \mathcal{S})$ be a pair where $T$ is a tree and $\mathcal{S}$ is a collection of subsets of $V$ which are in $1-1$ correspondence with the vertices of $T$. For a vertex $i$ in $T$  let $S_i \in \mathcal{S}$ be the subset that is assigned to the vertex $i$. The pair $(T, \mathcal{S})$ is called a ${\it clique}$ ${\it tree}$ of for $G$ if the following conditions hold:
(i)$\mathcal{S}$ is the set of maximal cliques in $G$
(ii)$\forall x\in V$, if $x\in S_i \cap S_j$, then $x\in S_l$ for every $l$ which lies on the path in $T$ from $i$ to $j$

In other words, a clique tree for $G$ is a tree of which the vertices represent the maximal cliques in $G$ such that for all $x\in V$, the maximal cliques that contain $x$ form a subtree of $T$.

Lemma: A graph $G=(V,E)$ is a chordal graph if and only if it is the intersection of a collection of subtrees of a tree, i.e., there exist a tree $T$ and a collection of subtrees $\{T_x : x \in V\},$ such that two vertices $x$ and $y$ of $G$ are adjacent if and only if $T_x \cap T_y \neq \emptyset$.

Finally, we are going to prove that the class of chordal graphs is closed under edge contractions:
Let $G=(V,E)$ be a chordal graph. Let $e = \{x,y\}$ be an edge of $G$ and let $G'$ be the graph obtained from $G$ by contracting the edge $e$ to a single vertex $xy$. We will prove that $G'$ is chordal. Since $G$ is chordal, we know from the above lemma that there exists a tree $T$ and a collection of subtrees $\{T_x : x\in V\}$ such that any two vertices $a$ and $b$ are adjacent if and only if $T_a \cap T_b \neq \emptyset$. 
Consider the subtrees $T_x$ and $T_y$. Since $x$ and $y$ are adjacent, we have that $T_x \cap T_y \neq \emptyset$. For the new vertex $xy$ define the subtree $T_{xy} = T_x \cup T_y$. Then $T_{xy}$ is a subtree of $T$. Any other vertex $z$ is adjacent to $xy$ in $G'$ if and only if $z$ is adjacent to $x$ or $y$ in $G$. Any subtree $T_z$ intersects $T_x$ or $T_y$ if and only if $T_z$ intersects $T_{xy}$.
Thus, the graph $G'$ is the intersection graph of a set of subtrees of a tree, and so, $G'$ is chordal.
A: It seems natural to prove the contrapositive.  I.e., we start with an arbitrary non-chordal graph $G$, and an arbitrary graph $H$ and edge $e \in E(H)$ for which $H/e$ (meaning $H$ contract $e$) is equal to $G$.  Then prove that $H$ is not chordal.
So we assume $G$ has an induced $k$-cycle for $k \geq 4$.
Let $v$ be the vertex in $G$ formed by contracting $e$.  If $v$ does not belong to this induced $k$-cycle, then $H$ also has this induced $k$-cycle, which shows that $H$ is not chordal.  So assume $v$ belongs to the $k$-cycle.
Let $a$ and $b$ be the two neighbors of $v$ in the $k$-cycle.  Let $e=\{e_1,e_2\}$.  We know:


*

*The $k$-cycle excluding $v$ forms an induced $(k-1)$-vertex path in $H$ with endpoints $a$ and $b$.

*Both $a$ and $b$ are adjacent to $e_1$ and/or $e_2$ (otherwise one or both would not be neighbors of $v$ in $G$).

*Neither $e_1$ nor $e_2$ are adjacent to a vertex $x$ in the $(k-1)$-vertex path, other than $a$ or $b$ (otherwise the $k$-cycle in $G$ would have a chord from $v$ to $x$).


Thus, we have the situation depicted below:

We check the remaining possible subgraphs induced by $\{a,b,e_1,e_2\}$ case-by-case, and we get either an induced $k$-cycle or an induced $(k+1)$-cycle in $H$.
