Equivalence between two definitions of Tree width
First, let me note that the claim, as stated in the question, is false:
Consider the following graph:


  $\begin{matrix}  &amp; &amp; 2&amp; - &amp; 3\\
&amp; \diagup\\
1 &amp; &amp; | &amp; &amp; |\\
&amp; \diagdown \\
 &amp; &amp;4 &amp;-&amp; 5
\end{matrix}
$


The complete graph on $5$ vertices is a chordal completion of this graph. However, this graph has treewidth $2$. (A decomposition is $\{1,2,4\}$, $\{2,3,4\}$, $\{3,4,5\}$ )



The correct claim is


  The treewidth of $G$ of is the size of the largest clique $\omega(G) - 1$ in a chordal completion with the minimum largest clique of the graph $G$.


Now, a proof is as follows: if the treewidth of $G$ by tree decomposition is $k$, then every tree decomposition has a bag of size at least $k$. We use the concept of a separator graph from a paper by Parra and Scheffler, in particular the fact that every maximal clique in the separator graph of $G$ is a tree decomposition of $G$ (This can be seen by comparing the definition of tree decompositions with those in the paper)

Then, by Theorem 4.7$^*$ from the same paper, every minimal chordal completion of $G$ has all the bags of some tree decomposition as a clique. This means every minimal chordal completion of $G$ has a clique of size $k$, so the chordal completion method also gives a tree-width of $k$. $\square$



*: Paraphrased in our notation, Theorem 4.7 states:


  A graph $H$ is a minimal chordal completion of $G$ if and only if $H$ is the graph $G$ with exactly the edges added such that all vertex sets in $\mathcal{S}$ are cliques. Here $\mathcal{S}$ is a maximal clique in the separator graph, which consists of minimal separators of $G$.


I attempted to find a proof using only elementary techniques, but I don't think an easy proof without any deeper theory would be easy to find.