TITLE: Diameter of a graph when removing a non-cut edge
QUESTION [13 upvotes]: It appears plausible to me that if we have a connected graph $G$ with diameter $d$ and we remove a non-cut edge $e$ from it, the diameter of the resulting graph $G_e$ will be at most $2d$.
By removing a non-cut edge I mean that the resulting graph is still connected. 
I was not able to find a proof or a counter-example.
Any suggestion?
Thanks.
EDIT:
I've tried to follow Perce's hint, and I'm quite sure I success in showing what I was looking for, but it was hard for me to explain. Then I ask for the correctness of my proof. 
Let $G$ be a graph with a certain diameter $d$, and let $$(u_0,u_d)=<u_0, u_1, ..., u_i, ..., u_d >  (0\leq i \leq d)$$ a shortest path from some node $u_0$ to some node $u_d$ such that its lenght is $d$. Let's add to $G$ an edge $e$. 
By adding $e$, maybe, there will be a shorter path $(u_0,u_d)'$ in the graph from $u_0$ to $u_d$ that differs from $(u_0,u_d)$ for some nodes $u_i$ (with $0<i<d$ ); it's easy to see that these different nodes are consecutive in  $(u_0,u_d)'$, that is they replace a substring of $<u_0, u_1, ..., u_i, ..., u_d >$.
Indeed, they form a new shortest path between some $u_j$ and $u_k$ using the new edge $e$ (because this new path $(u_j,u_k)$ wasn't the shortest before adding $e$); if there were not-consecutive $u_i$s in  $(u_0,u_d)'$, then $e$ would be used more than one time, and  $(u_0,u_d)'$ would not be a shortest path. 
EDIT2: fixed the following part of my solution
Hence, adding an edge we create at most one shorter route among the nodes in the path $(u_0,u_d)$, that in the best case is of lenght 1 (if we use $e$ itself).
Now, it's easy to see that, by adding a an edge between two nodes in a path, we cannot reduce its lenght $l$ below $l/2$.

REPLY [4 votes]: I unfortunately still can't comment, but this refers to Louis's answer below:
Louis, I don't see why the combinatorial structure of H' implies that each layer has at most two vertices. Suppose G is just a path $(u_0, u_1, u_2,u_3,u_4,u_5)$. Then $H = G$. Now let us add the edge $[u_0,u_4]$ to get $G'$. Now $H' = G'$ and the layers are:
Distance $1$ from $u_0$: $u_1,u_4$
Distance $2$ from $u_0$: $u_2,u_3,u_5$, which has three vertices.
In fact there is no vertex with distance $\geq 3$ from $u_0$, but $dist(u_2,u_5) = 3$.
EDIT: 
I believe this completes Louis's proof in a satisfactory manner:
We need to show that the diameter of $H'$ is at least $D/2$. Now from the construction it is clear that there are three possibilities for $H'$: it is either
(a) a cycle with $D+1$ edges; (b) a cycle with $D+1-i$ edges plus a path with $i$ edges starting at one of the vertices; (c) a cycle with $D+1-i-j$ edges plus two paths, with $i$ and $j$ edges respectively, starting at adjacent vertices in the cycle.
(a) and (b) clearly have diameter $\geq D/2$. For (c), suppose $i \leq j$ and let $u_j$ be the vertex of the cycle where the path with $j$ edges begins, and $v_j$ the respective endpoint. There exists $w_j$ in the cycle such that $dist(u_j,w_j) \geq \frac{D-i-j}{2}$. Therefore
$$dist(w_j,v_j) = dist(w_j,u_j) + dist(u_j,v_j) \geq \frac{D + j - i}{2} \geq \frac{D}{2},$$
which completes the proof.