The furthest vertex of half of the vertices in graph is also a vertex that defines the diameter - need proof So I was looking at graph algorithms and stumbled upon an interesting question:  https://stackoverflow.com/questions/1190543/good-algorithm-for-finding-the-diameter-of-a-sparse-graph
Now, an answer to post made by Karoly Horvath, posted by Josep Valls caught my interest -
the author said:
"By definition, one of the two vertices that define the diameter is also the furthest vertex to half of the vertices in the graph."
If this is really true, that really makes the diameter problem easier, but I was unable to come up with a proof, neither was I able to find an answer in the Internet - Is this just an observation that works for most graphs, or maybe there is an elegant proof?
Of course I consider fully connected undirected graphs.
Edit
please consider this graph:
Now clearly, vertices 13 and 11 form the diameter:
Vertex 13 is the furhtest vertex for vertices 11, 10, 9
Vertex 11 is the furthest vertex  for vertices 13, 1, 3, 4, 5, 2, 6, 7, 8
So for this graph the assumption holds, because wherever I start I am still going to get to the diameter endpoint.
Now, can I be sure, that this happens for every graph?

 A: It's not clear what the title of your question even means, but it sounds more like the converse of the actual question than the actual question itself.
I don't know what "fully connected" means, but I'm going to assume that it just means "connected".
It's not clear whether "$x$ is the furthest vertex from $y$" means "$d(x,y)\gt d(z,y)$ for all $z\ne x$" or whether it means "$d(x,y)\ge d(z,y)$ for all $z$". Under the first interpretation, any complete graph with more than two vertices is a counterexample. The graph described below is a counterexample under either interpretation.
$G$ is a graph with $8$ vertices and $21$ edges. The vertices are $x_0,x_1,x_2,x_3,x_4,u,v,w$. The edges are $x_ix_j$ ($i\ne j$), $ux_i$, $vx_i$, and $wx_0$.
$\operatorname{diam} G=2=d(u,v)$.
$u$ is a furthest vertex from the three vertices $x_0,v,w$; and $v$ is a furthest vertex from the three vertices $x_0,u,w$. However, only $w$ is the furthest vertex from each of the four vertices $x_1,x_2,x_3,x_4$, since for $i\in\{1,2,3,4\}$ we have $d(x_i,u)=d(x_i,v)=1\lt2=d(x_i,w)$.
Of course we can modify this example so that there are a zillion vertices for which $w$ is the unique furthest vertex, but still only three vertices for which $u$ (or $v$) attains the maximum distance.
A: 
Diameter is $2$, as for example witnessed by the top left and bottom right vertex. However, each vertex is furthest at most for one vertex.
So as stated, the claim is false. However, the context was: if you start with any vertex $a$, then pick a vertex $b$ that is furthest from $a$, then pick a vertex $c$ that is furthest from $b$, it holds for this graph as well that $a,c$ form a diameter.
