This falls slightly short of a proof but is too long for a comment.
The question inquires into the largest family of triangulations in which every pair of members shares some diagonal.
If we instead fix any single diagonal in common to every member of a family of triangulation of an n-gon, it is self-evident (from the noncrossing property of triangulations) that the number of triangulations sharing it is given by the product of the enumerations of the two families of triangulations of the ploygons so formed either side of it.
For any triangulatable n-gon, any given diagonal is a member of up to two fan triangulations, in which it can be assigned an integer the $d^{th}$ diagonal from the perimeter satisfying $1\leq d\leq \frac{n-2}{2}$. By symmetry we may choose either of the two fans with no effect upon the distance of the diagonal from the nearest edge.
$d=\frac{n-2}{2}$ shall be the distance of the central triangulation if $n$ is even and $d=\frac{n-3}{2}$ shall be the distance of the two "most central" diagonalisations if $n$ is odd.
The number of triangulations of any $n$-gon sharing the $d^{th}$ diagonal is the product of the number of triangulations $\cal T(d,n)$ either side of it:
$$\cal T(d,n)=C_d\times C_{n-d}$$
$$\cal T(d,n)=(d+1)^{-1}\binom{2d}{d}(n-d+1)^{-1}\binom{2n-2d}{n-d}$$
The ratio of one Catalan number to the next is greater than the ratio of the previous two Catalan numbers, and therefore in the choice of which diagonal to share, in order to maximise the size of the family, for any given choice $d$ of diagonal which is not at the edge, we may always identify a larger family by choosing the diagonal $d-1$ which is one position closer to the perimeter and by induction the outermost diagonal $d=1$ yields the largest family.
The size of the largest family of triangulations sharing the same diagonal is therefore given by:
$|\cal S| = C_{n-2-1}\times C_1=|\cal T_{n-1}|$
My claim which would complete the proof (and which remains to be proven) is that there is no family of triangulations in which every pair shares some diagonal, which is larger than the largest family of triangulations in which every triangulation in the family shares the same diagonal.
I suspect this is proven by the fact that the largest face on any associahedron has the same number of edges as the number of vertices of the associahedron one order smaller. In particular I believe that each face of the associahedron represents a family of triangulations which all share the same edge. These two statements alone would in fact be sufficient to answer the entire question in the affirmative.
Best Answer
Hi,
The question itself and all answers seem to be "practice-oriented" but as far as the complexity status of the problem is concerned then you are speaking about famous L.Lovasz's "polytope- polyhedron question". Namely, to check whether a polytope given by vertices coincides with a polyhedron given by facets. Note, that the lengths of the vertex and, respectively, the facet descriptions may differ exponentially and thus we should speak about incremental polynomial algorithms. Say, if a polyhedron is given by facets you should find a new vertex in polynomial time with respect to the input and the length of the list of already calculated vertices. In this format the question is still open for (bounded) polytopes but is proved to be NP-hard for (unbounded) polyhedrons (see, http://portal.acm.org/citation.cfm?id=1109640 and references therein)