Maximal families of equal length intervals consist of equilateral triangles

Question

My question is a follow up to How to find n points on a plane so that as many pair of points as possible have the same distance? -- see the conjecture at the bottom of this post.

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Let $\ n\ $ be a positive integer. Let $\ e(A)\ $ be the maximal cardinality of set $\ P\subseteq\binom A2\ $ such that distance $\ d(x\ y)\ $ is constant over all $\ \{x\ y\}\in P.$

Let $\ D_n\in\binom{\Bbb R^2}n\ $ be such that $$ \forall_{A\in\binom{\Bbb R^2}n}\quad e(D_n)\ge e(A) $$

Set $\ \Delta\subseteq\binom{D_n}2\ $ such that distance $\ d(x\ y)\ $ is constant over all $\ \{x\ y\}\in \Delta\ $ and $\ |\Delta|=e(D_n)\ $ is called critical.

CONJECTURE:  Let $\ n>2.\ $ Let $\ \Delta\in\binom{D_n}2\ $ be critical. Then for every $\ \{x\ y\}\in\Delta\ $ there exists $\ z\in D_n\ $ such that $\ \{x\ z\}\,\text{and}\, \{y\ z\}\,\in\,\Delta.$

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I am sure that the following weaker version is indeed true. Namely, only cardinalities $\ e(D_n)\ $ are defined uniquely but not sets $\ D_n.\ $ Thus I am confident that the conjecture is true for at least one $\ D_n\ $ (for every $n$) but -- who knows -- the conjecture fails for some $\ D_n.$

Answer 1

Following up on Gerry Myerson's suggestion, the following graph from Ágoston and Pálvölgyi's Improved constant factor for the unit distance problem may be a counterexample.

This is the $n=13$ example illustrating the maximum number (30) of unit distances among 13 points in the plane. Schade's 1993 thesis establishes that this is the unique maximal "unit distance graph" for 13 points. Looking at the 3 neighbors of the lefthand marked vertex $x$ and the 5 neighbors of the righthand marked vertex $y$, you can see that there is no vertex $z$ adjacent to both $x$ and $y$.