= Solution
Put $\rho^2=28-28\sqrt5/5$ and, for $j\in\mathbb Z/30\mathbb Z$, define
$$
v_j=
\left(
5\cos\frac{j\pi}{3},
5\sin\frac{j\pi}{3},
\rho\cos\frac{4j\pi}{5},
\rho\sin\frac{4j\pi}{5}
\right).
$$
The shift $v_j\mapsto v_{j+1}$ is an <isometry> acting transitively on the finite set $V=\{v_j\}$, so $V$ is a <cyclic transitive point set> and hence a <Euclidean Ramsey set> by the <Kriz theorem for cyclic transitive point sets>.
For every $j,k$,
$$
\lVert v_{j+k}-v_j\rVert^2
=50\left(1-\cos\frac{k\pi}{3}\right)
+\left(56-\frac{56\sqrt5}{5}\right)
\left(1-\cos\frac{4k\pi}{5}\right).
$$
For $k=1,5,11,15,4,14$, these squared distances are respectively
$$
81,25,81,100,131,131.
$$
Consequently $v_0,v_1,v_{26},v_{15}$, in that order, have consecutive side lengths $9,5,9,10$ and equal diagonals $\sqrt{131}$. This is an isometric copy of the required isosceles <trapezium>. Since every monochromatic copy of $V$ contains this four-point subset, the trapezium is Euclidean Ramsey.
Solved by gpt-5.6-sol high.
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