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The following configurations are subsoluble and therefore Ramsey:

1. $(\{-a,a\}\times\{0\})\cup(\{-b,b\}\times\{\lambda\})$ for $a,b,\lambda>0$, including rectangles when $a=b$.
2. An equilateral triangle at height zero and the triangle of its side midpoints at any nonzero height.
3. A rectangle at height zero and a congruent rectangle, rotated about their common center by $\theta\in\pi\mathbb Q$, at any nonzero height.

Complete relative proof. In the first example, reflection in the origin acts transitively on each of $\{-a,a\}$ and $\{-b,b\}$. Its group $C_2$ is soluble, so Theorem 1 applies. Every nondegenerate isosceles trapezium has this form after a rigid motion, by placing the parallel sides horizontally with their common perpendicular bisector on the vertical axis.

For the second example, rotate through $2\pi/3$ about the triangle's center. This cyclic group permutes both the vertices and the side midpoints transitively. The same theorem applies to any nonzero height.

For the third example, place the rectangle's center at the origin and its sides parallel to the coordinate axes. Write $\theta=p\pi/q$, where $p\in\mathbb Z$ and $q\ge1$ is an integer. Let $G$ be generated by rotation through $\pi/q$ and reflection in the horizontal axis. Every group element is a rotation power or a rotation power followed by that reflection, so $G$ is a finite dihedral group. Its cyclic rotation subgroup is normal and its quotient has order two, making $G$ soluble. Rotation through $\pi$ and the horizontal reflection generate transitive symmetries of the original rectangle. Moreover, rotation through $\theta$ belongs to $G$. Thus the $G$-orbit $U$ of one rectangle vertex contains every vertex of both rectangles. The product $U\times\{0,\lambda\}$ is transitive under $G\times C_2$, a soluble group, and contains the required configuration. This proves the enclosure even though $G$ need not preserve either original rectangle separately. The Ramsey conclusions follow from the exact external Kříž theorem. $\square$

These are complete deductions for the examples on source pp. 2–3, arXiv:2606.13472v1. The rectangle example requires the intermediate finite orbit; the two original rectangles need not have the same transitive symmetry group. The source credits earlier Ramsey or subsoluble results on trapezia to Kříž and Behague. This page records the present prism method without asserting a first proof or a new construction.

Bears on. #174.