Is there a nice proof for the following fact?
In a plane, there does not exist a square such that its vertices are at a rational distance from each vertex of some equilateral triangle.
What if we replace the square with a cyclic quadrilateral?
Well, it seems the following idea works for the first part of the question: It is well known that there is no equilateral triangle in the Cartesian plane with rational coordinates. Let $(x,y)$ be at rational distance from the square with vertices, $(0,0), (0,a), (a,a), (a,0)$. Then it is easily seen that $2a(x-a)$ is rational. Hence if the coordinate system is scaled by a factor of $a$ and the origin is shifted to $(a,a)$, the coordinates become rational.