Convex hexagon $ABCDEF$, is the graph of the sides and 3 diagonals a rigid graph?
I attempted to apply Menger's theorem: $V(ABCD)=0\iff$ $$\det {\begin{bmatrix}0&d(AB)^{2}&d(AC)^{2}&d(AD)^{2}&1\\d(AB)^{2}&0&d(BC)^{2}&d(BD)^{2}&1\\d(AC)^{2}&d(BC)^{2}&0&d(CD)^{2}&1\\d(AD)^{2}&d(BD)^{2}&d(CD)^{2}&0&1\\1&1&1&1&0\end{bmatrix}}=0$$ and similar equations for $ABCE,ABCF$. We get a system of 3 equations in 3 variables $AD,BE,CF$. The question becomes, whether it has unique solution (on positive real numbers)?
In general a planar embedding of the complete bipartite graph $K(3,3)$ is rigid unless the vertices lie on a conic.
That covers the convex cases when the vertices are not on an ellipse. Even in that case the graph will be determined up to rigid motions, but will have an infinitesimal deformation. Think of the vertices of a regular hexagon. You can move every other vertex infinitesimally in while the others move out without momentarily increasing the length of any edge or diagonal.
See WHEN IS A BIPARTITE GRAPH A RIGID FRAMEWORK at https://msp.org/pjm/1980/90-1/pjm-v90-n1-p04-s.pdf