Drytime wrote: Given two (identical) polygonal domains in the Euclidean plane, it is not possible in general to superpose the two using only translations and rotations. Prove that this can however be achieved by splitting one of the domains into a finite number of polygonal subdomains which then fit together, via translations and rotations in the plane, to recover the other domain. First, we have to use the fact that: The two identical polygonal domains cannot be superposed by only translations and rotations without splitting into parts if and only if one of them is the image of the other by reflected with respect of a line. Then we can splitting one of the domains into a finite number of triangles by cutting it through some diagonals of the polygon. Therefore we only need to prove that for every triangle, we can split it into some parts which can be fitted together to form reflected triangle. This time, we need another fact that isosceles triangle can be superposed using only translations and rotations. Actually this is obviously true. And finally, we need to prove that every triangle can be split into four isosceles triangle. Let the triangle be $\triangle ABC$ and $\angle A$ is the largest angle of the triangle. Let $D$ be feet of altitude with respect of $A$ and $E$, $F$ are the mid-point of $AB$ and $AC$ respectively. It is easy to check that $\triangle AED$, $\triangle BED$, $\triangle AFD$, $\triangle CFD$ are isosceles triangles. The proof is complete.