Congruent triangles/Same lengths/Fact/Proof

Translations and isometries preserve lengths; therefore, congruent triangles have the same lengths of edges. Let now two triangles ${\displaystyle {}(A,B,C)}$ and ${\displaystyle {}(A',B',C')}$ be given, having the same edge lengths. After renaming, we may assume that the edge lengths fulfill the relation

${\displaystyle {}d(A,B)\geq d(A,C)\geq d(B,C)\,,}$

and that the same holds for the second triangle. We can assume ${\displaystyle {}E=\mathbb {R} ^{2}}$, and we can also assume after a translation that ${\displaystyle {}A=A'=0}$ holds. After rotations of the triangles around the origin, we may further assume that ${\displaystyle {}B}$ as well as ${\displaystyle {}B'}$ lie on the positive ${\displaystyle {}x}$-axis. Because of the length equality, we have ${\displaystyle {}B=B'}$. The points ${\displaystyle {}C}$ and ${\displaystyle {}C'}$ have, on one hand, the same distance to ${\displaystyle {}0}$, and, on the other hand, the same distance to ${\displaystyle {}B}$. That is, they lie on the intersection of a circle about ${\displaystyle {}0}$ and a circle about ${\displaystyle {}B}$. Since there are only two intersection points, we have either ${\displaystyle {}C=C'}$, or ${\displaystyle {}C}$ and ${\displaystyle {}C'}$ can be transformed to each other by a reflection at the ${\displaystyle {}x}$-axis.