Hilbert's Foundations: Segment Congruence Is an Equivalence Relation

Examples of equivalence relations include lines being parallel (often denoted by $\parallel$), triangles being similar (denoted by $\sim$), and shapes being congruent (and denoted by $\cong$).

It's obvious that segment congruence is an equivalence relation. To be rigorous, let's prove it.

Proof. First we show transitivity. Let $\overline{AB}$, $\overline{CD}$, $\overline{EF}$ be three segments such that $\overline{AB} \cong \overline{CD}$ and $\overline{CD} \cong \overline{EF}$. Then by definition of congruence, we see that $AB = CD$ and $CD = EF$. Since equality is transitive, we conclude that $AB = EF$ and thus $\overline{AB} = \overline{EF}$. Reflexivity and symmetry can be proven similarly by using the definition of congruence. This will be left as an exercise to the reader. $\Box$