Axiom:Five-Segment Axiom

Axiom
Let $\equiv$ be the relation of equidistance.

Let $\mathsf B$ be the relation of betweenness.

Let $=$ be the relation of equality.

This axiom asserts that:
 * $\forall a, b, c, d, a', b', c', d':$


 * $\left({\neg \left({a = b}\right) \land \left({\mathsf B a b c \land \mathsf B a' b' c'}\right) \land \left({a b \equiv a' b' \land b c \equiv b' c' \land a d \equiv a' d' \land b d \equiv b' d'}\right)}\right)$


 * $\implies c d \equiv c' d'$

where $a, b, c, d, a', b', c', d'$ are points.

Intuition
Note that the following section does not cover degenerate cases.


 * Tarski's Five Segment Axiom.png

Let $a, b, c$ and $a', b', c'$ be collinear.

Let $a c d$ and $a' c' d'$ be triangles.

Draw a line connecting $d$ to $b$ and $d'$ to $b'$.

Suppose that every corresponding pair of line segments so constructed have been confirmed to be congruent except for segments $c d$ and $c' d'$.

Then $c d$ and $c' d'$ are also congruent.

Also see

 * Triangle Angle-Side-Angle Equality
 * Triangle Side-Angle-Angle Equality