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:
 * $\forall a, b, c, d, a', b', c', d':$


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


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

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 $acd$ 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 $cd$ and $c'd'$.

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

Also see

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