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.

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 and Side-Angle-Angle Equality

Sources

• June 1999:  Alfred Tarski and Steven Givant: Tarski's System of Geometry (The Bulletin of Symbolic Logic Vol. 5, no. 2: 175 – 214) : Page 178 : Axiom $5$
  Illustration courtesy of Steven Givant.