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:

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

$\paren {\neg \paren {a = b} \land \paren {\mathsf B a b c \land \mathsf B a' b' c'} \land \paren {a b \equiv a' b' \land b c \equiv b' c' \land a d \equiv a' d' \land b d \equiv b' d'} }$

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

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

Intuition

Note that the following section does not cover degenerate cases.

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 Congruence
• Triangle Side-Angle-Angle Congruence

Sources

• June 1999: Alfred Tarski and Steven Givant: Tarski's System of Geometry (Bull. Symb. Log. Vol. 5, no. 2: pp. 175 – 214) : p. $178$ : Axiom $5$
  Illustration courtesy of Steven Givant.