Axiom:Five-Segment Axiom
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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
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.