Axiom:Uniqueness of Triangle Construction

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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:

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

$\paren {\neg \paren {a = b} \land ac \equiv ac' \land bc \equiv bc' \land \mathsf{B} bdc' \land \paren {\mathsf{B} adc \lor \mathsf{B} acd } }$

$\implies c = c'$

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

Intuition

Construct a triangle $abc$.

Pick a point $c'$ such that:

The length of the line segment $ac$ is the same as that of $ac'$.

The length of the line segment $bc$ is the same as that of $bc'$.

Draw a line segment connecting point $b$ and point $c'$.

Let $d$ be some point on segment $bc'$.

If points $a,c,d$ are collinear, then points $c$ and $c'$ are the same point.

Note that this axiom still holds in degenerate cases.

For example, one can consider a line segment as a degenerate triangle.

In such a case, this axiom might be interpreted as "if two points are the same distance away from the left and right endpoints of a line segment and are on that line segment, they are the same point".

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Also see

• Two Lines Meet at Unique Point

Sources

• June 1999: Alfred Tarski and Steven Givant: Tarski's System of Geometry (Bull. Symb. Log. Vol. 5, no. 2: pp. 175 – 214) : p. $187$ : Axiom $20$
  Illustration courtesy of Steven Givant.