Axiom:Axiom of Segment Construction

Axiom
Let $a,b,q,x$ be points.

Let $\equiv$ be the relation of equidistance.

Let $\mathsf{B}$ be the relation of betweenness.


 * $\forall a,b,c,q,: \ \exists a,b,c,q \implies \exists x: \mathsf{B}qax \land ax \equiv bc$


 * AxiomOfSegmentConstruction.png

Intuition
Let $bc$ be a line segment.

Let $a$ be a point on a ray with endpoint $q$.

One can create a line segment congruent to $bc$. This line segment is $ax$.