Hilbert's axioms from Grundlagen der Geometrie involve notions of incidence, between-ness, segment congruence and angle congruence.

Consider the sub-theories of either Euclidean or hyperbolic geometry involving only the notions of incidence and between-ness. Of course, a priori, notions of congruence may occur in *proofs* (though not statements) of theorems solely about incidence and between-ness.

As a matter of fact, consider any 2-dimensional compact convex body $B$, with every boundary point extreme, say, but $B$ not affinely equivalent to a round disk. One can associate to $B$ an incidence geometry after the manner of Beltrami-Klein (with "lines" equal to interiors of chords). Such a geometry will *not* have an isomorphism to hyperbolic or Euclidean incidence geometries even though Hilbert's axioms about incidence and between-ness will hold! Indeed the incidence theory of such a quasi-Beltrami-Klein model actually determines $B$ up to an affine transformation. (**Has this fact been recorded in the literature?**)

**Main Question: Can one formulate a finite, or at least an elegant set of incidence and between-ness axioms, extending Hilbert's, so as to capture the theory of Euclidean and/or hyperbolic incidence?**

Added later: In light of Will Jagy's comment I'll sketch very briefly why I think the incidence geometry determines $B$ as much as it does, in case I'm obviously wrong. Given $B$, one can decorate it in many way with three families of evenly spaced parallel chords with triple intersections wherever intersections occur. The infinite of maximal configurations of this sort constitute approximations sufficient to recover $B$ up to affine.

allstraight chords to straight chords. – David Feldman Sep 2 '12 at 5:09notdetermine $B$ up to affine equivalence in general, but only up to projective equivalence. There are projectively equivalent convex bodies that are not affinely equivalent (of course, this cannot happen for ellipses, which are all affinely equivalent), and incidence and betweenness relations generated by chords will be the same for both bodies. Thus, for example, any two convex quadrilaterals are projectively equivalent, but generally, they won't be affinely equivalent. – Robert Bryant Sep 3 '12 at 18:40