Absolute geometry is any one that satisfies Hilbert's axioms of plane geometry without the axiom of parallels. It is well-known that it is either the Euclidean or a hyperbolic plane. For an elementary version we also drop the (Cantor's) axiom of continuity, Greenberg calls such geometries Archimedean H-planes in his survey paper. We can still define a metric on them in the usual way. Repeatedly bisecting a picked "unit segment" and laying its pieces off of any other gives a binary fraction (possibly infinite) that is assigned as the segment's length. The distance between two points is defined to be the length of the line segment connecting them, unique by incidence. Incidence, order and congruence imply that perpendicular is shorter than oblique, so the triangle inequality holds. Without the continuity however Archimedean H-planes may not be metrically complete. Will their completions still always be Euclidean or hyperbolic? In other words, if we remove the axiom ensuring completeness, and then take the completion, do we end up with what we started with?

Here is why I am having doubts. It is easy to check by limit arguments that most axioms still hold in the completion, but not that two lines intersect at no more than one point, for example. With extra points added perhaps some lines intersect more than once. Also, there is an algebraic classification of Archimedean H-planes due to Pejas described in the Greenberg's paper (p.760). One of them, called the semi-elliptic plane, is quite peculiar. It satisfies the Lambert's hypothesis of the acute angle (in any quadrilateral with three right angles the fourth angle is acute), but any two non-intersecting lines in it have a unique common perpendicular. In other words, no two lines are asymptotically parallel. If the semi-elliptic plane is isometric to a subset of a hyperbolic one then we should be able to obtain it by removing some points from the latter. But if we remove all lines asymptotically parallel to some line in the hyperbolic plane, nothing will be left other than that line itself. On the other hand, if the completion is elliptic, as the name suggests, then how does it square with the hypothesis of the acute angle?

Is the metric completion of an Archimedean H-plane always Euclidean or hyperbolic? If so than what subplane of a hyperbolic plane is semi-elliptic? If not then what is the completion of the semi-elliptic plane?

Unfortunately, Pejas's original works weren't translated from German. Here is the passage from Greenberg describing the semi-elliptic plane.

"*Pejas gave the following example of an Archimedean H-plane in which the acute angle hypothesis holds but which is not hyperbolic; it is an example of a semi-elliptic plane, defined by the property that any two parallel lines have a unique common perpendicular (in a hyperbolic plane, two asymptotically parallel lines have no common perpendicular, whereas two divergently parallel lines have a unique one): Let $K_0$ be an Archimedean ordered field with two distinct orderings $<$ and $<'$ (for example, $K_0 = \mathbb{Q}(\sqrt{2})$). Let $L$, $L'$ be the real closures of $K_0$ with respect to these orderings within a given algebraic closure. Set $K = L\cap L'$. Then $K$ is
Pythagorean, Archimedean, and contains an element $k$ such that $k<0$ and $0<'k$. We take $k$ as metric constant and the points of $\mathcal{K}$ to be all $(x,y)$ in the affine plane over $K$ for which $k(x^2 +y^2)+1>0$. $\mathcal{K}$ is the interior of the "absolute conic" $x^2 +y^2=-k^{-1}$, which is empty because $\sqrt{-k}\notin K$. Since $\mathcal{K}$ is maximal, that conic is the locus of all ideal points, so asymptotic parallels do not exist and the plane is semi-elliptic.*"

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