Model of Hilbert's plane ordered geometry in which the elliptic parallel property holds Elliptic parallel property: Given a line $L$ and a point $a\notin L$, there exist no lines parallel to $L$ passing through $a$.
Euclidean parallel property: Given a line $L$ and a point $a\notin L$, there exists exactly one line parallel to $L$ passing through $a$.
Hyperbolic parallel property: Given a line $L$ and a point $a\notin L$, there exist at least two lines parallel to $L$ passing through $a$.
Note that each of these properties states existence of particular number of parallel lines for every line $L$ and every point $a\notin L$. It means that negation of one of these properties does not imply one of the other two without additional assumptions.
As you know, the whole set of Hilbert's axioms describes Euclidean geometry. If we replace parallel postulate with it's negation we get hyperbolic geometry. In other words, assuming Hilbert's axioms for neutral geometry (i.e. without parallel postulate or its negation) we can prove that euclidean or hyperbolic parallel property holds. However, the proof relies on congruence and continuity axioms. What if we omit congruence and continuity axioms and restrict only to incidence and order? Can we then prove the existence of at least one parallel line to a given line passing through a given point? Or there is a model of ordered geometry in which there exist a line $L$, point $a\notin L$ and no lines parallel to $L$ passing through $a$ ? We can ask even more: Does there exist a model of ordered geometry in which elliptic parallel property holds?
EDIT. Apparently the answer to my question is in Veblen's paper "A system of axioms for geometry". On page 370 he proves the existence of at least one parallel line using only order and continuity axioms (no congruence and parallel axioms). On page 348 he provides a model for ordered geometry (without continuity axioms) in which elliptic property holds.
You can find the paper here: https://archive.org/details/jstor-1986462
EDIT. I'm still waiting for someone to expand on the model from Veblen's paper if it is indeed the answer to my question. The model is rather unclear to me, it uses some concepts from projective geometry which I wish someone could explain. Also I'd like to know how to modify it for plane model. Any other models are welcome, too.
 A: Yes, the example in the Veblen's paper gives a model of ordered geometry (only axioms of incidence and order) with the elliptic parallel property.
If I understood his example correctly, he just takes the projective plane $\mathbb{P}^2(\mathbb{Q})$ and defines the ordering of points there. In case of the affine plane there is no problem to define the ordering on points. When adding the points in infinity, one has troubles how to order such a point within a line it belongs to. Veblen's solution is 


*

*to pass to $\mathbb{P}^2(\mathbb{R})$,

*choose another line to be a line in infinity,

*in the new affine chart order the points accordingly,

*restrict to the original set of points.


Cleverly he chooses a line containing no point of $\mathbb{P}^2(\mathbb{Q})$ (e.g. the line passing through $(\pi,0)$ and $(0,\pi^2)$) to be a new line in infinity. Then after passing to the original set of points no point is omitted and therefore all points of the plane are ordered.
So actually it is NOT true that projective planes are no models to ordered geometry: a fact I thought to be true (because I found it mentioned in Greenberg's book, for example) but could not figure out the the proof.
A: Yes
without congruence, the frame is so rough that It has enough room to have such model as you mentioned.
with congruence (even without continuity), such model dose not exist.
In Hilbert plane (Euclidean plane without any form of parallel postulate and continuous), the parallel lines do exit. You can always use double-perpendicula to do so. Namely, drop a perpendicular , say M, from point A to L and then erect a perpendicular ,say N, at point A to M. N is always parallel to L (parallel means they do not intersect)
thanks to Gupta’s construction, to drop a perpendicular (and from that, erect perpendicular s and get midpoint of any segment) can be always done without help from any any form of parallel postulate and continuous. This construction relys on the facts about the base midpoint of isosceles triangles and congruence of triangles so it indeed needs congruence axioms
So, to eliminate the elliptic property, you need either (incidence+order+congruence) or (incidencr+order+continuity)
