Prove: "if three points are on a straight line, at least one point is between the other two." In Wikipedia, the third order axiom of Hilbert's axioms states that

"Of any three points situated on a line, there is no more than one which lies between the other two.

Note: The existence part ("there is at least one") is a theorem.
So if there are three points on a line, then there should be one and only one point that is between the other two. However the axiom states only that there can't be more than one such point, and does not tell anything about the existence of the middle point. How do you prove the existence of the point?
The whole order axioms are

*

*If a point B lies between points A and C, B is also between C and A, and there exists a line containing the distinct points A, B, C.


*If A and C are two points, then there exists at least one point B on the line AC such that C lies between A and B.


*Of any three points situated on a line, there is no more than one which lies between the other two.


*Pasch's Axiom: Let A, B, C be three points not lying in the same line and let a be a line lying in the plane ABC and not passing through any of the points A, B, C. Then, if the line a passes through a point of the segment AB, it will also pass through either a point of the segment BC or a point of the segment AC.
I do think the proof can be done with only these axioms (and of course with the obvious incidence axioms) without concerning congruency and continuity axioms, however I can't come up with one.
 A: Let $A,B,$ and $C$ be three distinct points on a line $\ell$, and suppose neither $A$ nor $C$ is between the other two points.  We will show that $B$ must be between $A$ and $C$.
Take a point $E$ which is not on $\ell$, and take a point $D$ such that $E$ is between $D$ and $B$.  Then line $CE$ passes through segment $BD$, and so by Pasch's axiom it must pass through segment $AB$ or $AD$.  But it cannot pass through segment $AB$ since $C$ is not between $A$ and $B$, so it meets segment $AD$ at some point $F$.  Similarly, line $AE$ meets segment $CD$ at some point $G$.
Now line $CE$ meets line $AD$ at $F$, line $DG$ at $C$, and line $AG$ at $E$.  Note that $F$ is between $A$ and $D$, but $C$ is not between $D$ and $G$ (since $G$ is between $C$ and $D$).  So, by Pasch's axiom (for line $CE$ and triangle $ADG$), $E$ must be between $A$ and $G$.
Finally, consider line $BD$ and triangle $ACG$.  Line $BD$ meets segment $AG$ at $E$, and meets line $CG$ at $D$ which is not on segment $CG$ (since $G$ is between $C$ and $D$).  Thus by Pasch's axiom, line $BD$ passes through segment $AC$.  But the intersection of line $BD$ and line $AC$ is $B$, so $B$ is between $A$ and $C$, as desired.
(In each application of Pasch's axiom, I have omitted verifying that the line does not pass through any of the vertices of the triangle, since these verifications are rather tedious.  They mostly boil down to concluding that all the points would be on the same line, contradicting our choice of $E$ as not being on line $\ell$.)
