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Prove directly from the axioms $(I1)-(I3)$ and $(B1)-(B4)$ that for any two distinct points A, B, there exists a point C with $A * C * B$.

My original thought was to use B2 to place a point D on the line AB s.t ABD holds then use it again to place a point E so that EBD holds but there is no guarantee doing this that E lies on the correct side of of A.

My next try was to use I3 to place a point F not on the line segment AB then connect AF and extend it with B2 to get G s.t $A*F*G$ now connect BG and we can use Paschs axiom on a line through the point F to say there is some point H either between BG or AB but this also become cyclic cause we can never guarantee that we place the point in between AB...

Incidence axioms

I1)For every two points A and B there exists a line a that contains them both.

I2)There exist at least two points on a line.

I3)There exist at least three points that do not lie on the same line.

Axioms of betweeness

B1) 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.

B2) If A and B are two points, then there exists at least one point C on the line AB such that B lies between A and C.

B3) Of any three points situated on a line, there is no more than one which lies between the other two.

B4) 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.

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Your second try should work:

  1. use I3 to place a point $F$ not on line $AB$;

  2. use B2 to extend $AF$ to a point $G$ such that $A∗F∗G$;

  3. use B2 to extend $BG$ to a point $N$ such that $G∗B∗N$;

  4. apply B4 to triangle $ABG$ and line $FN$: there exists some point $H$ on line $FN$ either between $BG$ or $AB$;

  5. but point $H$ cannot be on $BG$, otherwise line $FN$ would have two points in common with line $BG$ and thus $F$ would be on $BG$, which is impossible by construction as $A\ne B$;

  6. hence $H$ is on segment $AB$.

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