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In the figure above, $A'C'$ is parallel to $AC$. It is obvious, using similar triangles, that if $B$ is the midpoint of $AC$, then $B'$ is the midpoint of $A'C'$.

I would like to know how easily this can be proved without using the theory of ratio, similar triangles or area. Loosely speaking, what this means is that I don't want to use any theorems derived from considerations that involve products of lengths or irrational ratios of lengths.

I will state the problem more formally in terms of Hilbert's axiom system. I would like a proof relying only on the axioms of incidence, order and congruence, as well as the parallel axiom, but not using Archimedes' axiom and its consequences. I would also like the proof to be purely geometric. I add this requirement because I already know that it is possible, based on the stated axioms, to construct an ordered field $F$ such that the plane is isomorphic to $F^2$, and the problem then becomes trivial. But that construction is relatively involved, and what I would like is the most direct geometric proof possible.

I would prefer a proof not using any assumptions about the intersection of two circles or of a circle and a line, but a proof using these would be better than no proof.

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  • $\begingroup$ Are we allowed to use Desargues' theorem (which can be thought as an axiom in projective geometry)? $\endgroup$
    – Zilin J.
    Mar 15, 2016 at 20:03
  • $\begingroup$ @ZilinJ. In principle, Desargues' Theorem is a consequence of the axioms I stated, but I think--correct me if I'm wrong--that its proof from those axioms is via coordinatization. (I guess I should have pointed out that I'm only using Hilbert's axioms in the plane. You're not permitted to consider the plane as being embedded in space.) If you do have a proof relating this statement to Desargues' Theorem, then I would find that fascinating, but my preference would still be for a proof that doesn't use that. $\endgroup$
    – David
    Mar 15, 2016 at 21:02
  • $\begingroup$ I understand. And I think you are right about the Desargues' theorem (using only Hilbert's axioms). I was just trying to "cheat" by working in a different set of axioms which characterizes projective geometry. $\endgroup$
    – Zilin J.
    Mar 15, 2016 at 21:13
  • $\begingroup$ Since I've received a downvote, please let me know if you think the question needs improvement. $\endgroup$
    – David
    Mar 17, 2016 at 2:49

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