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.