Why are the medians of a triangle concurrent? In absolute geometry

Below is a summary of Hjelmslev's argument as it is explained in Bachmann's book, see the references to German and Russian editions in Misha's answer. The goal of this answer is at first to help people not speaking Russian or German, and at second to summarize quite a long theory (when you start to read this proof, you are referred to previous lemmata and definitions, then again and so on.)

The main idea is to identify any line or point with reflection in this line/point. So we multiply the lines and points as elements of the group of isometries. Also, let $[AB]$ denote a line through points $A$ and $B$.

This viewpoint allows, in particular, to define the pencil of lines: it is the set of lines for which the composition of any three reflections is again a reflection in some line. In particular, the set of lines which share a common point is a pencil: if $a,b,c$ are three lines through point $P$, then $\mathcal{R}:=bc$ is a rotation with centre $P$, there exists a line $d$ through $P$ for which $ad=\mathcal{R}$ too, thus $abc=a\mathcal{R}=aad=d$ is a reflection in $d$.

Another important example of a pencil is the set of lines orthogonal to a given line $p$. Indeed, let $a,b,c$ be three lines orthogonal to $p$, let them meet $p$ at points $A,B,C$ respectively. It is easy to find a point $D\in p$ such that $ABCD$ (it is a product of reflections) is identical on $p$. Let $d$ be a line through $D$ perpendicular to $p$. Then $abcd$ is identical on $p$ where it is the same as $ABCD$ and preserves the half-plane bounded by $p$, thus $abcd=\rm{id}$ and $abc=d$. Analogous argument shows that $AbC=d$ in this example.

To show how this stuff works, define an isogonal conjugation. Let $ABC$ be a triangle with side lines $a,b,c$ (respective as usual). Let $a_1,b_1,c_1$ be lines through $A,B,C$ respectively belonging to some pencil (for example, sharing a common point). Then we may find the lines $a_2,b_2,c_2$ through $A,B,C$ respectively such that $a_2b=ca_1$ (that implies $a_2c=a_2bbc=ca_1bc=(ca_1b)^2ba_1=ba_1$, we used that $ca_1b$ is a reflection in some line), $b_2c=ab_1$, $c_2a=bc_1$. It yields $a_2c_2b_2=(ca_1b)(bc_1a)(ab_1c)=c(a_1c_1b_1)c$ is a reflection in a line, since so is $a_1c_1b_1$. Thus $a_2,b_2,c_2$ belong to a pencil, which is called isogonally conjugate to the pencil $(a_1,b_1,c_1)$. If both pencils correspond to points, these two points are isogonally conjugate with respect to the triangle $ABC$.

A more general result which is proved in the same spirit is the following

9 Lines Theorem. If $x_1,x_2,x_3$ are distinct isometries and so are $y_1,y_2,y_3$ and $x_iy_j$ is a reflection in a line for 8 pairs $(i,j)\in \{1,2,3\}\times \{1,2,3\}$, then all these lines are from the same pencil and the ninth product is also a reflection in a line from this pencil.

It implies the following

Coupling theorem. Let $a_1,b_1,c_1,\ell,a_2,b_2,c_2$ be seven lines from the same pencil such that $a_2=\ell a_1\ell,b_2=\ell b_1\ell, c_2=\ell c_1 \ell$ are symmetric to $a_1,b_1,c_1$ with respect to $\ell$ and $a_1,b_1,c_1$ pass through the vertices $A,B,C$ respectively of $\triangle ABC$. Then the pencils $(a,a_2)$, $(b,b_2)$, $(c,c_2)$ share a common line. In particular, if there exist the points $a_2\cap a, b_2\cap b, c_2\cap c$, they are collinear.

This is a general stuff, and now consider the midpoints $A_1,B_1,C_1$ of sides of a triangle $ABC$. Let $c_1\ni C_1$ be the perpendicular bisector to $AB$. We start with

Midline lemma. The midline $q$ through $A_1$ and $B_1$ is perpendicular to $c_1$.

Proof. Let $h$ be a perpendicular from $C$ to $q$. The composition of reflections $A_1qB_1$ is a reflection via some line $r\perp q$ (proved above), on the other hand it maps $B$ to $A$, so $r=c_1$.

Now we prove that the medians of $ABC$ are concurrent. Let the points $P,Q$ be symmetric to $A,C_1$ with respect to $A_1$, respectively, $k\ni A_1$ be the perpendicular bisector to $AP$. By Midline Lemma applied to $\triangle AA_1C$ we have $[QB_1]\perp k$. Consider the pencil $\Phi$ of lines orthogonal to $k$. It contains the line $\ell=[AP]$, $r_C\ni C, r_B\ni B, [QB_1]=r_Q\ni Q$, $r_1\ni C_1$. Note that the projections of $B$ and $C$ onto $k$ are symmetric with respect to $A_1$. It implies that $r_B$ is the image of $r_C$ under reflection in $\ell$ (in other words, $r_B=\ell r_C \ell$), analogously $r_Q=\ell r_1\ell$. The lines $r_C,\ell, r_1$ of the pencil $\Phi$ pass through the vertices of $\triangle AC_1C$. Thus there images under reflection in $\ell$ meet the opposite sides in concurrent points. These images are $r_B,\ell,r_Q$ respectively, they meet corresponding opposite sides of $\triangle AC_1C$ in the points $B,[CC_1]\cap [AA_1],B_1$ respectively. Thus by Coupling Theorem these three points are collinear as we need.

There is such a proof given by Hjelmslev; it is based on a clever application of central symmetries (point-reflections). You can find it on pages 102-104 of

Ф. Бахман, Построение геометрии на основе понятия симметрии. (Russian) [The development of geometry based on the concept of symmetry] Translated from the German by R. I. Pimenov. Edited by I. M. Jaglom. With an appendix by the author. Izdat. "Nauka'', Moscow 1969, 379 pp.

The proof builds upon earlier results and, thus, a self-contained proof is too long to be reproduced here.

See also the German original, pages 74-75 of

Friedrich Bachmann, Aufbau der Geometrie aus dem Spiegelungsbegriff. (German) Zweite ergänzte Auflage. Die Grundlehren der mathematischen Wissenschaften, Band 96. Springer-Verlag, Berlin-New York, 1973. xvi+374 pp.

Those who would like to read the proof in English are, alas, out of luck.