This fact holds true in absolute geometry, and I would like to see an elementary synthetic proof not using the classification of absolute planes (Euclidean and hyperbolic planes) and specific models. Actually I know such a proof (from Skopets - Zharov book), but it uses the third dimension which has to be justified itself. 


UPDATE. Here is the aforementioned proof using stereometry, maybe somebody may see how to get pure 2d-proof.


Let triangle $ABC$ lie in a horizontal plane $\pi$. Draw unit vertical segments $AA', BB'$ above $\pi$ and $CC'$ below $\pi$. The segments $AC$ and $A'C'$ pass through the midpoint $K$ of $AC$, $BC$ and $B'C'$ through the midpoint $M$ of $BC$. Let the medians $BK$ and $AM$ meet at $G$, the segments $BA'$ and $AB'$ at $P$. The planes $BA'C'$ and $AB'C'$ have three common points $C',P,G$ which are therefore concurrent. Thus their projections to $\pi$ are concurrent. But $C'$ projects to $C$, $P$ projects to the midpoint $N$ of $AB$ (by the symmetry of the quadrilateral $A'ABB'$). Hence $G$ belongs to the third median $CN$ and we are done.



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


