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.
