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.

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    The title is cute, but perhaps it is a good idea to make it more clear what is being asked so that homework-seekers don't get the wrong idea of what sorts of questions belong on this site. – LSpice Oct 10 at 19:01
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    @LSpice fixed, but I am afraid they do not look for previously asked questions. At least they behave as they do not. – Fedor Petrov Oct 10 at 19:09
  • The following might be useful and could be proven quite easily in absolute geometry: for three points $A,B,C$, letting $A',B',C'$ be the midpoints of the segments $BC,CA,AB$ correspondingly, the perpendicular to $AB$ at $C'$ is also perpendicular to $B'A'$. – Uri Bader Oct 14 at 8:51
  • Another remark, though not in the spirit of what you're asking and also not checked: consider the operator that take a triangle ABC to its medial traingle A'B'C', then I think that iterating this operator will converge at the centroid of ABC. – Uri Bader Oct 14 at 9:06
  • @UriBader no, in general it converges to something different, since $A'B'C'$ and $ABC$ have different centroids. – Fedor Petrov Oct 14 at 11:14

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