This is a curious problem that is relatively easy to prove in Euclidean geometry, but has stumped me a good while in neutral geometry.
For a given triangle, how can one show that the line joining the midpoints of two legs is orthogonal to the perpendicular bisector of the third?
Suppose you're given a triangle $ABC$ with midpoints $D$, $E$ and $F$. Armed with the parallel postulate, it is fairly easy to prove that all the smaller triangles $ADE, DBE, FEC, DEF$ are congruent, and so $D$ and $DE\parallel BC$, from which it follows that the perpendicular bisector of $BC$ is also perpendicular to $DE$.
But in neutral geometry, there is no parallel postulate, so I don't see a way to make a similar argument. I think I lose any information about $DE$, $DF$ and $EF$ concerning congruences. How could one then show that $FG\perp DE$?
At best, I was able to prove it in the special case that $ABC$ is equilateral using repeated side-side-side arguments, but I can't find a proof for any arbitrary triangle. Thank you for any thoughts.