In neutral geometry, the line connecting midpoints in a triangle is orthogonal to the perpendicular bisector? 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.
 A: The key is to focus on the line DE (call it $L$), which is (provably) in the middle of the picture.
We won't even think about line BC until the very end.  If we consider curves of constant distance from $L$, one passing through A, and one passing through B and C (we will prove this), the picture has a nice symmetry, bringing us so close to the solution that the final step is easy.
Proof:
Point A is at some distance from $L$.  If we drop perpendiculars from A and B to $L$, we get congruent triangles (AAS) and so B is the same distance from $L$ as A.  Similarly for A and C, so by transitivity B and C are the same distance from $L$.
Say the perpendiculars from B and C meet $L$ at M and N.  Now define GF to be the perpendicular bisector of MN, with F on BC (but we do not yet know where or at what angle GF hits BC).  Triangles MBG and NCG are congruent (SAS), so BG=CG and ∠BGM=∠CGN, giving us ∠BGF=∠CGF, which means ▵BGF and ▵CGF are congruent (SAS).  This finally gives us that GF, which we already know to be perpendicular to DE, is indeed the perpendicular bisector of BC.  Q.E.D.
