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 $DE$.

enter image description here

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 $FG\perp DE$?

enter image description here

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.

  • $\begingroup$ I think you need the intercept theorem or something similar. I think it should still hold in neutral geometry. See number E27-28. $\endgroup$ – Raskolnikov Apr 18 '11 at 9:25
  • $\begingroup$ You can find this result as lemma 13.1 of W. Schwabhäuser, W Szmielew, A. Tarski, Metamathematische Methoden in der Geometrie. $\endgroup$ – Julien Narboux May 22 '15 at 14:33

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.


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.

  • $\begingroup$ How timely that I just logged on! Thanks for your answer Matt, it seems reminiscent to the proof that for any triangle there exists a Saccheri quadrilateral with summit angles equal to the angle sum of the triangle. Much appreciated! $\endgroup$ – yunone Apr 18 '11 at 21:06

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