Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
I think you need the intercept theorem or something similar. I think it should still hold in neutral geometry. See number E27-28. –  Raskolnikov Apr 18 '11 at 9:25

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
    
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! –  yunone Apr 18 '11 at 21:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.