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.

Let $A, B, C, D$ be four points with the following properties:

  1. Three of those points are never collinear.
  2. The angles $\angle ABC$ and $\angle DAB$ are right angles.
  3. $C$ and $D$ are located on the same side of the line running through $A$ and $B$.
  4. $AD \cong BC$.

I want to prove without use of the parallel postulate that the line running through $C$ and $D$ is parallel to the line running through $A$ and $B$. However, whenever I try to prove this, I really want to use the fact that the sum of the inner angles of every triangle equals two right angles (which of course assumes the parallel postulate). I think that it is pretty easy to show that $AD$ and $BC$ are parallel to each other, however that does not seem to be of a big help.

How should I approach this exercise? Should I prove it directly or by contradiction? Do I have to use the fact that $AD$ and $BC$ are parallel to each other? I really only need a hint.

Thanks for any answers in advance.

EDIT: I don't know whether this is relevant, but all of the other Hilbert axioms are given.

share|improve this question
    
Are you sure this can be proven without the parallel postulate? Is it possible to model 1-4 on a sphere? –  Bill Cook Nov 11 '11 at 21:13
    
Can you prove it without using the parallel postulate? If you are working on the sphere (points are antipodal points, lines are great circles), you can find four points with all four properties, but neither is $AD$ parallel to $BC$, nor is $CD$ parallel to $AB$. –  Arturo Magidin Nov 11 '11 at 21:13
    
I highly assumed that it was possible to do so, as it was a homework exercise of a geometry class at my university. Do all other of Hilbert's postulates hold on the sphere too? (if not: I should probably add that to the question) –  Huy Nov 11 '11 at 21:16
    
Do you have a diagram? –  Emmad Kareem Nov 11 '11 at 21:28
    
Ah... the "same side of the line" could be the issue that disqualifies spherical geometry. In spherical geometry, you lose the notion of "in between", so you may also lose the notion of "on the same side". –  Arturo Magidin Nov 11 '11 at 21:29
show 1 more comment

1 Answer 1

up vote 3 down vote accepted

One ought to emphasize that "parallel" means the two lines under consideration never meet. On the sphere, all lines (great circles) meet, there are never any parallel lines.

Euclid's Proposition I.27 holds in a Hilbert plane, if you have a transversal with alternate interior angles equal, you have "parallel" lines. In your case, you just draw a line segment between the midpoint of AB and the midpoint of CD.

This is very similar to Exercise 10.10, on page 103, of Hartshorne, Geometry: Euclid and Beyond. Euclid's I.27 is Theorem 4.1, page 162 in Greenberg, fourth edition, Euclidean and Non-Euclidean Geometry.

share|improve this answer
    
It is quite late and maybe I have not thought about it properly, but how do I obtain equal alternate interior angles in this case after having drawn a transversal between the midpoint of AB? –  Huy Nov 11 '11 at 23:11
    
Go ahead and prove for me that "I think that it is pretty easy to show that AD and BC are parallel to each other" and then we can discuss this some more. –  Will Jagy Nov 12 '11 at 0:24
    
Let $AB$ denote the line running through $AB$ (not the line segment from $A$ to $B$). As $\angle DAB$ is a right angle, the adjacent angle $\angle BAP$ for some $P$ such that $A$ is between $P$ and $D$ is a right angle as well. Furthermore, $\angle ABC$ is a right angle and thus we have equal alternate interior angles. Hence, $AD$ and $BC$ must be parallel to each other. –  Huy Nov 12 '11 at 10:15
    
I believe I have found out how to prove it: Let $M_1$ denote the midpoint of the line segment $AB$. Then, by Hilbert's sixth congruence axiom, the triangles $\triangle AM_1 D$ and $\triangle M_1 BC$ are congruent. Now, I draw a line through $M_1$ which intersects $CD$ in $M_2$ such that $\angle M_2 M_1 A$ is a right angle. Thus, $\angle C M_1 M_2 \cong D M_1 M_2$ and again by the congruence axiom, we have that $\triangle M_1 C M_2$ and $\triangle M_1 M_2 D$ are congruent. Thus, $\angle M_1 M_2 D \cong \angle M_1 M_2 C$ and as these are adjacent angles, they are right angles. Hence, $AB \| CD$. –  Huy Nov 12 '11 at 10:45
1  
Yes, that is fine. On page 163 of Greenberg (fourth edition), we have Corollary 1, Two lines perpendicular to the same line are parallel (never meet). I think it a worthy exercise to prove this, carefully, from Hilbert's axioms. Note that a corollary is that no triangle has two right angles. On the unit sphere (not Hilbert), it is easy to have triangles with two, or three, right angles, take the edges of a triangle to be the coordinate planes, with $x,y,z \geq 0.$ Finally, in a hyperbolic (Hilbert) plane , it is possible to have lines that never meet but have no mutual perpendicular. –  Will Jagy Nov 12 '11 at 20:20
show 1 more comment

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.