# Aristotle's Axiom in Hyperbolic Geometry

I need to prove that Aristotle's Angle Unboundedness Axiom holds in hyperbolic geometry and I don't really know where to start. The problem says that we can take a segment parallel to one of the legs of an angle and make some construction based on that, but I don't really understand what that means. Any suggestions?

Thanks!!

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What can you use for that? How is that proved in euclidean geometry, exactly? –  Berci Nov 1 '12 at 21:23
What exactly is Aristotle's Angle Unboundedness Axiom? –  Will Jagy Nov 1 '12 at 21:45
I have found it by google: For any length $AB$ and any given angle $UOV\angle$, there exists an $Y$ on one side of the angle that orthogonally projected to the other angle gives $X$ and that $XY>AB$. –  Berci Nov 1 '12 at 21:50
@Berci Yes, that is the Angle Unboundedness Axiom in Euclidean/Neutral geometry, but I need to figure out a way to prove that statement holds in Hyperbolic geometry. –  roboguy12 Nov 1 '12 at 23:29
What elementary operations are you allowed to use? I don't yet see how a general parallel line will be of any use. On the other hand, the problem would be fairly easy to solve if you were allowed curves of equal distance. In hyperbolic geometry, these are not parallel lines, not even geodesics. –  MvG Nov 6 '12 at 15:17

The "Aristotle's Angle Unboundedness Axiom" establishes that given any segment $AB$, and an acute angle $\alpha$ (see figure), there exists a point $E$ on any branch of the angle such that if $F$ is the foot of the perpendicular from $E$ to the other side of the angle, $EF > AB$. In other words, the perpendicular segments from one side of an acute angle to the other are unbounded (Aristotle's Axiom is implied by the Archimedean Axiom).
It can be proved using the fact that "for any acute angle $\alpha$, there exists a line that is parallel to one arm of the angle and orthogonal to the other arm of the angle. In particular, there is a segment whose angle of parallelism is equal to $\alpha$." (image source)
Given a segment $AB$, and an acute angle $\alpha$ with arms $l$ and $m$, let $n$ be the parallel to one arm $m$ that intersects the other arm perpendicularly at a point $D$ (this line "$n$" always exists as established by the previous theorem, and is the hint mentioned by the problem).
Let $C$ be a point on $n$ such that $CD \cong AB$. Draw a perpendicular to $n$ at $C$, and let it intersect $m$ at $E$. Drop a perpendicular from $E$ to $l$ with foot $F$. Now we have a Lambert quadrilateral $\square CDFE$.
The angle at $E$ ( $\measuredangle CEF$ ) must be acute (as this is the fundamental assumption of the Hyperbolic geometry). Therefore $EF > CD \cong AB$.