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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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    All theorems that hold in neutral geometry also hold in hyperbolic geometry.2016-05-27

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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 $\alpha$." Aristotle's axiom holds in hyperbolic planes - demonstration(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 $EF > CD \cong AB$.

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If you understand the way distance, parallelism, and perpendicularity are modelled in the Klein disk model of hyperbolic geometry then you would understand very well this answer. Take a look at the following figure.

enter image description here

Every segment (within the circle) of a Euclidean straight through the the crossing point of the tangent lines (white) is perpendicular, in the hyperbolic sense, to the hyperbolic straight determined by $AB$. The figure depicts that for any $\alpha$ there are segments (for example $AB$) that cannot be exceeded by perpendiculars dropped from the hypotenuse of a given triangle.