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 $\alpha$."
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 $EF > CD \cong AB$.