I want to prove the following theorem without using Dedekind's axiom(i.e. only with axioms of hyperbolic plane)
Given arbitrary line $\mathcal l$ and a point $\mathcal P$ which does not lie on $\mathcal l$, let $\mathcal Q$ be the foot of the perpendicular to $\mathcal l$ from $\mathcal P$. Then for any segment $\mathcal A\mathcal B$ such that $\mathcal A\mathcal B$ $\le$ $\mathcal P\mathcal Q$, there exists a line $\mathcal m$ which satisfies following three conditions. : (1) $\mathcal m$ goes through $\mathcal P$ (2) $\mathcal l$ is divergently parallel to $\mathcal m$ (3) For a common perpendicular segment $\mathcal M\mathcal N$ between $\mathcal l$ and $\mathcal m$, $\mathcal M\mathcal N$ $\cong$ $\mathcal A\mathcal B$
(divergently parallel means that two parallel lines have a common perpendicular line)