Prove that, when two parallel lines are cut by a third line, they make congruent angles.
I'm not using Euclid's axioms, but instead I'm using Hilbert's. This is Theorem 19 of Hilbert's "The Foundations of Geometry" (PDF link via berkeley.edu).
Theorem 19. If two parallel lines are cut by a third straight line, the alternate-interior angles and also the exterior-interior angles are congruent. Conversely, if the alternate-interior or the exterior-interior angles are congruent, the given lines are parallel.
The definition of parallel lines is simply two lines who don't meet.
The definition of angle is a bit long it is on page 9. I think the important bit is that there is a bijection between angle and rays from a certain point.
And we have the (Euclid's) Axiom of Parallelism (page 7): Given a line $r$ and a point $A \notin r$ we can always draw one, and only one, line through $A$ parallel to $r$.
We can use that angles opposite on a vertex are congruent.
My attempt was this:
Given two concurrent lines $r$ and $s$, s.t. $r \cap s = A$ let's take a point on $s$ different from $A$ and draw the one parallel line to $r$ from it, call it $h$.
supose $\angle (h,s) < \angle (r,s)$ and let $h'$ be the ray (line) such that $\angle (h',s) = \angle (h,s)$ can we prove that $h'$ is another parallel line or that it is line $s$?
I think if we assume the angles formed are different than we would have two parallel lines through $B$ but I'm out of ideas.