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.
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The definition of parallel lines is simply two lines who don't meet.
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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 there is a congruence relation between them.
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And we have the (Playfair'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 $r$.
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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 $h$.
suppose $\angle (h,s) < \angle (r,s)$ (this is not defined but it is not hard to come up with a definition either) 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. Please don't use Tarski's axioms, I don't know what a sum of angles is.
Best Answer
Proving this in full detail from Hilbert's axioms takes a lot of work, but here is a sketch. Suppose $\ell$ and $m$ are parallel lines and $n$ is a line that intersects both of them. Say $n$ intersects $m$ at $P$. Now let $m'$ be the line through $P$ which forms angles with $n$ that are congruent with the the angles that $n$ forms with $\ell$ (using axiom IV,4). If we can prove that $m'$ is parallel to $\ell$, then we must have $m=m'$ by axiom III.
So, suppose $m'$ was not parallel to $\ell$. Then the lines $\ell, m',$ and $n$ would form a triangle. Since $\ell$ and $m'$ form the same angles with $n$, this triangle will have two angles (the two angles on $n$) which add up to a straight angle. Now you can prove this is impossible by essentially the same argument as in Euclid; see https://mathcs.clarku.edu/~djoyce/java/elements/bookI/propI17.html, for instance. (Note that Euclid's argument relies on the existence of midpoints, which Euclid proves by constructing equilateral triangles by intersecting circles. Doing this with Hilbert's axioms requires the use of the completeness axiom and is pretty complicated. Alternatively, without the completeness axiom, it is still possible to construct an isosceles triangle with a given base, which is enough to obtain the midpoint of the base.)