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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.

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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.)

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  • $\begingroup$ thanks! But I got to ask if you are sure this is the way to prove it, because it would be weird for Hilbert to say it is easy to deduce this theorem from the previous axioms he gave in the book. $\endgroup$ – hellofriends Oct 28 '20 at 17:33
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    $\begingroup$ I would guess that Hilbert would expect his audience to be quite familiar with Euclid and with more recent work on axiomatizing geometry, and so would be able to fill in the gaps with arguments that would already be familiar to them. In particular, this line of argument long predates Hilbert and was well-known as being essentially how you prove the equivalence of Euclid's parallel postulate and Playfair's axiom (which is the axiom of parallels that Hilbert uses). The only new nontrivial work would be how to formulate all the details in the context of Hilbert's axioms. $\endgroup$ – Eric Wofsey Oct 28 '20 at 19:20

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