 Welcome back to our lecture series, Math 3130, Modern Geometries for Students at Southern Utah University. As usual, I'll be a professor today, Dr. Andrew Missildine. In this video, we're going to conclude our discussion of Ordered Geometry that is talking about the betweenness axioms of Hilbert. We've previously introduced the incidence axioms. We've been talking about the betweenness axioms for the last few lectures. In the next lecture of this lecture series, we'll start talking about the congruence axioms. In lecture 13, we want to hit the pinnacle, the climax of the betweenness theory, and talk about the so-called crossbar theorem. Now, before we state and prove the crossbar theorem, there's a new definition I want to introduce to us. It's the idea of being between rays. We've previously come up with the idea of betweenness of points. In fact, this is one of the undefined terms of betweenness theory of ordered geometry. The idea is that if we have three collinear points, there's a betweenness relationship that holds. This would be an illustration to suggest that B is between the points A and C. Now, we can also do the same thing with rays. Imagine we have three rays that have a concurrent vertex, so we have a picture that looks something like this. Let's say that B is the common vertex of these three rays, and then our rays, we need some other point on them, so we have A, we have C, and we have D, something like that, so we have these three rays. And so what we want to find is what does it mean for a ray to be between two other rays? So we say a ray, B, C, in an ordered geometry is between two other non-opposite rays, B, A, and B, D. So the fact that the rays, B, A, and B, D are non-opposite means that B, A, D, this is not a collinear set, it's a non-collinear set, okay? So they're not just two halves of the same line. So we say that ray B, C is between the rays B, A, and B, D. If C is an interior point to the angle A, B, D. So the angle A, B, D would be this region right here, everything in the middle, and so this point C would be an interior point. So we'd say that the ray B, C is between B, A, and B, D. And we use the same notation here. We say that B, A dash B, C dash B, D, something like that. We'll say some more about betweenness of rays in the second video here, but I just wanna define it right now so we can use it when we talk about the crossbar theorem. Now we're ready to state and prove the so-called crossbar theorem. So the statement's the following. If we have a ray B, P in some order geometry so that B, P is between two other rays, B, A, and B, C, in other words, the ray B, P is interior to the angle, particularly the point P is interior to the angle A, B, C. Under those assumptions that it must be true that the ray B, P intersects the line segment AC and it happens at some point, let's call it D, that rest between A and C. So I wanna draw this picture for us and I also wanna do it in such a way that we can see the picture for the entirety of this proof. So we're gonna draw something like this. So let's label it here where you have an angle ABC, B is the vertex, we have some other point A right there, we have some other point C right here and let's say we have some interior point P that's inside the angle and then consider the ray that emanates from B and passes through P. So we get this ray like the following, okay? And so we have this interior ray, then we are gonna consider the so-called crossbar. So between these two points A and C that determine the angle, the line segment between them is commonly referred to as the crossbar of this angle. Of course, with different points A and C you'll get different crossbars but this is a crossbar of the angle and this is why this is called the crossbar theorem. The crossbar theorem tells us that if we have a ray which is inside of an angle, that ray must intersect the crossbar and it'll happen at this point called D. So we wanna prove the existence of this point D. So let's work through the details of this proof then. So the first thing we're gonna do is we're gonna extend the line BC so that we have this other point C prime in consideration here. Now use the extension axiom to get that point C prime. So by the extension axiom, it's guaranteed that the point B is between the point C and C prime. So by construction, the line segment that goes from C all the way down to C prime will intersect the line BP. It happens at the point B of course. And let's call the line, let's call the line here L, okay? So this line L is gonna be of consideration for what's going on here. So C and C prime are on opposite sides of L. That's something to remember. The next thing I want us to consider is consider the angle BPC prime. So BPBC prime would look something like this. We claim that A is an interior point to this angle right here. Now when you look at the picture, it seems clear that's the case, but we can't base our proof upon an illustration because I'm drawing a Euclidean picture. I need to be proving this in ordered geometry, not in Euclidean geometry. So I'm in a much broader theory where things might not look how they appear. Now the good news is we've already taken care of this in a previous proposition in our lecture series. We did this at the end of lecture 12, by the way. It was a very, very long proof, but we proved exactly this situation that the point A would be in the interior of this angle BPC. It seems intuitive, it seems clear from the picture, but we had a really long proof to prove this. So I would have to argue that if it's a really long proof, then it's not intuitive. And it took a long time to do. But in particular, because A belongs to the angle BPC prime, and since an angle is an intersection between two half planes, a half plane, of course, if you have an open half plane, that means that the two points are on the same side of the same line. This implies that the points A and C prime are on the same side as L. So notice what we have here. A and C are on the same side as L, but we also have C and C prime on opposite sides of the line L. So applying the plane separation theorem, it has to be the case that A and C are on opposite sides of L. Now notice here that A, C is exactly the crossbar we had considered before, okay? So we've now proven that the crossbar intersects the line L at some point, we're gonna call that point D, okay? Now that feels like we're almost done, but the problem is we now know that L is the intersection between A, C, the segment and L, the line. We have to prove that the intersection between, well, we have to prove that D is on the ray BP, all right? Which a line is basically two rays put together. Could D actually be like something weird? Could this intersection actually be over here somewhere on the opposite ray? That seems weird by the drawing, but the drawing isn't the proof. How do we know that? Okay, so let's prove that D does not live on the opposite side of the line. It doesn't live on the opposite ray. Now by assumption, A, B, C are non-colinear points because they're three points of an angle, they're necessarily non-colinear. As such, because these are not collinear, B can't be between A and C. A between this relation implies that they're collinear, they're on the same line. This was our first axiom of betweenness. Now since D is between A and C and B is not, it must be the case that B is not equal to D. That seems like a silly thing, but it does have to be said. Also, again, by that previous proposition, there was three parts of that proposition, we used the third part. The second part tells us that the opposite ray of an interior ray to an angle only intersects the angle at the vertex B. So by proposition, part B of the previous proposition, if we look at the opposite ray, so this is what we call negative BP from before, the intersection between negative BP and the angle ABC, this intersection has to be the point B. So if D were on the opposite ray, since it's, because it's on the opposite ray, what I'm trying to say over here is that it would have to be B, right? Because D belongs to the angle, D belongs to the angle there, the between cross lemmas coming into play right there, right? Because D is between A and C on the cross bar, the between cross limit tells us that D is an interior point to the angle. So it's here in the angle, but if we're on the opposite ray, the opposite ray intersects the angle only at the vertex B, but D is not equal to B. So that means it's not on the opposite ray. Now, since every line is a union of a ray and it's opposite, if D is not on the opposite ray, then ding, ding, ding, ding, it has to be on the ray BD, which then is exactly, that puts it exactly where we want it to be. D is the intersection between the line segment AC and the ray BP. The cross bar theorem is a very important theorem, not just for a betweenness theory and order geometry, but also in geometry in general, because the cross bar theorem gives us something that we really much want. It gives us intersections. It'll give us intersections of lines, specifically gives us the intersection between a line segment and a ray, but those rays and line segments can be extended to lines. So anything that guarantees the intersection between two lines is a very powerful, very useful theorem, which is why the cross bar theorem is one of the most important theorems of ordered geometry.