 In this video, we are going to prove our longest and most technical theorem associated to order to geometry. Imagine that a point P is contained inside the interior of the angle ABC. So it's not on the rays B, A, or B, C. It's inside the angle. And what we're going to prove is actually three things. We're going to keep it all as one proposition because they're very much related to each other. And these are all in preparation to prove the so-called crossbar theorem, which is probably the most important theorem for order geometry. And like I said, all of these statements are going to be quite technical. It takes a little bit of time to go through it tedious, but it's very important we establish these things because this will prove our intuitive notion of how rays and points and angles interact with each other. So we've defined previously the notion of a ray, right? We have this ray BP. So this is defined to be the points B and P. And then you take the union of that with all the points C on the line determined by BP such that C is between B and P. So that's how we can define it using between this, right? But I want to also point out that we've introduced the idea of orderings of lines. We did that in lecture 11. And so this right here, we can really think of it in the following manner. This is the interval B towards infinity, where of course we're assuming that B is less than P in this situation. If B was greater than P, this would actually be negative infinity to B. But that's what this is defined to be. So we're looking for the set of all, we're looking for all the points C that are on the line determined by BP, such that your point C is greater than or equal to B, something like that. So those are two ways of describing rays. This could be called a closed ray because it includes the boundary point B. So it's natural to also talk about the idea of an open ray. We're not going to talk a lot about these, but at least it's important in our current conversation. We could talk about the open ray BP, which it's just the ray BP, but you remove the boundary point, which for our purposes basically means we take C is greater than but not equal to B. So that's what's going on there. We're just removing the boundary point. So the first condition that we're going to prove right here is that an open ray BP is this open ray is contained in the interior of it. So if you have an interior point to an angle, then the open ray that emanates from the vertex of the angle to that interior point, that entire ray is contained inside of, that's entirely inside the interior of the angle there. So think about it from a drawing. There's two other statements we're going to prove, but we'll prove those in just a second. So we'll state them in just a moment as well. And I should put this in such a way that we still see the statement we're trying to prove right now. So imagine we have an angle and so hypothetically it looks something like this. The vertex is B, let's say this is point A, this is point C, and we have some point P that's interior to it. So what we claim is if we take the ray going from B to P, that whole rays in there, which clearly the point B is on the boundary of the angle. So if you exclude that point, the entire open ray is inside. That's all we're trying to prove here, okay? So how are we going to do this? So the idea is we have to take an arbitrary point Q that is inside of the open ray BP. And we have to argue that the point Q is interior to the angle ABC. Now by the construction of these things, right, by trichotomy of the order, we have because, but again, by assumption here, we're assuming that B is less than P. Well, Q has to be comparable to that. We have a couple options here. We have that Q is less than P. We have that Q is equal to P, or we have that Q is greater than P. If we turn those into statements of betweenness, that says that Q is between B and P, P and Q are equal, or P is between B and Q. So again, this is the trichotomy we're going here, we can do betweenness, we can do ordering, the two are equivalent at this, at this venture here. Now one of these options is a lot easier than the other. In particular, if Q is equal to P, we by assumption have that P is in the interior, so Q would be in the interior. So that case is super easy. All right? So we then need to look at the other two cases. So in the other two cases, we can conclude that B is not on the interval P to Q, or Q to P, whichever direction you want to go. So if you look at this right here, B is not in them, because if B were in there, that would then imply that we have Q dash B dash P, Quebec dash Bravo dash Papa, or you could say that Q is less than B which is less than P, something like that. That's not compatible with the statements we saw earlier, right? The original betweenness one here, and particularly when you look at the ray, the way we've constructed the order, B is less than or equal to everything on the ray, and it's strictly less than everything on the open ray, right? So we can't have B being in there, that would violate the betweenness trichotomy or the ordering, it doesn't matter which one you use, you can use either, I keep on talking about both, you can use ordering or betweenness here. Ordering often feels a little bit cleaner, and it's probably more intuitive for many of us here, so feel free to use that. We can't have B being in there, because that would imply that Q is less than B, which is less than, less than or equal to P, which that's not possible based upon the assumptions on B right here. So that happens if you're in this case, or this case, B is not in there, okay? For which also it's important to point out here that the line PQ intersects the line AB at a unique location, that if there was some other point, right? Like we know B is not in here, but we also know there's no other point on AB that would be on this line segment right here, because if we extended this line segment into the line, it would intersect the line AB at B, but B is not on the line segment. So the only point of intersection between the two lines is B. This interval is a subset, so if they were to intersect, that's really what we're trying to summarize here, B is not in here, okay? And so because this line segment does not intersect the line AB, because if it did, it would contain B and B is not on there, because of that, we see that P and Q are on the same side of the line AB. Now by a similar argument, if you look at the line BC, we know since B is not on this interval right here, we know that the line segment PQ cannot intersect the line BC, because if it did, it would happen at B, and it can't be B, therefore P and Q, the line segment doesn't intersect BC either. So we've now established that Q, it's on the same side as P of the line AB. We also know that Q is on the same side of P that BC is, so therefore Q is in the intersection of these two half planes associated to the lines AB and BC. And so be aware, this set right here is the same as the half plane of BC associated to the point A. And this half plane is the same one as the open half plane associated to the line AB, C right here. And so since Q is in both of its intersection, and we see that Q is then inside of it. And so we actually took care of both of these situations simultaneously. It didn't matter which the ordering was, we get that Q is in that half plane, it's in that half at plane intersection. So we see that Q is in the intersection, it's in this open angle ABC, the interior. So that was the first property that took a little bit just to do that one. It's not too complicated, but we just have to familiarize ourselves with what's going on here. The next one, property B, I haven't even stated yet, let's talk about it right now. Well, whenever you have a line, it's, well, let's say this, whenever you have a ray, pictorially it might look something like this. You have your point B, which is the vertex of the ray. You have some point P right here. Well, B and Q, they determine a ray yet, but they also determine a line. And so we can go the other way around. By extension, there's some other point over here called like, we're going to call it negative P for a moment. There's some other point on the other side. So that B is between P and negative P. If you look at the ray, B, negative P, which we're going to commonly denote that as negative BP, this gives us the so-called opposite ray of BP. So in other words, every line is the union of a ray with its opposite. We claim that the opposite ray, again, denoted negative BP intersects the angle ABC only at the point B. So let's consider what that diagram would look like. So reconstructing the picture, since it's no longer on the screen, of course, we have the line AB. We have the line BC. So labeling it, we get something like B right here, A right here, C right here, P with some interior point, like so. So we can construct the ray from B to P. That would look something like that. Then the opposite ray should then be going the other way around, something like this. Again, we have this point, negative B might be a little bit confusing here. Let's call P prime to match off also with what I say right here. But we have this, we claim that these two rays, BP and BP prime, which is just negative BP, they inters, this, well, we know they form a line, but we know we're trying to prove that the opposite ray is only going to, it's only going to interact with this angle at B. But again, that's how the diagram suggests it. But it's because I'm drawing an Euclidean diagram, which Euclidean geometry isn't or geometry, it's going to satisfy it. How do we know it doesn't do something like weird like this? This is the opposite ray. This is B prime, P prime, excuse me. Could P and P prime both be inside the line? Well, you know, if you had some type of like weird wraparound feature, it could be that if you extend past some line at infinity, it wraps around the other side. And so P prime is here. It might seem like there's nothing to prove here, but there's a lot to prove here. The fact that we want our geometry to behave the way we expected, that's exactly what we're trying to prove in this moment. So let P prime be a point so that B is between P and P prime. That is, P prime is on the line determined by P B, but it's on the other side. So then the opposite ray of B P is determined by this point P prime. So the ray B P prime is just the opposite ray. So by construction, we have the following thing. We want to show that the intersection of this ray with the angle ABC. So this is the closed angle. We include the boundary here is just the point B. Okay, I'm going to slide this down a little bit. So we can read the rest of it and I'll relabel some things. Here's P. C itself doesn't really matter. It's just multi-pointed in direction. So we can kind of slide it around and pick a different point so it fits still on the screen right here. So let's then consider it. Let's take the intersection of the angle ABC with the opposite ray B P and call that point Q. We want to argue that Q has to be just B. That's the only possibility here. And so let's consider that point. Okay. So let's take this point. So it's on the opposite ray. So we have our point Q right here. I want us to consider the line segment Q P. Consider that right here. Well, this line segment is going to intersect the line BC. You can call it L for the moment. It's going to intersect the line BC at the point B. Why is that? Well, by construction, we have that P and Q are on opposite sides of L. How do we know that? Well, P and P prime are on opposite sides of L because B is between them. We have that Q and P prime are on the same side because Q is inside the opposite ray, like so. This is what we proved. This is what we basically proved in part A about this open right here. So unless Q is B, of course, then there's no problem whatsoever. So if Q is some other point, we're going to get that the interval Q P prime is going to be on the same side. So then by plane separation, we get that Q and P are on opposite sides of L. I was trying to mention right now. So we have that. So that's because Q is on the opposite ray. Well, we also know that P is an interior point to the angle ABC. So as such, we know that A and P are on the same side of L. That's what it means to be the angle, right? Because the angle itself, the interior of the angle is the intersection of two open half planes, which means that A and P are going to be on the same side of the line L. So then by plane separation, what have we established so far? P and Q are on opposite sides of L. A and P are on the same side. So by plane separation, we get that A and Q are on opposite sides of the line L. So we can think of that in the following manner. If we look at the line segment between P and Q and A, excuse me, then we anticipate there's going to be some point of intersection right here. All right. Well, if A and Q are on opposite sides of the line L, that implies that Q is not an interior point to ABC, the angle ABC. Because after all, the angle, this angle is the intersection of two half planes. So in particular, to be in the angle, you have to be on the same side of L that A is, but these ones aren't. So that gives us a problem. All right. Thus, we have that Q is on the ray BA or it's on BC. So think about what we had there earlier, right? Because we're assuming that Q is inside of the angle AB, but we just ruled out the possibility that it can't be in the interior. So it's got to be one of the boundary rays, which then the rays are BA and BC. So without the loss of generality, let's say that Q is on the ray BA. It really does make much of a difference and I'll show you why in just a second. So if we assume that Q is on the ray BA, then that means that Q belongs to the ray BA. It also belongs to the opposite, the opposite ray of BP for which these are rays. It's like half of a line. So in particular, this is a subset of the intersection of the line BA with the line BP like so, which these two lines only intersect at a single point. It has to be B. So the two rays BA and the opposite ray of BP, they have to intersect at B. And so if Q is on the angle, which means it has to be one of these rays and it's on the opposite ray, then it has to be at B, right? Which enforces B to be Q, which is what we have said it had to be. Now, of course, if you assume that Q was actually on BC, same thing, BC and the opposite ray of BP intersect at B. So Q has to equal B and that proves as the second condition, which as a reminder, the second condition told us that the opposite ray of an interior ray to an angle intersects only at the vertex. All right, now the third one. Like I said, this is a lengthy proof. There's a lot of parts to be doing for this proposition, but it establishes many important principles that we need when we work with order to geometry. So then this third principle that we're going to prove here is that if B is between C and D, then we have that A belongs to the interior of PBD. What's D here? So we're just assuming that we have some other point D, right? If B is such a point that it's between C and D, then that tells us that A actually belongs to the interior of the angle PBD. So again, drawing a picture will help us a lot here, understanding what's going on here. So we're going to first start off by drawing the angles we keep them in drawing. So we have BC, excuse me, BA and BC, something like that. So B will be the vertex right here. We have some point A here. We have some point C right here. P is some interior point. And what we're saying is D is such a point that B is between B and C. So D has to belong to the line determined by B and C. So we're going to get something like the following extending the line. We have some point D over here. So B is between C and D. That's what we're trying to go for. So then when you look at the angle, right, so look at the angle, whoops, the angle, the angle D or PBD, like so, our claim is that A is going to be interior to this angle, okay? So note, because we can't just use the picture, even though the picture seems to suggest it, we've said that like twice already, I don't even say it again. We want to show that A is interior to the angle PBD. So to show that you're inside of an angle, you have to be inside of half planes. So that's what we're going to try to do right now. So look at A and P, right? They're on the same side of the line BD because BD is the same line as BC. And since P is interior to the angle ABC, that means that A and P are on the same side of the line BC. Right? BC and BD are the same line. And so since P is interior to the angle AB, we get that A and P are on the same side of the line BD. So A belongs to the open half plane determined by the line BD and the point P. So that's half of it. That's the first half. Let me scroll down a little bit so we can see more of the proof here. All right. So the next thing we need to argue is that we need to argue that A and D are on the same side of the line as BP. Like so. That's what we're going to try to do. So if A and D are on the same side of the line PB, then we would be done because they now imply A is inside this half plane. That's what it means to be inside of the interior of the angle. So what we're going to do is we're going to proceed by contradiction. So it's hard to prove directly that A and D are on the same side. So we're going to assume they're on opposite side. So for the sake of contradiction, assume that A and D are on opposite sides of the line PB. Then there exists a point E on the intersection of PB, the line, and AD, the segment. And in particular, E is going to be between the points A and D. And so we could try to draw such a thing. So we're going to draw this line segment. There's some point E that's between them. But E is also on the line PB. So something weird is happening like so. And let me use a different color to help emphasize this. The line PB, it apparently turns and does something like this. So we have our line PB like so. All right. Could we have something bizarre like this happening? I mean, it doesn't seem to make sense, but that's what we're trying to prove right now. So to work out this contradiction, we're going to use the between cross lima. So the between cross lima, because we've satisfied the conditions of the between cross limit here, tells us that E is inside the interior of the angle ABD. So it's inside this angle right here. Why is that? Well, the assumptions, well, what the between cross limit tells us is that if you have a point between A and D, like E is echo, then that tells us that echo is inside of the angle. So the between cross limit relates being interior to the angle to be between points. So there's that relationship there. What does that have to do with anything? Well, if E is inside of the angle ABD, in particular, that will tell us that the line segment AE doesn't intersect the line. AE won't intersect the line here BD. And so A and E are on the same side of the line BD, which of course is the same as the line BC. So we have this right here. These guys are on the same side. Let me highlight this for later. Oh, boy, without scratching it out. AE are on the same side of the line BD. But here, we already know that A and P are also on the same side of the line BC. BC and BD, of course, are the same line. AP is on the same side of BC because P is interior to the angle ABC. So then we can conclude that P and E are on the same side of BC by plane separation. So let's remind ourselves what we have here. P and A are on the same side of BC. A and E are on the same side of BC. So that implies by plane separation that P and E are on the same side of line BC. But also remember by contradiction, we're assuming that A and D are on opposite side of PB. We'll get to that in a moment. Let me zoom out a little bit so we can still see the picture, but then we can see the rest of the proof. Where were we again? These things get a little bit long. So B is not between P and E. B is not between P and E. Because after all these, this is one weird curvy line. And the way that I drew it, it looks like B is between P and E, but it turns out that can't be the case. Because if B was between P and E, that means the line segment PE would intersect the line BE at the point B. Why is that a problem? Well, we just said that PE doesn't intersect the line BC, which is the same thing as the line BD, which if it did intersect, it would happen at B. Because P and E are on the same side of the line. So B is not between them. So that then tells us, since B is not between B and E, so my picture doesn't seem to make any sense anymore, which is understandable. I'm developing an inconsistent theory right now because we're assuming something that's actually false, but we don't know that yet. We have to conclude instead that E is on the open ray BE. Because remember, what is this open ray doing again? The open ray, we're going to gather all of the points, all the points on the line BP, such that we are greater than BE, like so. But in particular, since B is not between P and E, what's going to have to happen is we have B-P-E or we're going to have B-E-P. One of those between the statements has to happen. Both of these would imply that our point E is greater than BE with that ordering there. So we get that E is inside of this open ray, like so. So what do we have here so far then? Well, if E belongs to the open ray BE, then this actually tells me the open ray is the same thing as the open ray BE, right? The vertex matters, but the other point doesn't matter as much. I could interchange P and E in terms of defining this ray. So since E is on the open ray BP, this tells us that P is on the open ray BE. But BE as a ray is a subset of the interior of the angle ABD, right? How do we know that? Well, remember by the between cross lima, E was interior to the angle ABD. And so by the first part of this proposition we're currently proving part A, we know that because E is an interior point, this entire open ray is inside of the interior of the angle. And as P belongs to this ray, that would imply that P is interior to the angle ABD, all right? And as such, since the interior of the angle is the intersection of two half planes, in particular, P and D are going to be the same side of the line AB, because that's one of the half planes that determines it. I'm going to move it so we can see the last paragraph. Some of my pictures obscured, but admittedly at this point, the diagram doesn't seem really reliable because it's really like E. It's not over here. It's like really over here, but things are getting really weird because, again, we're going to find a contradiction just a moment. So we now know that P and D are on the same side of the line AB. Also, P and C are on the same side of the line AB, since P is interior to the angle ABC, right? The fact, remember, angles are intersections of two half planes. Since P is interior to the angle ABC, that means with respect to the line AB, P and C are on the same side. So notice what we have here. We have the same line A and B. So P and D are on the same side. P and C are on the same side. So by plane separation, we have that C and D are on the same side of the line AB, which this finally gives us our contradiction, because the line segment CD intersects the line AB at the point B. Remember, by construction, B was between C and D, like so, right? So we can't have that C and D be on the same side because then they shouldn't intersect AB, but they do at B. So take a deep breath there. We found our contradiction. So then that shows that A and D were, in fact, on the same side of PB, which that implies that A was an interior point to the angle. And that then finishes this proposition. So that brings us to the end of Lecture 11. Lecture 11, at least this proposition we just proved in three parts, was very, very technical. But it takes care of all of the hard work we had to do with order geometry. All these intuitive ideas about interiors of angles and rays have now been established. And so in Lecture 13, we will prove the famous crossbar theorem, which will give us one of those important results for order geometry, because we laid the groundwork in this video right here.