 In previous lectures of this lecture series Math 3130, particularly as we studied finite geometries, we often see phenomenon like the following. Take for example, fan geometry. For example, the five point fan geometry is now illustrated here on the screen. One thing that was very interesting with these finite incidence geometries is you often had the case where you had different types of lines in your geometry. For example, there's this one line that has four different points upon it. But then you have these four other lines that only have two points on them. So these lines are not really the same because of the, you count the number of points incidence to each of the lines. And we could come up with several mini examples that we've considered, of course, in this lecture series. Another one I'm gonna provide here on the screen right here, for which we had this geometry which had these two lines that had three points on them. And these other lines which had two points, same type of phenomenon here that these lines we might say are incongruent with each other. They're not congruent lines. They're different lines in the same geometry. Now for incidence geometry, we have no problem with this whatsoever. When it comes to ordered geometry because of the extension axioms and then later on the notion of density, all lines had an infinite amount of points on them. And so this idea of congruence of lines becomes a little bit more muddled there, is that, well, okay, they all have infinitely many points which of course, could there be cannibly many points, uncannably many points? We didn't really pursue that very much. But also with this idea of congruence that we've now introduced with regard to a congruence geometry, even if we have points and infinitely many points like there's a congruence of this segment, there's a congruence of the segment and the segment, how does that affect things? So do we have a situation like an incidence geometry where you could have completely different types of lines in your geometry, or is there more uniformity? And so in this video, I wanna prove this so called congruence of lines theorem in a congruence geometry, which is gonna essentially give us the statement that any two lines in a congruence geometry are the same. That is, there's a congruence there. So how does one make that explicit? What do we really mean that say two lines are congruent? We can talk about line segments, so what does it mean for two lines to be congruent? Well, let's say what that means. So suppose we have two lines, two distinct lines in a congruence geometry. So L and L prime. And by congruence geometry, I mean we have the four axioms of incidence, the four axioms of betweenness and the six axioms of congruence. So given these two lines, L and L prime, there exists a one-to-one correspondence AKA a bijection. We will call it F as it goes from L to L prime. You wanna think of this as a function between the points of L and L prime because in an incidence geometry, the points incident to a line characterize the line. And therefore, we can define these functions between lines. Literally, this is a function between the points on L and the points on L prime. There's no ambiguity there. But this bijection is not just a bijection that says the two lines have the same cardality of points. There's further statements we can say here that if you take a line segment AB, where AB or points on L, this bijection will preserve the congruence of the line segment. So F of A is the image of the point A, F of B is the image of the point B, which therefore F A and F B are necessarily points on L prime here. We're saying that this bijection will preserve congruence that a segment in the domain will be congruent to a segment in the image of this function F. But it also is gonna preserve betweenness that if B is a point on A, excuse me, ABC are all points on L because of course with betweenness, you can't be between another point unless you're collinear. So AB and CR are points on L. If it's the case that B is between A and B, then the image of B will be between the images of A and C right there. So this bijection preserves congruence, it preserves betweenness and of course also has to preserve incidence the way that we've defined this function. And so when you put this together, we then come to this conclusion that all lines in a congruence geometry are congruent to each other in a fixed congruence geometry because of these type of bijections. As the lines L and L prime are arbitrary, this construction works for any pair of lines. And so by transitivity, all lines are gonna be congruent to each other in a congruence geometry. Okay, so how are we gonna accomplish this? Well, because we are talking about lines, we're in a congruence geometry, in particular a congruence geometry is an order to geometry that has the congruence axioms. So each of these lines has an ordering to them. And so for convenience, we're gonna use the ordering of the line to help us construct these things right here. So you do have to make a choice though. So pick a point P on the line L, pick a point P prime on the line L prime, that's a little slight typo there, sorry about that. So we pick P on L, P prime on L prime. And then we require that the, well I should say we require, we're gonna assign F of P to B, P prime. So we've picked these two lines, excuse me, we've picked two points on the lines. So let me try to sketch a little of a picture going on right here. So we have our two lines here, the yellow line, we will call that one L listed right here. We have the line L prime that we're gonna list right here. There's some point P that's on there. There's some other point P prime over here. And so our function F that we're constructing right now will associate to the point P, this point P prime. That's a matter of choice. We've also chosen an order on these lines so that we could say something like this, the arrow is pointing in the upward direction with regard to this line. All right, so then going from here, let's consider the point Q, where Q is some point on the interval P towards infinity. Well, since we have an order of line, P towards infinity would be this interval right here. If Q lives on that interval, notice this interval is just the ray that goes on that direction of the line here. Likewise, if this is P prime towards infinity, that also gives us a ray right here. So for any point Q that falls on side of this ray, then by segment translation, there's gonna be some point, call it Q prime, that's on the ray P prime towards infinity, such that the segment PQ is gonna be congruent to the segment P prime Q prime. So that's given by segment translation there. So we are going to then define the function F of Q that it's gonna map over to this unique point Q prime, because segment translation does guarantee that this point is unique. This point Q prime is unique on the ray P prime towards infinity. All right, so we make that assignment like so. So if you're over here, you're going to get some unique point over here as well. And be aware P actually is one of the points we could be considering here, which of course we associate that to P prime. This actually is agreement with what we've already talked about. So no ambiguity whatsoever. On the other hand, what if we have a point R that belongs to the ray negative infinity P? That interval is an array, it's an array over here. So if we have some point R, then again by segment translation, there's gonna be some unique point R prime that lives in the ray negative infinity to P prime, such that PR is congruent to P prime, R prime. And so by segment translation, there's some point over here so that this segment is now congruent to this segment. And by the uniqueness of segment translation, we know that this point, whoops, I just erased my point, sorry about that. Let's put it back on the screen. By the uniqueness of segment translation, this is the only point in that ray, we're of course talking about this ray down here and this ray right here. This is the only point in that ray that'll have that congruent statement. So we will then define the function F to associate R to the point R prime. Because of the uniqueness condition of R prime, this is a function and this will be well defined. Every point on L belongs either to the interval P to infinity or to negative infinity to P. P is included of course in both of them, but again by the uniqueness of segment translation, there's no disconnect. P will go to this point P prime. No worries with that whatsoever. So we have this function, this well-defined function. F goes from L to L prime, all right? We constructed a function from L to L prime, but we could also have gone the other way around. We could define a function, we'll call it F prime, that connects P prime to P and then by analog, any point in the interval P prime to infinity, we can identify it by segment translation to some point on L and then again, any point that's in negative infinity to P prime, we can associate that to a unique point in the interval negative infinity to P again by segment translation. Another function that goes the other direction, we'll call that one F prime, it goes from L prime to L by analog, F prime sends P prime to P and all the other assignments followed by segment translation. I want you to be aware that both of these things exist and if we look at their composition, if we compose F prime with F right here, this would be a map from L back to itself. And so if you follow this process, you go from Q to Q prime and then back from Q prime to Q, what happens here is we have that P prime was congruent to, excuse me, QP was congruent to P prime, Q prime, then we translate that back. So we took a copy of the segment PQ and we drop it back onto PQ when we compose these functions. So we've translated a segment back onto itself by the uniqueness of segment translation, this segment we translated has to be equal. That is what I'm trying to say here is that F prime of F of Q, so this has to go to Q prime right here, it has to go back to Q, this is the exact same thing, there's not some other point here. The composition is going to give us Q in the end. Therefore, these functions are inverses of each other, this F prime is actually F inverse. And so this is a bijection. So that's the easiest way to get this bijection, you can produce an inverse to the function. So that gives us the bijection that we're looking for, why does it do the other things? I wanna keep my picture on the screen a little bit longer, so we'll have to squint a little bit here, but imagine now we have two points A and B that are on the line. And so consider of course the following segments. So between the two lines, you have these fixed points P and P prime, you can kind of think of it like it's the origin of that line. If this was the x-axis, this is x equals zero or something like that. So we just have these two fixed points. So look at the segment, refer the points A and B on the line, look at the segment AP by construction. A prime, P prime are gonna be congruent to each other because A prime is just the image of A under the map F and P prime is just the image of P, right? So this tells us the segment AP is congruent to F of A, F P as this segment here. Cause again, how is A prime chosen? Depending where it was, we're just going to, we chose these points so that these segments are congruent to each other. So by construction, we get that. Likewise with P B, whatever that line segment turns out to be, it'll be congruent to P prime B prime, which of course is the same thing as this one right here. This is by the construction of F. We constructed the assignment of the function using segment translation. Now, we have that A prime, excuse me, AP is congruent to A prime P prime. We also have that P B is congruent to P prime B prime. If, because we have these three points now A, B and P, because they're collinear, there has to be some between this relationship between them. If P is between A and B, then we can use the segment addition axiom from congruence and we're going to get that A B is congruent to A prime B prime as segments, which of course is the segment F of A, F of B. Now, that's if P is between A and B. It could be that B is between A and P, in which case then we're going to use what we call segment subtraction, which then will give us that A B is congruent to A prime B prime, which is what we're looking for. Now we have an improvement segment subtraction in this lecture series. We took addition as a axiom. Segment subtraction, I'm going to leave as an exercise to the viewer here, for which what does segment subtraction even mean? We have two lines, we have three points on each of those lines. So you have some A, some B, some C. You're going to have some point over here, A prime, some B prime, some C prime, like so. And so segment subtraction we have as the following. We're going to assume that the segment AC is congruent to the segment A prime C prime. So those are congruent to each other. We're also going to assume that the segment AB is congruent to A prime B prime. So under those assumptions, segment subtraction then will give us that the segments BC and B prime C prime are congruent to each other. It's similar in nature to segment addition, but it does have to be proven. And like I said, I'm leaving it as an exercise to the viewer here. The basic idea is you are going to use segment addition, but you have to also combine it with segment translation. You translate this piece over here and do some segment addition and translate it back and using the uniqueness of translation. You then should be able to get the congruence that you're seeking. So because of segment addition, segment subtraction we have that our function F does preserve, that's not the one we wanted. It is going to preserve segment congruence, no matter what situation you're in. Because I talked about PB between A and B, what if B is between A and P? We have an argument. Well, there are other possibility is that A is between B and P, but that's similar to this argument using segment subtraction. So no further consideration is necessary here. The last thing that we have to consider is what happens when the point B is between the points A and C on the line L. What happens with regard to between the statements? Well, remember that A prime is just F of A, C prime is just F of C here. And so there exists a unique point, we're going to call it B double prime for the moment on L prime, that's between these points. So if we try to draw this picture real quick just to give some understanding here of what's happening, we will have our original line right there. So this of course, this yellow one is L, we have A, we have B, we have C like so, we have A over here, A prime I should say, we have C prime. Now what we're saying here is by between this preservation, there's going to exist some point B double prime such that because the segment AC is congruence to A prime, C prime, by between this preservation, there's some point we can carry over, we'll call it B double prime so that the segment AB is congruent to A prime, B double prime and the segment BC is congruent to B double prime, C prime. So these statements we have listed over here, I guess I can leave those on the screen, that's not a problem. We have by between this preservation that these congruences are happening, okay? And again, the point A is associated to A prime by F and the point C is associated to C prime given by F as well. The image of B is gonna be B prime but we're not yet saying that's the case over here. Then we get back to this idea here, AB is gonna be congruent to A prime, B prime. So there is some point over here, some point B prime, right? So where B prime is the image under F of that, we have by what we previously proved about the congruence going on here that the segment AB is gonna be congruent to the segment A prime, B prime. Well, by uniqueness of segment translation, this is gonna force that B prime and B double prime are equal to each other. And so that then gives us that B double prime which, well, excuse me, B prime, which is B double prime is then between A prime and B prime. And so the last little bit is the part that I'm really trying to emphasize here that the segment translation brings over the congruence of AB onto A prime, B prime. But how do I know that it's between these other points? You do have to combine it with the betweenness preservation. So between this preservation combined with segment translation shows us that this function F will preserve the betweenness. And so we really can think of this function F as a niceomorphism, it's a congruence between the lines. It preserves the incidence relationships because it's a bijection, but it also preserves segment congruence and betweenness of points. And therefore every notion we have about a line right now is translated from one line to another as these two lines are arbitrary. We see that every line in a congruence geometry is congruent to one another. And that of course brings us to the end of lecture 14 as we introduced the congruent axioms of line segments. In lecture 15, we will finish listing the congruence axioms and we'll talk about the congruence axioms of angles. So take a look at that video to learn more about these congruence axioms. I do appreciate you for watching. If you learned anything in this lecture, like the videos, subscribe to the channel to see more videos like this in the future and post any questions you might have in the comments below and I'll be glad to answer them.