 In this video, I want to prove a proposition of congruence geometry, which we are going to call betweenness preservation. So we define what a congruence geometry was previously. A congruence geometry is an order geometry that satisfies the six axioms of congruence given by Hilbert here. And the betweenness preservation theorem here is going to essentially solidify the connection between congruence and betweenness. The betweenness comes from this notion of an order to geometry and congruence, then we add it on top of that. And congruence and betweenness are actually closely related to each other. Now when we talked about the axioms of segment congruence, the only axiom that specifically you talked about betweenness was segment addition. So it's going to come as no surprise that segment addition is going to be the main ingredient in this proof here. But what is the statement here? Imagine that we have a segment, AC, that's congruent to another segment, A prime, C prime, such that the point B sits in between A and C. So then betweenness preservation is going to guarantee that there exists some point, a unique point B prime, that's between A prime and C prime, such that A B is congruent to A prime, B prime, and B C is congruent to B prime, C prime. So in picture, we're thinking of the following situation. We have this segment, which has, of course, three points on it. So we're going to have A, B, and C. We then have this other line segment, maybe down here or something like this, for which we then have A prime and C prime on it. And the idea is that this line segment right here is going to be congruent to this line segment right here. So then those are the assumptions here. So what the preservation theorem is going to give us is that there's some third point B prime over here so that these segments are congruent and these segments are congruent. So we're going to show how to do exactly that right here. So the first thing is we're going to use segment translation to get us, so this is axiom C1, we're going to use segment translation to get the point B prime into the game here. Because after all, we could think of the ray, the ray that's determined by the points A prime, C prime. On that ray by segment translation, there is going to be some point B prime so that A B is congruent to A prime, B prime. Okay, so that's really what we want to do here is we can translate this segment down over here and so we get some point B prime like so. So by construction of the point B prime, we're going to have that A B is congruent to A prime, B prime. Now I have to be a little bit careful and this is always something we should be cautious about when we draw diagrams here. How do I know that B prime is between A prime and C prime? I actually don't know that yet. So we have to be very, very careful. I'm going to draw my picture in that manner but I cannot assume yet that I know B prime is between A prime and C prime. I know that B prime is on the ray, A prime, C prime but it could be over here. Maybe it's equal to C prime for all we know. We don't know that yet. Okay, so we use segment translation once. We're actually going to use it a second time here. So what we're going to do next is that again, using segment translation, there's going to be a point P such that B C is congruent to B prime P. And so we then will know that B prime is going to be between A and P. So this is the little trick that we're going to try to do here. As we look at this segment, I'm going to again draw the ray back on the screen because we don't know where all these live necessarily. But the idea is we're going to take the segment B C now and translate it down over here. So there's some point P that lives over here. P could be past C prime. Maybe it's between B prime and C prime. Maybe it's C prime itself. I mean that's what's going to happen. We don't know that yet. But by translation this time, we know that B C is going to be congruent to this one right here, B prime P. And we do know by construction that B prime will be between A prime and P. So we translated twice. We translated this one down. We translated this one down. And so we have to then determine how to C prime play into all of this. That's really what matters here. So now this is where segment addition is going to come into play here. So by what we know, A prime is congruent to, excuse me, A B is congruent to A prime B prime. We also know that B C is congruent to B prime P. So by segment addition, since these two pieces, those two pieces add up to be C prime. Let me put C. Let's put still C on the board there. So we know that this is congruent to that. We know this is congruent to that. So their unions must be congruent as well. So we're going to get that A C is congruent to A prime P. Like so. But I should also mention that we already know by assumption that A C is congruent to A prime C prime. Like so. So by transitivity of segment congruence, we have to have that A prime P is congruent to A prime C prime. Like so. And then we're going to come back to segment translation here. We have two segments on the exact same rate. Both of these segments start at A prime. We have a congruence between two segments on the same rate. By the uniqueness condition of segment translation, it's got to happen that C prime and P are actually the same point. Like so. So really, we can just remove this point from consideration here. And C prime is the one we are looking for. Okay. So therefore A B is congruent to A prime B prime by construction. We're going to have that BC is congruent to B prime C prime because P and C prime are the same thing. And then B prime is going to be between A prime and C prime because again P is C prime. That's proving the betweenness preservation theorem here. So this is a very important thing that if we have two congruent segments, we can preserve the betweenness relationship from one segment on to the other. And that's going to be a very, very useful skill. Now from that, we can actually define what it means for, well, we can define an ordering on segments. So for segments A B and C D, if there exists some point E that is between C and D, then we say that and A B is congruent to C E. Then we say that A B is less than C D. And so the idea of course is the following. We have our two segments, something like this. So we have A, we have B, we have C, we have D, right? And so then the thought is if I translate segment A B on to the segment C D such that because I should say you translate on to the Ray CD, then there's going to be some point E over here, right? Where does E live? Is E between C and D? If so, we say that A B is less than C D. What if it's, what if E is equal to D itself? Well, then then you'd get congruent in that situation. What if it's past the point though, where D is between it, then you get, it's greater than, right? But those observations are going to have to be proven, right? I'm going to leave this up to exercise to the viewer here. So if E is between C and D, we say the segment A B is less than C D. We can say that A B is less than or equal to C D if either they're congruent as segments or that one's actually backwards right there. Sorry about that. A B would be less than C D. Not sure how that thing got twisted around. But yeah, less than or equal to has the usual interpretation that you're either congruent or less than. So it's not equal. It's not less than or equal to it should be really less than or congruent to but we'll still just use the less than or equal to symbol here. Like I said, I'm going to leave this as an exercise to the viewer here to prove that in fact this this relationship we've defined is in fact a total order. If you look of course at this one less than or equal to here, it should be a reflexive that is your A B is congruent to that is A B would be less than or equal to A B will since congruence is reflexive that's pretty easy there is an anti symmetric. If A B is less than or equal to CD and CD is less than or equal to A B that they're actually congruent that there's something has to be proven there. Is it transitive? Well, we know that congruence is transitive but is this is this definition transitive. You can prove that there's some details there is a totally ordered. Yeah, the trichotomy in between this preservation is going to be coming to play in all these. So again, I'm going to leave it as an exercise to the viewer to prove this, but this is something we're going to use in future videos that. When we discuss segments, we can then put a total order on segments based upon how they relate to each other with regard to congruence and between us.