 Welcome back to our lecture series Math 3130, Modern Geometries for Students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Missildine. In lecture 19, our main goal is to prove the so-called midpoint theorem, which we'll do in this video in fact, which proves that every line segment has a midpoint. But related to midpoints is also the notion of bisectors for which we will try to define what those things are right here and now. For our first definition here, we're in a congruence geometry. Remember a congruence geometry is a geometry that satisfies the four axioms of incidence, the four axioms between this and the six axioms of congruence. Suppose we have two points A and C. Let B be a point that's between A and C, such that the line segments AB is congruent to the line segment BC. Pictorially, we're thinking of something like the following, we have our point A, we have our point C, and we have some point B that sits between A and B, such that the line segment AB is congruent to BC. In this situation, we call B a midpoint of the line segment AC. It's, you know, we don't have a notion of measure, so we don't really have a notion of distance, but we can use congruence here. If this segment is congruent to that segment, we can say that B is midway between A and C, so-called midpoint. Furthermore, we say that a line, a line is a perpendicular bisector to the segment A, if the line L is perpendicular to AC as a line, and then B goes through that point. So the perpendicular bisector being something like the following, that's not quite right, there we go. Like so, so we would say that this line L is the perpendicular bisector to the segment AC, if, first of all, the line L intersects the line determined by A and C as a right angle, and then it passes through this midpoint, so a perpendicular bisector. So like I mentioned a moment ago, the main purpose of this lecture, in particular in this video for this lecture, we're going to prove the so-called midpoint theorem that says that every line segment has a unique point. Now, if you have a notion like measure, like if we can measure the length of an interval, and that's a real number, well, then sure, we can cut that distance in half and go from there. But in a congruent geometry, we don't have any notion of distance, we don't have any notion of measure. That's something that we can inherit from continuity, but at this point in our construction of geometry, we have no measures of segments or angles, but we can prove the existence of midpoints, and this will be a consequence of things like the exterior angle theorem and other results that we've developed already for congruence geometry. All right, so this one's a very fun argument, very lengthy, surprisingly, it's difficult to actually construct midpoints. And so what we're gonna do is we're gonna start off with a segment, we'll put it here on the screen. So this will be the segment BC. And so we're gonna prove that BC has in fact a midpoint. And so to start off with, we're gonna pick some point A that is not on the line BC. The existence of such a point is quite taken for granted by now, it's a consequence of incidents. If I take the line determined by A and C, that line can be extended so that there's a point D such that C is between A and D on that line. This is a consequence, of course, of the extension axiom and such. And so I then want you to consider the triangle ABC that you see right here. Then the angle DCB would be an exterior angle to this triangle. In particular, if we look at angle B over here, because angle BCD is an exterior angle, DCB will be larger than angle B by the exterior angle theorem, which the exterior angle theorem was a consequence of the alternate interior angle theorem. So that's really who's proving this theorem, the existence of midpoints. Some people actually use the midpoint theorem to prove things like exterior angles or alternate interior angle theorem. We haven't proven the existence of midpoint yet. So we use those theorems to prove midpoints that we're doing right now, I should say. Now, because angle DCB is larger than angle B, that gives us that there has to be some point, there's got to be some point in the interior of the angle DCB, maybe some point over here or something. We're gonna call that point E. There's some point so that if we take the ray like so, where E is on this ray, then this angle BCE is then congruent to the angle ABC. Because again, it's interior to the angle like so. And then also by segment translation, we can assume that there's a point along this ray over here so that this segment is congruent to the segment AB. And so without the loss of generality, we can assume that E actually does that point like so. So the point E exists so that the angle ABC is congruent to the angle BCE and that the segment AB is congruent to the segment CE. So we have the following construction now on the screen here. Now since angle E, excuse me, since point E is contained inside the angle DCB, we know that E is on the same half plane that D is with respect to the line BC. So if we were to like extend this line a little bit further here at the line BC, then of course the point D and E are on the same side of the half plane they're on the same side of the line BC as each other. On the other hand though, we know that A and D are on opposite sides of this line BC because in particular, since C is between them, it's right here in our diagram, since C is between A and D, the interval AD intersects the line BC. So since AD are on opposite sides, excuse me, and since DE are on the same side of the line BC, by the plane separation theorem, it then holds that the line segment AE is gonna intersect the line BC. In other words, A and E are gonna be on opposite sides of the line BC by plane separation. Let's call that point of intersection between the line segment AE and the line BC, we're gonna call that point F. By construction, we know that F is between A and E. Now my diagram seems to suggest that F is between B and C, but we don't quite know that yet. We'll get that in just a moment. But I wanna pause here for a moment and just point out that it also kinda looks like these two segments are congruent, they are going to be, right? This point F is our candidate for the midpoint of the line segment. We have to prove that, of course, in just a moment, but that's what our goal's gonna be going forward here. We have to prove that F is this midpoint. So how are we gonna do that? Well, okay, let's actually see if F is between B and C, because if F's not between B and C, then it's not a midpoint. So that's the first thing. It has to be on the line segment BC. So consider the point E, right? E is on the same side of the line DC, which of course we're now focusing on this line right here. B and E are on the same side of that line, because after all, the line CD is the same as the line AC. So the half plane that contains B is the same in both situations, right? And how did we get the point E in the first place, right? We put it over here in this half plane by angle translation. So we do get that B and E are on the same side of the line AC. Again, this comes from the fact that E is an interior point to DCB. That's how we came up with E in the first place. Angle translation, we put it inside of this angle, all right? Now next, since AB, the angle and the angle ECB are congruent ulterior, excuse me, they're congruent alternate interior angles to the pair of lines, A, B, C, E. We can conclude that the lines A, B, and C, E are parallel to each other by the alternate interior angle theorem. So let's unravel that sentence for a second, right? How do we know that we have these congruent angles? Well, okay, yeah, remember, angle ABC was congruent to ECB, right? We use angle translation to do that very thing here. And so when we look at these lines, the line AB, which is this line right here, and the line CE, like so, because we have these alternate interior angles, it's transversed by the line BF. We have that AB and CE are parallel lines. So even though it seems like they might connect eventually, that's just a crudeness of my diagram right here. These are in fact parallel lines by the alternate interior angle theorem, all right? So then, sorry about that typo right there. So in particular, the line AB is gonna be parallel to the segment CE. So I want you to think about that, for example, for a moment, right? If the lines AB and CE don't intersect each other, if I replace the line with just the line segment, you can't get any new intersection points by taking a smaller set. So the line AB doesn't intersect the segment CE. So from that, we can conclude that the point E is on the same half plane as C with respect to the line AB. That is to say this line segment doesn't touch the line AB. So CE are on the same side of the line AB, okay? Which then shows for us that the point E belongs to the angle BAC, right? Belongs to the angle BAC. Because to be in the angle BAC, you need to be inside this half plane and you need to be inside of that half plane. So it belongs to the angle. E is an interior point. If we apply the between cross lemma in this situation, since E is this interior point and since F is on the line BC, that also implies, well, let me say it this way, since E is an interior point, the ray that comes out of A going towards E, that's an interior ray to the angle. So every point on that ray is likewise interior other than the vertex A itself. So F is an interior point to the angle ABC, so excuse me, the angle BAC. And we know that BFC are collinear because of F was the intersection that came from the crossbar theorem that we mentioned before, right? And so because F is interior to the angle and it's collinear with BC, by the between cross lemma we have to have that F sits in between. So okay, so now we have our point which is in the right location. So F is between, it's between B and C. So the next thing we have to argue is that the segment BF is congruent to the segment FC. I'm gonna zoom out a little bit so we can continue to see our diagram, but we can also see the next paragraph here, all right? So we have to show now that the segment BF is congruent to CF. We're gonna try to use some type of congruent triangles to do exactly that. I want you to notice what we already have going on here. Notice that if we look at the triangle AFB and we look at the triangle CFE, we wanna show that those two triangles are congruent to each other. They already have two congruent sides. They already have two congruent angles. And so with the triangle congruences we've done previously, if we can get one more side congruence, particularly if we could show those were congruent, we'd have side angle side, but we can't do that because that's what we're trying to prove here. If we prove that those ones were congruent, those sides, that side side angle, that's the ambiguous case. We can't quite work that one. If we could prove that these angles were congruent, oh, angle side angle would be appropriate, but turns out the direction we're gonna go is this one right here. Look at the angle BFC or AFB if you prefer, Air Force Base, AFB, and then look at the corresponding angle EFC. These two angles are gonna be vertical angles to each other. And as such, as we proved in the homework, vertical angles are in fact congruent to each other. So in this triangle, we have an angle angle side. The AAS triangle condition applies and this shows us that the triangle AFB is congruent to the triangle EFC for which then corresponding parts of congruent triangles are congruent. So we then get that BC, excuse me, BF is congruent to FC. So this then proves that the midpoint BC has a, yeah, the segment, excuse me, the segment has a midpoint, this point F, okay? It's important also to establish that this point is unique. So we've proven that midpoints exist. Why are they unique, okay? That's what we're gonna try to prove right now. So I don't need the previous picture, but we're gonna take our segment BC again. So what if it had two midpoints? What if the line, what if there were two midpoints to the line segment BC right here? So let's say there's one F right here and there's some other one F prime for which there would have to be some type of between this relationship that looks something like to follow with my notes, we'll do it this way. There'd have to be some between this relationship where B dash F prime dash F dash C prime happens. And we could do this by limiting your ordering of some kind. We can put an ordering on the line BC so that B is less than C. And so because F and F prime belong to the segment, the line segment, F and F prime, they're between B and C, but there has to be some ordering between them. So we can just assume without velocity generality that F prime is less than or equal to F, thus giving this between this statement we have right here. So if that's the case, we get the following. The line segment B F prime is gonna be less than the line segment BF because F is further along the line. It's farther away from B. But we're assuming these things are midpoints, right? So if F is in fact a midpoint, this would mean that the line segment BF is congruent to the line segment FC because it's a midpoint. So we then get that BF is congruent to FC. But FC is this line segment right here. If we were to go further on, F prime is farther down the line there. So we'd get that the line segment F prime C is actually larger than FC. But like we said, F prime is also by assumption a midpoint right here. And since it's a midpoint, that means that F prime C is congruent to BF prime. So we get this. And so notice what happened here. We started with BF prime, we ended with BF prime. So the only way these two things can be congruent is that every single inequality that showed up in the middle must actually be a congruence. In particular, the one we really care about was the very first one. We end up with BF prime, the segment is congruent to the segment BF. But then by uniqueness of segment translation, because both of these points live on the ray emanating from B going in the direction of F. If you have two congruence segments on the same ray by the uniqueness of segment translation, those segments have to be equal, which means their endpoints have to be equal, F and F prime are in fact the same point. So two different midpoints actually have to be one and the same thing. So this then proves the midpoint theorem, which says that every segment has a unique midpoint. And so like I said, this was a consequence of the exterior angle theorem, which came from the alternate interior angle theorem. So in closing this video, I also want to provide the definition of an angle bisector. We talked about a perpendicular bisector, but that really has to do with segments. You could think of like a perpendicular bisector it's the angle bisector of a flat angle. You can make that connection there, but we're gonna properly define these things because the idea of a midpoint and a bisector are very much related to each other. The midpoint gives you half of a segment. The angle bisector gives you half of the angle. Imagine we have points ABD in a congruence geometry and suppose that C is some point in the interior of the angle such that the angle ABC is congruence to CBD, okay? And so then we get a picture that looks something like the following. So we label these things, B is the vertex, we have the point D over here, we have the point A, right? So, and so then if there's some interior point C, like so, so that the angle ABC is congruent to the angle CBD, we call such a thing a bisector. So the ray BC is then the bisector of the angle ABD sometimes called an angle bisector. And that's because we don't wanna confuse it with a perpendicular bisector, which really is like a special case of what we talked about, what we're talking about right now. So similar to the midpoint theorem, there is the so-called bisector theorem for which the first statement is analogous to what the midpoint theorem says. The bisector theorem tells us that every angle has a unique bisector. So bisectors exist and they're in fact unique. In addition to that, I'm also gonna give us a second part to the bisector theorem that says every segment has a unique perpendicular bisector. Now the first part right here, every angle has a unique bisector. We can go through a proof of this, very similar to what we saw before, but it turns out because of what we did with the midpoint theorem, okay? If we were to look at the crossbar AD, and that crossbar has a midpoint, call that midpoint M. If we take the ray going from B through M right there, I claim that that ray is the bisector of the angle. If the ray passes through the midpoint of the crossbar, then that's gonna give us the angle bisector. Now that does require proof. I'm gonna leave it as an exercise to the students here to prove that this angle, that excuse me, that this, the ray passing through the midpoint gives you the angle bisector, but if you play around with congruences of triangles, you can get that this is in fact the angle bisector, and then by uniqueness of midpoints, you can also show that these bisectors are unique. Basically, it's gonna be an angle bisector if and only if it passes to the midpoint of every crossbar. That's an interesting result. I'm gonna leave it for the student to do that. Also, I'm gonna leave it as an exercise to the viewer here to prove the second half of the bisector theorem that every segment has a unique perpendicular bisector. But given what we've already done, this is pretty easy if you have the midpoint theorem. So given any point, any segment A and B, right? It has a midpoint, M right here. Whoops, M, there you go. We've proven previously, when I say we proven, I mean, this was left as an exercise to the student as well, but it was proven through previous homework assignments that a perpendicular can be erected out of any line through any point on that line. So the line AB has the point M. There's some perpendicular that can be erected out of it. And then one just has to claim that that's the perpendicular bisector. That should be fairly straightforward, which is why I leave it as an exercise to my viewer.