 Welcome back to our lecture series, Math 3130, Modern Geometries for Students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Misseldine. We are in our last and final lecture, lecture 36 in our series. That's only because this series is originally based upon our traditional 15-week semester. Of course, in the future, I might decide to add some new topics to this series, just for some fun topics that we could continue on with geometry, especially if schedules change or things like that. But for now, we'll consider this our last lecture in the series here. And so what we've been talking about recently is elliptic geometry, some of the consequences of elliptic geometry. And so I wanna talk a little bit about quadrilaterals in elliptic geometry, particularly Sikeri and Lambert quadrilaterals. We'll also talk a little bit about triangles. And actually that brings us to our first theorem, our first proof for this lecture video here. The thing is in elliptic geometry, we lose a lot of the things that came from between this, things like the alternate here, angle theorem, the exterior angle theorem, and to name a couple other things. And so there's sort of like this is devastation that occurred because we lost some critical theorems and therefore the theory that was developed on top of that. Included in that list is gonna be opposite, the angle opposite side property or AOS that we've referred to many times in this series right here. AOS we lose because the proof of it originally depended upon the exterior angle theorem. And so if we wanted to keep AOS, we have to provide a new proof of that. And although we're not gonna prove a full blown version of AOS for elliptic geometry, we can prove a right triangle version of AOS, something similar to the best approximation theorem that is gonna be useful for our considerations here. So in a right elliptic triangle, each of the other two angles has a measure that's less than equal to or greater than a right angle depending on whether the side opposite has a measure that is less than equal to or greater than a polar distance. So if we have a right triangle, we can compare the sides of a triangle and we can say that they're either large, they're either acute right or up to just based upon how the opposite side of that angle measures to the polar distance which is a fixed number for the entire geometry. So the argument is gonna go similar to the following. Let's take an elliptic right triangle. So we might get something like this. So we have angles A, B and C. And let's suppose C is the right angle in this consideration, okay? And since it's a right angle, what we wanna do is we wanna find the pole associated to the line AC. So we have this line AC right here, consider that our line L. And if we extend potentially this line here, the line BC, there's gonna be some point up here which we call P, it's the pole, the pole associated to the line L, the line AC there. And because C has a right angle there, we know that the pole will drop onto L at C and form this right angle. So we get a picture much like the following. So in particular, the line AP which I'm gonna add that to the picture as well. If we drop that picture down, right? The line AP will be perpendicular to AC because P is the pole to L which is AC, right? And therefore the line drop from P to any point on L is always gonna be a perpendicular like so. So there are some possibilities, what about the point B, right? It could be the possibility that B and P are exactly the same point. And so if B is a pole to the line L, that means the length from B to C, this line right here would be the pole or distance. And therefore that distance would be the pole or distance. Then the segment B, or the length of the segment P to A, since that's B, that would also be the pole or distance right there. And so that would give us the situation we were looking for before. That would tell us that the angle CAB is a right angle from what we had there. So okay, that's great if B is the pole. So let's suppose that B is between C and P. And of course, when I say between, I mean in a relative sense. So there's some point at infinity, some ideal point we could use to make this statement. But look at the picture we have right here in a relative sense B is between C and P. Now in that situation, we're gonna get that the segment BC would be less than a pole or distance because again, B is between C and P right there. So B is less than a polar distance. But we also get that the angle right here is less than a right angle. Basically like the picture is telling us right here. So less than a polar distance gives us an acute angle. So those things are corresponding to each other. The other possibility of course is if things were reversed, if P was between C and B in some relative sense, then you'd have something larger than a polar distance and you'd get something larger than a right angle. And so this result follows from that. And so let's think a little bit about, let's think a little bit about quadrilaterals in elliptic geometry. For the most part, the proofs we start off with with quadrilaterals are gonna work very similar than we did before. So the notion of a security quadrilateral still makes sense in elliptic geometry. Although the summit seems to curve away, but it seems to curve the other way around. So if you have your security quadrilateral right here, this will be assumed we have two right angles and two congruent legs. If we consider the diagonals of these things, we can still do a side angle side argument to show that the diagonals are congruent. And then to show that the summit angles are congruent, well, you look at this triangle right here where you do a side angle side argument as well to show that those two triangles are congruent. So therefore the congruent angles would still, the summit angles would still be congruent. If we introduce the altitude into this thing, the altitude is the line that connects the two midpoints of the summit and base midpoints exist. And then you get essentially just a side angle side angle side argument to show that the two quadrilaterals you created here, which are course lumber quadrilaterals. We don't know that APOR though, but we're gonna show that these two quadrilaterals are congruent by side angle side angle side, which is still true in elliptic geometry because of side angle side. And so therefore the corresponding angles would be congruent. These two angles would be congruent as they are supplementary that gives us right angles. So the existence of altitude, all that jazz, so it's gonna work out really very much the same. The issue that we have to be concerned about is these summit angles, right? What's the measure of these summit angles? Are they gonna be right angles, acute angles or obtuse angles? What are the possibility? The comparison, and also comparisons we make about like, is the altitude longer equal to or shorter than the legs? Is the summit longer equal to or shorter than the base? We have to make some comparisons about these things. And in elliptic geometry, it's gonna be always different than what we saw in neutral geometry, right? We're gonna see that the altitude is actually longer than the legs in this quadrilateral. And more importantly, we're gonna see that the summit angles are obtuse. The so-called obtuse angle hypothesis. The proof of this, you can see in front of us right here. In elliptic geometry, the summit angles of a security quadrilateral are obtuse. Although I am going to use some progressive slides made by a former student. You can find a link to this in the comments below if you want to. And that's just because the students have already created the pictures for us. And so let us, oh, that's right. I need to open this up in an outside page. It doesn't quite work right here. So let's see if I can do this in real time. Let me try this again. Alrighty then. Nope, not in Desmos. Let's go to the website. All right, hopefully you can see this okay. Continuing on with this thing. We started with the picture. So take a security quadrilateral, A, B, C, D, and let EF be the altitude. E is on the summit, and E is on the base, and F is on the summit right here. And again, the congruences of these things are valid from before. Because we are in elliptic geometry, this security quadrilateral is not a parallelogram. We don't have the alternative angle theorem anymore. In particular, the summit and base have to intersect each other because it is elliptic geometry. Say that the intersection between the summit and base is some point P. Now I want to admit that in the projective model of elliptic geometry, this point P would actually be the intersection on both sides of the line because there's really only one side. But in spherical geometry, this point P, there would be an antipode over on the left side of negative P there, but we don't have to worry about that necessarily. So, and we'll stick with the projective model here. In a relative sense, we do get that vertex B is between F and P. We also get that vertex C is between E and P as well. And so look at the triangle we formed here, the PEF or FEP for the FEP triangle. This is gonna be a double right triangle because it has the two angles associated to the altitude of the security quadrilateral. And so because this is a double right triangle, the segments FP and the segment EP are gonna be polar distances. That's gonna be important here. And so let's look at the smaller, another right triangle, BCP because angle C is associated to a right angle of the security quadrilateral. It's supplement is also a right angle BCP. And that's gonna be a right angle. And so if we use the elliptic version of AOS, the side opposite of the right angle, I mean that BP right there, I guess what I'm trying to say is since we have a right angle in this triangle, it's a right triangle, elliptic right triangle, that tells us that the other two sides, if they're shorter than an elliptic distance, they're gonna correspond to acute angles. And so consider this right here, we have a right angle C. So the other angles, since we're on a right triangle, angle B and angle P, I only care about angle B right here. So angle CBP, it's opposite of a side that's shorter than a polar distance. The segment CP is shorter than EP, which is the polar distance. And since CP is shorter than a polar distance, the angle B, that is CBP is an acute angle. There's a typo in the proof right there. It's just at the bottom line there, it should say angle CBP, sorry about that. Anyways, so that angle B is an acute angle. Therefore, its supplement is a up to single. And that supplement is, of course, a summit angle of the securit quadrilateral. Angles A and angle B are congruent to each other. And this proves the obtuse angle hypothesis. Notice we essentially proved this theorem right here in the same argument right there, the fourth angle of a Lambert quadrilateral is obtuse and elliptic geometry. Again, this exact same argument applies right there. And I should also mention that this shows that in elliptic geometry, there are no rectangles. So there's no rectangles in hyperbolic geometry or elliptic geometry. They only exist in Euclidean genre. And their existence is actually equivalent to that Euclidean parallel postulate. I wanna talk a little bit more about triangles in the next video, stay tuned for that. And other consequences of this obtuse angle hypothesis.