 Welcome back to lecture 28 from our lecture series, Math 3130, Modern Geometries at Southern Utah University. As usual, I will be your professor, Dr. Andrew Missildine. Thanks for watching our video here today. This lecture is going to be loosely based upon section 4.2, the Parallel Apostlet and some implications from the textbook Roads to Geometry by Wallace and West. We'll call it, we'll title the lecture, Equivalence to the Euclidean Parallel Apostlet. And it's actually a little bit of bookkeeping. We have to actually finish a lecture that we had started previously in lecture 27. So with respect to our series, theorem 3.4.4 stated that the Euclidean Parallel Apostlet is equivalent in neutral geometry to the angle sum of a triangle always equaling 180 degrees. What we have proved last time is that if we assume the Euclidean Parallel Apostlet, then every triangle's angle sum will equal 180 degrees. What we want to prove today is the converse of that statement. So that is we're going to assume that the angle sum of triangles is always 180 degrees. And then we're going to prove the Euclidean Parallel Apostlet from that. So whenever you're trying to prove the Euclidean Parallel Apostlet always start off with a line L and take a point P that's not on that line. And so we want to show there's a unique parallel line that goes through P and that's parallel to L. Now we have the guaranteed parallel line M. I kind of missed my point P a little bit. If that ever happens to you, no big deal. Just make your point just a little bit bigger. And that way you hit the mark there. So let M be the guaranteed parallel line. So L and M are parallel to each other. And also let the line T be the guaranteed transverse, the perpendicular transversal, sorry, to L and M. So T is parallel to L and M, both right there. And so what we're going to do is prove this by contradiction. So assume we had a different parallel line. So let's say we had another line M' so that M' is parallel to L but M' is different from M. So how we're going to get a contradiction here is since M is the only line perpendicular to the line T at the point P, that means M' – let me make my prime more obvious there – the only way – so M' intersecting T has to be at angles other than right angles. So if we look at the two supplementary angles right here, one of those has to be acute and one of those has to be obtuse. By the diagram that appears to be the left side. So we're going to work with that. So let's work with this acute angle side of T and take S to be a point that's on M at that side of the line. Let R be a point on M that's on the same side of T that S is. And let's actually call the intersection of L and T the point Q. So we have some angles RPS and SPQ. You're going to be complementary angles because they combine to give us RPQ, which is a right angle. So this is a right angle right here. We'll use that complementary relationship later. Let's see. So in particular, we do know that S is going to be inside the angle RPQ because as SPQ is an acute angle and RPQ is right, it's going to be interior this right here. So the next thing we're going to do is we're going to create a new point using the Archimedean principle. We're going to call that point T. So it's going to be some point on the left side of the line L called T so that if we take the point Q, sorry, the point P and connected to T here, I want that this angle is called angle one is less than the angle two. So as we can choose a point T so that PQ, sorry, PTQ is less than SPR. And the idea is the farther along we choose the point T down this ray, the smaller angle one will get. And eventually this can be done so that angle one is smaller than angle two. So I want us to kind of keep track of this for later information. So the angle QTP is less than the measure of SPR. Now next let's consider how the point S interacts with this new ray PT that we brought in here. Now by trichotomy of rays, either the ray PT is between PS and PQ like we see in the diagram, or it could be that PS is actually between, what if PS was between PT and PQ? Well if PS was between PT and PQ, that would mean that S is an interior point to QPT. And using the crossbar theorem that actually would guarantee there's some intersection between the ray PS and the line L. Well if there was an intersection, since PS is part of the line M' this would imply that M' is not parallel to L which is a contradiction. So this situation can't happen, let me erase it from your memory, we never ever considered it. This isn't a possibility here whatsoever. So in fact it's got to be that like the diagram suggests PT is between PQ and PS. And so we get this inequality that well because PT is between PS and PQ that would suggest that T is inside the angle QPS like the diagram suggests. And so that gives us this inequality that T PQ is less than SPQ. When we put these two inequalities together we get the following, T PQ is less than SPQ and we get that QTP is less than SQR. So the sum of those two angles on the left, so let's label these things right here. QTP was angle one, SPR, so this right here is angle one, this is angle two. T PQ, T PQ, T PQ, movie this one right here, this is angle three. And lastly, this is angle three, and then SPQ is this one right here, this is angle four. So we have these angles right here, so angle one and angle three are less than angle two and angle four. Now like we mentioned earlier, angle two and angle four, these are complementary angles so they add up to be 90 degrees. So we get that angle one plus angle three are less than 90 degrees. But wait a second, if we look at the triangle PQT, we know that angle Q is the right angle. And by assumption, all triangles has an angle sum of 180 degrees. So we have 90 degrees already accounted for, but these other two angles, one and three, their angle sum is strictly less than 180 degrees. What the junk? That actually is a contradiction, you can't have, this isn't a 180 triangle. And like we've seen before, if one triangle is 180, they all have to be. But in fact, actually that was our assumption that all triangles have an angle sum of 180 degrees. And so this gives us a contradiction. We've contradicted our assumption here. And so we have to go back to the original assumption. Because of the contradiction, because of the contradiction, we couldn't have a second parallel line. So it turns out, sorry, m prime must have intersected L. And so we get a unique parallel line which gives us the Euclidean parallel postulate. So with that theorem now proved, what I want to do is I kind of summarize some of these parallel postulates. Sorry, equivalents to the Euclidean parallel postulate that we've done so far. So what do we know so far? So our original version was play fairs version, uniqueness of parallel lines. So if you have a unique parallel, if you have a line and a point off the line, there's only one parallel line going through that point. We've proven that in an affine geometry, if we assume EPP with the incidence axioms, we proved an affine geometry that one implies two, proclaslima. Remember, proclaslima says that if you have two parallel lines and a line intersects one of those two parallel lines, it must intersect the other one as well. And then in affine geometry, we prove that proclaslima implies transitivity of parallelism. And so although there's an argument to be done, one can actually show that in incidence geometry, transitivity of parallelism, transitivity means that parallelism is a transitive relationship there, transitivity of parallelism actually implies play fairs version, that there's a unique parallel line. In incidence geometry, these three statements one, two, and three are all logically equivalent to each other, which means in neutral geometry they'll be equivalent to each other as well. But some other things to mention here, the angle of parallelism is always right. That's equivalent to the Euclidean parallel postulate. When we talked about the angle of parallelism a couple of lectures ago, we showed that if the angle of parallelism was acute, you can get multiple parallel lines. That would imply not EPP. That actually gives us a hyperbolic parallel postulate. And then therefore if you have unique parallel lines, you have to get right angles, free angles of parallelism. So we didn't do all the details of this one right here, but you can kind of fill in the gaps there, connect the dots, to show that one is equivalent to four in neutral geometry. In the previous lecture, we showed that EPP was equivalent to Euclid's fifth, and we also showed that EPP is equivalent to the converse of the Alternative Angle Theorem. And then half last time, half this time, we showed that the 180 condition is equivalent to EPP, that all triangles angle sum is 180 degrees. Now if you combine the 180 condition with the existence of right triangles, that is the all or nothing theorem we did, was that two lectures ago? Yeah, I think in lecture 26 we did that one. So if rectangles exist, all triangles have an angle sum of 180 degrees, and that's equivalent to the existence of rectangles. So you get rectangles and 180 triangles one together. Well, since the 180 condition implies is equivalent to EPP, that gives the existence of rectangles is equal to EPP as well. Again, some dots to connect there, but we basically have those ones. The right angle hypothesis says that the summit angles of a security quadrilateral are always right angles. That's actually equivalent to saying the fourth angle of a Lambert quadrilateral is always the right angle as well. Well, okay, why is that equivalent to EPP? Well, if your security quadrilateral has right angles, a summit angles, that means it's a rectangle. Right angles exist, you're going to get EPP. On the other hand, if you have EPP, you get rectangles. Rectangles are security quadrilaterals, and therefore you get that the angles have to be right. So that's a pretty quick one to connect there. These last two, I'm not going to say too much about them. We'll talk a little bit more about these in the future. Parallel lines mean equidistant. This issue is we haven't even defined what equidistant even means. Although this is the way that people usually think about parallel lines. Think of like you have train tracks. So your two parallel lines are the sides of your train tracks, and then you have these bars that cross them over and over and over again. And so these bars are common perpendiculars, and the length of these bars doesn't change no matter where we are along the parallel lines. So think of train tracks here. That's how people often think about parallel lines as equidistance. And we'll show sort of in the future that this is in fact an equivalence. We're not going to do that right now. We'll talk some more about equidistance when we talk in hyperbolic geometry, parallel lines. Because this idea of multiplicity of equidistance is in fact something that's equivalent to EPP, right? In hyperbolic geometry, we'll see that if you have a common perpendicular, it's unique. There's not multiple ones. And so this will be more clear in future lectures. As for the Pythagorean theorem, this is a result we're quite common with. If you have a right triangle, and we'll label the sides A, B and C, so that the angle opposite the side C is the right angle, we get the usual equation A squared plus B squared equals C squared. This is the Pythagorean theorem. If you have a right triangle, then you get the equation A squared plus B squared equals C squared. The converse of the Pythagorean theorem is actually if a triangle has side lengths ABC and their lengths add up to be A squared plus B squared equals C squared, then that's a right triangle and the right angle is opposite the side C. That's the converse of the Pythagorean theorem. Turns out those two statements are logically equivalents to each other, the Pythagorean theorem, and it's converse, and they're both equivalent to EPP in neutral geometry. With the Pythagorean theorem, you can actually start to develop the usual notions of analytic geometry and trigonometry that one sees in high school geometry or pre-calculus. When you take a high school or freshman college, it doesn't really matter. You can develop the usual analytic geometry trigonometry using the Pythagorean theorem, these other things. The Pythagorean theorem is actually a Euclidean result. It's kind of hard to believe that at first glance, though, but right triangles in hyperbolic geometry do not satisfy the Pythagorean relationship. There's a ton of other equivalents to Euclidean parallel postulate, which we won't discuss in detail in this lecture or really much in future lectures as well. There's a whole ton of them listed in Road to Geometry. I encourage you to read those in this section if you want some more discussion about those. In fact, going forward, we're not really going to talk about these equivalents as much more. Although there's numerous equivalents to Euclidean parallel postulate, we're not going to prove these equivalences going forward. Instead, we'll be more interested in improving theorems of Euclidean geometry. Many of these theorems will in fact be genuine equivalents to the Euclidean parallel postulate, but we're not going to bother approving the converse. Like I was alluding to before, assuming EPP, we were able to prove Proclus' lemma and transitivity of parallelism. Although there is a proof that goes the other direction, I'm not going to supply that one. I'm not too interested in that right now. What we're going to do is we're actually going to prove theorems in Euclidean geometry, which are equivalent, but we're not going to supply the converse of those things. In fact, the next two theorems are going to be exactly that type of situation. One only needs to know if two statements are logically equivalent if we're trying to prove a specific interpretation of undefined term, terms in a model of a certain axiomatic system, and we believe it's easier to prove an equivalent to an axiom instead of the original or if we want to study a more broad or more narrow axiomatic system, that is, if we want to create independent axioms. The first one I'm trying to say is that if we're interested in the broader meta-geometry type statements, like what axiomatic system should we create? Is this a model? Does this model give us the axioms? Then we care about equivalencies. But if we're just like, hey, I'm trying to develop Euclidean geometry, I only need to see these things as theorems. So we're not going to prove the converses of these things, but these next two theorems are equivalencies, and you'll prove some equivalencies in the homework here. So here's a theorem. In Euclidean geometry, the opposite sides of parallelograms are always congruent. This is equivalent to EPP, but I'm only going to give you the one direction. EPP implies the opposite sides of parallelograms are congruent. So imagine we have a parallelogram. By definition, a parallelogram means we have a quadrilateral so that opposite sides correspond to parallel lines. So if we take the parallelogram A, B, C, D, what we can assume is that A, B is parallel to C, D, and that A, D is parallel to B, C. So we have these parallel lines, and this is going to be useful for us because we're going to use the converse of the alternative angle theorem. So before we do that, we're going to dissect the quadrilateral of the two triangles. So we get the triangle A, B, D, and the triangle C, B, D, like so. Well, because the side A, D is parallel to C, B, we're going to get that the angle A, D, B is congruent to the angle D, B, C by the converse of the alternative angle theorem. We also are going to get, using the parallel sides, D, C, and A, B, that angle A, B, D is congruent to angle C, D, B. Again, using the converse of the alternative angle theorem. And so if we put this all together, so we get that these angles, this angle together, I'm sorry, we're not there yet, we have the blue angles are congruent, the green angles are congruent, and we also have this common edge, D, B, is congruent. So these triangles A, D, B, and the triangle C, D, B are congruent to each other by angle side angle. And so that gives us that the corresponding sides, A, D is congruent to B, C, and likewise, D, C is congruent to A, B. So we get very quickly by the converse of the alternative angle theorem, which is a theorem of Euclidean geometry. We get that opposite sides of a parallelogram are congruent. Again, using, we can show this as an equivalent to EPP, but we're not going to do that. We'll just think of this as a theorem of Euclidean geometry. And as a last theorem right here, we're going to make a statement here. In Euclidean geometry, three non-colonial points determine a unique circle, or we could call this circle determination. This one kind of seems surprising to us that we're so used to this fact because we're so biased to Euclidean geometry because that's the only geometry we're taught when we're little youngsters. But three non-colonial points do not necessarily form a circle. We'll see some examples of three non-colonial points that cannot be inscribed in a circle. Later on, we'll talk about hyperbolic geometry, but this right here is a theorem of Euclidean geometry. In fact, it's logically equivalent, but we're only going to look at one direction here. EPP implies circle determination. As often happens, I do have some lectures where a previous student, Hailey John, had created some slides for us I'm going to use. So we're going to switch over to the web browser here for a second. If you look at the transcripts of this lecture, you can actually find a link to these slides. There's also a link posted in the video description here. So theorem 4-2-2 in Euclidean geometry, three non-colonial points determine a unique circle. So let's start with our three non-colonial points, A, B, and C. We're going to form some lines associated to that. So connect the dots between A and B and B and C right there. So we connect these dots right here and we connect these dots right here. So form the perpendicular bisectors to the segment AB, call that line L, and the perpendicular bisector to BC, call that line M. And so we have these perpendicular lines, L and M, going along right there. All right. Oh, let's see. I think I made a goof by turning... Oh, here we go. Sorry, I was afraid that since I turned on the laser pointer, I can't finish the slides here, but I can. I'm learning still how to use these videos here, how to make these videos. So with the lines L and M, I claim that they're going to intersect, and the intersection of L and M is going to be the center of our circle that's forthcoming. But for the moment, let's assume they didn't. Let's assume they didn't. Well, if L and M don't intersect each other, they would be parallel. And so since they're parallel, you might get some funky picture going on like this, right? Well, since L and M are parallel, and the line AB intersects the line L, by Procos Lemma, which is a theorem of Euclidean Geometry, M has to intersect AB somewhere, as the diagram seems to suggest. And so then we get... I'm going to kind of rearrange it so that like L and M actually look like they're parallel lines now, in which case AB is acting like this transversal here. So AB transverses L and M. And these points of intersection right here. And so now, since the lines L and M are parallel, we can use the alternate-tier angle theorem. We get that M is going to be perpendicular to AB, because L is. So we probably should draw a little right angle right here or something like that. That's a little messy. I'm going to get rid of it. All right, so since AB is parallel to BC, we get that... Well, so sorry, let's back up for a second. So M is perpendicular to AB, L is perpendicular to AB. AB is perpendicular to L, and since M is parallel, AB would have to be a common perpendicular. And so this implies that AB is parallel to BC. By the alternate-tier angle theorem. Since AB and BC now have a common perpendicular, a.k.a. L. But this is a problem. The alternate-tier angle theorem says that AB is perpendicular to B, sorry, AB is parallel to BC by the alternate-tier angle theorem, but they clearly both have the point B on them, so they can't be parallel. This is a contradiction, which in fact gives us that L and M must have intersected at some point. Let's call that point of intersection O. Again, this is going to center of a forthcoming circle. So using the side-angle-side argument, we're going to get that OA is congruent to OB, which is congruent to OC. So how are we doing that? How are we getting those type of situations right here? So we have L intersects M at the point O. How are these going to be radii of all the same circle? How is that going to come into place here? Let's see. So to give you a little bit better explanation there, so we got side-angle-side. So what are the angles we're talking about right here? Well, remember L was chosen to be the perpendicular bisector of A and B. So we get that this segment and this segment are congruent because L bisects that. This segment is shared, and since right angles are congruent to each other, we get that this triangle right here is congruent to this triangle right here. So that's the side-angle-side argument we were talking about just a moment ago. So those triangles are congruent to each other, and therefore the corresponding segment's AO is congruent to OB. But by doing the same argument, we get that OC is congruent to OB, and by transitivity congruence AO, OB, and OC are all congruent to each other. So we in fact do have that these things are all congruent to each other. And so then take the circle whose center is O and whose radius is OA. Because all of those three segments are congruent, they're all going to be points on the circle. Let's see. And so I claim that this circle, so we have now constructed a circle, but this circle is going to be unique. Since a basic side-side-side argument shows that if you had a different circle, if we had some different circle over here, some point P, and you connected these things together, so if we had some different circle trying to do this, you could do a side-side-side argument of the associated triangles that we would form, and this is going to get us a contradiction so that this circle is in fact unique. Alright then. That actually finishes our lecture for today. And the homeworking approved some other equivalencies that we didn't do right now. So let me know if you have any questions for all students, either those in my class or those, either those whether in my class or whether you are just watching these videos. It's good to have you here. Post your comments down below if you have any questions, and I'll be glad to answer them or other other commenters can help you as well. I will see you next time. Bye.