 Welcome back everyone to our lecture series for Math 3130 Modern Geometries at Southern Utah University. This lecture is lecture 27 in our series entitled Euclidean Geometry, and it is loosely based upon section 3.4, the place of parallels in the textbook Roads to Geometry by Wallace and West. Welcome back to all my students and also anyone else who wants to be following our lecture series. Good to have you here. If you have been following our series so far, you'll know that I've been very hesitant to do anything in Euclidean Geometry. But given past lectures, it's kind of unavoidable right now. We have to say something about Euclidean Geometry. I mean, and after all, it's certainly the most famous and popular geometry out there. But because students and viewers actually have a strong understanding of Euclidean Geometry, I don't want to say much about it. But here, finally, drum roll please. We're actually going to have an official definition of what Euclidean Geometry is. As a reminder, what do we mean by Neutral Geometry? Neutral Geometry is any geometry that satisfies the Hilbert axioms of incidents, betweenness, congruence, and continuity. Neutral Geometry has no assumptions about parallelism whatsoever. Euclidean Geometry, we finally do accept as the axiom of parallelism, the Euclidean Parallel Apostolate. And remember, we're using the play fair version of the Euclidean Parallel Apostolate, which means that if we have a line L and a point P that is off the line L, there exists a unique parallel line M that goes through P and is parallel to L. So that's our version of the Euclidean Parallel Apostolate. And it turns out that in Neutral Geometry, parallel lines always exist because of the following argument. You can take the unique perpendicular that's dropped from P, and so it's perpendicular to L, call this line T, and so we get that T is perpendicular to L. And then you can take the perpendicular that's erected out of P, but perpendicular to T, that's our line M, so T is perpendicular to M. And by the Alternative Angle Theorem, because we have congruent alternate angles, because they're both right angles, this one and this one right here, we can get that line M and L are always going to be parallel to each other. Whoops, they're always parallel to each other. And so this is what we call the guaranteed parallel line. And this line T associated will call the perpendicular transversal. In our constructions today, this construction will show up a lot. I should say in the proofs we're going to talk about today, this guaranteed parallel line in this perpendicular line as well. Now, I should mention that in the search for a proof of the Euclidean Parallel Apostolate. I mean, people were obsessed with this thing for a couple hundred years, right? Some mathematicians of prominence in this regard. I mean, we should first mention Giovanni Sicari, an Italian mathematician. He lived from 1667 to 1733. Sicari was an expert in his day in the Reducto Aud Assertum. R-A-A-R will be more commonly referred to as proof by contradiction nowadays. Sicari was like the expert of at the time. It was like state-of-the-art technology in the 17th century there. And so Sicari's approach to the Euclidean problem was that he wanted to assume the negation of the Euclidean Parallel Apostolate and then arrive upon a contradiction. If he could construct an inconsistent geometric system, then that would imply that we actually have, we must assume the Euclidean Parallel Apostolate as truth. And because of the Sicari-Legender theorem, we know that if you take the negation of Euclide's fifth postulate, that would actually be equivalent to triangles having an angle sum of strictly less than 180 degrees. But he was never actually able to do it. Adrienne Mariez Le Genre, a fringe mathematician who lived about 100 years after Sicari, Le Genre lived from 1752 to 1833. He developed basically the same type of logical system, independent of Sicari, doing basically the same thing, trying to build an inconsistent logical system to prove the Euclidean Parallel Apostolate. Johann Lambert kind of did the same thing as well. He was a contemporary of Le Genre. He lived from 1728 to 1777. Lambert quadrilaterals are named after, of course, Johann Lambert. And Sicari quadrilaterals are naturally named after Giovanni Sicari. And then the Sicari-Legender theorem also named after these people as well. And so they try to develop this inconsistent logical system by negating the Euclide Parallel Apostolate. But they could never do it. It turns out they actually developed a lot of theorems of hyperbolic geometry, not knowing it. But as geometers extensively tried to prove the Euclidean Parallel Apostolate as a theorem of neutral geometry, they consistently failed. And this is due to the fact that Euclidean Parallel Apostolate is independent of the neutral axioms. This is evidenced by the consistent models of hyperbolic geometry, which we've talked about before, and we'll talk about these in future lectures as well. Although the Euclidean Parallel Apostolate could not be proven from the neutral axioms, many important equivalents to the Euclidean Parallel Apostolate were discovered throughout this journey. Also, theorems of hyperbolic geometry were proven along this way as well. And so these useful theorems we're going to talk about in this lecture, the next one. And these are going to be theorems of Euclidean geometry. And you could also phrase them as theorems of neutral geometry by saying these things are equivalent to EPP. So let's talk about those. And maybe the one we should talk about first is, in fact, that the Euclidean Parallel Apostolate Playfairs version is equivalent to Euclid's fifth postulate, which stated here is that if two lines intersect, so we have two lines, say M and L, if they intersect a transversal T, and by transversal here we mean the line T intersects both L and M, but they don't intersect at some concurrent point, if the angle is associated to this transversal, if on one side of the line the consecutive interior angles add up to be less than a flat angle, that is less than 180 degrees, then we actually get that these two lines intersect somewhere on that side of the line. Now this version of Euclid's fifth postulate, I have modernized the language a little bit and also put in English, Euclid didn't speak English. Anyways, so it's not perfectly how Euclid would have written it, but this is basically the same thing. And so we want to prove that Euclid's fifth postulate is equivalent to the Euclidean Parallel Apostolate that we're using due to Playfair. And it might not be too surprising that Sicari and others thought they could prove Euclid's fifth postulate as a theorem in neutral geometry. After all, Euclid was not very good at stating his assumptions and to give Euclid some credit, right? I mean, this was done thousands of years ago. He was really good at sort of starting the art of axiomatic geometry, but we've done better since then. And also to Sicari and others benefit, Euclid's fourth postulate, we could in fact prove as a theorem of the neutral axioms. So why not, why not the fifth postulate? Well, we'll see why you can't do that. I mean, this is forthcoming, right? So when you're trying to prove that two things are logically equivalent, there is basically two proofs for the price of one. So let's first assume Euclid's fifth postulate and we're gonna prove the Euclidean parallel postulate from that. So to prove Euclidean parallel postulate, we start off always with a line, L, it's a generic line, and we take any whole point P that's not on L. And we wanna show that there's only one, exactly one parallel line that goes through P. It's parallel to L. So let's start off with the guaranteed parallel line, M. And let's take the perpendicular transversal T. So T is perpendicular to M and L there. And suppose we had a different line. Any other line, take M prime. So M prime goes to the point P, but it doesn't form right angles with T. So because this is distinct from M, the angles with respect to T and M prime here, one of them has to be acute and the other one has to be obtuse. Focus on the side of the line that has an acute angle. Well, because one angle is acute and the other angle is right, this says that the angle sum on that side of the line is gonna be strictly less than 180 degrees. And so if we use Euclid's fifth postulate, which we're assuming right now, this would say that on the right side of the line here, we have an angle sum less than 180 degrees. So the postulate then gives us that these two lines have to intersect somewhere. That is L is not parallel to M prime. These are intersecting lines. This would apply to every line except for the guaranteed line and therefore we're gonna get a unique parallel line to L through P and that gives us the Euclidean parallel postulate. So that's pretty quick, pretty quick argument there. Let's go the other way around. Let's assume this time we have the Euclidean parallel postulate and let's say we have the picture we had from before. So we have two lines, L and M, they're cut by a transversal T and let's assume that the angle sum on one side is less than 180 degrees. How do we form a, how do we get our intersection right here? So for some labeling purposes, let's say the following. Let's say that B is the intersection of L and T and let the intersection of T and M be the point D. On the side of T for which the angle sum is the consecutive interior angles is less than 180, we'll call C a point on L on that same side. On the opposite side, we'll call that point A and then on the same side of M that C is, Charlie, let's take the point echo, that is E right here. And so we can talk about these angles right here. So just for future discussion here, we have angle one, angle two, and angle three. All right. Now, if angle one and angle three were congruent to each other, I'm sorry, what can we say so far? So angle one and angle two have an angle sum less than 180 degrees. That we have by assumption. We also that have angles two and three as their supplementary angles. If we add them together, we get 180 degrees. If you put those two arguments together, you're gonna see that the measure of angle one is less than the measure of angle three. And so therefore, we can translate this angle over here and so that this new angle is angle three. It's bigger angle and let F be a point in this direction right here. So now we get that this new line associated to DF, it has an alternative angle congruent to angle ABD. And so by the alternative angle theorem right here, we get that this line DF, we'll call this line M prime here. We get that M prime is parallel to L, right? So we constructed a parallel line to L here. Now, because we're assuming that you're including parallel postulate, this is the only line that can be parallel to L that goes through D. Therefore, the line M must intersect L somewhere. And so we get the intersection point that we are looking for. And in fact, because of the Sicari-Legionner theorem, the intersection must be on the same side that the angle sum was less than 180 degrees because on the other side, if this side's angle sum is less than 180 degrees, then the left side's angle sum would have to be larger than 180 degrees, which couldn't have a triangle that big by the Sicari-Legionner theorem. So we get the intersection is on the side of the smaller angles like we expected for Euclid's fifth. And so this shows that the two are logically equivalent to each other. And so it's probably useful to know that Playfair's version of Euclidean parallel postulate is equivalent to Euclid's version of Euclidean parallel postulate. All right, let's look at another alternative here. In the previous theorem, we use the alternative angle theorem and be aware that we can use that because the alternative angle theorem is a theorem of neutral geometry. What we wanna prove now is that the Euclidean parallel postulate is actually equivalent to the converse of the alternative angle theorem, which the converse of the alternative angle theorem says if two lines are parallel, then the alternative angles are congruent. What we used in the previous proof was that if alternative angles are congruent, that the two lines are parallel. One has to be very careful because we don't make mistake in these things because using the converse of alternative angle theorem is equivalent to the Euclidean parallel postulate, but you can actually use alternative angle theorems and neutral geometry without any problem. So again, this is an equivalence. So let's prove one implies the other. So let's assume the Euclidean parallel postulate, Playfair's version. And so we're gonna improve the converse of alternative angle theorem. So we start off with parallel lines. So we have lines L and M. Whoops. And they're cut by a transversal, T. And let's assume that the associated alternate interior angles are congruent to each other. So the forsake of labeling, let's call the intersection of A and, sorry, the intersection of T and L is A. On one side of the line, we have the point B. On the other side of the line, we have the point C. We'll say the intersection, I'm sorry, not C. I'm gonna call this point right here B prime. So A sits between A and B prime. The intersection of M and T, we're gonna call that C. And then on the same side as, on the same side as the point B, we're gonna put the point D. So D is on M and it's on the same side of T that B is. And likewise on the same side of T, we're gonna put the point D prime. It's on M, it's on the same side with B prime. So we get this picture right here. So we have parallel lines and by assumption, we're gonna have alternate interior angles are congruent. No, I'm sorry, I'm sorry, I take that back. We're proving the converse here. See, it can be confusing at times. So we assume that L is parallel to M and we wanna show that alternate interior angles are congruent to each other. That's what our goal is right here. So what we wanna do is we have that M is parallel to L by assumption, we're assuming the alternate interior angle. I'm sorry, we're assuming the Euclidian parallel postulate. And I mean, I should say that we were assuming B and D are on the same side of the line. We can also get that D prime and B prime are on the same side of the line as well. So we're only assuming D is on the same side that B is. But using things like the crossbar theorem and other theorems of betweenness, we can guarantee that these points are on the same side. I'm sorry, I take that back. There's no issue. We just use the extension axiom right here. We can get D prime and D are on opposite sides of C and we get B prime and B are on opposite sides of A. And then so it's really just plain separation that gives these things are on the same side, these same side and these are on the same side. Sorry about that right there. So what we're gonna do is we are going to, if these angles were not congruent, if these were not congruent, then we can actually construct an angle, we can construct an angle that is congruent to angle CAB prime, this angle above right here. So these are both congruent to each other, angle one there. But then because alternate to your angles will be congruent to each other, this new line called M prime would be parallel by the alternate to your angle theorem. But, so now we get that L is parallel to M prime. But by the Euclidean parallel postulate, there can only be one line parallel to L that goes to the point C. And as these were different angles, we would get a contradiction right there, right? So what that actually implies to us is that that line doesn't exist. The original two alternate to your angles must in fact be congruent to each other. And so this proves the converse of the alternate to your angle theorem. We have to prove the other way around that assuming the converse of the alternate to your angle theorem implies the Euclidean parallel postulate. But I'm gonna leave that as a homework question to my readers and to my students and the viewers of this video right here. But it can be proven by a similar type of argument. All right, so another equivalence to the Euclidean parallel postulate, we get is that the Euclidean parallel postulate is equal to the angle sum of every triangle being 180 degrees. That is every triangle is a 180 triangle if you assume the Euclidean parallel postulate. So how does that argument go right here? Well, again, there's two directions, one has to go. So we're gonna first assume the Euclidean parallel postulate and then we're gonna prove that we have a 180 triangle. So let's take an arbitrary triangle in neutral geometry we're assuming EPP. So actually we're in Euclidean geometry right here. So we have a triangle ABC and I'm gonna prove that the angle sum of ABC is 180. So because we can assume the Euclidean parallel postulate constructs the unique parallel line that's parallel to AB that goes through C. This line is gonna be unique and then choose points on this line. We can call this line L right here. Choose points on L. We're gonna label these points A prime and B prime. We want that A prime is on the opposite side of A associated to the line AC and we want that B prime is on the opposite side of B. I'm sorry, I said that backwards, try that again. A prime, no, that's right. A prime is on the opposite side of CB associated to A so A and C prime are opposite sides there and we do want that B and B prime are opposite sides of the line AC. And again, we can place these using just usual between this arguments of extension, the crossbar theorem. We can guarantee that this picture we have is accurate right here. And so let's consider the triangles, the angles associated to this triangle right here. So if we take angle one to be angle A, angle two to be angle C and angle three to be angle B of the original triangle, then we actually get using the converse of the alternate to your angle theorem. Notice because we're using, we're assuming EPP, we get everything that's equivalent to EPP that we've already proven. And in the previous theorem, we showed that the converse of alternate to your angle theorem is equivalent to EPP. So that's actually what I'm gonna use right here. So because L and AB, the line, are parallel to each other, so L is parallel to the line AB. We get by the converse of the alternate to your angle theorem that this angle right here, the B prime CA is congruent to A, B, C because we're using the transversal, not that transversal, sorry. Let's see, am I drawing the wrong picture again? It's one or the other, I always have to keep track of these things. Let's see. So we want to use the converse of the alternate angle theorem, so it's one or the other, let's just draw one and see which angles are right here, right? Oh, I'm sorry, this is the issue. This is angle one. There you go, fixed it right there. So by the converse of the alternate to your angle theorem, because AB and L are parallel to each other and they're transversed by the line AC, the angle B prime CA is congruent to the angle BAC. So those are both angle one. And then if we look at the transversal BC, this transverses the line AB and L so that angle A prime CB is congruent to angle ABC. So we get, those are congruent right there. All right, sorry for that little brain fart there, but we got all the pieces together now. So angles one, two, and three add up together to be the angle sum of ABC, but also angles one, two, and three add up to be a flat angle, which means their angle sum is 180 degrees. So this shows that the triangle has an angle sum of 180 degrees as well. So if we assume EPP, we get that all triangles are 180 triangles. We're actually gonna prove the other direction in our next lecture. So that'll be lecture 20, which one are we on right now, 27 right now. So in the next one, we're gonna do lecture 28. We'll finish this one. But as a homework exercise, we actually get that Euclidean parallel poshia is equivalent to the existence of triangles. And so the basis behind this proof is that if you take the previous lecture, 26, we prove the all or nothing theorem that if you have one 180 triangle, all triangles 180, in the process of proving that, we show that 180 triangles imply the existence of rectangles. So use the previous theorem with the all or nothing theorem, we get that EPP is equivalent to the existence of rectangles. And again, that's left as a homework exercise. We'll actually complete the proof of the previous theorem, this theorem, 344 next time. So please look for lecture 2028 for that right there. As usual, the transcripts of this lecture, you can find online, just click the link that's in the description to find those. If you want to read those along, those are available to my students. In fact, anyone who wants to watch these videos. And if you have any questions or comments, please put them in the comments on this video below. I'll try to answer any questions or other people viewing can help you answer some questions as well. Thanks for coming in. Stay tuned for next time. Bye.