 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 17, we're going to prove two very important theorems for congruence geometry. In particular, we're going to talk about, in this video, the ultimate interior angle theorem, and in the next video, we'll talk about the exterior angle theorem for triangles. Now to do the so, we need to first clarify some important definitions of what these things mean. So what does it mean to be alternate interior angles in the first place? So imagine we have two lines, L and M, that we can put here on the screen like so. L will put on top, and M will put on the bottom like so. And these are distinct lines. And suppose that we have a third line, T, that so to speak transverses the two lines. And this can actually take place in an ordered geometry. We don't need any notions of congruence yet, just to define the notion of alternate interior angles. Of course, the alternate interior angle theorem will say something about congruence, and so therefore we'll have to be in a congruence geometry. So we have our line T, which is referred to as the transversal of these two lines. So the transversal intersect the two lines at two different places. And now I don't necessarily claim that the lines L and M are parallel to each other, although the way I've drawn them on the screen, it seems to appear that way. But this idea of a transversal just hits the two lines, but it's not at some concurrent location. In fact, the point of intersection between the line L and the transversal T, we're going to call that P. And we're going to call the point of intersection between the line M and the transversal T, we're going to call that Q. And so to be a transversal, we do call those different points. So those is not just one common point there. And then also for description of angles here, we're going to add some points here. We're going to say A is a point on L and B is a point on M. And we're going to say that A and M are, excuse me, A and B are on the same side of the transversal T. We're going to make it so that the point A prime is another point on L, but it's going to be the case that P is between A and A prime. Therefore, A and A prime are on opposite sides of the line T, but both on the line L. And then similarly, we're going to add the point B prime onto our diagram here, where just like before, Q is going to sit between B and B prime. So this is going to give us that B and B prime are on opposite sides of the line T, but both of them are on the line M like so. And so by labeling these points on our diagram here, this creates four angles that we want to talk about. So there's going to be the angle APQ, which of course is this angle right here. There's the angle A prime PQ, which would be this angle right here. There's going to be the angle BQP, which is this angle right here. And then there's going to be the angle B prime QP like so, which is this angle right here. So these angles. And so I'm going to label them one, two, angle three, angle four. These four angles that you see labeled now on the screen are what we refer to as the interior angles of the transversal T with respect to the lines L and M, which it transverses interior because it sits in between the two lines L and M. And so naturally, we can also talk about exterior angles in this situation. So we can talk about this angle, this angle, this angle like so. And so there will be analogous statements we can say about the exterior angles as well. We're going to mostly just focus on the vocabulary for the interior angles here. So we say that two interior angles are consecutive if they're on the same side of the line. So angle one and angle three are considered consecutive interior angles because they're on the same side of the of the transversal T. So AQ, excuse me, APQ and BQP are consecutive interior angles. And likewise, A prime PQ and B prime QP are consecutive interior angles. We say that I mean, clearly these two angles right here, one and two, they're supplementary angles to each other, three and four, they're supplementary interior angles. So if they need a label, we can call them that. Then the next one, the main one we're going to go for here is the notion of an alternate interior angle. So we would call angles one and four, alternate interior angles because they're on opposite sides of the line, but they're also not the supplements of each other. So you go down over here. So they're kitty corner from each other with regard to this diagram. Angles one and four, which, of course, are angles APQ and B prime QP. Those are considered alternate interior angles. And likewise, angles two and three, that is A prime QP and BQP. These are also considered alternate interior angles. All right. With regard to the exterior angles, we can say similar things as well. Like if we had this angle right here, this exterior angle would be its supplement. This angle right here would be its consecutive, excuse me, consecutive exterior angle. And this one right here would be its alternate exterior angle. So you can do some analogous statements like so. You can also connect together, of course, exterior angles with interior angles, because these angles are vertical angles with each other. These angles are supplement angles with each other. But like I said, our emphasis is going to be placed upon these alternate interior angles. In particular, that leads to the so-called alternate interior angle theorem, which says that if two distinct lines, if they're cut by a transversal, then the pair of alternate, if the alternate interior angles are congruent, excuse me, then we're going to have that the two lines are parallel. So the alternate interior angle as a theorem in congruence geometry, just so we're aware here, that if we have congruent alternate interior angles, we have to have a notion of congruence. So that happens exactly in, of course, an alternate congruence geometry. So alternate interior angle theorem is a theorem of congruence geometry. If the alternate interior angles are congruent, then this guarantees that the two lines that were transversed were actually parallel to each other. Now, I have to caution you that as we prove the alternate interior angle theorem in this video, we are not proving the converse of the alternate interior angle theorem, which the converse would tell us that if two lines are parallel, then alternate interior angles are congruent to each other. That's not what we're proving here. The converse of a statement and the statement itself are not logically equivalent in general. That would only happen if we have a bi-conditional situation, which in neutral geometry, or I should say in congruent geometry, we do not have that notion. We will see later on in this lecture series that the converse of the alternate interior angle theorem is actually equivalent to the Euclidean parallel postulate. So what we're going to do is we're going to reconstruct the diagram we had on the screen just a little bit ago. Sorry about that. Let me put it right here and right here, like so. And so this was our line L on top, our line M on the bottom. We had this transversal T, something like this. And again, we're labeling all the points. So the point between L and T, we call it P. We had A on one side of the line. We had A prime on the other side of the line. The intersection between B, excuse me, the intersection between T and M was Q. B was on that line on the same side of A. And B prime was on the same side that A prime was. So we had this picture right here. And by assumption, we're assuming that the alternate interior angles are congruent. So we're going to assume the angle APQ is congruent to the angle B prime QP. So that's the assumption that we have in this situation. We have these congruent alternate interior angles like so. Now we have to prove that the lines L and M are parallel to each other. So we're going to do this by contradiction. So we're going to assume for the sake of contradiction that L and M intersect each other. And we're going to call that point of intersection R, in fact. So we're going to extend our lines. How's that going to happen? We're going to extend our lines to be something like this. It's kind of weird, but we claim the lines are intersecting. So there's going to be some point of intersection R right there. And without the loss of generality, since L and M are intersecting each other, we can assume that the point of intersection is going to be on the same side of the line T that A and B are, because again, if it was on the other side, we could just relabel A with A prime, etc. We could change this label. So without the loss of generality, we can assume that R is on the same side as A and B. OK, well, by segment translation, what we can do is we can take the segment PR right here and we can translate it onto the ray B, excuse me, QB prime. So in particular, we can we can add a point R prime onto the ray QB prime so that the line segment, the line segment PR is congruent to the line segment QR prime like so. I should also mention, of course, that the line segment PQ is, of course, congruent to itself, of course, by just congruences in equivalence relation. You're congruent to yourself right there. So I want you to notice what we have here. We have a side angle side situation. In particular, the segment PQ is congruent to itself by assumption, the alternate to your angles AP, Q and B prime QP are congruent to each other. And then by segment translation, the segment PR is congruent to the segment QR prime like so. So if I were to connect the dots like so, then we have a triangle, believe it or not, we have this triangle PQR. It'll be congruent to the triangle PQR prime like so by side angle side. The side angle side axiom applies right here. So in particular, since the triangles are congruent to each other, we're also going to get that will corresponding parts of congruent triangles are congruent. So if we look at the angle PQB, which is this one right here, it'll be congruent, which I should say PQB, of course, is the same thing as the angle PQR, it'll be congruent to the angle QPR prime, which is this angle right here. So again, corresponding parts of congruent triangles are congruent like usual. So then I want us to investigate this angle right here, right? Because notice that angle RQP, this is the supplement to B prime QP like so. And likewise, the angle A prime PQ is the supplement to the angle APQ. And since we have that these alternate to your angles are congruent, their supplements must be congruent as well like so. So in particular, we have to infer that the angle A prime PQ is congruent to the angle BQP like so, because supplements of congruent angles are congruent to each other. So what we have here is we have two angles. We have the angle QPR prime and the angle QPA prime. They rest on in the same half plane of the line PQ on the same ray PQ, but they're congruent to each other. So by uniqueness of by uniqueness, well, I should first of all say that by transitivity of congruence as both of these angles are congruent to this angle down here, BQP, they're congruent to each other. And so then by uniqueness of angle translation, it's got to be that the angle A prime PQ is in fact congruent. I should say not just congruent, it's equal to the angle R prime PQ. So it turns out my diagram is actually not quite correct. Let's correct it so slightly. What we actually have instead is that this line is in fact the line P A prime is the line PR prime like we saw before. And so that these were the angles that were congruent to each other. The supplement was in fact the case there. So we have two triangles over two congruent triangles. So we have the triangle PQR. We also have the triangle PQR prime like so those were congruent to each other by the side angle side like so. And so notice notice what's happening here. We have the ray P A that intersects the ray P B at the point R. We also have the ray P A prime that intersects the ray Q B prime at the point R prime like so. But if you take the ray P A and P A prime and put them together, that's a single line. That's the line L that we talked about earlier. And if you take the ray if you take the ray Q B and combine it with the ray Q B prime, those two rays together form the line M. And so you'll notice that the lines L and M if they intersect, they in fact have two different points of intersection. They intersect at R and R prime, which that's a contradiction to line determination, which guarantees that when lines intersect, they do it at a unique location. And that gives us the contradiction we were seeking here. That if the lines L and M were parallel, if they're not parallel to each other, then they have too many intersections and that's not possible in a congruence geometry, thus proving the ultra interior angle theorem. I also want to very briefly mention a corollary to the ultra interior angle theorem. This would be the so-called consecutive interior angle theorem, which says that if two distinct lines cut by a transversal have a pair of consecutive interior angles, which are such that one interior angle is congruent to the supplement of the other. So basically, if the congruent interior angles are supplements of each other, then the two lines have to be parallel as well. So I want you to think of the picture we had from before, something like this for which we had Q, we had P, and suppose that we have, well, again, this is L, this is T, this is M. We have A over here, we have B over here. Let me just throw all the points on there. B prime like so, A prime like so. So suppose we have some point, OK? So let P be between some point Q and R, which Q we already know, R would have to be something over here, right? Then look at the angle A, P, R, this angle right here. This is the supplements to the angle A, P, Q, right? So notice that these two angles are going to be supplements to each other. That's what we're saying here. Suppose that the angle B, Q, P is congruent to A, P, R. So this is what we're trying to say that by assumptions, we have these two, we have these two consecutive interior angles, angle one and angle two. We're assuming the angle two is congruent to the supplement of angle one. So call that angle three just for simplicity here. All right. But I also want to point out that A, P, R and A prime Q, A prime P, Q, which would be over here. This is angle four. These are going to be congruent to each other as they are vertical angles. OK. And so by transitivity, we have angles two and angle four are congruent to each other. But these are alternate to your angles. So then by the alternate to your angle theorem, we get that the two lines are parallel and so we get that immediate consequence. Thus proving the consecutive interior angle theorem. This alternate to your angle theorem is very important. I do want to make a comment about it that we're going to see in the not too distant future of this lecture series that because of the alternate to your angle theorem, there's going to exist a so-called guaranteed parallel line given any point, excuse me, given any line and a point off of that line. And that guaranteed parallel line will be an authentic parallel line, which actually shows us that the alternate to your angle theorem is inconsistent with the elliptic parallel postulate. And therefore elliptic geometry cannot be realized inside of congruent geometry, a topic we will talk about in the future.