 The previous video for this lecture, we defined the notion of a supplement to an angle. That is, two angles are supplementary if their union forms a half plane. With the notion of a supplement now defined, we can define what it means to be a right angle. So we say that an angle is right if it is congruent to its own supplement, okay? So we don't have any notion of like 90 degree angle or angle measure whatsoever. So we can't really talk about, oh, an angle is 90 degrees. We don't have that notion in congruent geometry, but what we can say is the following. If we have some angle, A, B, C, something like this, then we can say like, oh, its supplement would be D, B, C. And if the angle is congruent to its supplement, then we call that a right angle. And typically when we have a right angle, we'll denote the angle using this little square icon here. This is to suggest like it's the corner of a rectangle. After all the word rectangle itself means right angle, okay? So we can define the notion of a right angle inside of congruence geometry. I wanna prove a few things about right angles in this video here. The first thing is any angle that's congruent to a right angle is itself a right angle. Now I wanna caution you, I wanna point out what this does not say as opposed to what it does say. So this says that if you're congruent to a right angle, then you're a right angle. What this doesn't say is that if you have two right angles that they are necessarily congruent to each other. That last statement is actually Euclid's fourth postulate that he wrote in the elements. And we will prove it as a theorem very, very soon, but not in this video right here. So we're just gonna show that if two angles are congruent and one of them is right, then the other one must be right as well. So it's important that we have that distinction. So let me scroll down here so that we can try to put a diagram on the screen. So let's take two angles. So we have one angle there in red, then we'll have some other angle, maybe something like this. So we have these angles A, I guess I should do that one in red. That's my color coding, isn't it? So we have this angle here, A, this, let's just excuse me, this point, A, B and C. Like so, we also have the angle A prime, B prime, B can be looking a little bit better there, and then C prime, like so. And then we're also, we're gonna assume that these angles are congruent to each other. So let's make a little mark about that. So these angles are congruent, but we're also going to assume that the angle ABC is a right angle. So it's congruent to its supplement. A, B, well, ABC is congruent to its supplement. So we need some point to describe this supplement here. So let's take the point D, so that DBC is the supplement of ABC, and therefore their ABD is congruent to DBC. Now, that's what it means to be a right angle. So be aware that if ABC is congruent to DBC, as we have right here, and we also have by assumption that ABC is congruent to A prime, B prime, C prime by transitivity of segment congruence, we also have that A prime, B prime, C prime is congruent to CBD. So it's important also that we note that these angles, they're all congruent to each other. These two angles are congruent to each other as well. And so what we can of course do is we can insert, you know, a point over here, D prime, so that we can talk about the supplement of A prime, B prime, C prime. That supplement of course would be C prime, B prime, D prime. And so we have supplementation going on right there. Okay, so in a previous video when we introduced supplementary angles, we proved that supplements of congruent angles are congruent to each other. And so notice what happened here. We have that angle A prime, B prime, C prime is congruent to ABC. And then, so that means since these angles are congruent, this will mean that their supplements have to be congruent to each other as well because supplements of congruent angles are congruent. But we also have that A and B here, excuse me, ABC is congruent to DBC because they're both right angles. So they're congruent to each other. So these two angles are congruent by transitivity, but these two angles are congruent. So if we follow all of this transitivity here, so this line gives us congruence, this line gives us congruence, this line gives us congruence. So by transitivity of congruence, these two angles have to be congruent as well. If we go through it just one more time just to be very clear what we're talking about. By assumption, these two angles are congruent because we said they were. These two angles are congruent because this angle over here is a right angle. And then the supplements of congruent angles are congruent a previous theorem right there. And so following this congruent statement by transitivity, these two angles are congruent. They're supplements. Therefore the angle A prime, B prime, C prime is a right angle. Thus proving the very first statement we wanted to do. Now that we've introduced the idea of a right angle it also makes sense to talk about perpendicular lines, okay? So let's take two lines L and M. These are two distinct lines in a congruence geometry. And as such, their intersection, let's assume they intersect, they're not parallel. So these are intersecting lines and let's say that their unique point of intersection is B, okay? And so I'm gonna draw a diagram here to help us keep track of what's going on here. So we have one line right here, another line like so. And so let's say that this line is L, this line is M and I said that their common point of intersection will be B, okay? So on L, let's place points A and D so that B is between A and D. On M, let's place some points C and E so that B is between C and E. The axioms of incidence and betweenness guarantee that such a thing is gonna happen here, okay? So with these labels in play here, we say that the line L is perpendicular to M to note it as L per P M if the angle ABC is a right angle. So we want that this angle right here is a right angle. Now be aware that the supplement of ABC is the angle DBC. It's also the angle ADE, those are both supplements. Those are both supplements in that situation here. And this is something we're gonna, well, I typically leave this as an exercise to my students here, the idea of vertical angles. Vertical angles are angles that are across the, they're on the opposite side of these two intersecting lines right here. Vertical angles are gonna be congruent to each other because they are supplements of the same angle. And as the angle is congruent to itself, supplements of congruent angles are congruent. So vertical angles are gonna be congruent to each other as these are both supplements of a right angle, they're gonna be right angles themselves. And then likewise, this over here, these are vertical angles as well, they're gonna be congruent. So it's gonna, I should say these angles, I have to be careful here because we haven't proven that right angles are congruent. But as these two angles are gonna be congruent and this is a right angle, we just prove that this is a right angle as well. So it actually doesn't matter which of these four angles you talk about, if one of them is right, all four of them are gonna be right. That's why we can describe lines being perpendicular without referencing which angle because it doesn't matter which of the four angles you get when two lines intersect each other, they form four different angles. But if one of them is right, all of them are gonna be right. And so when that happens, we say that the lines are perpendicular to each other. Perpendicular, of course, is a symmetric relationship. If L is perpendicular to M, then M is perpendicular to L. It doesn't really make sense for a line to be perpendicular to itself. So we don't allow for that. And we also don't claim anything about transitivity because that actually is not gonna be the case typically. If two lines are perpendicular to some common line, we'll actually see in congruence geometry that they're gonna be parallel to each other. So the only thing we can guarantee right now for perpendicular lines is that it's a symmetric relationship. And so to finish this video in particular, I wanna prove a very important theorem with regard to perpendicular lines. We will call this the dropping of a perpendicular to a line. This will be very different from another theorem which we will talk about later about erecting a perpendicular out of a line. So there's dropping versus erecting out of a line. The direction does matter in this situation. So the statement of the theorem is the following. We have a line L. We have a point P that is not on L, okay? Then we claim that there exists a line M that contains P but is perpendicular to L. So when you look at the surface of this, this sounds a lot like the Euclidean parallel postulate. If you have a line and a point not on the line, then there exists another line that contains the point but is parallel to L. In congruent geometry, we have no assumptions about parallel lines. So we're not doing that, but this theorem is very similar but it doesn't guarantee the existence of a parallel to L. It actually guarantees a perpendicular to L. And so let me draw a picture that'll help us explain the proof, of course, that you can read over here. I hope you do read along with these things. So we have a line L and we have some point P which is a point not on L. What we're trying to then construct is a line that passes through P and is perpendicular to the line L. That's what we're trying to do. We're trying to construct this line M right here, okay? Now we don't have that yet, we're gonna construct it. So what we do have is the following. There do exist at least two points on the line L that they're not equal to each other. This is just the secancy axiom for line determination, excuse me for incident geometry, line determination was another axiom there. So we have two points A and B and they're distinct points on side of L. By angle translation, there exists a point C that's on the opposite side of L that P is. So P's over here, so there's some point over here, call it C and we can have it so that the angle BAP is congruent to the angle BAC. So I'm gonna move my picture up a little bit, covering up the proof. And so we have this angle, this angle BAP, but we also know there's some other point down here, call it C, so that the angle BAP is congruent to the angle BAC. This is a consequence of angle translation. We also know that by segment translation, we could take the segment AP and we could translate that onto the ray AC so that we have some point, call it C prime, so that AP as a segment is congruent to the segment AC prime. Now without the loss of generality, we could assume that point was C in the first place because after all, C is just giving us the direction, it's just giving us the location of the ray on that side of the line. So again, by swapping C with C prime, we can assume that the segment AP is congruent to C. And in fact, just to make the correspondence even more available to us, we're gonna call this point P prime. So we had this point C earlier, but we don't need it anymore. So P prime was the point on the ray AC so that AP prime is congruent to AD, okay? And so that gives us the two points we're very interested in. And I'm gonna actually adjust my point one more time. Let's see, let me clean this up because I want the diagram to look perpendicular. So I kind of actually need to get rid of this point right here because actually if I were drawing this in Euclidean geometry, the point would be about right here. So here's our point P prime. Like I said, we don't actually care about C anymore so I don't mind erasing it. We get the following, it's not perfect but we'll do this trick, right? I want my diagram to look a little bit better. I'm gonna take the line that goes through P and is perpendicular to L, I'm just gonna draw it. I don't know that they're perpendicular yet and then I'll fix my diagram using that. So here's P prime and so that fixes my picture there. So this new line we just constructed, what is it? We're gonna call it M. This line is the line determined by P and P prime. We have that line right here. I don't claim yet it's perpendicular. We're gonna have to prove that but that's the bulk of the remaining proof here. So we have our candidate for the line M. Clearly by construction it contains the point P. We didn't have to argue that it is perpendicular to L. The first thing we kind of have to do is of course argue that they actually intersect each other. I took some great detail to draw the picture more accurately but nonetheless, how do we know that these points actually intersect each other? Well, there's a couple of options we have to consider, two cases really. So we have these points A, P and P prime. What if they happen to be co-linear, right? We could have something like this happening where here's P, here's A and here is P prime and then B's over here somewhere. Well, if that's the case, then it would have to be that A is between P and P prime since P and P prime are on opposite sides of L and that's the consequence of plane separation, posh's axiom, right? So if that's the case, then okay, then we have the intersection. The intersection would be some point called A in that situation. Now, if I were to guess randomly, that's probably not the case here. I mean, that would have only been if A was a fantastic guess, but that's okay. No big deal. It's really not such a big deal whatsoever but I guess I actually still want this picture on the screen here. So if A is in fact between them, should have called that P prime. So we know that these angles P, A, B and P prime, A, B are supplementary to each other and that's because P, A, and P prime are all on the same line, that's that supplement. But also by construction, the angles B, A, P is congruent to the angle B, A, P prime because we translated the angle from one side of the line to the other side. So these angles are congruent and they're supplements so that makes them right angles. In that situation, we would then have the line M does in fact intersect the line L and then they're gonna be perpendicular. The point of intersection is A in that situation. So we get that M and L are perpendicular. So that's the first case that we have to consider. The second case is now suppose that A, P and P prime are not collinear. So let Q be the point of intersection with M and L which we would label that right here. Okay, just a little bit of vocabulary here. This point for a perpendicular line intersects the other one or I should say you have this point P where you drop down the perpendicular line to get this point of intersection. This is commonly referred to as the foot of the perpendicular but maybe I'm gonna a little ahead of ourselves. We don't yet know that these angles are perpendicular but also how do we even know these lines intersect each other, right? This actually is a consequence of the crossbar theorem. If you look at the angle P, A, P prime and then it's crossbar P, P, P prime there then of course the interior angle B is gonna be interior there. So this ray has to hit the crossbar somewhere. So it's gonna be at this point we're calling that point Q, all right? And of course B is an interior points to the angle of because of consequences we had proven previously in order to geometry. So we can conclude that Q is between P and P prime. Now of course Q is on the line L by construction. So it has to be the case that A, B and Q since they're co-linear there has to be some between this relationship going on there. Let me zoom out here. So where does Q live relative to A and B? It could be like the picture is illustrated on the screen right now where Q is between A and B. It could also be that B is like over here maybe B could be between A and Q. That's a possibility. Now in this situation if that were the case the ray, the ray A, B is identical to the ray A, Q. And so in that situation the angle Q, A, P would equal the angle B, A, P for either of those two situations. There is a third possibility over here could be where B is. Maybe A is between B and Q. And I did misspoke earlier I had said something like that B was an interior point to the angle P, A, P prime. We don't necessarily know that. We do know that if B is not an interior point like in this situation there is an interior point on the other side of it, call it B prime or something. So we can still use the crossbar theorem of course to get this point of intersection Q. The location of B was somewhat irrelevant. So I misspoke in that situation. But I want you to consider this possibility. What if A was between B and Q like in this situation? Well if that happens then it turns out the angle B, A, P is a supplement to the angle Q, A, P like so. But then their supplements would be congruent to each other. And so what we want to conclude here is whether my diagram is correct or not the angle Q, A, P is congruent to the angle Q, A, P prime even if the betweenness relationships aren't what I have here on the screen. And so that's what I'm trying to argue here. Let me scooch down the diagram now and let's zoom out. So again, I wanna see the words and the diagram happening right here. So this is a pretty good vantage point. So we know that whatever assumptions we have about A, Q and B either all of them lead to the same conclusion that angle Q, A, P is congruent to angle Q, A, P prime. And since A, Q was congruent to A, Q prime and Q, A, P is congruent to Q, A, P prime. And lastly, A, P is congruent to A, P prime by the side angle side assumption axiom we get that Q, A, P is congruent to Q, A, P prime. So I want you to think of these triangles right here and these triangles, they have to be congruent to each other. So notice here that A, P was congruent to A, P prime by assumption. We just argue that these angles were congruent to each other. And of course the segment A, Q is congruent to A, Q. I mean, that's just reflexive. So these triangles are gonna be congruent to each other. So in particular as congruent parts of corresponding, excuse me as corresponding parts of congruence triangles are congruent, we get that the angle A, Q, P is congruent to angle A, Q, P prime. So this angle right here is congruent to this angle right here. There's supplementary angles by construction and therefore they have to both be right angles. So we did construct a perpendicular line, M to the line L right there. I'm gonna zoom in to the text in case you wanted to copy it down. Pause the video if you need to. It might've been harder to see on the previous resolution but that does bring us to the end of this proof here about the existence of perpendicular lines. In particular, what we said here is if you have a line and you have a point off the line you can always drop a perpendicular onto that line. This point of intersection here, we call the foot of the perpendicular and this is a construction that we will use many times in the future of congruence geometry.