 In this video, we want to prove Euclid's fourth postulate, and we're going to prove this as a theorem of congruence geometry. Now, it might seem weird. How can you prove a postulate? Isn't this something like an axiom? You just assumed to be true. Well, it's called Euclid's fourth postulate because in the book The Elements written by Euclid, Euclid took this as his fourth postulate of geometry, which of course then later on became known as Euclidean geometry. He didn't name it after himself. It is, of course, in homage to the work that Euclid did. Now, we took a different axiomatic approach to Euclidean geometry. In particular, we're taking the axioms of David Hilbert to axiomatize Euclidean geometry, and we're taking one group of axiom at the times. In our lecture series, we took the incidence axioms, proved some stuff about incidence geometry. We took the betweenness axioms and then proved stuff about order geometry. Now, currently, we've added onto those the axioms of congruence, and now we're proving things about congruence geometry. As such, we took a slightly different statement for our axioms. We have angle and segment translation, angle and segment transitivity of congruence. We have angle and segment addition. We have the side angle, side axiom. This is our bedrock for congruence geometry. Now, we defined right angles as those angles which are congruent to their own supplement. Now, we're in a position where we can actually prove that all right angles are congruent to each other. I should point out that we've already proven that if an angle is congruent to a right angle, then it must also be right. This sort of gives you the other side of the coin that all right angles are congruent to each other. So, what we have to do is we have to take two arbitrary right angles. So, we're going to take angles A, B, C, and A prime, B prime, C prime to B are right angles. So, let's say we have one of them like so. There's a right angle, and then we're going to have another right angle, maybe something like this. And they're not necessarily related to each other in space whatsoever. So, we'll call this one angle A, B, C. Let me lower my B here to be something like that. A, B, C, like so. Let's take this one over here, same basic idea. We have A prime, B prime, and C prime, like so. And we know that both of these are right angles. So, I'm going to draw a little rectangle here to indicate that we have two right angles. Now, what does it mean to be a right angle? It means that you're congruent to your supplement. So, it makes sense to introduce the supplement in this conversation here. So, there is going to exist some point D so that B is between A and D. And so, these two angles are congruent to each other because they're both right. The right angle is congruent to its supplement by definition. And we can do the same thing over here. We can extend the line so that there's some point D over here. And it's just, and it will then be the case that D, B prime, C prime is congruent to A prime, B prime, C prime. Oh, sorry, not call this point D. It should be called D prime, like so. And so, by assumption, because these are right angles, A, B, C is congruent to its supplement, C, B, D. And then, the angle A prime, B prime, C prime is congruent to its supplement, C prime, B prime, D prime. And so, we know that these angles, these angles are congruent to each other. And these angles are congruent to each other. Our goal is to prove that these two angles are in fact congruent to each other. And so, this is what we're going to do. We are going to do angle translation. So, we're going to translate one of these angles onto the other one. And in particular, we're going to translate the angle A, B, C onto this angle over here. So, what does that exactly mean? It means there's some point P which lives in the open half plane determined by the boundary line A prime, D prime, and is on the same side as the point C. So, there's some point over here. Let's just for the sake of it, say it's right here. There's some point P over here that when we construct this ray, B prime to P prime to P, then we get that the angle A, B, C will be congruent to the angle A prime, B prime, P, like so. And so, this seems alien, but that's because of course we're sort of considering the possibility that maybe that's not the case. We're then assuming that A, B, C is congruent to the angle A prime, B prime, P. So, I don't want your diagram to have any, lead to any confusion here. I'm not saying that P is interior to the angle A prime, B prime, C prime. It could be, we don't know that. It could be over here. It could be on this ray. There are some possibilities, but for the sake of drawing, I had to draw one of them. Now, what else can we say here? Now, since CBD and P, B prime, D prime are supplements to congruent angles, they are also congruent to each other as well. This angle right here, so this notice here is D prime, B prime, P. This has to be congruent to the angle D, B, C. Because if they're congruent to, if your supplements to congruent angles, then you're likewise congruent with each other. Now, P is on one side of B prime, C prime. Without the loss of generality, we can assume that P is in the same half plane that B prime, C prime, excuse me. We can assume that P is in the same half plane bounded by the line B prime, C prime that A prime is. Basically, what I'm saying here is I can assume, because if you look at the line B prime, C prime, right? I can assume that P is inside this angle. That's what I'm doing right here. I'm saying without the loss of generality, it looks like my picture because if it was over here, if this is where P was, we could be like, oh, let's make this become A prime and oh, we'll make this become D prime. Because after all, this angle is a right angle, but also this angle is a right angle. And so we can swap them around and there's no loss in doing that whatsoever. So we can assume that my diagram is actually the one we're okay with. The diagram we have is acceptable here. P belongs to the interior of the angle A prime, B prime, C prime. Because if not, we could relabel things or we could just provide the exact same argument twice, move in the appropriate parts. Not a big deal. So because P belongs to the interior of this angle, this means that the ray B prime P lives between the ray B prime, A prime and B prime, C prime. Which then implies that the angle A prime, B prime, P is less than or equal to the angle A prime, B prime, C prime. Just like the diagram seems to suggest right here. We do have a total order on angles, just like we have a total order on segment congruence rate, on segments. So again, our diagram is then correct, up to, without the loss of generality. Now on the other hand, if you look at the points A prime and C prime, like so, A prime and C prime, they're going to be on opposite sides of the line A prime, B. So if you think about that right there. Okay, if you look at the crossbar, for example, this ray B prime P must intersect the crossbar somewhere between A prime and C prime. This is the crossbar theorem. And as such, A prime and C prime need to be on opposite sides of the line B prime P. And similarly, the points A prime and D prime are on opposite sides of this as well. Because after all the line segment between A prime and D prime, it contains B, B is obviously on the line determined by B prime and P right there. So we get that B prime sits between A prime and D prime like so. So since A prime and D prime are on opposite sides of the line B prime P, and since A prime and C prime are on opposite sides of the line B prime P, then by plane separation, we have that C prime and D prime are actually on the same side of the line B prime and P. Let me move my diagram a little bit, zoom out. This is actually a really nice argument what we're going to do here, believe it or not. So let's see where were we. So this of course gives us that the segment B prime C prime sits between B prime P and B prime D prime. Basically just justifying that the diagram I have in front of me is in fact valid. So since this ray sits in between the other two rays, that means that the angle, the angle C prime B prime D prime is less than the angle A prime B prime P. So this angle right here is bigger than this angle right here like so. So what we're trying to do is establish a bunch of inequalities. We have this one right here, we have this one right here. And remember this symbol less than or equal to is really less than or congruent to. So we have a bunch of congruences because right angles are congruent to their supplements and we had some other assumptions about congruence. We have these inequalities as well. And so I want you to notice here that we get this chain of inequalities here. The angle ABC was congruent to the angle A prime B prime P. The consequence of angle translation now just we just observed that a prime B prime P is less than or equal to a prime B prime C prime. As we track down where C prime lives relative to the other points, you know, other points A prime B prime D prime and P and things like that. Also, the angle A prime B prime C prime is congruent to the angle C prime B prime D prime because their supplements and right angles are congruent to their supplements. But then this angle C prime B prime D prime, we just argued was was less than or equal to the angle A prime B prime P because again P is the interior point to A prime B prime C prime. But then this angle right here P D P B prime D prime, this is congruent to the angle CBD. Why is that? Because these angles are, we said this one earlier, these angles are supplements of congruent angles, therefore they're congruent as well. And finally CBD is the supplement of the angle ABC. These are right angles so they're congruent to each other. So I want you to notice what we had going on here. We have the angle ABC is ultimately congruent to ABC. So we have this we have this string of congruences and less than or equal to is right. We start and stop with ABC the angle. So some of the times that's congruent that's okay but some of the times you have less than or equal to the angle ABC is not strictly less than ABC. So the only way we could have this string of inequalities that all of the inequalities actually had to be congruences. So these two congruences excuse me these two less than or equal to is actually had to be congruences. That's what they had to be for which then if we track down what's in the middle of the list, we have ABC, which of course is what we start with. But we also have a prime B prime C prime is in there as well. So those two angles are congruent to each other by this transitivity of congruence. And as these are two arbitrary right angles, this then proves Euclid's fourth postulate that all right angles are congruent to each other. And so that brings us so proving Euclid's fourth which kind of sounds like it's a symphony or something like that. Proving Euclid's fourth then finishes lecture 16 in this lecture series so I appreciate you watching it. If you learned anything about angle congruence and things like that please like these videos subscribe to the channel to see more videos like this in the future. And always post your questions in the comments below and I'll be glad to answer them when I can. Thank you everyone. Bye.