 Welcome back to our lecture series Math 3130, Modern Geometries for Students at Southern Utah University. As usual, I'll be a professor today, Dr. Andrew Missildine. In lecture 16, we're going to continue our discussion of the congruence axioms and focus on the idea of a congruence geometry. Just as a reminder, in lecture 14, we introduced three axioms about congruence of line segments. We had segment translation, transitivity of congruence for segments, and then we had segment addition. But then in lecture 15, when we talked about congruence of angles, well sure, we had an axiom about angle translation, we had an axiom about transitivity of angle congruence, but we didn't take angle addition as an axiom. We actually took something better, the side angle side axiom. But I then also argued, or at least mentioned in the video at the time, that all of the properties about segment congruence we had would also be true for angle congruence. And in particular, in this video, we're going to prove that angle addition is in fact a theorem of the axioms we have set up here. Now, we're going to basically use many of the axioms for congruence, particularly the segment side of things like segment translation, what have you. But there's this connection between angle congruence and segment congruence. And this is where we're going to utilize these connections to actually prove angle addition here. So what does it mean to add to angles? Because we don't have any operation, we don't have measurement here. What does it mean to add angles? Well, imagine we have some rays, BA, BC, BD, these are clearly concurrent arrays, because they all have the same vertex of B. And let's say that BC is between BA and BD, with our usual notion of betweenness of angles here. So we might get something like the following. So we'll have like BA, BC, BD, something like this, and I'll label the points so we can remember. The vertex was, of course, B. These are rays, we have some point A, we have some point C, and then we have some point D. So these are just the first of our assumptions here. Okay, then we also have some other, some other ones, some other rays that is we have B prime, A prime, B prime, C prime, and the ray B prime, D prime. These are also concurrent arrays as they have the common vertex of B prime in all three situations. And let's likewise assume that B prime, C prime is a ray sitting between B prime, A prime, and B prime, D prime. So I'm going to scooch up the screen just a little bit, so we have a little bit more space to draw. But we're basically going to get the exact same picture that we had before. We have these three rays, like so they are rays because they do extend beyond what we see potentially right here. So we have B prime, we have some A prime like so, some C prime like here, and some D prime like so. So we have this setting, so B prime, C prime is in fact between the other two. So that's just the basic setup now for the addition part. Suppose that the angle ABC is congruent to the angle A prime, B prime, C prime. So we have these two angles right here are going to be congruent. And let's also suppose that the angle CBD is congruent to the angle C prime, B prime, D prime. So CBD would be this one right here. And we're going to assume it's congruent to this one. So these are the assumptions here for the angle addition. The conclusion is that the angle ABD is congruent to the angle A prime, B prime, D prime. So we're going to prove that this angle is congruent to this angle. And this is the analogous statement when we consider like angle addition versus segment addition. So the first thing we want to do because in order to get angle addition, we're going to have to use side angle side. And as such, we have to form triangles with congruent sides. Segment translation can help us out here. So because of segment translation, without the loss of generality, we can assume that the segment BA is congruent to the segment B prime A prime. Because if we take the segment BA, we can translate it down here. And that's going to give us some point A double prime, so that B prime A double prime is congruent to the segment BA. And so given that the point A prime was just giving the direction of the ray and A prime, A double prime is another point on the array. We can swap A prime with A double prime. And so without the loss of generality, we can assume the segment AB is congruent to A prime B prime. By similar reasoning, we can also assume without the loss of generality that the segment BC is congruent to the segment B prime C prime. And the segment BD is likewise congruent to B prime D prime. Again, this is a consequence of segment translation. Next, we're going to play around with the crossbar theorem. So by the crossbar theorem, we can assume that the point CD is between, excuse me, that the point C is between A and D. So I'm actually going to fix my picture a little bit because if we look at the angle ABD, by the crossbar theorem, there's going to be some point of intersection between the ray BC with the segment AD. And so without the loss of generality, we can assume that C is that point itself. So actually, let me clean this thing up a little bit to match up with what I'm suggesting here. Filling my gaps a little bit. It's a little bit crude, but I hope it's okay. So we can actually assume that C is the point that, so that these three points A, C and D are all collinear and C is between them. But then again, from our previous assumptions, we have that the segment BC is congruent to the segment B prime C prime. Now I want you to be aware, this isn't actually part of the proof, but this is more of just a word of caution. Be aware that we can't assume yet that C prime is between A prime D prime because that would be assuming too much. We can assume the betweenness of A, C and D, but because of segment translation, we're not ready to say that C prime is between D prime and A prime. I mean, that will be the case, but we don't know that yet. And so we can't go beyond the mark there. So we can assume congruence of these segments by segment translation, and we can assume betweenness of the points ADC because of the crossbar theorem. So now we're in a situation for which we can talk about triangle congruences. So what are the triangles that are going to be congruent here? So I first want us to consider the triangle A, B, D, excuse me, ABC like so. So I'm actually going to add a little segment here. So look at the triangle ABC versus A prime B prime C prime. So when we look at what we have here, we have a side. We have an angle. We have a side. So by side angle side, these triangles are going to be congruent to each other. And that's what's mentioned right here. In particular, because corresponding parts of congruent triangles are congruent, we have that the angle ACB is congruent to the angle A prime C prime B prime. Sorry about the typo there, how to fix it in the middle of the lecture. Whoops a daisy. Anyways, but let's also look at another triangle congruence. If we look at the triangle CBD, so we connect these dots right here, connect these dots right here. Again, side angle side is going to come into play here because we assume that these sides are congruent, these angles are congruent, and these sides are congruent. So again, by side angle side, we're going to get that this triangle here is congruent to this triangle here. And so again, as corresponding parts of, excuse me, corresponding parts of congruent triangles are congruent, we get that the angle DCB is going to be congruent to the angle D prime C prime B prime. Like so. Now in the situation of these ones right here, we do know that ACB, ACD are collinear. As my diagram suggests, we don't necessarily know about collinearity between A prime C prime and D prime. That's an important part of this situation. Now, but because we because we know that C is between A and D, we do have that these two angles right here are in fact, supplements of each other. But the problem is, again, we don't know we have collinearity between A prime C prime and D prime. And so that's actually the thing we want to focus on next. Let me zoom out a little bit and so we can look at more of the proof here. So we know that those angles ACB and DCB are in fact, supplements to each other. But let's take a point D double prime, alright, so that we can extend the line A prime C prime so that D double prime is there. So in particular, what I'm suggesting is by by the extension axiom, we do something like the following, we extend this line. And so there's this point D double prime right there. Let me write that again D double prime, like so. And this will give us that C prime sits between A prime and D double prime. In this situation, then, then the angle, the angle A prime C prime B prime has as its supplement, the angle right here, D double prime C prime B prime, like so. And so then we have that remember these angles are congruent right here. Their supplements are located here and here. It turns out as we've proven previously that the supplements of congruent angles are congruent. So we can conclude that this supplement right here is congruent to this supplement right here, which I'm going to use the color scheme we had from before. We're going to get something like this. Okay, now this picture seems a little bit weird, right? Because when you look at the ray, the ray C prime B prime, we have two angles on the same side of the line that are congruent to each other, that by the uniqueness of angle, by the uniqueness of angle translation suggests that these angles actually must be one and the same thing. Because after all, they're congruent to each other because they're both congruent to the angle DCB angle transitivity comes into play here. So these two angles are congruent to each other, they're living on the same right in the same half plane. Therefore, it must be that D prime and D double prime are actually one and the same point because they're on the same side of the line, which then actually means we can correct the picture since these two things are the same, then it actually means that D prime was always collinear with C prime and A prime like we had originally suggested. We did have to provide an argument of that. Okay, so by the uniqueness of angle translation, D prime and D double prime are the same point. Therefore, the segment C prime D prime and the segment C prime and C double prime are actually the same segment. We have A, we have A, C, D and D double prime are all collinear because again D prime and D double prime are the same point. So we get that C prime is between A prime and D prime because D prime and D double prime are the same point. So we have a four-stat collinearity happening there. So once we have the collinearity, then and only then can we use the angle addition to get us that A D and A prime D prime are the same point. So as we try to fix this diagram, it turns out that the segment B prime D prime really looks more, that's just not to say the segment, but the ray A prime D prime really looks something more like this and I'm going to color this stuff in. So this looks just like a single prime now. We can sort of ignore these things right here. We did have this congruence here like so. In particular, we have that the segment, oh, what are we on right now in our Roman numerals? So the segment A, C was congruent to the segment A prime C prime and we also have that the segment, the segment D, C is congruent to D, excuse me, the segment D C, yeah, is congruent to D prime C prime, which gives you that one. And so then finally, segment addition happens here that the segment AD is congruent to the segment A prime D prime. The big hiccup was were these things collinear? Now we force that they are collinear. So we can make that statement there coming back down here. So we then have this statement AD is congruent to A prime in the segment A prime D prime. So finally, we're going to have one more side angle side here. So we have that AB is congruent to A prime D prime. We have the angle B AC is congruent to the angle B prime A prime C prime. And we have the segment AD is congruent to A prime D prime. So this is going to get us that the triangle ABD is congruent to the triangle A prime B prime D prime by side angle side, like so. And so looking at our messy picture right here, the idea is we've then proven by side angle side that the triangle ABD right here is congruent to the triangle A prime B prime D prime like so. Therefore, their corresponding angles are going to be the same for which these angles right here A prime B prime D prime is congruent to ABD. So it took a lot more effort to accomplish it, but we are able to prove angle addition using side angle side translation of segments, but in particular segment addition was used in there. What we had to do basically is we had to turn angle addition into segment addition using side angle side for which then we invoked angle addition to show that two angles were excuse me that we invoke segment addition so that two segments were congruent to each other and then we push it back onto the angle. So it was a little bit round about, but side angle side allows us to connect together angles and the congruence of angles and the congruence of segments. Now I want to point out here that because we now have angle addition, we essentially have all the same axioms for angles congruence that we do for segment congruence. We have segment translation. We have angle translation. We have transitivity of segment congruence. We have transitivity of angle congruence. And then the other axiom for segment congruence was segment addition. Now that we have angle addition, then anything we can prove about segment congruence, we can essentially prove the same thing for angle congruence using the exact same proof. I mean, remember when we first introduced angle congruence, we argued that because we had angle translation, because we had transitivity of angles, we had that angle congruence was an equivalence relation. We use the exact same proof that we did for segment congruence. You just change the appropriate part from segment, from segments to angles. And so as such, all of the theory we've developed about segment congruence is going to translate to angle congruence because we have the same three properties. This is the beauty of the axiomatic method. If you have the same axioms, you get the same theory. Now, yes, there is some slight subtle points that are different between segments and angles, but as we have essentially the same axioms, we'll develop essentially the same theory here. And so I'm going to leave many of these statements as exercises to the viewers, of course. So for example, we have segment subtraction. We also have angle subtraction. You can use the same proof that you do for segment subtraction, which I admittedly also left that as an exercise to the viewer here. Because of that, we can also get angle subtraction. What about linear order? We proved that you can, well, I guess I didn't prove that one either. I stated that you can prove that there's a linear order on segments in a congruence geometry. I left that as an exercise. But by that same proof, you can argue that there's a linear order on angles based upon congruence. Speaking of the total order on angles, let's actually provide the definition, be very explicit about it. Suppose we have two angles, angles ABC and angle DEF. If there exists a ray EP such that P is an interior point to the angle DEF and the angle ABC is congruent to DEP, then we say that the angle ABC is less than the angle DEF. All right, that's a lot to unravel here. So let's try to be very specific here. So we have our two angles, ABC. So here is B, here is A, here is C. And then we have our second angle, maybe somewhere else in the plane who knows where it is. But we're drawing some picture of it. So these are our three points right here. So we'll call this one D, we'll call this one E, and we'll call this one F. So we say the angle ABC is less than DEF. If there exists a point P that's interior to the angle, for which case if we think of the ray, like so in particular with the ray EP is interior to the angle, that's the first thing. The next thing of course has to do with congruence. If angle ABC is congruent to the angle DEF, then we say that the angle ABC is less than the angle DEF, like you see right here on the screen. Now, we can also say that the angle ABC is less than or equal to DEF. We really should say less than or congruent to, because this symbol right here will mean either ABC is less than DEF in this notion right here, or that the two angles are actually congruent to each other. I'll again leave you to prove that one. I don't want to provide one or the other because that's the same proof. And again, I want to leave that as an exercise for the students here. But what I will do is we did prove this one in the lecture videos. We proved but the between this preservation of segment congruence, I want to show you that this is the same proof. This is the same proof for angle congruence as well. So suppose we have two congruent angles, ABD and A prime B prime D prime. And let's suppose that the ray BC sits in between the rays BA and BD, then there is going to exist a unique ray B prime C prime, that's between B prime A prime and B prime D prime, such that the angles ABC are that's congruent with A prime B prime C prime and the angle CBD is congruent to the angle C prime B prime D prime. So the exact same notion of between this preservation happens there as well. So let's try to sketch a picture to make it easier to follow along. So we're going to have our two angles like so, although one of them has a third ray inside of it. So let's do that one first. We have our three rays, they're all concurrent. It's vertex of B. So we'll call this the ray BA. Here's the ray BD. And then we have some other ray BC like so. And then we have a second angle. Let's put it down here somewhere like so. I'll put little arrowheads on them so that we know they're rays. They continue on. So we have the point B prime here. We have the point A prime here. And we have the point D prime like so. And so by assumption, remember, we have that the angle ABD, which is this angle right here, this is congruent to the angle A prime B prime D prime, which is right here. So those are the assumptions of the problems. And then we have this interior ray right here. So the basic argument, just like we did with segment preservation between us, is we're going to translate this thing over heat down here and then make an argument from there. All right. So using angle translation, we can copy this angle down below here. In particular, there exists some ray that emanates from B prime that's on the same side of the line A prime B prime that D is on. So there's some ray interior to the angle. There's some point C prime associated to that will give us that the angle ABC is congruent to the angle A prime B prime C prime like so. So angle translation gives us exactly that. We did that with the between this preservation of segment of segment congruence. Now again, there is going to be a unique ray such that a unique ray B prime D, excuse me, B prime P such that the angle CBD is congruent to the angle C prime B prime D, P, excuse me. So in this situation, there's some point, I'm going to put it over here. There's some point P like so. So when we consider the ray, it's going through P, try that again, like so that this angle right here, the angle C prime B prime P is congruent to the angle CBD like so. And now I'm calling this point P because I don't know how it's necessarily connected to the point D prime yet, but angle addition, excuse me, angle translation allows us to do that exact same thing. Now this is where angle addition is going to come into the play here. Now since these two angles are congruent to each other, and since these two angles are congruent to each other, for which I should add an extra label so it's clear there, we have we can use angle addition, which we just proved in this video. And this is going to give us that the angle A B D here, let me use a different color here. So we're going to say the angle A B D is congruent to the angle A prime B prime P, like so. Now we already know by assumption, so we already know by assumption that the angle A B D is congruent to the angle right here, A prime B prime D prime, right? So by transitivity of congruence here, we have that the angle A prime B prime D prime is congruent to A prime B prime P because they're both congruent to the same angle. But by uniqueness of angle translation, if I have two angles on the same right in the same half plane, they have to actually be the same equal the ray B prime D prime is actually equal to the ray B prime P so that these two are actually equal to each other. So this this is all happening here. This is actually the ray that we have for all of them. So therefore, we have that a the angle ABC is congruent to the angle A prime B prime C prime, the angle CBD is congruent to C prime B prime D prime. And we have that the angle, excuse me, that the ray B prime C prime sits between B prime A prime and B prime D prime. So putting that all together here, we did in fact put this ray in between it so that these two angles we broke up were congruent to the ones we started with. So between this preservation happens, it was the exact same proof that we did for segment between this preservation. So this is the illustration I want to you want to offer you. So whenever you consider a proof about congruence of segments, because we have the same axiomatic framework for congruence of angles, you can use the exact same proof. And so we can develop basically one of the two notions, because the other one is going to develop the same way. So mostly what we want to then do is develop the interconnections between segment congruence and angle congruence, which particularly means we want to develop the notion of triangle congruence, because that's where you marry together the segment and angle congruence notions.