Common Core Geometry Unit 2 Lesson 8 Congruence Reasoning About Triangles

Hello and welcome to common core geometry by E math instruction. My name is Kirk Weiler and today we're going to be doing unit number two lesson number 8 on congruence reasoning about triangles. Now, this is sort of the culmination of everything that we've been doing in this unit. We do have one less and left after this, but it's kind of a fun lesson on symmetry. Within this lesson, what we're going to be doing is we're going to be using properties of rigid motions and the definition of congruence in terms of rigid motions. In order to show that two triangles are congruent. And remember, what that's all about is it's about taking a rigid motion like a reflection and rotation or a translation and showing how we map the three vertices of one triangle onto the three vertices of another. So let's make sure that we've kind of reviewed all of the essential pieces. The first thing, triangle congruence. Two triangles in the planar congruent. If a sequence of rigid motions can be found that make the vertices of one triangle, coincide with the vertices of the other, or lie on top of them. So in other words, I've got these two triangles, right? I want to show that they're identical. So what I do is I find some kind of a rigid motion or perhaps a sequence of them, reflections rotations translations, and I show that ABC maps onto DEF. All right? And even though a triangle is composed of an infinite infinite number of points, I just have to do it for the three vertex points because all rigid body motions map line segments to line segments, right? So if I've got those three vertices the same, then all the points in between them will be the same. All right, so let's jump into it. If I can advance the slide. Here we go. Let's take a look at our first rigid body motion proof that two triangles are congruent. Let's take a look. Exercise number one in the diagram to note below, we know the following. Now this is exceptionally important. These are what are called givens. Givens in geometry, given ADC, BD, perpendicular to AC, and AD congruent to CD. Prove that triangle BDC is congruent to triangle, BDA. All right, so this is the first time we've really done this. Based on these givens, all right? We're able to prove that those two triangles have the same shape and the same size. All right? Now, letter a based on the givens and the diagram, what rigid body motion do you think will be used to map triangle BDC onto triangle BDA, try the transformation with tracing paper. All right. So take a moment now. Pause the video, use some tracing paper and think about what rigid body motion we're going to use to map this triangle onto that one. All right. In many of these, it's going to be fairly obvious and specifically what we're going to be looking to do. Is we're going to be using a reflection. A reflection across segment BD. All right? We can even take a look at what that looks like visually. By flipping it, left, right. All right? So right there, what I did was I reflected triangle BDC across line segment BD and it fell on triangle BDA. And interestingly enough, of course, I can reflect it back. Okay. So now, how do I actually prove that? In other words, what is it about all of this information? That allows me to conclude that underneath that reflection, the three vertices of BDC will get mapped to the three vertices of BDA. Pause the video for a moment if you need to and think about this for a bit. Everything that we need has to be contained in those givens. Okay, so let's talk about it. All right. So. Under the reflection. That's supposed to be of of triangle BDC. In BD. Points B. And D. Get mapped. To themselves. Let's pause just for a moment and think about this. So under that reflection, B and D don't go anywhere. Let me get rid of this period there. Because we should give a reason because they lie. On the line of reflection. All right. Now remember, in order to prove that that triangle is congruent to that one, we just have to find a rigid body motion, and it's going to be this reflection that maps the three vertices of that triangle. Of this triangle onto the three vertices of that triangle, but we already have two of the three done. All right? In other words, we've mapped B to B and D to D so now the real question is, why is it? With these givens, what is it about these givens that make me know that under this reflection C will get mapped to a? What is it? Because that's the last little piece of our explanation. All right. Well, let's talk about it. You see, BD. Is the perpendicular. Is the perpendicular bisector. Of AC. Again, let me step back. Let's talk about this a little bit. Why is this the perpendicular bisector? Why is BD the perpendicular bisector of a C? Well, for two reasons. Number one, because we've been told that BD is perpendicular to AC. And number two, because we were told that AD is congruent to D.C., right? And therefore, the two conditions of being a perpendicular bisector are met. IE its perpendicular and number two, it bisects that line segment. It cuts it in half. So. By the very definition. By the definition of a reflection. Point C. Let me bring that up. We'll get mapped. To point a. Under a reflection. In. BD. All right. Let's talk about that just a little bit, that kind of line of reasoning, because we're going to see it come up a fair amount. See, by the very definition of a line reflection, what a line reflection says is that any point that is not on the line of reflection, like point C if it gets reflected in BD, wherever it ends up landing, when you connect that point with where it gets where it lands, where it gets mapped to, then the line of reflection is the perpendicular bisector of that segment. Therefore, because BD is the perpendicular bisector of AC, then C will get mapped over to a now one little piece to finish off this proof, right? Since let me extend the page. Since a reflection. Is a rigid motion, wow, that was weird. I love the fact that that turned into red. Since a reflection is a rigid motion and the three. Vertices of triangle BDC. Got mapped. To the three. Vertices. They must be congruent. And that's it. Wow. A little bit weird. Get rid of that. There's a little bit better. So understand the line of reasoning here. Based on the givens, we can say when we reflect triangle BDC in BD, B and D go nowhere. And that's okay. Because BD is the perpendicular bisector of AC, C has to get mapped over to a and now because a rigid motion reflection maps the three vertices of BDC onto the three vertices of BDA, the two triangles have to be congruent because rigid motions preserve shape and size. And that's it. Again, a lengthy explanation. I think sometimes you'll find in the middle of geometry, you feel like it's more like an English class than a math class because you're writing a lot. And I know that that can be challenging for people, especially when they understand the math, and they understand the reasoning, but getting it all down into words is a little bit more difficult. Still, I encourage you to keep at it because quite frankly, explaining your reasoning is more powerful than being able to do the reasoning itself. Or put another way, you can do all the reasoning you want. But if you can't explain it or articulate it in writing, then it's not worth as much. All right, let's keep doing these types of problems. Exercise number two, let's take a look at this one. In the diagram below, we know the following. Given ACD and BCE with C being the midpoint of AD and BE. Prove using rigid motions that triangle ABC is congruent to triangle DEC. As a little side note, one of the reasons that I'm tossing this phrase in right now is that later on in the course, we're going to see ways to prove in this problem that these two triangles are congruent, not using rigid body motions, but some other piece of mathematics that we've developed it. Same thing with the last exercise. We'll have a different way of proving those two triangles are congruent later on. Right now we simply want to use the properties of rigid motions to do it. So what I'd like you to do is take a few minutes now and think about what rigid body motion were even going to use. You may want to use tracing paper. I'm going to do a tracing paper movement over a triangle DCE, or DEC. And then once we have the rigid body motion, then we'll try to understand how the givens. Tell us that the mapping can be done. All right, take a few minutes. I'll give you my best thinkers pose. All right, should we go through it? Let's do it. The key here is that we don't have a reflection. What we have is a rotation. Specifically, what we're going to be looking to do is rotate this triangle. A 180° about point C so let's do it. And let's talk about why this is going to work. Now, let's take a look at these givens just for a moment. All right? First things first, let's talk about these two. These two givens just tell us both. That three points. Are collinear. Linear collinear. That three points are collinear. This, though, C is the midpoint of both a D and B while being the midpoint of AD tells me that these two segments are congruent. And being the midpoint of this segment tells us that these two are congruent. All right? So let's talk about why we can now use this rotation by a 180° about point C to map all the vertices of one triangle onto the other. A rotation. Of 180° of triangle. Let's go with DEC since that's the way I did it. About C. We'll map it. Onto triangle ABC. All right, so I'm just kind of making an opening statement. Ah, this is the rigid body motion I'm going to use. So a rotation of a 180° of triangle DEC a PowerPoint C will map it on a triangle ABC. And now I need to explain why. I need to take each vertice in turn and show how to be mapped. And I'm going to start with the easy one. Point C. Will get mapped to itself. Why? Well, that's because. It is the center. Of rotation. Very similar to when we had points on the line of reflection and we reflect, they didn't go anywhere. Same thing here when we have a point that is the center of rotation, right? When you rotate it about that point, whoops. You rotate it about that point. Then that point doesn't go anywhere. Now for the harder two, right? And those are points D and E, why does D get mapped to a, why does E get mapped to B? Now, the good news is, both of them are getting mapped to eventually the right spot, for exactly the same reason. Or I really should say reasons. So first, why a 180°? The 180° rotation is important because BCE and ACD are collinear sets of three points. Okay? If they weren't collinear, if we didn't know these two givens up front, then we wouldn't be able to say a rotation of a 180° would do the job. So, all right? B, C, E, and D, CA, R, collinear. Sets a points. But that's not enough, right? It's not enough that their code linear. The collinear helps us get just the fact that the 180° rotation is what we want. Just think about point E right here. The fact that they're collinear, let's try to do that again. The fact that they're collinear, allows this segment to lie on top of that segment. But why does that point get mapped exactly to be? Well, it gets mapped exactly to be because EC and BC are the same length. Okay? As well, EC is equal to BC. And AC is equal to D.C.. So a 180° rotation. About C. What? Maps E. To B. And D two, I lost it. There it is to a let's do that. Final conclusion. The final conclusion is really stating the fact that we've got a rigid body motion here. So because. A rigid motion. Mapped all the vertices. Of triangle whoops. No, not a no way there. Triangle D, CE two triangle. DCE to triangle a, CB. Must be congruent. Wow. There is no way I could make that our red even if I wanted to. It just magically turned red. Seriously, do you see it? Now watch this. I'm going to erase it. And I bet it'll become blue. Who knows? Technology. It's awesome. Again, the idea here is really kind of cool. All right? Which is that using a rigid body motion. Okay, specifically a 180° rotation about point C all right? We can show that C maps the C, E maps to B, D maps to a and because all three vertices of this triangle map to all three vertices of that triangle using a rigid body motion, the two triangles have to be identical. The same size, the same shape. All right. Let's do one more. The last one is going to be the most challenging one we do. But let's take a look at it. All right, exercise number three. Okay? The last really kind of tough reasoning work here. We'll have one more problem after this, but it's going to be fairly easy. Let's take a look at it. In the following diagram, it's known that AB bisects both angle CAD and angle CBD. This will be enough using rigid motions to show that triangle ACB is congruent to triangle ADB. All right, letter a asks, what angle pairs must be congruent based on what is known, explain. Mark these pairs on the diagram. All right, so what I'd like you to do is take a moment to think about this. Based on this given, all right. We can talk about congruent angle pairs. Why don't you write them down? All right, let's talk about them. Now, one little tip. Okay. We can definitely go with the three letter naming system, and there's absolutely nothing wrong with that. Okay? But to speed things up, you can oftentimes number angles, okay? And as long as it's clear from the diagram, what angles you're talking about, the numbering system works quite well and is completely acceptable. So I'm going to label this angle one and this angle two. I'm going to call this angle three in this angle four. And I'm going to claim based on the given that AB bisects both angle CAD and angle CBD that angle one must be congruent to angle two and angle three must be congruent to angle four. Now if on the other hand, you said angle CA B is congruent to angle D, a, B, right? That would be these two. And that's, of course, completely okay. But a lot of people like using the one and the two and the three and the four, because it's just simply easier than writing down all three letters. By the way, it does ask, oops. I guess I can't get rid of that. It does ask for us to explain. And the explanation, which I have to now sneak in is because angle bisectors. Divide an angle into two congruent angles. I have to go a little bit onto my diagram. That's okay. So keep in mind, angle one congruent to angle two. An angle three congruent to angle four. Note how I use a single arc to show that these two angles are congruent and double arcs to show these two are congruent. Sometimes what people will also do is instead of having something like that and this again depends on your teacher, sometimes they'll put a single dash mark through these angles. And a double dash mark through these. We just want to make sure we know which angles are congruent to which. Okay. Now, for the reasoning part, that's just interpreting the given. That's it. Letter B, if ADB, the triangle on the bottom was reflected in AB, what two points would not move called fixed points and why wouldn't they? So in other words, if I take triangle ADB and I flip it across AB. What two points won't move? We've answered this question a few times. So take a moment. All right. Well, hopefully very quickly. A and B. Because they are on because they on. How about because they are on. The line of reflection. All right. So you probably can already see where we're going with this. If we reflect ADB over and I want to prove that ADB is congruent to ACB, right? That reflection gets two out of the three points done. Now we've seen a problem like this in previous lessons. The question now becomes. Why does D get mapped to C? And why is knowing that these angles are the same enough to do so? So let's take a look at letter C all right, now I've got the diagram here again. I'm going to extend the page a bit. So we can really properly frame it. I'm going to write down one and two, three, four, and I'll go with maybe the double arcs right now. All right? Letter C, if ADB was reflected in AB, why must the image of point D D prime fall on both rays AC and BC? Think about this for a moment. Because this kind of really, really goes back in our work a little ways, right? Don't worry about it over here. If I flip this up, why does D, why does the image of D have to lie somewhere? On ray AC. Do you remember the reasoning behind that? Or hopefully, because angle one is congruent to angle two, D prime must Y on AC. Again, pause for a minute, because we talked about this a few lessons ago, specifically when we were investigating isosceles triangles and other things. If these two angles are congruent and they are, and they share a common ray, and they do AB, right? Then when I reflect this ray, AD, across that line, it has to fall on ray AC because reflections preserve angle measures, right? And again, this is just all about rays. If I reflect this ray ray AD across AB, it's got to land on AC or these two angles wouldn't be congruent. Likewise, because angle three is congruent to angle four. D prime must Y on ray BC. Again, simply put, right? If I'm thinking about reflecting this array, this one, and I flip it over, its image must land. Some are on that ray. Otherwise, these two angles wouldn't be congruent. Ah, letter D since D prime must fall on both AC and BC, where does this tell you that it must lie? Explain. This is awesome now, right? Wherever D goes, wherever its image goes, it has to go on this ray, while at the same time it has to go on this array. So therefore, D prime must fall on C because whoops. And we'll just write out the whole word because. C. Is the only point. On both. And that's key so key I'm going to underline it on both AC and BC. All right, it's kind of cool, right? Just based on the fact that I know these two angles are equal, I know D has to lie somewhere on that ray. Based on the fact that these two angles are equal, I know that D has to fall somewhere on that ray. And the only point that's on both of those rays is letter C or point C and therefore the image of D D prime has to go to C so finally, wrapping it all together in a nice neat package, explain why we now know that triangle ACB is congruent to triangle ADB. A reflection. Mapped. Three vertices. Of triangle ADB onto. The three vertices. Of triangle ACB. Sensor reflections. Are rigid motions. The two triangles, the two triangles. Must be congruent. Kind of cool. All right. In traditional geometry, this is not the way we would prove these two triangles are congruent. And in fact, in the next unit, we're going to give you getting back to that more traditional geometry that relies on some other theorems and things like that. But it's as plain as simple as that. Because we've got these two pairs of angles, we can prove using a reflection that the three vertices of this triangle fall in the three vertices of that one. All right. Let me kind of step back a little bit in case you need to write any of this down. Okay, let's move on to the last problem. All right, pretty nice and simple. Multiple choice problem. You know, one of the I think questions that a lot of both students and to be honest, teachers have about all of this kind of material is how will it be tested? What depth will it be at on any kind of a standardized exam? And quite frankly, the answer is, I don't know. We can look at past exams. We can see what they've asked before. And sometimes this kind of thinking even comes up. In multiple choice problems. So let's take a look at one of those. Exercise number four, given that line R is the perpendicular bisector of both BC and AD, which of the following would be used to justify that triangle ABE is congruent to triangle DCE. All right. Well, why don't you take a shot at this? Like which one of these four choices is the correct one for how we would justify it? We're not going to actually write out a long form proof or anything. But which one would it be? All right, let's go through them. So choice one said a reflection of a, B, E, across line AD. Well, if we took ABE and we reflected it across AD, it would end up kind of looking well, sort of like this. I'm not doing a great job drawing it, but well, this triangle isn't sitting on top of that one. So definitely not. All right. A 180° rotation of a, B, E, about point E well, honestly, that would come a little bit closer. It would actually, if we did do that, if I can actually select this, one more time, there we go. It would end up being something like that. Well, at least it matched up two of the three vertices, but still incorrect. Let me bring it back down. All right. So no. Three, a translation of a, B, in the direction of BC, by a distance of BE. Well, that would mean I would be moving this in the direction of BC and moving at a distance of BE. Well, again, not too bad, it matches up two out of the three vertices, but it doesn't get all three. So our final choice ends up being the correct one, which is a reflection of ABE across line R and that exactly gets the job done. All right, when we reflect cross line R, the three vertices of one triangle match up with the three vertices of another. Now, again, that's primarily because we know that R is the perpendicular bisector of BC and AD. And again, just by the definition of what a line reflection is, if R is the perpendicular bisector of this segment and this segment, well, then that means that when I flip this point across that perpendicular bisector, it has to land there. And likewise, this one would have to land there. All right. So hopefully simple enough choice for the correct answer. All right, let's wrap this up. So rigid body motions. Transformations that don't change the size nor the shape of an object can be used in a variety of different ways. In this lesson, what we did is we used the very, very formal properties of these rigid body motions. To justify that two triangles are congruent by mapping the three vertices of one triangle onto the three vertices of the other. And what we repeatedly saw is that oftentimes we would find a transformation that would leave some points unchanged, fixed, and other points would get moved in some way according to the transformation. As long as we can find that transformation and we can properly explain why all the vertices move to where they move, then we've got a rigid body motion proof. All right, we'll be moving on to other types of proof in the next unit. For now, I just want to thank you for joining me for another common core geometry lesson by E math instruction. My name is Kirk weiler, and until next time, keep thinking. And keep solving problems.