Hello and welcome to common core geometry by E math instruction. My name is Kirk Weiler and today we'll be doing unit three lesson number four CP CTC. If you haven't figured it out, CPC TC is an acronym, and it's the most important acronym in all of geometry. Well, obviously describe what that acronym means what all the letters stand for. We're going to be using it a lot, and we're going to really want to understand the reasoning behind it. So let's jump right into it. Otherwise, it's just a bunch of letters. All right. CPC TC. Corresponding parts of congruent triangles are congruent. Corresponding parts of congruent triangles are congruent. Now, that's a mouthful, and it can be kind of confusing.
Obviously, there's many words in this that you understand, like congruent triangles, congruent, right? But corresponding parts, that's a little bit confusing too. So I like to talk about CPC TC as a type of reasoning that we use. And let me illustrate it this way. So you've probably all been by one of those housing developments where every single house looks the same, right? You might even live in one of those things. The houses can be absolutely gorgeous. These two houses are absolutely identical. Except for maybe the color, right? So you could say that these are congruent houses. The idea behind CPC TC is if I know that these two houses are congruent. Then any parts of the houses that correspond to one another must be congruent as well. So let's say I was trying to prove for whatever reason, right? That this window and this window were identical. Well, I could use a line of reasoning that kind of went something like this.
These two houses are in a housing development where all the houses are identical. Because these two houses are in that location, they must be identical as well. Now, every time I say the word identical, you think congruent. And because these two houses are identical to Windows, which are in the same place on the two houses, must also be identical. Now, I would never try to make the argument, I would never try to make the argument that this window and this window are identical. Because they're not in the same place on the two houses. They're not corresponding. So the idea of CPC TC reasoning is actually pretty easy. Things on two geometric objects that are in the same place, those are corresponding parts. Of two geometric objects that are identical, like triangles, congruent triangles must also be identical. Now, a 100% of the time, a 100% of the time in this course, when we use the idea of CPC TC, it will either be to prove that two side lengths are congruent. Or to prove that two angle sizes are the same. Or congruent. Okay? Every single time.
So what we're going to do now is jump right into the idea of using CPC TC reasoning. To prove one of the most important facts in all of geometry. So let's get right into that. All right. Actually, not quite yet. My apologies. Before we get into proving something using CPC TC, I want to make sure that you understand what it means to have corresponding parts. So let's take a look at exercise number one. Exercise number one, in the diagram below, it is given that triangle EAC is congruent to triangle FDB. State the corresponding side and angle pairs. All right, so I'm telling you right now that these two triangles that overlap are congruent. And that can be a little bit confusing. Especially when we're trying to identify corresponding parts. That's where it wouldn't hurt if we knew that these two triangles were congruent. To let's say redraw this one in a manner where it's oriented the same way. Now I know that FDB and EAC are congruent to one another. So if I were to label this, I would want to make sure wherever E is, that's where I put F and wherever a is, that's where I would put D and wherever C is that's where I would put B all right, now honestly, that's just really this triangle after it's been sort of reflected over some kind of a vertical line. But now it's pretty easy, right? If I want to list corresponding sides, well, I could say that EA. Is congruent to FD.
Now, it says just corresponding sides so I could maybe put a comma in between them. But the whole idea of CPC TC is if I know that this triangle on this triangle are congruent, then I know these, this pair, of corresponding sides, is also congruent. Likewise, right? I could say that EC is congruent to FB. And I could also say that AC is congruent to DB. Right? Those are my three pairs, of corresponding sides. And they must be congruent to one another because the triangles are. As well, we've got pairs, of corresponding angles. Maybe a little bit less important than the corresponding sides, but important nonetheless. So for instance, and I'm going to use the three letter naming system right now because these triangles are overlap and it could be confusing what's angle C what's angle B et cetera but one thing I can tell is that angle a E C that's the one right up here must be congruent to angle D, FB. This guy right here. I can also say that angle E, a, C that's this one. Must be congruent. I'm going to run out of room to angle F whoops, F D, B, down here. And finally, use three arcs, angle E, C, a, must be congruent to angle. F B D all right.
Now, again, it's important. If I know that two triangles are congruent, it's important to be able to orient them in any way I need to in order to identify corresponding sides or corresponding angles. Now again, this line of reasoning, right? Is all about establishing that two segments have equal lengths or two angles have equal measures based on two triangles being congruent. And now, let's take a look at a very, very important result that you can get from CPC TC. All right. So one of the things that we studied back when we were looking at transformations in rigid body motions was the following geometric fact. A point that lies on the perpendicular bisector of a line segment. We'll lie equidistant from its endpoints. A point that lies on the perpendicular bisector of a line segment will lie equally distant from its endpoints. And again, I'd like to illustrate this for a second with a geo gebra widget. And let's take a look at it. In this widget, what we see is we see that line segment AB has a perpendicular bisector drawn in red. Okay? And again, if I move one of these points around a little bit, we can see that it is the perpendicular bisector because that angle remains now 90°. And we see that it must be passing through midpoint M because a M is 3.1 and MB is 3.1. But we can see right away that these that point P, which lies on the perpendicular bisector, is equidistant from the endpoints of the line segment. 3.76, 3.76.
And it doesn't matter where I put that point P anywhere on the perpendicular bisector. That point is the same distance from one endpoint as it is from the other. Now we actually proved this result back with backed by using rigid motions in unit two. But now we're going to use Euclidean geometry proof to do it along with CPC TC reasoning. Oh, that's a mouthful. Let's take a look at how. And we're going to kind of hang out on the same slide. It changed back over to blue. All right. Exercise number two. Let's read through it. In the diagram below, line M is the perpendicular bisector of AB. Point P lies on M we want to prove that P will always be equidistant from a and B no matter where it is on M so again, understand what we're given in this problem. We have a line segment. And we're given that line M is the perpendicular bisector of segment AB. Point P is just some arbitrary point sitting on the line segment. I'm sorry, sitting on the perpendicular bisector, my apologies. What we ultimately want to prove is that point P is equidistant from a and B all right, let's take a look at what the problem asks us to do. It says draw in AP and BP.
What are their lengths and inches? All right, so take your straight edge out. And let's draw in this segment. And we might as well measure it at the same time. That is one and a half inches. Likewise, if I come up here to be. And I draw the line segment in. And I look at its measurement, it's also 1.5 inches. Okay, so at least experimentally, point P does lie equidistant from the two end points of this line segment. Now, the 1.5 isn't very important. Those could have both been two inches, they could have been three miles. It doesn't really matter. Now what we want to do is we want to prove that point P lies equidistant from the endpoints. Let's take a look at what the problem asks us to do. Letter B, given that M is the perpendicular bisector of AB. Prove that triangle APC is congruent to triangle B, P, C all right. In the last lesson, we actually kind of did this proof. But we're going to do it again right now. And we're going to use a paragraph proof form. So I'm not going to do a statement and a reason T table proof. I'm going to use a paragraph proof. And again, don't worry about the 1.5 inches.
In fact, I'm going to delete that. And hopefully, maybe not. I guess not. I can't get the 1.5 inches without taking away the line segment. So I'm going to leave them both. But remember, it has nothing to do with the 1.5 inches. We want to write up a proof for why this triangle must be congruent to that one. And let's kind of outline that proof. Remember, we have only three ways of proving that triangles are congruent right now. Side side side, side angle side, and angle side angle. So what do we know so far? Well, we know that M is the perpendicular bisector of AB, which means that this will be a 90° angle, and this would be a 90° angle. We also know that AC must be equal to BC because it's a bisector. Finally, we also know that the two triangles share side CP. And therefore, we've got a side equal to a side. An angle equal to an angle, and another side equal to a side. Of course, that side equal to a side is equal to itself. So do you notice what we're going to be using now is side angle side. We essentially have done the proof, we just need to write it out. This gets to be the portion where you got to write a little bit. So let me go ahead and do that. We'll have you do some on your own in a little bit.
All right, so what do I want to say? I'm going to say. Because M is the. Perpendicular. Bisector. Of segment AB. What do we know? Well, we know that AC must be congruent to BC. Also. We know angle a, C, P, and angle B, CP, are both right angles. And R therefore. Congruent. Finally. CP is congruent to CP. By the reflexive property. So triangle a, P, C is congruent to triangle. B, P, C, by the side angle, side. There. Or just by SAS. That's okay too. So again, make sure that you understand why these two triangles are congruent. We know that AC is congruent to BC because it's the bisector. We know that these two angles are right angles because it's the perpendicular bisector. And therefore, these two angles are congruent because, well, every 90° angle is congruent to every other 90° angle. And finally, they share a side PC, so psi equal to a side, angle equal to an angle, psi equal to a side. We have the side angle side theorem. These two triangles must be identical. Great. How does this now prove that these two points or at that point P is equidistant from a and B? Let's take a look at that in the final part of the problem. It says why now, does it make sense that AP, right? Is equal to BP. P is equidistant from a and B well, think about this. I know that these two triangles are identical, right? And if I draw them in this orientation, what I can see is that AP and BP are in the corresponding places on those two triangles.
And therefore, AP must be equal to BP because corresponding parts of congruent triangles are congruent. And ultimately, because those are the same length, P is equidistant from the endpoints of that line segment. All by itself, that result is amazingly important. We're going to be using it time and time again in the next unit on constructions. But for now, it's a wonderful result of CPC TC reasoning. In other words, if I go into this proof, and I just say to you, hey, look, I've got the perpendicular bisector of a segment, explain why that endpoint or sorry, explain why that point must be equidistant from the endpoints of the line segment. Well, it's questionable how we do that. But if I can establish that these two triangles are identical, then these two links have to be identical. And if those two links are identical, well, then P is equidistant from a and B all right? Let me step back here for a second. I'll give you a chance to take a look at this text right down anything you need to. All right, likewise. There's not that much to write down here, but it's helpful, right? They have to be equal to each other because they're corresponding parts on two congruent triangles.
Almost all really interesting results of geometry. Are a product of CPC TC reasoning. It's very rare that I actually care that two triangles are congruent. It's almost always the case that I care about two parts of two triangles being congruent, but I first have to prove the triangles are congruent before I can say the parts are congruent. All right. Let's keep moving on. Do a little bit more CPC TC reasoning. All right. Let's take a look. Exercise number three. In the diagram below, it's given that AB is congruent to CB. And BD bisects angle ABC. We would like to prove that BD must also bisect angle ADC. All right. Now, let's start to mark up this diagram a bit. Okay? Just based on what we know. So we're told that AB is congruent to BC. We're also told that BD bisects angle ABC. Well, that's going to tell me that this angle is congruent to this angle. All right? Now finally, and maybe do this in another color. Just to distinguish what we're trying to prove is we're trying to prove that BD also must bisect angle ADC. Isn't that remarkable? That has to be the case just based on the fact that that length is equal to that length, and that angle is equal to that angle.
But let's take a look at why. Letter a what would need to be proven to show that BD bisects angle ADC. What would we need to show in order to show that BD bisects that angle? Think about that for a moment, pause the video and write down what you think will have to prove. An angle bisector means that we've split an angle into two equal angles. So we need to prove. Or show however you want to put it that angle three is congruent to angle four. If I can show that angle three is congruent to angle four, then BD must bisect that angle because that's the definition of an angle bisector. So letter B, what triangles should we show congruent in order to show letter a? So I want to somehow show I'm working the proof backwards. You see, I need to show that angle three is congruent to angle four in order to show that BD bisects ADC. All right, but to do that, to get two angles congruent, I want them to be corresponding parts of congruent triangles. And let her be just says, well, which triangle should we show congruent in order to prove that three is congruent to four? Write down a congruent statement.
All right. So we would need to show that triangle, let's say, B, a, D, bad triangle, bad, bad triangle, triangle, BAD, is congruent to triangle. B, C, D now, you might say that that's obvious because quite frankly, there's only two triangles in the picture. So what else would I do? It's going to be trickier later on. There might be three or four different triangles in a picture, and you really have to kind of figure out based on what you're trying to prove, which two triangles you need to prove congruent, or you might have to look at all your givens and just say, which two triangles can I prove congruent? But in this case, I'm only got that trying on that drawing all so if I can prove the one on the top is identical to the one on the bottom, then that will tell me three is congruent to four, and that will tell me I have an angle bisector. So let's go through it. This time with a T table proof. All right? So I've got the picture reproduced down here. A little bit trickier because we don't have the given sitting on the board anymore. But remember what our givens were, right? And let me actually go back a little bit to blue. All right, I know that this is congruent to this. That I was told, and I was also told that BD bisects angle ABC.
Now ultimately, that will tell me that those two angles are congruent. So what's the first thing I want to put down in my proof? Well, I'm going to just put down the given AB is congruent to CB. Why? Because it was given. And that's one of the sides inside angle side. Isn't that what I'm going to have to use again? Sine equal to a side, angle equal to an angle. And psi equal to a side. Watch for those shared sides. Eventually, we may even have some shared angles. Those are trickier. Anyway, AB is congruent to CB. Wonderful. Let's get that next given down. BD bisects angle ABC, right? BD bisects angle ABC. But what conclusion can we draw on that? What can we get from that? Well, from that, we can get the fact that angle one is congruent to angle two. Why? Because angle bisector is divide and angle into two congruent angles. Now again, this is where your teacher might say, oh, it's completely okay to just write down definition of an angle bisector. If so, go for it. It's a lot less you have to write. Okay? Now at this point, I've got a side equal to a side, an angle equal to an angle. And to get that last side equal to a side, I get that shared side. BD is congruent to itself. By the reflexive property.
All right. That's going to come up in a lot of our proofs. But now, we essentially have enough to conclude, hopefully, that triangle ABD is congruent to triangle CBD. And we certainly do have enough because we have a side angle side situation. Now, keep in mind what we now have. We now have enough information to say that this triangle and this triangle are identical. That means that any corresponding parts in the two triangles must also be identical. For instance, line segment AD would have to be congruent to line segment CD. Angle a would have to be congruent to angle C but for our purposes we only care about one thing. And that's the fact that angle three is congruent to angle four. Why? Because corresponding parts of congruent triangles are congruent. Angle three must be congruent to angle four. These two. All right? Now the reason that I said that goes all the way back up to letter a what would we need to prove, what would mean to be proven to show that BD bisects ADC? Well, three would have to be congruent to four.
So what is the final thing I'm going to say? Well, I'm going to say what I set out to prove. Let me extend the page a little bit so you might be able to see this a little bit better. I'm going to say BD bisects angle ADC. Now what's my reason for that? Here, I'm going to write down definition of an angle bisector. Now, I've been talking a lot about when should you put that down when should you not put that now? And it may almost seem a little bit a hypocritical of me to write down this now. But what else would I write down? I could write something convoluted down like if a line segment divides an angle into two angles of equal size, then we call it an angle bisector. But here, I actually like using just that phraseology. Definition of an angle bisector. In other words, that's just what it is. If I know these two angles are congruent, three and four, then by the definition of what makes an angle bisector an angle bisector, that thing has to be an angle bisector. If somebody comes and puts a fire out at my house, there are firefighter. That's what they are by the definition of what a firefighter is, right? All right, so what great reasoning, you know? Going into it, we're just given a couple very easy pieces of information, AB is congruent to BC. BD bisects angle ABC, we need to prove that it also bisects angle ADC. How do we do that? We first establish that these two triangles are congruent.
Based on that, we then say these two angles are congruent, and based on that, we now know that BD must bisect that angle. Let me bring up the lines of reasoning in this proof so you can take a look at them. I'll pause for a minute to give you a chance to write down anything you need to. And then we'll go on and do one more CPC TC proof. All right, let's move on. Always have to give you a little bit of time to write down the steps and approve they can be long, and as I'm going through them, you might not be quick enough to write them down. Always take advantage of the fact that you can pause the video at any point you need to write down information that you weren't quick enough to write down in the first place. Let's take a look at the last problem. In the diagram below, if it is known, if it is known that AD bisects angle BAC and is perpendicular to BC, is that enough to conclude that triangle ABC is isosceles, explain your thinking. All right, this is kind of cool. So what I'd like you to do is I'd like you to take everything that I've told you in this problem.
Mark up that diagram and see if we have enough to then conclude that sort of the larger triangle ABC is isosceles. Take a little bit of time with this. All right. Let's take a look. Well, AD bisects angle BAC. What does that tell me? It tells me that these two angles are congruent. Great. Also, it's perpendicular to line segment BC. Which means both of these angles are right angles. All right, wonderful. Now remember, the whole idea here is CPC TC reasoning, right? And in order to have CPC TC reasoning, we have to establish the fact that two triangles are congruent. Well, the two triangles that we're looking at here is this right triangle, and that right triangle. Do we have enough to know that they're congruent? Well, there is this shared side right here. So right now I could say that triangle a DB is congruent to triangle a, D.C. by side. No, not by side angle side. But by angle side angle. All right? Now, if these two triangles are identical, then any two parts of them that correspond are also identical.
Well, going back to thinking about what this problem has asked us, do we have enough information to know that sort of this larger triangle is isosceles, well, what is an isosceles triangle? When I saw these triangles one that has two or more sides that have the same length. Well, it sure looks like this and this are the same length. But do we have enough information to conclude that? And the answer is absolutely. Yes. Right? And we do have enough information because a, B must be congruent, or the same length as AC, because they are corresponding parts of congruent triangles, and therefore they are congruent. We have enough. And this is kind of cool. Again, what we're saying is we're saying if I take any random triangle, right? And I have a line drawn that bisects this angle, and it's perpendicular to the side, then this thing, which doesn't look like an isosceles triangle. Is an isosceles triangle. That's enough to know it. And again, it's as simple as knowing that with that information I can prove that these two right triangles are congruent because they're congruent, these two side lengths, which are the hypotenuses of the right triangle, are the same length, and because of that, I have a triangle, the overall triangle that has two sides of the same length, so it must be isosceles. Okay? Let's wrap it up.
Today, we looked at one of the most important ideas in all of geometry. Which is if we can establish, if we can prove, given information we're told, that two triangles are identical, then any parts that correspond on those two triangles must also be identical. Corresponding parts of congruent triangles are congruent. We will see this line of reasoning again and again and again. In future lessons. In fact, it's going to be rare that any lesson where we do a geometric proof. Especially a Euclidean geometry proof doesn't involve CPC TC. But we'll see that in future lessons. For now, I 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.
