Common Core Geometry Unit 3 Lesson 5 Proofs with Partitioning
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Learning Common Core Geometry Unit 3 Lesson 5 Proofs with Partitioning by EmathInstruction
Hello and welcome to another common core geometry lesson by E math instruction. My name is Kirk weiler. And today we'll be doing unit number three lesson number 5 on proofs with partitioning. This lesson is going to involve many of the techniques we've already seen in many of the triangle congruence theorems we've already seen. Along with the work that we did with the axioms of equality, just a few lessons ago. We're going to try to put it all together and by the end, we're going to see some proofs that are fairly lengthy, but that you have absolutely all the tools that you need for need. Now before we get into those though, I think it would be helpful if we reviewed those axioms of equality.
Let's get right into it. All right. Now remember, an axiom is something that's fairly obvious. Something that's so obvious that we have no real way to prove it we're just going to use it as a fact. So let's review the four important axioms of equality. And instead of equality, you could also have congruence, that would be fine too. But let's review them right now. All right. The first axiom is the whole is the sum of its parts. And that's exactly what it says it is, right? If we add all the parts together of some geometric object, we get the hole. The substitution axiom equals can be substituted for equals. Should make all the sense in the world. The addition axiom equals added to equals result in equals. And finally, the subtraction axiom is exactly the same as edition, but just with subtraction.
In other words, equals subtracted from equals result in equals. Again, all of these should be obvious. And what we're going to do now is we're going to start by doing some very simple partitioning proofs. Proves where you either have to add things together or sort of break them apart and subtract them. And then we'll get into proofs that involve triangles and triangle congruence. So let's start with a very, very simplistic. Addition proof. Let's say that. All right, let's take a look at exercise one and what we're given. An exercise one, it says in the following diagram, AB is equal to D.C.. And we want to prove that AC is equal to BD. All right, so let's think about what we know. Right now we know that this segment is congruent to this segment. And what we're trying to prove is that this segment is the same length as this segment. All right? So pretty easy, pretty standard. It's going to be harder when this gets wrapped up in a larger proof.
But for right now, let's make a very simple observation, right? When we think about what we're trying to prove, the fact that AC is equal to BD, what we should notice is that this segment BC is part of both of them, right? And in fact, if I add BC onto AB, I'm going to get AC. And if I add BC on the CD, I'm going to have BD. And that kind of essentially outlines how this proof is going to go. Now, since we've done a proof like this before, I'd ask you to pause the video now and make an attempt at doing this proof and then we'll take a look at how to do it in just a minute. All right. Let's go through it. First, let's get our little T table up. Now, sometimes on a standardized exam, they'll actually have these laid out for you. Sometimes you'll have to write them for yourself. But here we go. So let's get into this proof. All right, no great surprise. We're going to start off with our given AB is equal to D.C.. And there it is. Given. Now ultimately, we want to establish that AC is equal to BD.
In order to do that, though, we've got to add BC onto both sides. This may seem silly, but that means that we've got to write down BC is equal to BC. And remember, any time you state that two or that a geometric object is equal to itself, or congruent to itself, that's called the reflexive property. All right? But now we're all set, right? So let's keep going. If we add these two equations together, we can do that based on the addition axiom or the addition property or possibly you just wrote down addition. That would be okay too. All right? Now, think about it. We've done this, and that could be obvious. We could have had a, B equals D.C., and two equals two, and we could have added those two equations together. That would have been fine. But we did this for a very specific reason. We did it now because AB plus BC is going to be AC. And BC plus D.C. is going to be BD.
Now why can we say that AB plus BC is AC and BC plus D.C. is BD? Well, that's because the whole is the sum of its parts. In other words, a B and B C make the whole AC and BC and D.C. sounds like I'm talking about a 1980s rock band. Makes BD. All right? Putting it all together. You want to look for that in larger proofs. And we'll see one of those later on in this lesson. Okay? Before though, we get into those longer proofs what we should do is also a proof that involves subtraction instead of addition. So let's take a look at one of those. All right, exercise number two. I'm kind of already spoiled the surprise here. We know it's going to involve subtraction, but let's take a look at the overall problem. Exercise number two in the diagram below, ray SU is perpendicular to ray SW. And ray ST is perpendicular to ray SV, prove that the measure of angle one is equal to the measure of angle three.
All right. Now, of course, it's extremely important that you know what perpendicularity means. Right? It means that this ray in this ray form a 90° angle, and this ray. And this ray form a 90° angle. We can even mark that up on our diagram. So we know that that is a 90° angle. And that is a 90° angle. And I think I prefer to go back to working in blue. All right, now ultimately, we want to prove somehow that three I stayed in green, that's funny. That three and one are equal to each other and measure. Or the two angles are congruent. And what we can see is that angle two is sort of part of both of those two right angles. So somehow, if we can remove angle two from those two right angles, we'll be left with angle three and angle one. It's a classic subtraction situation.
So let's see how this thing is going to play out. All right? I'm going to get my statement and reason column up there. And again, since we've done a proof like this before, what I'd ask is that you pause the video now and take a few moments with it and see where you can get. All right, let's take a look. So no great surprise. Let's start with those two givens. All right? SU is perpendicular to SW, ST is perpendicular to SV. Simple enough. But what do we get from that? Well, what we immediately get from perpendicularity is that we've got two right angles. Angle WSU is a right angle. And angle VST are right angles. And the reason for that, well, is that perpendicular lines or rays or segments form right angles. It's really as simple as that. Now the fact that they're right angles isn't nearly as important as what comes next. Which is the fact that their measures are equal to each other.
That's what's important about the fact that they're right angles. Not so much that they're both 90°, but the fact that they're equal to each other. And the statement is simple. All right angles are equal in measure. Okay? Now, how does that help us? Remember, ultimately, we want to get angle three and angle one equal to each other. And yet, the right angle WSU, WSU is made up of angle three and angle two. And Ryan VST is made of angle to an angle one. So let's break these angles up. In other words, WSU is measure of angle three plus measure of angle two, and VST is measure of angle two plus measure of angle one. Why can we say that? Because the hole is the sum of its parts, right? All I'm doing is taking this hole and breaking it up into its parts, and this hole and breaking it up into its parts. Okay. But now you probably see it, right? We're trying to prove that the measure of angle three is equal to the measure of angle one, and on both sides of this equation, angle two is showing up, or at least the measure of angle two.
So we need to get rid of it. And of course, we get rid of something from both sides of an equation by subtracting it. One of the unfortunate things, you know, one of the things that probably seems a little like, why do I have to do that? Is the fact that I have to actually state what I'm going to subtract from both sides. And that's the measure of angle two. So I have to say, the measure of angle two is equal to the measure of angle two, but that's easy enough. That's the reflexive property. And now I can subtract this angle from both sides to get the measure of angle three equal to the measure of angle one, and that's because of subtraction. All right? Now, the good thing is, we're going to be doing some lengthy proofs in just a moment that involve both addition and subtraction, and we'll see how to recognize it when we get there. All right?
Take a moment and pause the video now if you need to to write any of the steps and the reasons down in this proof. All right, let's move on. Okay. So now we're going to get into a real proof. Now don't get me wrong, you could certainly have some of those simple partitioning proofs on a standardized exam. I would certainly expect in your classrooms for you to have simple proofs like that. But more often they are wrapped up in larger proofs, like the one that we're going to see next. So let's take a look at it. Exercise number three in the diagram, B, C, D, F, remember what a given like this means, we're going to take a shot at changing that to blue one more time. What a given like this means is that those three points are co linear. We also know that BC is congruent to FD. I'm going to highlight that on my picture because that's going to be very, very important. These two segments are the same length. We also know that angle one is congruent to angle two. And AD. Is congruent to EC.
Now, I don't know about other states so much, but I do know that in New York, where there's a common core geometry regions exam. That highlighters are allowed on that exam. And I would highly suggest using them. You know, even color coding them, so that like here, I've outlined in blue, all the things that I know or I am given are true. And now in a different color, I'm going to highlight what I'm trying to prove. Maybe I'll do it in red. So it asks me to then prove that AB. Is congruent to EF. All right. It's very, very easy when you're doing a geometry proof, especially a more lengthy one, to kind of get lost in all the details, all the givens, the confusing picture, and things like that. All right? So let's really try to understand this, right? Ultimately, I want to get this line segment congruent to that one.
Now this should be a big red flag for you, especially because it's a more complicated diagram that we're going to have to use, what's known as CP CTC reasoning. Remember what that means? It means congruent parts of congruent triangles are congruent. All right? In order to use CPC TC reasoning, we first must prove that two triangles are congruent. And the question is, which two triangles do we want to prove congruent in order to get those two side lengths congruent? Well, there's not a lot of choice in this diagram, which is actually kind of nice. We need to somehow get triangle, draw it out. We need to get triangle ABD, ABD, somehow congruent, to triangle, EF, C all right? Now, think about what I already know. About these two triangles. I know that AD is congruent to EC. Highlighted in blue. I also know that this angle, which is angle two, is congruent to this angle, which is angle one. Okay? But that's not enough.
Remember, at my disposal right now, I have side side side, side angle side, and angle side angle. Okay? Now I'm not going to get the job done with side side side. All right, it's either going to be side angle side or angle side angle. Now, in order to use angle side angle, the other angle would have to be angle a and angle E, the other angle pair, I apologize. Because that side has to be sandwiched in between them. And there's really nothing in this problem, or nothing in the givens that will help me figure out that angle a and angle ER congruent. On the other hand, to use side angle side, I would have to figure out that BD is congruent to CD. Right? I'd have to be able to figure out that this is congruent to this. As if that didn't make my diagram way more confusing. And I'm kind of halfway there. I already know that this segment length is equal to that segment length. Got to do a little raging there. I just somehow have to add in CD to both of them to establish that entire length is congruent. So I'm more or less have my strategy, right? I got to figure out I've got to prove that BD and CF are congruent by using some kind of an addition proof.
Once I have those two congruent, I can then get the fact that these two triangles are congruent by side angle side. And once I know those two triangles are congruent by side angle side, I can then get the fact that a, B is congruent to EF by CPC TC. Notice how I've kind of scoped out almost the entire proof before ever doing any kind of writing. All right, that's a good, good idea. So let's get into the proof. All right, first things first, we've got our statements. I'm going to move this up here. Okay. First thing, let's get down the fact that BC is congruent to FD. That's given. All right? Second thing I'm going to do is I'm going to say CD is equal or congruent to CD. And that's the reflexive property. So what I'm doing immediately is I'm going to do the hardest part of this proof, which is the addition part. I want to get BD and CF congruent. The order of this proof, this doesn't have to come first. But it has to come before side angle side and side angle side has to come before CPC TC. So so far, I've got those two small segments congruent to each other. And I've got CD congruent to itself.
Now I'm going to add them together. All right? I'm going to add these two segments together and I can do that obviously because of the ignition property. Of congruence or just addition. All right, I'm going to bring this up a little bit, so it might be easier to see. Okay. And finally, I'm going to then say that BD, what I really want BD is congruent to FC, why can I say BD is congruent to FC? Well, because the whole is the sum of its parts. Okay? I'd really like to get this thing off of there, but I don't know how to do it. Oh, and then of course I ended up moving my little reason there. All right. I guess X marks the spot. So we've now done the hardest part of the proof. Because everything else is essentially given to us. But we had to do a little bit of work to get BD and CF to be. Congruent, or the same length. But now that's one of our sides inside angle side. So our next step, angle one is congruent to angle two and AD is congruent EC.
Notice how I lumped those two givens into one. Those just complete our side angle side kind of argument. The angle angle one congruent to angle two being the angle, AD congruent EC being the other side. We can now say that those two triangles ADB and ECF are congruent to each other. Let's see if we can get the picture into the frame. There we go. We can say that because of the side angle side theorem. And then finally, we can say that AB is congruent to EF using CPC TC. Okay. Now, as a little commentary, okay. One of the biggest mistakes that students make in proof, especially proof that involves CPC TC. Is that they'll actually put CPC TC in before ever establishing that two triangles are congruent. Now, in terms of a logic flaw, that's about as big as you can get, right? I don't want to claim that two parts of congruent triangles are congruent.
Unless I've actually proven or are given the fact that those two triangles are identical. So be careful about that. Most often CPC TC, if it's not the final part of the proof, is pretty darn close to the end of it, because you have to establish two triangles or congruent in order to use it. Let me step out of the way a little bit. And let you write down some of these reasons. If you need to. Okay, let's move on to one final proof for the day. Where do I? My advancer got there is. Figuring out a good place for that. All right, another complicated diagram with some overlapping triangles. I don't know about U I see at least three triangles four triangles, 5 triangles in this picture, I'm counting as I go. So a little bit more complicated of a diagram than before. Let's take a look at what we're asked to prove. And what we're given. All right, exercise number four, I'm going to go back to my blue. We're given that angle M NP and NP is congruent to angle RPN. So M and P, which is this angle, and RPN, which is this angle. We also know that angle one is congruent to angle three. These two.
All right, great. And finally, we know that QN. Right there is congruent to QP. Great. Now, eventually, one of these givens will be a redundant. We'll be a redundant. We'll be redundant, all right? So if you're if your math teachers have said, well, he kind of over specified he gave us more than what we need. I completely agree. But we'll have to wait until we do a little bit more work with isosceles triangles to get to that point. Still, let's keep going. Take a look at what we're asked to prove. All right, I'm going to change over to red. And it asks us to prove that MQ. Is congruent to RQ. Okay. Now again, we're asked to prove that two segments are congruent to each other. Besides very simple partitioning proofs like what we started this lesson with. In that kind of situation, you're pretty much a 100% of the time talking about. Right? And it's essential that as quickly as possible, you figure out what two triangles you have to prove congruent in order to get those two sidelines congruent to each other.
Now in this case, those two segments are part of triangle M, Q, N, this guy. MQ N and part of triangle. Sorry about it getting a little crowded here. R Q, P, the one that's right over here. Like that. That's it. So somehow I have to prove that those two triangles are identical. How much of it do I know so far? Well, I know that the sides congruent to this side. That's nq is congruent to PQ. I think that's it. Pause for a minute. See, and take a little bit of time to think. About what else we can figure out about these triangles to allow us to prove that they're congruent. Take a minute. All right, well, one thing is that we're going to eventually be able to say that this angle is congruent to this one. Up on the picture, I don't know that I really want to put it in red or blue. So I'll go with green. And I'm going to use three arc marks, all right? Maybe I'll come back over here. Eventually, I can say that those two angles are congruent. Because they're vertical angles. And remember vertical angle pairs are always congruent to each other. So that gets me closer. I've got an angle on a side, congruent to an angle and a side.
Now, of course, we were told all this business about angles being congruent to each other down here. It'd be very, very nice if somehow I could then prove that angle two is congruent to angle four. If I could somehow do that and I don't want to put all the arcs I'd need there and then four different arcs. If I can prove that those two angles are congruent, then I'll be able to say these triangles are congruent by angle side angle. And then based on those triangles being congruent by angle side angle, I can then say that those two side lengths are congruent by CPC TC. So I want you to pause for a minute and think about how you might prove that angle two is congruent to angle four. Hopefully, hopefully you've said ah, it's going to be a subtraction proof. Okay? And it's going to be a subtraction proof, because we know that this larger angle M and P is congruent to this larger angle, RPN. Those two angles are made up of one and two and three and four respectively. But since we know that one in three are congruent, we'll be able to subtract them away and be just left with two and four.
We've got these larger angles congruent to each other with smaller portions of each being congruent to each other. If we get rid of those smaller portions, what's left over will also be congruent. So let's dive into the proof. All right? So there's going to be a subtraction one going on in here. All right, here we go. Get up my statement bars. Statement reason. First things first, let's get those big angles congruent to each other. That given. Is congruent to RPN. All right. Now, I want to break those things up into pieces and then subtract off the pieces that are equal. So let's break those angles up into pieces. And then P is one and two, RPN is three and four. The whole is the sum of its parts. Anytime you take a hole and you break it up into its sum or take a sum and put it back together into its whole, it's because the hole is the sum of its parts. What's next? Well, we know angle ones congruent to angle three. And that's just a given. Hold on. It's not even there. That's too funny. Let's just write it in. Who knows where that given what? I don't know.
Anyway, I ain't going to congruent to angle three, and that's a given, right? It's right up here. Anyway now, what can we do? We can subtract one and three from both sides of this equation and we're left with angle twos congruent to angle four. And that's just subtraction, right? That's the subtraction property. And again, it's nice because once we have two congruent to four, those are these two angles up here. Here, and here. All right. The rest of the proof should be relatively easy. Let's go through it. We now say that QN is equal to QP. Or congruent, and that's a side. Given. Right? Now we can say that MQ N is equal to R QP. This is important. MQ N is congruent to R QP. Those are these two angles up here. Right? And that's that vertical angle business. What do I say? Vertical angles are congruent. If you had said that the two things were equal to each other and then said vertical angles are equal, that's okay as well.
Equality and congruence when it comes to line segments and angles, right? Those two are kind of hard to distinguish. When it comes to triangles, you should never say that two triangles are equal. That's weird. It's like saying an elephant is equal to an elephant. It just doesn't make a lot of sense. All right, line segment equal to a line segment and angle equal to an angle, especially angles. That makes a lot of sense. Anyway, look what we have. We have an angle, a side, and an angle of these two triangles. We can now say that those two triangles are congruent by the angle side angle theorem. And then, because we have these two triangles congruent, what we wanted to prove, MQ is equal to R Q, right? By CPC TC. That's it. What makes this proof challenging? What makes this proof challenging is the fact that there is a subtraction mini proof in the middle of it.
All right? And again, that's not going to be surprising. We'll see some proofs eventually that are quite long. Proofs where maybe you have to prove two triangles are congruent in order to prove two other triangles are congruent. Okay? Proof within proof, argument within argument. Anyway, pause the video now if you need to, to write down any of this. Okay. Let's wrap up. Now, there are partitioning proofs that are small. And simple. And we spend an entire lesson just working on those. And then there are partitioning proofs that show up in the middle of much larger proofs. Those were the last two. Essentially, all partitioning, though, kind of works the same. Either you're taking small pieces and adding on pieces that are equal to make the larger pieces equal. Or you're taking larger quantities that are congruent or equal, breaking them apart and subtracting out smaller parts that are congruent or equal to get other two smaller parts can grow it.
In one case, you're talking about addition. And the other case you're talking about subtraction. All right? And we'll see these things pop up and proofs from time to time. For now though, I'd like 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.