Common Core Geometry Unit 9 Lesson 7 Tangent and Secant Proofs and Practice

Hello and welcome to another common core geometry video by E math instruction. My name is Kirk weiler, and today we'll be doing unit number 9 lesson number 7. Tangent and secant proofs and practice. So before we get into this, let me just remark about the fact that this truly is a lesson where all we're going to be doing is proving some theorems that we saw in lesson number 6. So in lesson number 6, we saw a bunch of different theorems that were obviously going to be reviewing today about how angles that are created by tangents and chords, tangents, and secants to engines and tangents. How those angles relate to the arcs that all of those intercept. All right? So today we're going to actually prove those theorems. The practice part of the lesson really comes from the homework. So the homework, you're not going to really be doing these proofs. Maybe a little bit here and there. But mostly you're just going to be practicing more with the theorems. So if you haven't watched lesson number 6 yet, this one could be a little bit more dicey. But hey, join me in it anyway. Let's jump right into it. All right, here we go. Exercise number one, angle formed by a cord and a tangent. Angle formed by a cordon tangent. So if you recall, from the last lesson, that if we've got a tangent like BE here, and we've got a cord like AC here, or you could even claim that that's a secant. Then the measure of angle BAC BAC will be one half of the measure of that intercepted arc. One half of arc AC. And what we're going to do in exercise one is we're going to prove that. And that's very important because this theorem will then get wrapped up strangely enough in some of the other ones that we try to prove. So let's jump right into it. All right, to derive the measure of angle BAC is one half the measure of arc AC. So letter a simple enough asks us, what is the measure of angle B, a D? How do we know? So what's the measure of angle BAD? Pause the video now and write that down. All right, well, let's do it. Hopefully, because this wasn't that many lessons to go. Let me get rid of my little black dot there. The measure of angle BAD is 90°. Okay. And the reason why is that tangents. Are always perpendicular. To radii. At the point of tangency. Tangents. At the point of tangency. My red pen, I think we'll leave it that time. Tangents are always perpendicular to radii at the point of tangency. So let me actually draw that in. We've got this nice 90° angle there. All right. Great. All right, letter B, what is the measure of angle CAD in terms of arc measurements? So now I'd like to know the measure of angle CAD, CAD in terms of arc measurements. Yet again, pause the video, pause the video and write something down here. All right. Well, this one should be easy. The measure of angle CAD is going to be one half of the measure of arc CD, right? So we learned and proved we learned and proved a few lessons ago that the measure of any inscribed angle, and look, CAD is an inscribed angle, will always be one half the measure of its intercepted arc. Okay? So we're working with that as well. Now, let us see. It says fill in the missing angles and arcs. This should be pretty easy. See if you can pause the video now and fill in the blanks here. All right. Well, here we go. The measure of angle BAC plus the measure of angle CAD, this is simple enough. That's going to be the measure of angle. BAD. Now, if you actually said that that was equal to 90°, that's okay. Or if you just said 90, or if you just said BAD, that's all good as well. This is the hole is the sum of its parts, right? So angle BAC plus angle CAD has to be angled BAD. Bad bad angle. All right, and also hole is the sum of its parts. The measure of arc AC plus the measure of arc CD has to be equal to the measure of arc AD. Which also is a 180° arc, okay? That's important all by itself. Arc AC plus arc CD is our AD and that is a 180° arc. And now we have everything we need to prove that the measure of angle BAC is one half the measure of arc AC. And let's take a look at that. Letter D use a, B, and C to derive the fact that the measure of angle BAC is one half the measure of our KC. So let's do this together. All right. What did we learn? And letter a, we recall that measure of angle BAD was 90°. In letter B, the measure of angle CAD is one half the measure arc CD. And let her see we had a few holes the sum of its parts. Now at the end of the day, what I'm trying to do though is I'm trying to solve for the measure of angle BAC. So one thing I can do is I can immediately rearrange that second or sort of that equation and letter C to solve for BAC. In fact, the measure of angle BAC will be the measure of angle BAD minus the measure of angle CAD, okay? So all I'm doing here is rearranging this equation to solve for the angle I'm interested in. BAC. Now let's do some other substitution. We know, right? We know that the measure of angle BAD is 90°. So I'm just going to do that. Okay? Likewise, we know that the measure of angle CAD is one half the measure of arc CD. Okay. But we want to somehow get AC involved because AC, well, as part of the theorem, but also AC is the arc that is intercepted by BAC. So I don't really want CD in the mix here. That's where this final equation comes in, right? Because let me extend the page a little bit. Oh, I extended it way too far. Because we can rearrange this equation and solve for the arc CD. So the measure of arc CD is the measure of arc AD minus the measure of arc AC and this is really right here is the one that we want to have involved. So let's keep going. The measure of angle BAC is 90. I'll drop the degree symbol for right now minus one half of the measure of arc AD minus the measure of arc AC. That's one beautiful looking arc symbol there. Let's try that again. Wow, while I'm getting closer, remember I want BAC to be one half the measure of arc AC, so I've almost got there, right? Okay, but what else can I do? Well, I know that the measure of arc AD, the measure of arc AD, all the way up here is a 180°. So the measure of angle BAC is 90 minus one half of a 180 minus measure of arc AC. And I essentially have it now. All right, and here's why. I'm now going to distribute the one half, and I'm going to get the measure of angle BAC is 90 minus one half times one 80, which is 90. Minus one half times minus measure of AC is going to be positive one half measure of our KC, and now what happens the 90s cancel. And the measure of angle BAC is one half the measure of arc AC. There it is. That's what I needed. That's the whole thing. Now you might look at this and think, wow, that is rather daunting. But it really uses things that we already know. It uses things about the perpendicularity of a radius to a tangent. The fact that an inscribed angle has a measure that's one half its intercepted arc uses the fact that these two arcs would have to add up to a 180°. All of which come together to tell us or to allow us to prove that the measure of angle BAC is one half of the arc that it intercepts. All right, let me step out of the way for a moment, write down anything you need to. And then we'll move on and prove another theorem. All right, let's do it. What we're really doing today are what are known as derivations. You could also call them algebraic proofs if you wanted to, right? They're not so much statement and reason, but they're using things like the whole as the sum of its parts and previous theorems that we've used or learned, along with the rules of algebra to come up with the results that we're looking for. So now we're going to get into all of those theorems about an exterior angle being created when two secants intercept a circle, things like that. So let's take a look at exercise number two. Angle formed by two secants. So remember, this is all about the fact that the measure of angle P will be one half the difference between arc CD and arc AB, okay? And I think I'm going to actually highlight those a little bit. All right. So it's all about AB. That's interesting. And about CD. And I'm kind of thinking that I'll have some mystery blue pen show up. But not likely. So it's all about those two arcs. Poorly highlighted though this one is arc AB and RCD. So let's jump into the proof. Okay. Here we go. Letter a using the exterior angle theorem, fill in the following. The exterior angle theorem, right? Remember what the exterior angle theorem says. Let me review that really quick. If I've got some triangle, and I extend one of the side lengths out, it creates what's called an exterior angle right here. And that exterior angle is always the sum of the two remote interior angles. So this angle plus this angle will be that one. We're going to use that theorem quite a bit right now. So let's do it. Okay. Now, notice this dashed line I have drawn in here in order to create a triangle, specifically triangle PAD. PAD. So let's take a look. Using the exterior angle fit theorem fill in the following, the measure of angle FAD, watch for that F, a, D, F, a, D the measure of angle FAD must be the measure of angle P plus what other angle. Well, it's an exterior angle. F a D is the exterior angle one exterior angle of this triangle. So one remote interior angle P and the other one down here. You could call it angle D, I'm going to call it angle BDA or PDA, right? So I'm going to say angle, yeah, we'll call PDA. Public display of affection for angles. So this one, down here. Simple enough. Letter B says, solve the above equation for the measure of angle P well that's easy enough. Go ahead and do that really quick. Just rearrange this equation so you have measure of angle P equals something. All right, it should be easy enough. All I have to do is take PDA, throw it on the other side. And the measure of angle P will be the measure of angle FAD minus the measure of angle PDA. All right? So what now? Well, letter C says, using your answer to B, along with inscribed angle relationships derive this fact. Well, look at how easy this is going to be. Watch. The measure of angle P is going to be the measure of angle FAD minus the measure of angle PDA. But what is extended a little bit? What is the measure of angle FAD? FAD, well, that's going to be one half of arc CD. The inscribed angle FAD intercepts arc CD. So this is going to be one half the measure of arc CD. Minus, and PDA, P, D, a, is an inscribed angle that intercepts arc AB. So that's going to be one half the measure of arc AB. And now look, this is what I'm trying to come up with. This is what I have. All I need to do is factor out that common one half. So the measure of angle P is one half the measure of arc CD minus the measure of arc. AB. And that's it. And I have it. That was even simpler than the last one, right? Because all I had to do was use the exterior angle theorem, which said that this angle is the sum of this one plus this one. And then I'm solving for this. This is the one that I care about, right? So I solve for this, which says this angle is simply going to be this one minus that one. But then both this angle, angle FAD, and this angle, PDA, are both inscribed angles, and therefore must be one half their intercepted arcs. And that gets me all the way home. All right, step back, copy down anything you need to, then let's go prove another theorem. All right, so exercise number three, an angle formed by a tangent and a secant. Let's take a look. So what we want to derive is that the measure of angle P that exterior angle is again one half the difference in the intercepted arcs, right? And we've got that larger arc, BCD, and we've got the smaller arc AB, and we want to show just like we did in the last exercise that the measure of angle P that exterior angle is one half the measure of the difference in the measure of those two arcs. And just like before, we have to actually draw in this little extra segment to help us out and for exactly the same reason we need to get the exterior angle theorem involved. So let's take a look. Let array, it says using the exterior angle theorem. Let me change back. Using the exterior angle theorem, fill in the following. The measure of angle fbd, FB D all right, is equal to the measure of angle P plus what exactly. Ah, well in this case, we need the measure of angle B, D, P, B, D, P, or PDB, whichever, or you could even say the measure of angle D but I'm going to say the measure of angle, B, D, P so in other words, that one right up there, right? The exterior angle is always the sum of the two remote and terror angles. And just like before it says, all right, go ahead and solve this for the measure of angle P why don't you go ahead and do that. All right, again, it should be piece of cake, right? The measure of angle P is simply the measure of angle F, B, D minus the measure of angle B, D, P all right? This should feel pretty familiar to the last one, right? Because then it says use the inscribed angle relationship and the chord tangent angle theorem to show that the measure of angle BCD are the measure of angle P is one half the measure of arc BCD minus the measure of arc AB. Well, let's just write this relationship down that we had up above. The measure of angle P is the measure of angle. F minus the measure of angle B, DP. All right, extend the page a little bit so that we've got some room to work. Okay, well, what do we have? Well, F, D, F, B, D, is not an inscribed angle. F all right? But it is an angle made by a tangent and a cord. And what we've already proven is that an angle made by a tangent and a cord will always be one half of the intersected arc, which in this case is arc B, CD. All right? So one half the measure of arc, B, C, D now, angle B, D, P, on the other hand, be DP, is an inscribed angle. It's an angle that has its vertex on the circle, and whose two rays intersect the circle, specifically at point a and point B therefore, the measure of angle BDP is just one half the measure of arc AB. And we're basically there. Just like before, we can factor that one half out. And we have the relationship we were looking for. In other words, the measure of that exterior angle is one half the difference in it to intercepted arcs. Simple enough. All right, go ahead and write down anything you need to on these. Let me step out of the way for a minute, and then we'll wrap up. All right. So in today's lesson, we looped back and we proved theorems that we introduced in the last lesson. Specifically, the theorem that says that the measure of an angle made by a chord in a tangent is one half the measure of its intercepted arc. And the measure of an exterior angle created by two secants or a secant and a tangent is one half the difference in the intercepted arcs. If you're really kind of on your game, you might notice that we've left one out, and that's the measure of an angle created by two tangents. And what you'll be doing is you'll be looking at that one on the homework, okay? We had to save one so that you actually got a little bit of practice on this. The other homework problems, though, for today's lessons, really specifically concentrate on just applying the theorems. Because they can be tricky all by themselves, especially because there's so many of them that are floating around. They all have one half in them. So you need a little bit of practice to really kind of reinforce them. For now, I'd like to thank you though for joining me for another common core geometry video by E math instruction. My name is Kirk weiler, and until next time, keep thinking. And keep solving problems.