Comon Core Geometry Unit 9 Lesson 8 Secant and Tangent Lengths

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 8 on secant and tangent segment lengths. So for the last few lessons, we've been concentrating on the angles that secants and tangents make with one another and make exterior to the circle and how they relate to those arcs that the secants and tangents intercept. Today though, we're going to be concentrating on the lengths of secants and tangents and how they relate to each other. Thankfully, there's basically one theorem that encompasses all of this material. And so we're going to kind of take a look at that in the first exercise, then we'll kind of formally introduce it and eventually will prove it. All right? But for right now, let's jump into it and see what we're even talking about. All right, take a look. Exercise number one. Encircle O shown below sequence PAC and PBD are drawn. It is also known that PA is three. Let's label these as we go. Oh, maybe with a slightly smaller pen. All right, I said that they will all fit. So PA is three. We've got a, C is 5. All right, PB is four. And BD is two. All right? And letter a says, find the following products, and it's just a bunch of different products. This should be easy for you to find. So pause the video now and go ahead and find these products. All right, let's go through them. It's not all that interesting. The first one asks us to find the product of PA and AC. That's three times 5. So that's 15. All right, we've got PB times BD. That's four times two. And that's a product of 8. Then we've got PA times piece C, PA times PC. Well, PA is three. PC, of course, is going to be 8, so then that's going to be 24. PB times PD that's going to be four times 6, and that's going to be 24 as well. AC times PC, well, AC is 5. Times 8, which is 40. And BD times PD that's going to be two. Whoops. There's my two times 6. Wow, this thing is just going crazy. Let's try that all the way over. Let's try two times 6, and that's going to be 12. All right. Now letter B, quite easy. Asks us which pair of products is equal. How would you describe the products in terms of the segments? So there's one pair of these products that's obviously equal, and that's this one. Right? In other words, trying to keep that up. PA times PC is equal to PB times P D okay, and how would we describe that? Let's take a look. PA is this length. Right? And maybe a different color. And PC is this length. All right? Then going back, PB is this length. And PD is that length. So how could I describe that? Well, we've got these two secants, and it looks like when I take the portion that's exterior to the circle. And I multiply it by the entire length, and then I take the portion that's exterior to the circle. And I multiply it by the entire length. It looks like that product is consistent, or stays constant. So I think we would describe it as exterior segment times whole segment. Is constant. So for every secant, there's a portion of it that lies outside of the circle, and then there's the entire length of that segment, we're assuming that the secant is a segment. It's not continuing on forever. And so it looks like and this will be the case that if we take the portion of the secant that's exterior to the circle. And multiply it by the entire length, that product stays constant. That product stays constant. Let's kind of summarize that. If we can move forward, there we go. C can't segment length theorem. Let's take a look. If two secants are drawn from a common exterior point, the product of the length of one segment with the length of its external segment is equal to the product of the length of the other secant with the length of its external segment. All right? So here we've got that picture. PA times PC is equal to PB times PD. And that's really the entire thing that we're going to be doing today. Even when we get tangents involved, it'll be the same idea. When we take the portion that's exterior to the circle, and then we multiply it by the entire length, we'll get the portion exterior to the circle, multiplied by the entire length. And we will eventually prove that. But first, let's take a look at using it. It's an exceptionally easy theorem to use. Let's take a look at exercise two. In the diagram of circle O shown, AB is equal to 6. All right, let's label that. BD is equal to two. AC is equal to four. And we want to know the length of CE. Now, of course, the incorrect way of using this theorem is to say, okay, 6 times two equals four times X and then solving for X that would give me X equals three, and it would be incorrect. What the theorem says is it says that AB times AD, right? AB times AD is equal to AC, AC times AE. Now, just kind of filling the blanks, while AB is equal to 6. AD, though, is equal to 8, right? I have the 6 plus two. AC is equal to four. And this will now let me solve for AE. Granted, I'm not actually looking for AE. I really want the length of CE, but I need a E to find it. This is simple enough. 6 times 8 is 48. Four times AE divide both sides by four, right? And what we find is that AE is equal to 12. And because we now know that the length of this entire segment AE is 12, we can just subtract off the four. And we find that CE must be 12 minus four CE's gotta be 8. That's it. Exterior segment times the entire length equals exterior segment times the entire length. So the whole deal. All right, let me step out of the way so you can write something down. And then we'll move on. All right, let's do it. Okay, now, what happens if we have a secant and a tangent? We won't even really talk about two tangents, because quite frankly, when you have two tangents, something that we learned before was that the length of the two tangents have to be equal. So there's no real fancy theorem for that other than the length of the two tangents have to be equal. But let's take a look at this one. If a secant and a tangent are drawn from a common exterior point, then the product of the length of the secant with the length of its external segment, so that's the same thing as we had before. All right, we're taking a look at the total segment times the external segment is equal to the square of the length of the tangent. Wow, another theorem to memorize. So we've got PA times PC is equal to PB squared. Now that almost seems like it's a different theorem than the one we had before. But keep in mind, the one that we had before said, if we take the external segment and multiply it by the whole thing, we'll get the external segment multiplied by the whole thing. But the external segment and the whole thing are the same for a tangent. And that's why we just get the PB squared. All right? So it's absolutely absolutely the same theorem. All right, it just happens that with a tangent, it's entire length is exterior to the circle. So when we take the exterior length and we multiply it by the whole length, we just get the length of the tangent squared. All right, let's use it. It's an easy enough theorem to use just like below. Let's take a look. Or before, exercise number three. Encircle O, a tangent is drawn from a to the circle at point B secant ADC is also drawn and has a length of 12 units. So we know that this is 12 units. If AD to D.C. is one to three, determine the length of a, B all right, so the theorem says that if I take a, B, and I square it, I'll get AD times AC. So this times this is equal to this times this. AB squared is AD times AC. The problem is, although I do know what AC is, it's 12. I don't yet know what AD is. And that's where I've got to do that piece with the ratios, right? I know that the ratio of AD to D.C. is one to three. Now, many of you can probably just kind of go, oh, all right, then AD must be blank, okay? But if you don't, then we go with one of these again. We'll let AD be equal to X will let D.C. equal three X that maintains that one to three ratio one X to three X. Then I can use sort of this hole as the sum of its parts to say X plus three X has to be 12. When I add the two together, I've got to get those 12 units. So for X must be 12 and X must be three. Which means a D is three and D.C. is 9. I don't really need the 9 though and I want to be careful not to use it. I really just needed that three. So now I can say that AB squared is equal to AD, which is three times AC 12 AB squared is equal to 36. Take the square root of both sides. And a, B is 6. All right? Not too bad, right? The exterior times the hole is equal to the exterior times the hole. That's the theorem. Let me step back right down I think you need to. Then we'll go on and we'll prove this thing. All right, let's do it. So the last exercise, we're going to prove the secant length theorem. So we're not going to do the tangent length. That's going to show up on the homework. We're going to prove the secant length theorem. And again, let me give you a little bit of a setup. What we've got right now is we've got these two secants drawn, PAC, and PBC. And again, I want to prove that PA times PC is equal to PB times PD. In order to do this, I had to draw these extra lengths in, okay? And let's kind of take a look. Okay, exercise number four. In circle O secants have been drawn from external point P, we will prove PA times PC equals PB times PD. But array explain why angle C must be congruent to angle D why does this angle angle C have to be congruent to angle D? Pause the video now and see if you can answer this question. All right, well, it's because they are both inscribed angles. That intercept. The same arc. Specifically, arc AB, but whatever. So they both are intercepting or intersecting arc AB. And again, since they have to be half the measure of arc AB, that means they both must be the same measurement. So if arc AB is 60°, C and D are both 30°. Ah, letter B, why must these two triangles be similar? Now, again, keep in mind, right, that what I've done is I've taken triangle PCB right here. And triangle PAD, and I've just kind of put them here. Why do those two triangles have to be similar? Well, let's take a look. Number one, we just proved or explained why angle C and angle D must be congruent to each other. But one of the beautiful things here is that they also share an angle. They share angle P and so they're similar. By the angle angle theorem of similarity. Hey, red pen. It's been a little lost since we've seen red pen. There, which is cross it out. Okay. They're similar by the angle angle theorem similarity. There they are. Angle equal to an angle, angle equal to an angle. Great. Now let's finish the proof. This is going to be simple, right? It says, using a proportion or using similar triangles, set up a proportion involving the segment lengths in the theorem and write it in product form. And remember what I'm looking to do, right? Let's write down what we really want. We want to get the fact that PA times PC is equal to PB times PD. That's at the end of the day what I want. I just want to be able to bring it up here. And now it's all about the similarity of these two triangles. Keep in mind, write those similarity proofs we did, where the final line was to prove some product. Well, that's really in geometry to set us up for proving something like this. So let's do that similarity piece, right? What we know is that we know that if I do PC divided by PD, all right, if I do PC divided by PD, I would have to get PB divided by PA. And if we were doing this in really proof form, our little reason there would be the ratio of corresponding sides in similar triangles or proportional. So alongside a PC of that triangle divided by long side of this triangle is equal to short side of this triangle divided by short side of that triangle. PC divided by PD is equal to PB divided by PA. But now I can write it in product form. And in fact, I can just sort of cross multiply if you will, and I get exactly what I want. PA times PC is equal to PB times PD. And the reason for that would be the product of the means is equal to the product of the extremes. So it's really kind of cool. This theorem that we were using today is a result of similar triangles. That similarity being caused by the fact that two angles two inscribed angles that intercept the same arc must have equal measures. Isn't that great how math kind of brings it all together in the end? All right. Let me pause now so that you can write down anything you need to, and then we'll wrap up. All right, let's do it. Wrap this lesson up. So this lesson was a fairly short one, because it really presented a fairly simple theorem, which is if two secants intersect outside of a circle. Then their segment lengths are related by being able to take the product of the exterior portion of the secant times the entire length of the secant, and that product remains constant. All right? So any circle, if you locate a point outside of it and you draw two secants, the products of those external segments with the entire segment stays constant. When it comes to a secant and a tangent since the entire portion of the tangent lies outside of the circle, it simply the length of the tangent squared is then consistent with that external segment times the entire segment of the secant. All right. Well, you'll get a lot of work on those on the homework. For now, I'd like to just 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.