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Did I do that right? Yep, that's 80 plus 15, yep, 95. And then if I subtract 95 from both sides, what am I left with? I'm left with Y is equal to 85 degrees. And so this is going to be equal to 85 degrees. And then this last triangle right over here, I have an angle that has measure 36, another one that's 59. So by the same logic, this one over here has to be 85 degrees.
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Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3
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I'm left with Y is equal to 85 degrees. And so this is going to be equal to 85 degrees. And then this last triangle right over here, I have an angle that has measure 36, another one that's 59. So by the same logic, this one over here has to be 85 degrees. So let's ask ourselves, now that we've figured out a little bit more about these triangles, which of these two must be congruent? So you might be tempted to look at these bottom two triangles and say, hey, look, all of their angles are the same. You have angle, angle, angle, and angle, angle, angle.
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Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3
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So by the same logic, this one over here has to be 85 degrees. So let's ask ourselves, now that we've figured out a little bit more about these triangles, which of these two must be congruent? So you might be tempted to look at these bottom two triangles and say, hey, look, all of their angles are the same. You have angle, angle, angle, and angle, angle, angle. Well, they would be similar. If you have three angles that are the same, you definitely have similar triangles. But we don't have any length information for triangle IJH.
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Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3
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You have angle, angle, angle, and angle, angle, angle. Well, they would be similar. If you have three angles that are the same, you definitely have similar triangles. But we don't have any length information for triangle IJH. You need to know at least one of the lengths of one of the sides in order to even think, start to think about congruence. And so we can't make any conclusion that IJH and LMK, triangles IJH and triangles LMK are congruent to each other. Now let's look at these candidates up here.
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Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3
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But we don't have any length information for triangle IJH. You need to know at least one of the lengths of one of the sides in order to even think, start to think about congruence. And so we can't make any conclusion that IJH and LMK, triangles IJH and triangles LMK are congruent to each other. Now let's look at these candidates up here. We know that their angles are all the same. And so we could apply, we could apply angle, I'll do this in a different color, angle, side, angle. 36 degrees, length six, 82 degrees.
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Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3
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A tiny but horrible alien is standing at the top of the Eiffel Tower, so this is where the tiny but horrible alien is, which is 324 meters tall, and they label that the height of the Eiffel Tower, and threatening to destroy the city of Paris. A men in black, or I guess a men in black agent, I was about to say maybe it should be a man in black, a men in black agent is standing at ground level, 54 meters across the Eiffel Square, so 54 meters from, I guess you could say the center of the base of the Eiffel Tower, aiming his laser gun at the alien, so this is him aiming the laser gun. At what angle should the agent shoot his laser gun? Round your answer if necessary to two decimal places. So if we construct a right triangle here, and we can, so the height of this right triangle is 324 meters, this width right over here is 54 meters, it is a right triangle, what they're really asking us is what is the angle, what is this angle, what is this angle right over here? And they've given us two pieces of information, they gave us the side that is opposite the angle, and they've given us the side that is adjacent to the angle. So what trig function deals with opposite and adjacent?
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Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3
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Round your answer if necessary to two decimal places. So if we construct a right triangle here, and we can, so the height of this right triangle is 324 meters, this width right over here is 54 meters, it is a right triangle, what they're really asking us is what is the angle, what is this angle, what is this angle right over here? And they've given us two pieces of information, they gave us the side that is opposite the angle, and they've given us the side that is adjacent to the angle. So what trig function deals with opposite and adjacent? And to remind ourselves, we can write like I always like to do, SOH CAH TOA. And these are really by definition, so you just kind of have to know this, and SOH CAH TOA helps us. Sine is opposite over hypotenuse.
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Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3
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So what trig function deals with opposite and adjacent? And to remind ourselves, we can write like I always like to do, SOH CAH TOA. And these are really by definition, so you just kind of have to know this, and SOH CAH TOA helps us. Sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. Tangent is opposite over adjacent. So we can say that the tangent, we can write that the tangent of theta, the tangent of theta is equal to the length of the opposite side, 324 meters, 324 meters over the length of the adjacent side, over 54 meters.
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Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3
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Sine is opposite over hypotenuse. Cosine is adjacent over hypotenuse. Tangent is opposite over adjacent. So we can say that the tangent, we can write that the tangent of theta, the tangent of theta is equal to the length of the opposite side, 324 meters, 324 meters over the length of the adjacent side, over 54 meters. Now you might say, well, okay, that's fine, so what angle gives, when I take its tangent, gives me 324 over 54? Well, for this, it will probably be useful to use a calculator. And the way that we'd use a calculator is we would use the inverse tan function.
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Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3
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So we can say that the tangent, we can write that the tangent of theta, the tangent of theta is equal to the length of the opposite side, 324 meters, 324 meters over the length of the adjacent side, over 54 meters. Now you might say, well, okay, that's fine, so what angle gives, when I take its tangent, gives me 324 over 54? Well, for this, it will probably be useful to use a calculator. And the way that we'd use a calculator is we would use the inverse tan function. So we could rewrite this as we're going to take the inverse tangent, and sometimes it's written as tangent, kind of this negative 1 superscript. So the inverse tangent of tan of theta is going to be equal to the inverse tangent of 324 over 54. And just to be clear, what is this inverse tangent?
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Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3
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And the way that we'd use a calculator is we would use the inverse tan function. So we could rewrite this as we're going to take the inverse tangent, and sometimes it's written as tangent, kind of this negative 1 superscript. So the inverse tangent of tan of theta is going to be equal to the inverse tangent of 324 over 54. And just to be clear, what is this inverse tangent? This literally says, this will return what is the angle that when I take the tangent of it, gives me 324 over 54. This says, what is the angle that when I take the tangent of it, gives me tangent of theta? So this right over here, this just simplifies to theta.
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Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3
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And just to be clear, what is this inverse tangent? This literally says, this will return what is the angle that when I take the tangent of it, gives me 324 over 54. This says, what is the angle that when I take the tangent of it, gives me tangent of theta? So this right over here, this just simplifies to theta. Theta is the angle that when you get the tangent of it, gets you tangent of theta. And so we get theta is equal to tangent, inverse tangent of 324 over 54. Once again, this inverse tangent thing, you might find confusing, but all this is saying is, over here it's saying tangent of some angle is 324 over 54.
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Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3
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So this right over here, this just simplifies to theta. Theta is the angle that when you get the tangent of it, gets you tangent of theta. And so we get theta is equal to tangent, inverse tangent of 324 over 54. Once again, this inverse tangent thing, you might find confusing, but all this is saying is, over here it's saying tangent of some angle is 324 over 54. This is just saying is my angle is whatever angle I need, so that when I take the tangent of it, I get 324 over 54. It's how we will solve for theta. So let's get our calculator out, and let's say that we want our answer in degrees.
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Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3
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Once again, this inverse tangent thing, you might find confusing, but all this is saying is, over here it's saying tangent of some angle is 324 over 54. This is just saying is my angle is whatever angle I need, so that when I take the tangent of it, I get 324 over 54. It's how we will solve for theta. So let's get our calculator out, and let's say that we want our answer in degrees. And actually they should, well, I'm just going to assume that they want our answer in degrees. So let me make sure my calculator is actually in degree mode. So I'll go to the second mode right over here, and actually it's in radian mode right now.
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Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3
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So let's get our calculator out, and let's say that we want our answer in degrees. And actually they should, well, I'm just going to assume that they want our answer in degrees. So let me make sure my calculator is actually in degree mode. So I'll go to the second mode right over here, and actually it's in radian mode right now. So let me make sure I'm in degree mode to get my answer in degrees. Now let me exit out of here, and let me just type in the inverse tangent. So it's in this yellow color right here.
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Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3
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So I'll go to the second mode right over here, and actually it's in radian mode right now. So let me make sure I'm in degree mode to get my answer in degrees. Now let me exit out of here, and let me just type in the inverse tangent. So it's in this yellow color right here. Inverse tangent of 324 divided by 54 is going to be 80 point, and they told us to round to two decimal places, 80.54 degrees. So theta is equal to 80.54 degrees. That's the angle at which you should shoot the gun to help defeat this horrible alien.
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Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3
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So pause this video and see if you can figure that out. Well, the key realization to solve this is to realize that this altitude that they dropped, this is going to form a right angle here and a right angle here. And notice, both of these triangles, because this whole thing is an isosceles triangle, we are going to have two angles that are the same. This angle is the same as that angle because it's an isosceles triangle. This 90 degrees is the same as that 90 degrees. And so the third angle needs to be the same. So that is going to be the same as that right over there.
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Pythagorean theorem with right triangle.mp3
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This angle is the same as that angle because it's an isosceles triangle. This 90 degrees is the same as that 90 degrees. And so the third angle needs to be the same. So that is going to be the same as that right over there. And since you have two angles that are the same and you have a side between them that is the same, this side, this altitude of length 12 is on both triangles, we know that both of these triangles are congruent. So they're both gonna have 13, they're gonna have one side that's 13, one side that is 12. And so this and this side are going to be the same.
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Pythagorean theorem with right triangle.mp3
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So that is going to be the same as that right over there. And since you have two angles that are the same and you have a side between them that is the same, this side, this altitude of length 12 is on both triangles, we know that both of these triangles are congruent. So they're both gonna have 13, they're gonna have one side that's 13, one side that is 12. And so this and this side are going to be the same. So this is going to be x over two and this is going to be x over two. And so now we can use that information and the fact and the Pythagorean theorem to solve for x. Let's use the Pythagorean theorem on this right triangle on the right-hand side.
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Pythagorean theorem with right triangle.mp3
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And so this and this side are going to be the same. So this is going to be x over two and this is going to be x over two. And so now we can use that information and the fact and the Pythagorean theorem to solve for x. Let's use the Pythagorean theorem on this right triangle on the right-hand side. We can say that x over two squared, that's the base right over here, this side right over here, we could write that x over two squared plus the other side, plus 12 squared, is going to be equal to our hypotenuse squared, is going to be equal to 13 squared. This is just the Pythagorean theorem now. And so we can simplify.
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Pythagorean theorem with right triangle.mp3
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Let's use the Pythagorean theorem on this right triangle on the right-hand side. We can say that x over two squared, that's the base right over here, this side right over here, we could write that x over two squared plus the other side, plus 12 squared, is going to be equal to our hypotenuse squared, is going to be equal to 13 squared. This is just the Pythagorean theorem now. And so we can simplify. This is going to be x, let me do that same color, this is going to be x squared over four, that's just x squared over two squared, plus 144, 144 is equal to, 13 squared is 169. Now I can subtract 144 from both sides. I'm gonna try to solve for x, that's the whole goal here.
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Pythagorean theorem with right triangle.mp3
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And so we can simplify. This is going to be x, let me do that same color, this is going to be x squared over four, that's just x squared over two squared, plus 144, 144 is equal to, 13 squared is 169. Now I can subtract 144 from both sides. I'm gonna try to solve for x, that's the whole goal here. So subtracting 144 from both sides. And what do we get? On the left-hand side, we have x squared over four is equal to 169 minus 144, let's see, 69 minus 44 is 25.
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Pythagorean theorem with right triangle.mp3
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I'm gonna try to solve for x, that's the whole goal here. So subtracting 144 from both sides. And what do we get? On the left-hand side, we have x squared over four is equal to 169 minus 144, let's see, 69 minus 44 is 25. So this is going to be equal to 25. We can multiply both sides by four to isolate the x squared. And so we get x squared is equal to 25 times four is equal to 100.
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Pythagorean theorem with right triangle.mp3
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On the left-hand side, we have x squared over four is equal to 169 minus 144, let's see, 69 minus 44 is 25. So this is going to be equal to 25. We can multiply both sides by four to isolate the x squared. And so we get x squared is equal to 25 times four is equal to 100. Now if we were just looking at this purely mathematically, you'd say, oh, x could be positive or negative 10. But since we're dealing with distances, we know that we want the positive value of it. So x is equal to the principal root of 100, which is equal to positive 10.
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Pythagorean theorem with right triangle.mp3
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And so we get x squared is equal to 25 times four is equal to 100. Now if we were just looking at this purely mathematically, you'd say, oh, x could be positive or negative 10. But since we're dealing with distances, we know that we want the positive value of it. So x is equal to the principal root of 100, which is equal to positive 10. So there you have it, we have solved for x. This distance right here, the whole thing, the whole thing is going to be equal to 10. Half of that is going to be five.
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Pythagorean theorem with right triangle.mp3
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So x is equal to the principal root of 100, which is equal to positive 10. So there you have it, we have solved for x. This distance right here, the whole thing, the whole thing is going to be equal to 10. Half of that is going to be five. So if we just looked at this length right over here, I'm doing that in the same color, let me see. So this length right over here, that's going to be five, and indeed, five squared plus 12 squared, that's 25 plus 144, is 169, 13 squared. So the key realization here is, isosceles triangle, the altitude splits it into two congruent right triangles.
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Pythagorean theorem with right triangle.mp3
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Half of that is going to be five. So if we just looked at this length right over here, I'm doing that in the same color, let me see. So this length right over here, that's going to be five, and indeed, five squared plus 12 squared, that's 25 plus 144, is 169, 13 squared. So the key realization here is, isosceles triangle, the altitude splits it into two congruent right triangles. And so it also splits this base into two. So this is x over two, and this is x over two. And we use that information and the Pythagorean theorem to solve for x.
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Pythagorean theorem with right triangle.mp3
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So this is a diameter, that's what they're telling us. That's SE, and they say, what is the measure of angle ISE? So what we care about, the measure of angle ISE, I-S-E. So we're trying to figure out this angle right over there. And like always, I encourage you to pause the video and see if you can work through it yourself. So there's a bunch of ways that we can actually tackle this problem. The first one that jumps out at me is, you know, there's a bunch of triangles here, and we can use the fact that the angles, the interior angles of a triangle add up to 180 degrees.
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Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3
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So we're trying to figure out this angle right over there. And like always, I encourage you to pause the video and see if you can work through it yourself. So there's a bunch of ways that we can actually tackle this problem. The first one that jumps out at me is, you know, there's a bunch of triangles here, and we can use the fact that the angles, the interior angles of a triangle add up to 180 degrees. So we could look at this triangle right over here. And so we know one of the angles already. We know this angle has a measure of 27 degrees.
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Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3
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The first one that jumps out at me is, you know, there's a bunch of triangles here, and we can use the fact that the angles, the interior angles of a triangle add up to 180 degrees. So we could look at this triangle right over here. And so we know one of the angles already. We know this angle has a measure of 27 degrees. If we could figure out this angle right over here, this is what would be angle SIE, then if we know two angles, two interior angles of a triangle, we can figure out the third. And this one, SIE, we can figure it out because it's supplementary to this 61 degree angle. So this angle right over here is just going to be this angle plus the 61 degree angle is going to be equal to 180 degrees because they are supplementary.
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Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3
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We know this angle has a measure of 27 degrees. If we could figure out this angle right over here, this is what would be angle SIE, then if we know two angles, two interior angles of a triangle, we can figure out the third. And this one, SIE, we can figure it out because it's supplementary to this 61 degree angle. So this angle right over here is just going to be this angle plus the 61 degree angle is going to be equal to 180 degrees because they are supplementary. Or we could say that this angle right over here is going to be 180 minus 61. So what is that going to be? 180 minus 60 would be 120, and then minus one would be, this is 119 degrees.
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Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3
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So this angle right over here is just going to be this angle plus the 61 degree angle is going to be equal to 180 degrees because they are supplementary. Or we could say that this angle right over here is going to be 180 minus 61. So what is that going to be? 180 minus 60 would be 120, and then minus one would be, this is 119 degrees. And so this angle that we're trying to figure out, this angle plus the 119 degrees, plus the 27 degrees, is going to be equal to 180 degrees. Or we could say that this angle is going to be 180 minus 119 minus 27, which is going to be equal to, so let's see, 180 minus 119 is 61, and then 61 minus 27 is going to be 34. So there you have it.
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Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3
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180 minus 60 would be 120, and then minus one would be, this is 119 degrees. And so this angle that we're trying to figure out, this angle plus the 119 degrees, plus the 27 degrees, is going to be equal to 180 degrees. Or we could say that this angle is going to be 180 minus 119 minus 27, which is going to be equal to, so let's see, 180 minus 119 is 61, and then 61 minus 27 is going to be 34. So there you have it. Measure of angle ISE is 34 degrees. Now I mentioned that there is multiple ways that we could figure this out. Let me do it one more way.
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Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3
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So there you have it. Measure of angle ISE is 34 degrees. Now I mentioned that there is multiple ways that we could figure this out. Let me do it one more way. So let me unwind everything that I just wrote. We already figured out the answer, but I want to show you that there's multiple ways that we can tackle this. So ISE is still the thing that we want to figure out.
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Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3
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Let me do it one more way. So let me unwind everything that I just wrote. We already figured out the answer, but I want to show you that there's multiple ways that we can tackle this. So ISE is still the thing that we want to figure out. Another way that we could approach it is, well, we know we have some angles, inscribed angles on this circle, and we know that if an inscribed angle intercepts the diameter, then it's going to be a right angle. It's going to be a 90 degree angle. So this angle right over here is a 90 degree angle, and we can use that information to figure out this angle, and we could also use that information if we look at a different, so if we look at this triangle, we could use 90 plus 61 plus this angle is going to be equal to 180 degrees.
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Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3
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So ISE is still the thing that we want to figure out. Another way that we could approach it is, well, we know we have some angles, inscribed angles on this circle, and we know that if an inscribed angle intercepts the diameter, then it's going to be a right angle. It's going to be a 90 degree angle. So this angle right over here is a 90 degree angle, and we can use that information to figure out this angle, and we could also use that information if we look at a different, so if we look at this triangle, we could use 90 plus 61 plus this angle is going to be equal to 180 degrees. So this angle right over here, another way to think about it, it's going to be 180 minus 90 minus 61, which is equal to, 180 minus 90 is 90, minus 61 is 29 degrees. So this one right over here is 29 degrees. 29 degrees.
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Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3
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So this angle right over here is a 90 degree angle, and we can use that information to figure out this angle, and we could also use that information if we look at a different, so if we look at this triangle, we could use 90 plus 61 plus this angle is going to be equal to 180 degrees. So this angle right over here, another way to think about it, it's going to be 180 minus 90 minus 61, which is equal to, 180 minus 90 is 90, minus 61 is 29 degrees. So this one right over here is 29 degrees. 29 degrees. And then we could look at this larger triangle. We could look at this larger triangle right over here to figure out this entire angle. To figure out this entire angle.
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Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3
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29 degrees. And then we could look at this larger triangle. We could look at this larger triangle right over here to figure out this entire angle. To figure out this entire angle. If we know this entire angle, you subtract 29, then you figure out angle ISE. And so this large, or what I've depicted as this kind of, this magenta, this measure right over here, of that angle, plus 90 degrees, plus 27 degrees is going to be equal to 180, because they're the interior angles of triangle SLE. So this angle right over here is going to be 180 minus 61 minus 27.
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Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3
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To figure out this entire angle. If we know this entire angle, you subtract 29, then you figure out angle ISE. And so this large, or what I've depicted as this kind of, this magenta, this measure right over here, of that angle, plus 90 degrees, plus 27 degrees is going to be equal to 180, because they're the interior angles of triangle SLE. So this angle right over here is going to be 180 minus 61 minus 27. Sorry, not minus 61, minus 90. Minus 90. It's 180 minus 90 minus 27 is going to give us this angle right over here, because the three angles add up to 180.
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Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3
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So this angle right over here is going to be 180 minus 61 minus 27. Sorry, not minus 61, minus 90. Minus 90. It's 180 minus 90 minus 27 is going to give us this angle right over here, because the three angles add up to 180. So minus 90 minus 27, which is equal to, so 180 minus 90 is 90. 90 minus 27, 90 minus 27 is 63. 63 degrees.
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Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3
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It's 180 minus 90 minus 27 is going to give us this angle right over here, because the three angles add up to 180. So minus 90 minus 27, which is equal to, so 180 minus 90 is 90. 90 minus 27, 90 minus 27 is 63. 63 degrees. So this large one over here is 63 degrees, and the smaller one is 29 degrees. And so angle ISE, which we set out to figure out, is going to be 63 degrees minus the 29 degrees. So 63 minus 29 is once again equal to 34 degrees.
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Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3
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63 degrees. So this large one over here is 63 degrees, and the smaller one is 29 degrees. And so angle ISE, which we set out to figure out, is going to be 63 degrees minus the 29 degrees. So 63 minus 29 is once again equal to 34 degrees. So the way I did it just now, a little bit harder. It really depends what jumps out at you. The first way I tackled it, it does seem a little bit easier, a little bit clearer, but it's good to see these different things, and at least here, we use this idea of an inscribed angle that intercepts a diameter.
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Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3
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So 63 minus 29 is once again equal to 34 degrees. So the way I did it just now, a little bit harder. It really depends what jumps out at you. The first way I tackled it, it does seem a little bit easier, a little bit clearer, but it's good to see these different things, and at least here, we use this idea of an inscribed angle that intercepts a diameter. And if you say, hey, how do we know? I mean, we prove it in other videos, but it comes straight out of the idea that an inscribed angle, the measure of an inscribed angle is going to be half of the measure of the arc that it intercepts. And notice, it's intercepting an arc that has a measure of 180 degrees.
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Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3
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Now, we call the longest side of a right triangle, we call that side, and you can either view it as the longest side of the right triangle, or the side opposite the 90 degree angle, it is called the hypotenuse. It's a very fancy word for a fairly simple idea, just the longest side of a right triangle, or the side opposite the 90 degree angle. And it's just good to know that because someone might say hypotenuse, like, oh, they're just talking about this side right here, the side longest, the side opposite the 90 degree angle. Now, what I want to do in this video is prove a relationship, a very famous relationship, and you might see where this is going, a very famous relationship between the lengths of the sides of a right triangle. So let's say that the length of AC, so uppercase A, uppercase C, let's call that length lowercase a, let's call the length of BC, lowercase b right over here, I'll use uppercases for points, lowercases for lengths, and let's call the length of the hypotenuse, the length of AB, let's call that C. And let's see if we can come up with a relationship between A, B, and C. And to do that, I'm first going to construct another line, or another segment, I should say, between C and the hypotenuse, and I'm going to construct it so that they intersect at a right angle. And you can always do that, and we'll call this point right over here, we'll call this point capital D. And if you're wondering, how can you always do that, you can imagine rotating this entire triangle like this, and this isn't a rigorous proof, but it just kind of gives you the general idea of how you can always construct a point like this. So if I've rotated it around, so now our hypotenuse, we're now sitting on our hypotenuse, this is now point B, this is point A, and if I rotated the whole thing all the way around, this is point C, you can imagine just dropping a rock from point C, maybe with a string attached, and it would hit the hypotenuse at a right angle.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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Now, what I want to do in this video is prove a relationship, a very famous relationship, and you might see where this is going, a very famous relationship between the lengths of the sides of a right triangle. So let's say that the length of AC, so uppercase A, uppercase C, let's call that length lowercase a, let's call the length of BC, lowercase b right over here, I'll use uppercases for points, lowercases for lengths, and let's call the length of the hypotenuse, the length of AB, let's call that C. And let's see if we can come up with a relationship between A, B, and C. And to do that, I'm first going to construct another line, or another segment, I should say, between C and the hypotenuse, and I'm going to construct it so that they intersect at a right angle. And you can always do that, and we'll call this point right over here, we'll call this point capital D. And if you're wondering, how can you always do that, you can imagine rotating this entire triangle like this, and this isn't a rigorous proof, but it just kind of gives you the general idea of how you can always construct a point like this. So if I've rotated it around, so now our hypotenuse, we're now sitting on our hypotenuse, this is now point B, this is point A, and if I rotated the whole thing all the way around, this is point C, you can imagine just dropping a rock from point C, maybe with a string attached, and it would hit the hypotenuse at a right angle. So that's all we did here to establish segment CD, where we put our point D right over there. And the reason why I did that is now we can do all sorts of interesting relationships between similar triangles, because we have three triangles here. We have triangle ADC, we have triangle DBC, and then we have the larger original triangle, and we can hopefully establish similarity between those triangles.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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So if I've rotated it around, so now our hypotenuse, we're now sitting on our hypotenuse, this is now point B, this is point A, and if I rotated the whole thing all the way around, this is point C, you can imagine just dropping a rock from point C, maybe with a string attached, and it would hit the hypotenuse at a right angle. So that's all we did here to establish segment CD, where we put our point D right over there. And the reason why I did that is now we can do all sorts of interesting relationships between similar triangles, because we have three triangles here. We have triangle ADC, we have triangle DBC, and then we have the larger original triangle, and we can hopefully establish similarity between those triangles. And first I'll show you that ADC is similar to the larger one, because both of them have a right angle. ADC has a right angle right over here. Clearly if this angle is 90 degrees, then this angle is going to be 90 degrees as well.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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We have triangle ADC, we have triangle DBC, and then we have the larger original triangle, and we can hopefully establish similarity between those triangles. And first I'll show you that ADC is similar to the larger one, because both of them have a right angle. ADC has a right angle right over here. Clearly if this angle is 90 degrees, then this angle is going to be 90 degrees as well. They're supplementary. They have to add up to 180. And so they both have a right angle in them.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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Clearly if this angle is 90 degrees, then this angle is going to be 90 degrees as well. They're supplementary. They have to add up to 180. And so they both have a right angle in them. So the smaller one has a right angle, the larger one clearly has a right angle. That's where we started from. And they also both share this angle right over here.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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And so they both have a right angle in them. So the smaller one has a right angle, the larger one clearly has a right angle. That's where we started from. And they also both share this angle right over here. Angle DAC or BAC, however you want to refer to it. So we can actually write down that triangle. I'm going to start with the smaller one.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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And they also both share this angle right over here. Angle DAC or BAC, however you want to refer to it. So we can actually write down that triangle. I'm going to start with the smaller one. ADC, and maybe I'll shade it in right over here. So this is the triangle we're talking about. Triangle ADC, and I went from the blue angle to the right angle to the unlabeled angle from the point of view of triangle ADC.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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I'm going to start with the smaller one. ADC, and maybe I'll shade it in right over here. So this is the triangle we're talking about. Triangle ADC, and I went from the blue angle to the right angle to the unlabeled angle from the point of view of triangle ADC. This right angle isn't applying to that right over there. It's applying to the larger triangle. So we can say triangle ADC is similar to triangle.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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Triangle ADC, and I went from the blue angle to the right angle to the unlabeled angle from the point of view of triangle ADC. This right angle isn't applying to that right over there. It's applying to the larger triangle. So we can say triangle ADC is similar to triangle. Once again, you want to start at the blue angle, A. Then we went to the right angle, so we want to go to the right angle again. So it's ACB.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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So we can say triangle ADC is similar to triangle. Once again, you want to start at the blue angle, A. Then we went to the right angle, so we want to go to the right angle again. So it's ACB. And because they're similar, we can set up a relationship between the ratios of their sides. For example, we know the ratio of corresponding sides are going to, well, in general for a similar triangle, we know that the ratio of the corresponding sides are going to be a constant. So we could take the ratio of the hypotenuse of this side, of the smaller triangle.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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So it's ACB. And because they're similar, we can set up a relationship between the ratios of their sides. For example, we know the ratio of corresponding sides are going to, well, in general for a similar triangle, we know that the ratio of the corresponding sides are going to be a constant. So we could take the ratio of the hypotenuse of this side, of the smaller triangle. So the hypotenuse is AC. So AC over the hypotenuse over the larger one, which is AB. AC over AB is going to be the same thing as AD, as one of the legs.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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So we could take the ratio of the hypotenuse of this side, of the smaller triangle. So the hypotenuse is AC. So AC over the hypotenuse over the larger one, which is AB. AC over AB is going to be the same thing as AD, as one of the legs. AD, just to show that I'm just taking corresponding points on both similar triangles. This is AD over AC. You can look at these triangles yourself and show, look, AD, point AD is between the blue angle and the right angle.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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AC over AB is going to be the same thing as AD, as one of the legs. AD, just to show that I'm just taking corresponding points on both similar triangles. This is AD over AC. You can look at these triangles yourself and show, look, AD, point AD is between the blue angle and the right angle. Sorry, side AD is between the blue angle and the right angle. Side AC is between the blue angle and the right angle on the larger triangle. So both of these are from the larger triangle.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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You can look at these triangles yourself and show, look, AD, point AD is between the blue angle and the right angle. Sorry, side AD is between the blue angle and the right angle. Side AC is between the blue angle and the right angle on the larger triangle. So both of these are from the larger triangle. These are the corresponding sides on the smaller triangle. And if that is confusing looking at them visually, as long as we wrote our similarity statement correctly, you can just find the corresponding points. AC corresponds to AB on the larger triangle.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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So both of these are from the larger triangle. These are the corresponding sides on the smaller triangle. And if that is confusing looking at them visually, as long as we wrote our similarity statement correctly, you can just find the corresponding points. AC corresponds to AB on the larger triangle. AD on the smaller triangle corresponds to AC on the larger triangle. And we know that AC, we can rewrite that as lowercase a. AC is lowercase a. We don't have any label for AD or for AB.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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AC corresponds to AB on the larger triangle. AD on the smaller triangle corresponds to AC on the larger triangle. And we know that AC, we can rewrite that as lowercase a. AC is lowercase a. We don't have any label for AD or for AB. Sorry, we do have a label for AB. That is C right over here. We don't have a label for AD.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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We don't have any label for AD or for AB. Sorry, we do have a label for AB. That is C right over here. We don't have a label for AD. So let's just call that lowercase d. So lowercase d applies to that part right over there. C applies to that entire part right over there. And then we'll call DB, let's call that length E. That will just make things a little bit simpler for us.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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We don't have a label for AD. So let's just call that lowercase d. So lowercase d applies to that part right over there. C applies to that entire part right over there. And then we'll call DB, let's call that length E. That will just make things a little bit simpler for us. So AD we'll just call d. And so we have a over c is equal to d over a. If we cross multiply, you have a times a, which is a squared, is equal to c times d, which is cd. So that's a little bit of an interesting result.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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And then we'll call DB, let's call that length E. That will just make things a little bit simpler for us. So AD we'll just call d. And so we have a over c is equal to d over a. If we cross multiply, you have a times a, which is a squared, is equal to c times d, which is cd. So that's a little bit of an interesting result. Let's see what we can do with the other triangle right over here. So this triangle right over here. So once again, it has a right angle.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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So that's a little bit of an interesting result. Let's see what we can do with the other triangle right over here. So this triangle right over here. So once again, it has a right angle. The larger one has a right angle. And they both share this angle right over here. So by angle-angle similarity, the two triangles are going to be similar.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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So once again, it has a right angle. The larger one has a right angle. And they both share this angle right over here. So by angle-angle similarity, the two triangles are going to be similar. So we can say triangle BDC. We went from pink to right to not labeled. So triangle BDC is similar to triangle.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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So by angle-angle similarity, the two triangles are going to be similar. So we can say triangle BDC. We went from pink to right to not labeled. So triangle BDC is similar to triangle. Now we're going to look at the larger triangle. We're going to start at the pink angle, B. Now we go to the right angle, CA.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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So triangle BDC is similar to triangle. Now we're going to look at the larger triangle. We're going to start at the pink angle, B. Now we go to the right angle, CA. From pink angle to right angle to non-labeled angle, at least from the point of view here. We labeled it before with that blue. So now let's set up some type of a relationship here.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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Now we go to the right angle, CA. From pink angle to right angle to non-labeled angle, at least from the point of view here. We labeled it before with that blue. So now let's set up some type of a relationship here. We can say that the ratio on the smaller triangle, BC over BA. Once again, we're taking the hypotenuses of both of them. So BC over BA is going to be equal to BD.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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So now let's set up some type of a relationship here. We can say that the ratio on the smaller triangle, BC over BA. Once again, we're taking the hypotenuses of both of them. So BC over BA is going to be equal to BD. Let me do this in another color. So this is one of the legs. The way I drew it is the shorter legs.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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So BC over BA is going to be equal to BD. Let me do this in another color. So this is one of the legs. The way I drew it is the shorter legs. BD over BC. I'm just taking the corresponding vertices. Over BC.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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The way I drew it is the shorter legs. BD over BC. I'm just taking the corresponding vertices. Over BC. And once again, we know BC is the same thing as lowercase b. BA is lowercase c. And then BD we defined as lowercase e. This is lowercase e. We can cross multiply here and we get B times B. I've mentioned this in many videos. Cross multiplying is really the same thing as multiplying both sides by both denominators. B times B is B squared is equal to CE.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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Over BC. And once again, we know BC is the same thing as lowercase b. BA is lowercase c. And then BD we defined as lowercase e. This is lowercase e. We can cross multiply here and we get B times B. I've mentioned this in many videos. Cross multiplying is really the same thing as multiplying both sides by both denominators. B times B is B squared is equal to CE. And now we can do something kind of interesting. We can add these two statements. Let me rewrite the statement down here.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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B times B is B squared is equal to CE. And now we can do something kind of interesting. We can add these two statements. Let me rewrite the statement down here. So B squared is equal to CE. So if we add the left-hand sides, we get A squared plus B squared is equal to CD plus CE. And then we have a C in both of these terms so we can factor it out.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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Let me rewrite the statement down here. So B squared is equal to CE. So if we add the left-hand sides, we get A squared plus B squared is equal to CD plus CE. And then we have a C in both of these terms so we can factor it out. This is going to be equal to C times D plus E. Close the parentheses. Now what is D plus E? D is this length.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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And then we have a C in both of these terms so we can factor it out. This is going to be equal to C times D plus E. Close the parentheses. Now what is D plus E? D is this length. E is this length. So D plus E is actually going to be C as well. So this is going to be C. So you have C times C which is just the same thing as C squared.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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D is this length. E is this length. So D plus E is actually going to be C as well. So this is going to be C. So you have C times C which is just the same thing as C squared. So now we have an interesting relationship. We have that A squared plus B squared is equal to C squared. Let me rewrite that.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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So this is going to be C. So you have C times C which is just the same thing as C squared. So now we have an interesting relationship. We have that A squared plus B squared is equal to C squared. Let me rewrite that. A squared. I'll do that in an arbitrary new color. I deleted that by accident.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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Let me rewrite that. A squared. I'll do that in an arbitrary new color. I deleted that by accident. So let me rewrite it. So we've just established that A squared plus B squared is equal to C squared. And this is just an arbitrary right triangle.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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I deleted that by accident. So let me rewrite it. So we've just established that A squared plus B squared is equal to C squared. And this is just an arbitrary right triangle. This is true for any two right triangles. We've just established that the sum of the squares of each of the legs is equal to the square of the hypotenuse. And this is probably what's easily one of the most famous theorems in mathematics named for Pythagoras.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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And this is just an arbitrary right triangle. This is true for any two right triangles. We've just established that the sum of the squares of each of the legs is equal to the square of the hypotenuse. And this is probably what's easily one of the most famous theorems in mathematics named for Pythagoras. Not clear if he's the first person to establish this, but it's called the Pythagorean theorem. And it's really the basis of, well, not all of geometry, but a lot of the geometry that we're going to do. And it forms the basis of a lot of the trigonometry we're going to do.
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Pythagorean theorem proof using similarity Geometry Khan Academy.mp3
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And I thought we would use this to really just get some practice with line and angle proofs. And what's neat about this, this even uses translations and transformations as ways to actually prove things. So let's look at what they're telling us. So it says line AB and line DE are parallel lines. All right. Perform a translation that proves corresponding angles are always equal and select the option which explains the proof. All right.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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So it says line AB and line DE are parallel lines. All right. Perform a translation that proves corresponding angles are always equal and select the option which explains the proof. All right. So let's see what they have down here. So they say perform a translation that proves corresponding angles are always equal. And then select the option that explains the proof.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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All right. So let's see what they have down here. So they say perform a translation that proves corresponding angles are always equal. And then select the option that explains the proof. So they've picked two corresponding angles here. And so you see this is kind of the bottom left angle, this phi. And then you have theta right here, is the bottom left angle down here.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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And then select the option that explains the proof. So they've picked two corresponding angles here. And so you see this is kind of the bottom left angle, this phi. And then you have theta right here, is the bottom left angle down here. So these are corresponding angles. Line FB is a transversal. And they already told us that line AB, and what did they call it, did they call it DE?
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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And then you have theta right here, is the bottom left angle down here. So these are corresponding angles. Line FB is a transversal. And they already told us that line AB, and what did they call it, did they call it DE? And line DE are indeed parallel lines. So we want to prove that these two things, that the measure of these two angles are equal. So there's many ways that you can do this geometrically.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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And they already told us that line AB, and what did they call it, did they call it DE? And line DE are indeed parallel lines. So we want to prove that these two things, that the measure of these two angles are equal. So there's many ways that you can do this geometrically. And we do that in many Khan Academy videos. But this one, they offer us the option of translating, of doing a translation. So let's see what that is.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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So there's many ways that you can do this geometrically. And we do that in many Khan Academy videos. But this one, they offer us the option of translating, of doing a translation. So let's see what that is. So I press the Translate button. And when I move this around, notice it essentially translates these four points, which has the effect of translating this entire intersection here. So this point where my mouse is right now, that is the point D. And I'm translating it around.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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So let's see what that is. So I press the Translate button. And when I move this around, notice it essentially translates these four points, which has the effect of translating this entire intersection here. So this point where my mouse is right now, that is the point D. And I'm translating it around. If I move that over to B, what it shows is, because under a translation, the angle measures shouldn't change. So when I did that, this angle, so it's down here. Theta is the measure of angle CDF.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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So this point where my mouse is right now, that is the point D. And I'm translating it around. If I move that over to B, what it shows is, because under a translation, the angle measures shouldn't change. So when I did that, this angle, so it's down here. Theta is the measure of angle CDF. And so when you move it over here, this right over here should be the same. This angle's measure is the same as CDF. I'm just translating it.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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Theta is the measure of angle CDF. And so when you move it over here, this right over here should be the same. This angle's measure is the same as CDF. I'm just translating it. And when you move it over here, you see, look, that's the same exact measure as phi. So this is one way to think about it. I just translated the point D to B.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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I'm just translating it. And when you move it over here, you see, look, that's the same exact measure as phi. So this is one way to think about it. I just translated the point D to B. And then it really just translated angle CDF over angle ABD to show that these have the same measure, or at least to feel good about the idea of them having the same measure. So let's see which choices describe that. So let me, I'm having trouble operating my mouse.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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I just translated the point D to B. And then it really just translated angle CDF over angle ABD to show that these have the same measure, or at least to feel good about the idea of them having the same measure. So let's see which choices describe that. So let me, I'm having trouble operating my mouse. All right. The translation mapping point F to point D. So point F to point D. We didn't map point F to point D. So this is already looking suspect. Produces a new line, which is a bisector of segment DB.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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So let me, I'm having trouble operating my mouse. All right. The translation mapping point F to point D. So point F to point D. We didn't map point F to point D. So this is already looking suspect. Produces a new line, which is a bisector of segment DB. OK, this doesn't seem anything like what I just did. So I'm just going to move on to the next one. Since the image of a line under translation is parallel to the original line, that's true.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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Produces a new line, which is a bisector of segment DB. OK, this doesn't seem anything like what I just did. So I'm just going to move on to the next one. Since the image of a line under translation is parallel to the original line, that's true. The translation that mapped point D to point B. That's what I did right over here. Maps angle CDF to ABD.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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Since the image of a line under translation is parallel to the original line, that's true. The translation that mapped point D to point B. That's what I did right over here. Maps angle CDF to ABD. And that's what I did. I mapped angle CDF to angle ABD. That's exactly what I did right over there.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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Maps angle CDF to ABD. And that's what I did. I mapped angle CDF to angle ABD. That's exactly what I did right over there. So this is to ABD. Translations preserve angle measures, so theta is equal to phi. Yeah, that looks pretty good.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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That's exactly what I did right over there. So this is to ABD. Translations preserve angle measures, so theta is equal to phi. Yeah, that looks pretty good. The translation that maps point D to E. I didn't do that. I didn't take point D and move it over to E like that. That didn't really help me.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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Yeah, that looks pretty good. The translation that maps point D to E. I didn't do that. I didn't take point D and move it over to E like that. That didn't really help me. Let's just keep reading it just to make sure. Produces a parallelogram. That actually is true.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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That didn't really help me. Let's just keep reading it just to make sure. Produces a parallelogram. That actually is true. If I translate point D to point E, I have this parallelogram constructed. But it really doesn't help us establishing that phi is equal to theta. So that one I also don't feel good about.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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That actually is true. If I translate point D to point E, I have this parallelogram constructed. But it really doesn't help us establishing that phi is equal to theta. So that one I also don't feel good about. And it's good, because we felt good about the middle choice. Let's do one more of these. So they're telling us that line AOB.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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So that one I also don't feel good about. And it's good, because we felt good about the middle choice. Let's do one more of these. So they're telling us that line AOB. And they could have just said line AB, but I guess they wanted to put the O in there to show that point O is on that line, that AOB are collinear. And COD is our straight lines. All right, fair enough.
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Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3
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