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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... | Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3 |
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 ... | Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3 |
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 an... | Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3 |
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... | Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3 |
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 ... | Calculating angle measures to verify congruence Congruence High school geometry Khan Academy.mp3 |
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 ... | Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3 |
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 righ... | Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3 |
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 oppos... | Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3 |
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. ... | Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3 |
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 ov... | Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3 |
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... | Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3 |
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 ... | Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3 |
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 ... | Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3 |
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 ... | Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3 |
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 ... | Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3 |
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 ... | Angle to aim to get alien Basic trigonometry Trigonometry Khan Academy.mp3 |
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 hav... | Pythagorean theorem with right triangle.mp3 |
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 sam... | Pythagorean theorem with right triangle.mp3 |
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 o... | Pythagorean theorem with right triangle.mp3 |
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 ove... | Pythagorean theorem with right triangle.mp3 |
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 equ... | Pythagorean theorem with right triangle.mp3 |
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 fr... | Pythagorean theorem with right triangle.mp3 |
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. A... | Pythagorean theorem with right triangle.mp3 |
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 mathe... | Pythagorean theorem with right triangle.mp3 |
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 equ... | Pythagorean theorem with right triangle.mp3 |
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 ... | Pythagorean theorem with right triangle.mp3 |
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 triang... | Pythagorean theorem with right triangle.mp3 |
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 your... | Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3 |
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 ... | Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3 |
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 degree... | Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3 |
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 t... | Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3 |
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... | Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3 |
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,... | Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3 |
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 ... | Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3 |
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, inscr... | Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3 |
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... | Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3 |
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, anot... | Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3 |
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... | Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3 |
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 a... | Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3 |
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. ... | Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3 |
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, w... | Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3 |
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 ou... | Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3 |
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 inscrib... | Inscribed shapes angle subtended by diameter High School Math Khan Academy.mp3 |
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 oppos... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 cal... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 righ... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 ang... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 t... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 sta... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 bl... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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. ... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 angl... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 c... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 ... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 triangle... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 co... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 ... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 A... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 ove... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 resu... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 ... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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. | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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. | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 ty... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 ... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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. | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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. | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 denominat... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 te... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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? ... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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. | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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. | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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. | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 hy... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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... | Pythagorean theorem proof using similarity Geometry Khan Academy.mp3 |
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 tra... | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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... | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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 th... | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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 ... | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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 o... | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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. Bu... | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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,... | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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 ov... | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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 ... | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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 th... | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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... | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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 tr... | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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 g... | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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 ov... | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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. | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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. | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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. | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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. | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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. | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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... | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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 C... | Line and angle proofs exercise Angles and intersecting lines Geometry Khan Academy.mp3 |
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