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And so from this point, you're gonna go straight as a right angle, and instead of going down seven, you're gonna go to the right seven. So you're gonna go to the right seven, just like this. And so the point I, or the corresponding point in the image after the rotation, is going to be right over here. So that green line, let me draw the hypotenuse now, it's gonna look, it's gonna look like, whoops, I wanted to do that in a different color, I wanted to do that in the green, I have trouble changing colors. All right, so there you go. It's gonna look, it's gonna look like that. So your new point I, if you rotate this triangle, this right triangle 90 degrees, your new point I, maybe I shouldn't say, I'll call it I prime, which is, what is the image of this point after I've done the 90 degree rotation, is going to be right over here.
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Points after rotation Transformations Geometry Khan Academy.mp3
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So that green line, let me draw the hypotenuse now, it's gonna look, it's gonna look like, whoops, I wanted to do that in a different color, I wanted to do that in the green, I have trouble changing colors. All right, so there you go. It's gonna look, it's gonna look like that. So your new point I, if you rotate this triangle, this right triangle 90 degrees, your new point I, maybe I shouldn't say, I'll call it I prime, which is, what is the image of this point after I've done the 90 degree rotation, is going to be right over here. And now we can do that for each of the points. We can do that for N here. Let's draw a right triangle.
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Points after rotation Transformations Geometry Khan Academy.mp3
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So your new point I, if you rotate this triangle, this right triangle 90 degrees, your new point I, maybe I shouldn't say, I'll call it I prime, which is, what is the image of this point after I've done the 90 degree rotation, is going to be right over here. And now we can do that for each of the points. We can do that for N here. Let's draw a right triangle. Let's draw a right triangle. And I could do it a bunch of different ways. I could draw it, I could draw a right triangle like this.
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Points after rotation Transformations Geometry Khan Academy.mp3
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Let's draw a right triangle. Let's draw a right triangle. And I could do it a bunch of different ways. I could draw it, I could draw a right triangle like this. Let me draw it like this. So that's one side of my right triangle. Actually, let me draw the hypotenuse first.
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Points after rotation Transformations Geometry Khan Academy.mp3
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I could draw it, I could draw a right triangle like this. Let me draw it like this. So that's one side of my right triangle. Actually, let me draw the hypotenuse first. So I have the hypotenuse, connects the origin to my point, just like that. And then I could, I could either draw it up here, or I could draw it down here, like this. I could draw it like this.
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Points after rotation Transformations Geometry Khan Academy.mp3
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Actually, let me draw the hypotenuse first. So I have the hypotenuse, connects the origin to my point, just like that. And then I could, I could either draw it up here, or I could draw it down here, like this. I could draw it like this. So if you rotate this 90 degrees, if you rotate this 90 degrees, this side, which is seven units long, and we're going seven below the origin, it's now going to be seven to the right, right? If I rotate it by 90 degrees, it's gonna be right over here. It's gonna be here.
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Points after rotation Transformations Geometry Khan Academy.mp3
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I could draw it like this. So if you rotate this 90 degrees, if you rotate this 90 degrees, this side, which is seven units long, and we're going seven below the origin, it's now going to be seven to the right, right? If I rotate it by 90 degrees, it's gonna be right over here. It's gonna be here. And this side, which has length, this, let me switch colors. This side, which has length two, it forms a right angle. So we're gonna form a right angle and have length two right over here.
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Points after rotation Transformations Geometry Khan Academy.mp3
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It's gonna be here. And this side, which has length, this, let me switch colors. This side, which has length two, it forms a right angle. So we're gonna form a right angle and have length two right over here. And so your, the image of point N is going to be, let me get the right color, it is going to be just like that. It is going to be, the image of point N is going to be right here. I'll call that N prime.
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Points after rotation Transformations Geometry Khan Academy.mp3
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So we're gonna form a right angle and have length two right over here. And so your, the image of point N is going to be, let me get the right color, it is going to be just like that. It is going to be, the image of point N is going to be right here. I'll call that N prime. So we know where the new, where the image of I is, the image of N is, so now we just have to think about where does the image of P sit? And once again, we can do our little right triangle idea. So let's draw a right triangle.
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Points after rotation Transformations Geometry Khan Academy.mp3
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I'll call that N prime. So we know where the new, where the image of I is, the image of N is, so now we just have to think about where does the image of P sit? And once again, we can do our little right triangle idea. So let's draw a right triangle. So just like that, I can draw that side, and I can do this side right over here. So if I were to rotate, if I were to focus on this, let me do this in a color I haven't. So if I were to focus on, I've already used that color.
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Points after rotation Transformations Geometry Khan Academy.mp3
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So let's draw a right triangle. So just like that, I can draw that side, and I can do this side right over here. So if I were to rotate, if I were to focus on this, let me do this in a color I haven't. So if I were to focus on, I've already used that color. If I were to focus on this right over here, and if I were to rotate it by 90 degrees, instead of going two to the right, it's gonna go two straight up. If I rotate this by 90 degrees, it's gonna be just like this. Now, this side on, let me pick another color.
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Points after rotation Transformations Geometry Khan Academy.mp3
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So if I were to focus on, I've already used that color. If I were to focus on this right over here, and if I were to rotate it by 90 degrees, instead of going two to the right, it's gonna go two straight up. If I rotate this by 90 degrees, it's gonna be just like this. Now, this side on, let me pick another color. This side right over here forms a right angle, and it has a length of three. So we're gonna form a right angle here and have a length of three. And just like that, we know where the image of P is going to be.
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Points after rotation Transformations Geometry Khan Academy.mp3
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Now, this side on, let me pick another color. This side right over here forms a right angle, and it has a length of three. So we're gonna form a right angle here and have a length of three. And just like that, we know where the image of P is going to be. It is going to be, it is going to be right over here. So this is P prime. And I know it's kind of confusing, but now we just have to, we now have to connect the P prime, the I prime, and the N prime to figure out what the image of my triangle is after rotation.
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Points after rotation Transformations Geometry Khan Academy.mp3
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And just like that, we know where the image of P is going to be. It is going to be, it is going to be right over here. So this is P prime. And I know it's kind of confusing, but now we just have to, we now have to connect the P prime, the I prime, and the N prime to figure out what the image of my triangle is after rotation. So let me do that. So if I connect these two, I get that. If I connect these two, I get that.
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Points after rotation Transformations Geometry Khan Academy.mp3
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And I know it's kind of confusing, but now we just have to, we now have to connect the P prime, the I prime, and the N prime to figure out what the image of my triangle is after rotation. So let me do that. So if I connect these two, I get that. If I connect these two, I get that. And if I connect these two, I connect those two, I have that. And there you have it. I have the image.
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Points after rotation Transformations Geometry Khan Academy.mp3
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If I connect these two, I get that. And if I connect these two, I connect those two, I have that. And there you have it. I have the image. And now I just have to input it on the actual problem. So let's see. Point negative three comma two.
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Points after rotation Transformations Geometry Khan Academy.mp3
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I have the image. And now I just have to input it on the actual problem. So let's see. Point negative three comma two. Let me get it out. So negative three comma two is there. I have seven comma two.
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Points after rotation Transformations Geometry Khan Academy.mp3
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Point negative three comma two. Let me get it out. So negative three comma two is there. I have seven comma two. So let me put that in there. Seven comma two. And then I have seven comma seven is there.
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Points after rotation Transformations Geometry Khan Academy.mp3
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I have seven comma two. So let me put that in there. Seven comma two. And then I have seven comma seven is there. So let me get this out. So seven comma seven. And then I'll draw it here again to connect the lines.
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Points after rotation Transformations Geometry Khan Academy.mp3
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And then I have seven comma seven is there. So let me get this out. So seven comma seven. And then I'll draw it here again to connect the lines. And I'm done. I've rotated it through an angle of 90 degrees. Or negative 270 degrees, which is what they originally, they originally asked me for.
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Points after rotation Transformations Geometry Khan Academy.mp3
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Where does the point P, which has a coordinates negative six comma negative six lie? And we have three options. Inside the circle, on the circle, or outside the circle. And the key realization here is just what a circle is all about. If we have the point C, which is the center of a circle, a circle of radius six, so let me draw that radius. So let's say that is its radius. It is six units.
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Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3
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And the key realization here is just what a circle is all about. If we have the point C, which is the center of a circle, a circle of radius six, so let me draw that radius. So let's say that is its radius. It is six units. The circle will look something like this. Remember, the circle is a set of all points that are exactly six units away from that center. So that's the definition of a circle.
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Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3
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It is six units. The circle will look something like this. Remember, the circle is a set of all points that are exactly six units away from that center. So that's the definition of a circle. It's a set of all points that are exactly six units away from the center. So if, for example, P is less than six units away, it's going to be inside the circle. If it's exactly six units away, it's going to be on the circle.
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Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3
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So that's the definition of a circle. It's a set of all points that are exactly six units away from the center. So if, for example, P is less than six units away, it's going to be inside the circle. If it's exactly six units away, it's going to be on the circle. And if it's more than six units away, it's going to be outside of the circle. So the key is is let's find the distance between these two points. If the distance is less than six, inside, distance equals six, we're on the circle, distance more than six, we are outside of the circle.
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Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3
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If it's exactly six units away, it's going to be on the circle. And if it's more than six units away, it's going to be outside of the circle. So the key is is let's find the distance between these two points. If the distance is less than six, inside, distance equals six, we're on the circle, distance more than six, we are outside of the circle. So let's do that. So if we wanted to find, and there's different notations for the distance, well, I'll just write D, or I could write the distance between C and P is going to be equal to, and the distance formula comes straight out of the Pythagorean theorem, but it's going to be the square root of our change in X squared plus our change in Y squared. So what is our change in X?
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Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3
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If the distance is less than six, inside, distance equals six, we're on the circle, distance more than six, we are outside of the circle. So let's do that. So if we wanted to find, and there's different notations for the distance, well, I'll just write D, or I could write the distance between C and P is going to be equal to, and the distance formula comes straight out of the Pythagorean theorem, but it's going to be the square root of our change in X squared plus our change in Y squared. So what is our change in X? So our change in X, if we view C as our starting point and P as our end point, but we could do it either way, our change in X, our change in X is negative six minus negative one. So negative six minus negative one, and we're going to square it. So what we have inside here, that is change in X.
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Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3
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So what is our change in X? So our change in X, if we view C as our starting point and P as our end point, but we could do it either way, our change in X, our change in X is negative six minus negative one. So negative six minus negative one, and we're going to square it. So what we have inside here, that is change in X. So we're taking our change in X squared and then plus our change in Y squared. So we are going, we're going from negative three to negative six. So our change in Y is negative six minus negative three.
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Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3
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So what we have inside here, that is change in X. So we're taking our change in X squared and then plus our change in Y squared. So we are going, we're going from negative three to negative six. So our change in Y is negative six minus negative three. Negative six minus negative three, and we're going to square everything. So that is our change in Y inside the parentheses, and we're going to square it. So this is equal to, this is equal to negative six, negative six plus positive one is one way to think about it.
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Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3
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So our change in Y is negative six minus negative three. Negative six minus negative three, and we're going to square everything. So that is our change in Y inside the parentheses, and we're going to square it. So this is equal to, this is equal to negative six, negative six plus positive one is one way to think about it. So this is negative five squared, and then this is negative six plus three. So plus negative three squared. And once again, you can see our change in X is negative five.
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Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3
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So this is equal to, this is equal to negative six, negative six plus positive one is one way to think about it. So this is negative five squared, and then this is negative six plus three. So plus negative three squared. And once again, you can see our change in X is negative five. We go five lower in X, and we're going three lower in Y. So our change in Y is negative three. So this is equal to the square root of 25, square root of 25 plus nine.
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Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3
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And once again, you can see our change in X is negative five. We go five lower in X, and we're going three lower in Y. So our change in Y is negative three. So this is equal to the square root of 25, square root of 25 plus nine. Square root of 25 plus nine which is equal to the square root of 34. Now, the key is, is the square root of 34 less than six, greater than six, or equal to six? Well, we know that six is equal to the square root of 36.
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Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3
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So this is equal to the square root of 25, square root of 25 plus nine. Square root of 25 plus nine which is equal to the square root of 34. Now, the key is, is the square root of 34 less than six, greater than six, or equal to six? Well, we know that six is equal to the square root of 36. So the square root of 34 is less than the square root of 36. So I could write the square root of 34 is less than the square root of 36, and so the square root of 34 is less than six. Square root of 36 is six.
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Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3
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Well, we know that six is equal to the square root of 36. So the square root of 34 is less than the square root of 36. So I could write the square root of 34 is less than the square root of 36, and so the square root of 34 is less than six. Square root of 36 is six. And so since the distance between C and P is less than six, we are going to be on the inside of the circle. If I somehow got square root of 36 here, then we'd be on the circle. And if I somehow got square root of 37 here, or something larger, we would have been outside the circle.
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Points inside outside on a circle Mathematics I High School Math Khan Academy.mp3
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Here's the center of it. And this is circle M. This circle right over here. And so it looks like at first she translates it. So the center goes from this point to this point here. And so after the translation, we have the circle right over here. And then she dilates it. And the center of dilation looks like it is point N. And so she dilates it with some type of a scale factor in order to map it exactly onto N. Alright, so that all seems right.
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Non-congruent shapes & transformations.mp3
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So the center goes from this point to this point here. And so after the translation, we have the circle right over here. And then she dilates it. And the center of dilation looks like it is point N. And so she dilates it with some type of a scale factor in order to map it exactly onto N. Alright, so that all seems right. Now, Brenda concluded, I was able to map circle M onto circle N using a sequence of rigid transformations. So the figures are congruent. So is she correct?
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Non-congruent shapes & transformations.mp3
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And the center of dilation looks like it is point N. And so she dilates it with some type of a scale factor in order to map it exactly onto N. Alright, so that all seems right. Now, Brenda concluded, I was able to map circle M onto circle N using a sequence of rigid transformations. So the figures are congruent. So is she correct? Pause that video, or pause this video, and think about that. Alright, now let's work on this together. So she was able to map circle M onto circle N using a sequence of transformations.
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Non-congruent shapes & transformations.mp3
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So is she correct? Pause that video, or pause this video, and think about that. Alright, now let's work on this together. So she was able to map circle M onto circle N using a sequence of transformations. She did a translation and then a dilation. Those are all transformations. But they are not all rigid transformations.
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Non-congruent shapes & transformations.mp3
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So she was able to map circle M onto circle N using a sequence of transformations. She did a translation and then a dilation. Those are all transformations. But they are not all rigid transformations. I'll put a question mark right over there. A translation is a rigid transformation. Remember, rigid transformations are ones that preserve distances, preserve angle measures, preserve lengths.
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Non-congruent shapes & transformations.mp3
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But they are not all rigid transformations. I'll put a question mark right over there. A translation is a rigid transformation. Remember, rigid transformations are ones that preserve distances, preserve angle measures, preserve lengths. While a dilation is not a rigid transformation. As you can see very clearly, it is not preserving lengths. It is not, for example, preserving the radius of the circle.
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Non-congruent shapes & transformations.mp3
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Remember, rigid transformations are ones that preserve distances, preserve angle measures, preserve lengths. While a dilation is not a rigid transformation. As you can see very clearly, it is not preserving lengths. It is not, for example, preserving the radius of the circle. In order for two figures to be congruent, the mapping has to be only with rigid transformations. So because she used a dilation, in fact, you have to use a dilation if you wanna be able to map M onto N because they have different radii, well, then she's not correct. These are not congruent figures.
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Non-congruent shapes & transformations.mp3
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And for our definition of congruence, we will use the rigid transformation definition, which tells us two figures are congruent if and only if there exists a series of rigid transformations which will map one figure onto the other. And then what are rigid transformations? Those are transformations that preserve distance between points and angle measures. So let's get to it. So let's start with two angles that are congruent, and I'm going to show that they have the same measure. I'm gonna demonstrate that. So they start congruent.
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Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3
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So let's get to it. So let's start with two angles that are congruent, and I'm going to show that they have the same measure. I'm gonna demonstrate that. So they start congruent. So these two angles are congruent to each other. Now, this means by the rigid transformation definition of congruence, there is a series of rigid transformations that map angle ABC onto angle, I'll do it here, onto angle DEF. By definition, by definition of rigid transformations, they preserve angle measure, preserve angle measure.
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Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3
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So they start congruent. So these two angles are congruent to each other. Now, this means by the rigid transformation definition of congruence, there is a series of rigid transformations that map angle ABC onto angle, I'll do it here, onto angle DEF. By definition, by definition of rigid transformations, they preserve angle measure, preserve angle measure. So if you're able to map the left angle onto the right angle, and in doing so, you did it with transformations that preserved angle measure, they must now have the same angle measure. We now know that the measure of angle ABC is equal to the measure of angle DEF. So we've demonstrated this green statement the first way, that if things are congruent, they will have the same measure.
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Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3
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By definition, by definition of rigid transformations, they preserve angle measure, preserve angle measure. So if you're able to map the left angle onto the right angle, and in doing so, you did it with transformations that preserved angle measure, they must now have the same angle measure. We now know that the measure of angle ABC is equal to the measure of angle DEF. So we've demonstrated this green statement the first way, that if things are congruent, they will have the same measure. Now let's prove it the other way around. So now let's start with the idea that measure of angle ABC is equal to the measure of angle DEF. And to demonstrate that these are going to be congruent, we just have to show that there's always a series of rigid transformations that will map angle ABC onto angle DEF.
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Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3
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So we've demonstrated this green statement the first way, that if things are congruent, they will have the same measure. Now let's prove it the other way around. So now let's start with the idea that measure of angle ABC is equal to the measure of angle DEF. And to demonstrate that these are going to be congruent, we just have to show that there's always a series of rigid transformations that will map angle ABC onto angle DEF. And to help us there, let's just visualize these angles. So draw this really fast, angle ABC. An angle is defined by two rays that start at a point.
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Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3
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And to demonstrate that these are going to be congruent, we just have to show that there's always a series of rigid transformations that will map angle ABC onto angle DEF. And to help us there, let's just visualize these angles. So draw this really fast, angle ABC. An angle is defined by two rays that start at a point. That point is the vertex. So it's ABC, and then let me draw angle DEF. So it might look something like this.
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Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3
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An angle is defined by two rays that start at a point. That point is the vertex. So it's ABC, and then let me draw angle DEF. So it might look something like this. D-E-F. And what we will now do is, let's do our first rigid transformation. Let's translate, translate angle ABC so that B mapped to point E. And if we did that, so we're gonna translate it like that, then ABC is going to look something like, ABC is gonna look something like this. It's gonna look something like this.
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Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3
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So it might look something like this. D-E-F. And what we will now do is, let's do our first rigid transformation. Let's translate, translate angle ABC so that B mapped to point E. And if we did that, so we're gonna translate it like that, then ABC is going to look something like, ABC is gonna look something like this. It's gonna look something like this. B is now mapped onto E. This would be where A would get mapped to. This would be where C would get mapped to. Sometimes you might see a notation A prime, C prime.
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Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3
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It's gonna look something like this. B is now mapped onto E. This would be where A would get mapped to. This would be where C would get mapped to. Sometimes you might see a notation A prime, C prime. And this is where B would get mapped to. And then the next thing I would do is I would rotate angle ABC about its vertex, about B, so that ray BC, ray BC coincides, coincides with ray EF. Now, so you're just gonna rotate the whole angle that way so that now ray BC coincides with ray EF.
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Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3
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Sometimes you might see a notation A prime, C prime. And this is where B would get mapped to. And then the next thing I would do is I would rotate angle ABC about its vertex, about B, so that ray BC, ray BC coincides, coincides with ray EF. Now, so you're just gonna rotate the whole angle that way so that now ray BC coincides with ray EF. Well, you might be saying, hey, C doesn't necessarily have to sit on F because they might be different distances from their vertices, but that's all right. The ray can be defined by any point that sits on that ray. So now if you do this rotation and ray BC coincides with ray EF, now those two rays would be equivalent because measure of angle ABC is equal to the measure of angle DEF.
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Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3
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Now, so you're just gonna rotate the whole angle that way so that now ray BC coincides with ray EF. Well, you might be saying, hey, C doesn't necessarily have to sit on F because they might be different distances from their vertices, but that's all right. The ray can be defined by any point that sits on that ray. So now if you do this rotation and ray BC coincides with ray EF, now those two rays would be equivalent because measure of angle ABC is equal to the measure of angle DEF. That will also tell us that ray BA, ray BA now coincides, coincides with ray ED. And just like that, I've given you a series of rigid transformations that will always work. If you translate so that the vertices are mapped onto each other, and then you rotate it so that the bottom ray of one angle coincides with the bottom ray of the other angle, then you could say the top ray of the two angles will now coincide because the angles have the same measure.
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Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3
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So now if you do this rotation and ray BC coincides with ray EF, now those two rays would be equivalent because measure of angle ABC is equal to the measure of angle DEF. That will also tell us that ray BA, ray BA now coincides, coincides with ray ED. And just like that, I've given you a series of rigid transformations that will always work. If you translate so that the vertices are mapped onto each other, and then you rotate it so that the bottom ray of one angle coincides with the bottom ray of the other angle, then you could say the top ray of the two angles will now coincide because the angles have the same measure. And because of that, the angles now completely coincide. And so we know that angle ABC is congruent to angle DEF. And we're now done.
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Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3
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If you translate so that the vertices are mapped onto each other, and then you rotate it so that the bottom ray of one angle coincides with the bottom ray of the other angle, then you could say the top ray of the two angles will now coincide because the angles have the same measure. And because of that, the angles now completely coincide. And so we know that angle ABC is congruent to angle DEF. And we're now done. We've proven both sides of the statement. If they're congruent, they have the same measure. If they have the same measure, then they are congruent.
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Angle congruence equivalent to having same measure Congruence Geometry Khan Academy.mp3
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So that's one line, and then let me draw another line that is parallel to that. I'm claiming that these are parallel lines. And now I'm gonna draw some transversals here. So first let me draw a horizontal transversal. So just like that. And then let me do a vertical transversal. So just like that.
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Proof parallel lines have the same slope High School Math Khan Academy (2).mp3
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So first let me draw a horizontal transversal. So just like that. And then let me do a vertical transversal. So just like that. And I'm assuming that the green one is horizontal and the blue one is vertical. So we assume that they are perpendicular to each other, that these intersect at right angles. And from this, I'm going to figure out, I'm gonna use some parallel line angle properties to establish that this triangle and this triangle are similar, and then use that to establish that both of these lines, both of these yellow lines have the same slope.
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Proof parallel lines have the same slope High School Math Khan Academy (2).mp3
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So just like that. And I'm assuming that the green one is horizontal and the blue one is vertical. So we assume that they are perpendicular to each other, that these intersect at right angles. And from this, I'm going to figure out, I'm gonna use some parallel line angle properties to establish that this triangle and this triangle are similar, and then use that to establish that both of these lines, both of these yellow lines have the same slope. So actually let me label some points here. So let's call that point A, point B, point C, point D, and point E. So let's see. First of all, we know that angle CED is going to be congruent to angle AEB because they're both right angles.
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Proof parallel lines have the same slope High School Math Khan Academy (2).mp3
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And from this, I'm going to figure out, I'm gonna use some parallel line angle properties to establish that this triangle and this triangle are similar, and then use that to establish that both of these lines, both of these yellow lines have the same slope. So actually let me label some points here. So let's call that point A, point B, point C, point D, and point E. So let's see. First of all, we know that angle CED is going to be congruent to angle AEB because they're both right angles. So that's a right angle, and then that is a right angle right over there. We also know some things about corresponding angles where a transversal intersects parallel lines. This angle corresponds to this angle if we look at the blue transversal as it intersects those two lines.
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Proof parallel lines have the same slope High School Math Khan Academy (2).mp3
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First of all, we know that angle CED is going to be congruent to angle AEB because they're both right angles. So that's a right angle, and then that is a right angle right over there. We also know some things about corresponding angles where a transversal intersects parallel lines. This angle corresponds to this angle if we look at the blue transversal as it intersects those two lines. And so they're going to have the same measure. They're going to be congruent. Now this angle on one side of this point B is going to also be congruent to that because they are vertical angles, and we've seen that multiple times before.
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Proof parallel lines have the same slope High School Math Khan Academy (2).mp3
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This angle corresponds to this angle if we look at the blue transversal as it intersects those two lines. And so they're going to have the same measure. They're going to be congruent. Now this angle on one side of this point B is going to also be congruent to that because they are vertical angles, and we've seen that multiple times before. And so we know that this angle, angle ABE, is congruent to angle ECD. Sometimes this is called alternate interior angles of a transversal and parallel lines. Well, if you look at triangle CED and triangle ABE, we see they already have two angles in common.
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Proof parallel lines have the same slope High School Math Khan Academy (2).mp3
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Now this angle on one side of this point B is going to also be congruent to that because they are vertical angles, and we've seen that multiple times before. And so we know that this angle, angle ABE, is congruent to angle ECD. Sometimes this is called alternate interior angles of a transversal and parallel lines. Well, if you look at triangle CED and triangle ABE, we see they already have two angles in common. So if they have two angles in common, well, their third angle has to be in common because this third angle's just going to be 180 minus these other two. And so this third angle's just going to be 180 minus the other two. And so just like that, we notice we have all three angles are the same in both of these triangles, or they're not all the same, but all of the corresponding angles, I should say, are the same.
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Proof parallel lines have the same slope High School Math Khan Academy (2).mp3
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Well, if you look at triangle CED and triangle ABE, we see they already have two angles in common. So if they have two angles in common, well, their third angle has to be in common because this third angle's just going to be 180 minus these other two. And so this third angle's just going to be 180 minus the other two. And so just like that, we notice we have all three angles are the same in both of these triangles, or they're not all the same, but all of the corresponding angles, I should say, are the same. Angle, this blue angle has the same measure as this blue angle. This magenta angle has the same measure as this magenta angle. And then the other angles are right angles.
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Proof parallel lines have the same slope High School Math Khan Academy (2).mp3
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And so just like that, we notice we have all three angles are the same in both of these triangles, or they're not all the same, but all of the corresponding angles, I should say, are the same. Angle, this blue angle has the same measure as this blue angle. This magenta angle has the same measure as this magenta angle. And then the other angles are right angles. These are right triangles here. So we could say triangle AEB, triangle AEB, is similar, similar, is similar to triangle DEC, triangle DEC, by, and we could say by angle, angle, angle. All the corresponding angles are congruent.
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Proof parallel lines have the same slope High School Math Khan Academy (2).mp3
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And then the other angles are right angles. These are right triangles here. So we could say triangle AEB, triangle AEB, is similar, similar, is similar to triangle DEC, triangle DEC, by, and we could say by angle, angle, angle. All the corresponding angles are congruent. So we are dealing with similar triangles. And so we know similar triangles, the ratio of corresponding sides are going to be the same. So we could say that the ratio of, let's say the ratio of BE, the ratio of BE, let me write this down.
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Proof parallel lines have the same slope High School Math Khan Academy (2).mp3
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All the corresponding angles are congruent. So we are dealing with similar triangles. And so we know similar triangles, the ratio of corresponding sides are going to be the same. So we could say that the ratio of, let's say the ratio of BE, the ratio of BE, let me write this down. This is this side right over here. The ratio of BE to AE, to AE, to AE, is going to be equal to, so that side over that side, well what is the corresponding side? The corresponding side to BE is side CE.
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Proof parallel lines have the same slope High School Math Khan Academy (2).mp3
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So we could say that the ratio of, let's say the ratio of BE, the ratio of BE, let me write this down. This is this side right over here. The ratio of BE to AE, to AE, to AE, is going to be equal to, so that side over that side, well what is the corresponding side? The corresponding side to BE is side CE. So that's going to be the same as the ratio between CE and DE, and DE. And this just comes out of similar, the similarity of the triangles, CE to DE. So once again, once we establish these triangles are similar we can say the ratio of corresponding sides are going to be the same.
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Proof parallel lines have the same slope High School Math Khan Academy (2).mp3
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The corresponding side to BE is side CE. So that's going to be the same as the ratio between CE and DE, and DE. And this just comes out of similar, the similarity of the triangles, CE to DE. So once again, once we establish these triangles are similar we can say the ratio of corresponding sides are going to be the same. Now what is the ratio between BE and AE? The ratio between BE and AE. Well that is the slope of this top line right over here.
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Proof parallel lines have the same slope High School Math Khan Academy (2).mp3
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So once again, once we establish these triangles are similar we can say the ratio of corresponding sides are going to be the same. Now what is the ratio between BE and AE? The ratio between BE and AE. Well that is the slope of this top line right over here. We could say that's the slope of line AB. Slope, slope of line connecting, connecting A to B. Or actually let me just use, I could write it like this.
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Proof parallel lines have the same slope High School Math Khan Academy (2).mp3
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Well that is the slope of this top line right over here. We could say that's the slope of line AB. Slope, slope of line connecting, connecting A to B. Or actually let me just use, I could write it like this. That is slope of, slope of A, slope of line AB. Remember slope is, when you're going from A to B, it's change in Y over change in X. So when you're going from A to B, your change in X is AE, and your change in Y is BE, or EB, however you want to refer to it.
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Proof parallel lines have the same slope High School Math Khan Academy (2).mp3
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Or actually let me just use, I could write it like this. That is slope of, slope of A, slope of line AB. Remember slope is, when you're going from A to B, it's change in Y over change in X. So when you're going from A to B, your change in X is AE, and your change in Y is BE, or EB, however you want to refer to it. So this right over here is change in Y, and this over here is change in X. Well now let's look at this second expression right over here, CE over DE. CE over DE.
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Proof parallel lines have the same slope High School Math Khan Academy (2).mp3
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So when you're going from A to B, your change in X is AE, and your change in Y is BE, or EB, however you want to refer to it. So this right over here is change in Y, and this over here is change in X. Well now let's look at this second expression right over here, CE over DE. CE over DE. Well now this is going to be change in Y over change in X between point C and D. So this is, this right over here, this is the slope of line, of line CD. And so just like that, by establishing similarity, we were able to see the ratio of corresponding sides are congruent, which shows us that the slopes of these two lines are going to be the same. And we are done.
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Proof parallel lines have the same slope High School Math Khan Academy (2).mp3
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And what I want to prove is that the sum of the measures of the interior angles of a triangle, that x plus y plus z, is equal to 180 degrees. And the way that I'm going to do it is using our knowledge of parallel lines or transversals of parallel lines and corresponding angles. And to do that, I'm going to extend each of these sides of the triangle, which right now are line segments, but extend them into lines. So this side down here, if I keep going on and on forever in the same directions, then now all of a sudden I have an orange line. And what I want to do is construct another line that is parallel to the orange line that goes through this vertex of the triangle right over here. And I can always do that. I can just start from this point and go in the same direction as this line, and I will never intersect.
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Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3
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So this side down here, if I keep going on and on forever in the same directions, then now all of a sudden I have an orange line. And what I want to do is construct another line that is parallel to the orange line that goes through this vertex of the triangle right over here. And I can always do that. I can just start from this point and go in the same direction as this line, and I will never intersect. I'm not getting any closer or further away from that line. So I'm never going to intersect that line. So these two lines right over here are parallel.
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Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3
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I can just start from this point and go in the same direction as this line, and I will never intersect. I'm not getting any closer or further away from that line. So I'm never going to intersect that line. So these two lines right over here are parallel. This is parallel to that. Now I'm going to go to the other two sides of my original triangle and extend them into lines. So I'm going to extend this one into a line.
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Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3
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So these two lines right over here are parallel. This is parallel to that. Now I'm going to go to the other two sides of my original triangle and extend them into lines. So I'm going to extend this one into a line. So I'll do that as neatly as I can. So I'm going to extend that into a line. And you see that this is clearly a transversal of these two parallel lines.
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Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3
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So I'm going to extend this one into a line. So I'll do that as neatly as I can. So I'm going to extend that into a line. And you see that this is clearly a transversal of these two parallel lines. Now, if we have a transversal here of two parallel lines, then we must have some corresponding angles. And we see that this angle is formed when the transversal intersects the bottom orange line. Well, what's the corresponding angle when the transversal intersects this top blue line?
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Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3
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And you see that this is clearly a transversal of these two parallel lines. Now, if we have a transversal here of two parallel lines, then we must have some corresponding angles. And we see that this angle is formed when the transversal intersects the bottom orange line. Well, what's the corresponding angle when the transversal intersects this top blue line? What's the angle on the top right of the intersection? Angle on the top right of the intersection must also be x. The other thing that pops out at you is there's another vertical angle with x, another angle that must be equivalent.
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Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3
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Well, what's the corresponding angle when the transversal intersects this top blue line? What's the angle on the top right of the intersection? Angle on the top right of the intersection must also be x. The other thing that pops out at you is there's another vertical angle with x, another angle that must be equivalent. On the opposite side of this intersection, you have this angle right over here. These two angles are vertical. So if this has measure x, then this one must have measure x as well.
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Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3
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The other thing that pops out at you is there's another vertical angle with x, another angle that must be equivalent. On the opposite side of this intersection, you have this angle right over here. These two angles are vertical. So if this has measure x, then this one must have measure x as well. Let's do the same thing with the last side of the triangle that we have not extended into a line yet. So let's do that. So if we take this one, so we just keep going.
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Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3
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So if this has measure x, then this one must have measure x as well. Let's do the same thing with the last side of the triangle that we have not extended into a line yet. So let's do that. So if we take this one, so we just keep going. So it becomes a line. So now it becomes a transversal of the two parallel lines, just like the magenta line did. And we say, hey, look, this angle y right over here, this angle is formed from the intersection of the transversal and the bottom parallel line.
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Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3
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So if we take this one, so we just keep going. So it becomes a line. So now it becomes a transversal of the two parallel lines, just like the magenta line did. And we say, hey, look, this angle y right over here, this angle is formed from the intersection of the transversal and the bottom parallel line. What angle does it correspond to up here? Well, this is kind of on the left side of the intersection. It corresponds to this angle right over here, where the green line, the green transversal intersects the blue parallel line.
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Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3
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And we say, hey, look, this angle y right over here, this angle is formed from the intersection of the transversal and the bottom parallel line. What angle does it correspond to up here? Well, this is kind of on the left side of the intersection. It corresponds to this angle right over here, where the green line, the green transversal intersects the blue parallel line. Well, what angle is vertical to it? Well, this angle. So this is going to have measure y as well.
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Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3
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It corresponds to this angle right over here, where the green line, the green transversal intersects the blue parallel line. Well, what angle is vertical to it? Well, this angle. So this is going to have measure y as well. So now we're really at the home stretch of our proof, because we will see that the measure, we have this angle and this angle. This has measure angle x. This has measure z.
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Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3
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So this is going to have measure y as well. So now we're really at the home stretch of our proof, because we will see that the measure, we have this angle and this angle. This has measure angle x. This has measure z. They're both adjacent angles. If we take the two outer rays that form the angle, and we think about this angle right over here, what's this measure of this wide angle right over there? Well, it's going to be x plus z.
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Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3
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This has measure z. They're both adjacent angles. If we take the two outer rays that form the angle, and we think about this angle right over here, what's this measure of this wide angle right over there? Well, it's going to be x plus z. And that angle is supplementary to this angle right over here that has measure y. So the measure of this wide angle, which is x plus z, plus the measure of this magenta angle, which is y, must be equal to 180 degrees, because these two angles are supplementary. So the measure of the wide angle, x plus z plus the measure of the magenta angle, which is supplementary to the wide angle, it must be equal to 180 degrees, because they are supplementary.
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Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3
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Well, it's going to be x plus z. And that angle is supplementary to this angle right over here that has measure y. So the measure of this wide angle, which is x plus z, plus the measure of this magenta angle, which is y, must be equal to 180 degrees, because these two angles are supplementary. So the measure of the wide angle, x plus z plus the measure of the magenta angle, which is supplementary to the wide angle, it must be equal to 180 degrees, because they are supplementary. Well, we could just reorder this if we want to put it in alphabetical order, but we've just completed our proof. The measure of the interior angles of the triangle, x plus z plus y. We could write this as x plus y plus z, if the lack of alphabetical order is making you uncomfortable.
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Proof Sum of measures of angles in a triangle are 180 Geometry Khan Academy.mp3
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I will now do a proof for which we credit the 12th century Indian mathematician Bhaskara. So what we're gonna do is we're gonna start with a square. So let me see if I can draw a square. I'm gonna draw it tilted at a bit of an angle, just because I think it'll make it a little bit easier on me. So let me, my best attempt at drawing something that reasonably looks like a square. You have to bear with me if it's not exactly a tilted square. So that looks pretty good.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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I'm gonna draw it tilted at a bit of an angle, just because I think it'll make it a little bit easier on me. So let me, my best attempt at drawing something that reasonably looks like a square. You have to bear with me if it's not exactly a tilted square. So that looks pretty good. And I'm assuming it's a square. So this is a right angle, this is a right angle, that's a right angle, that's a right angle. I'm assuming the lengths of all of these sides are the same.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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So that looks pretty good. And I'm assuming it's a square. So this is a right angle, this is a right angle, that's a right angle, that's a right angle. I'm assuming the lengths of all of these sides are the same. So let's just assume that they're all of length C. I'll write that in yellow. So all of the sides of the square are of length C. And now I'm gonna construct four triangles inside of this square. And the way I'm gonna do it is I'm going to be dropping, so here I'm gonna go straight down.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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I'm assuming the lengths of all of these sides are the same. So let's just assume that they're all of length C. I'll write that in yellow. So all of the sides of the square are of length C. And now I'm gonna construct four triangles inside of this square. And the way I'm gonna do it is I'm going to be dropping, so here I'm gonna go straight down. I'm gonna drop a line straight down and draw a triangle that looks like this. So I'm gonna go straight down here. Here, I'm gonna go straight across.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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And the way I'm gonna do it is I'm going to be dropping, so here I'm gonna go straight down. I'm gonna drop a line straight down and draw a triangle that looks like this. So I'm gonna go straight down here. Here, I'm gonna go straight across. And so since this is straight down this is straight across, we know that this is a right angle. Then from this vertex on our square, I'm going to go straight up. And since this is straight up and this is straight across, we know that this is a right angle.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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Here, I'm gonna go straight across. And so since this is straight down this is straight across, we know that this is a right angle. Then from this vertex on our square, I'm going to go straight up. And since this is straight up and this is straight across, we know that this is a right angle. And then from this vertex right over here, I'm going to go straight horizontally. I'm assuming that's what I'm doing. And so we know that this is going to be a right angle, and then we know this is going to be a right angle.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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And since this is straight up and this is straight across, we know that this is a right angle. And then from this vertex right over here, I'm going to go straight horizontally. I'm assuming that's what I'm doing. And so we know that this is going to be a right angle, and then we know this is going to be a right angle. So we see that we've constructed, from our square, we've constructed four right triangles. and in between we have something that at minimum looks like a rectangle or possibly a square. We haven't quite proven to ourselves yet that this is a square.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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And so we know that this is going to be a right angle, and then we know this is going to be a right angle. So we see that we've constructed, from our square, we've constructed four right triangles. and in between we have something that at minimum looks like a rectangle or possibly a square. We haven't quite proven to ourselves yet that this is a square. Now the next thing I want to think about is whether these triangles are congruent. So they definitely all have the same length of their hypotenuse. All of the hypotenuses, I don't know what the plural of hypotenuse is, hypotenai, hypotenuses, they have all length c. The side opposite the right angle is always length c. So if we can show that all the corresponding angles are the same, then we know it's congruent.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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We haven't quite proven to ourselves yet that this is a square. Now the next thing I want to think about is whether these triangles are congruent. So they definitely all have the same length of their hypotenuse. All of the hypotenuses, I don't know what the plural of hypotenuse is, hypotenai, hypotenuses, they have all length c. The side opposite the right angle is always length c. So if we can show that all the corresponding angles are the same, then we know it's congruent. If you have something where all the angles are the same and you have a side that is also the corresponding side that's also congruent, then the whole triangles are congruent. And we can show that if we assume that this angle is theta, then this angle right over here has to be 90 minus theta because together they are complementary. We know that because they combine to form this angle of the square, this right angle.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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All of the hypotenuses, I don't know what the plural of hypotenuse is, hypotenai, hypotenuses, they have all length c. The side opposite the right angle is always length c. So if we can show that all the corresponding angles are the same, then we know it's congruent. If you have something where all the angles are the same and you have a side that is also the corresponding side that's also congruent, then the whole triangles are congruent. And we can show that if we assume that this angle is theta, then this angle right over here has to be 90 minus theta because together they are complementary. We know that because they combine to form this angle of the square, this right angle. If this is 90 minus theta, we know this angle and this angle have to add up to 90 because we only have 90 left when we subtract the right angle from 180. So we know this has to be theta. And if that's theta, then that's 90 minus theta.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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We know that because they combine to form this angle of the square, this right angle. If this is 90 minus theta, we know this angle and this angle have to add up to 90 because we only have 90 left when we subtract the right angle from 180. So we know this has to be theta. And if that's theta, then that's 90 minus theta. I think you see where this is going. If that's 90 minus theta, this has to be theta. And if that's theta, then this is 90 minus theta.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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And if that's theta, then that's 90 minus theta. I think you see where this is going. If that's 90 minus theta, this has to be theta. And if that's theta, then this is 90 minus theta. If this is 90 minus theta, then this is theta. And then this would have to be 90 minus theta. So we see in all four of these triangles, the three angles are theta, 90 minus theta, and 90 degrees.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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And if that's theta, then this is 90 minus theta. If this is 90 minus theta, then this is theta. And then this would have to be 90 minus theta. So we see in all four of these triangles, the three angles are theta, 90 minus theta, and 90 degrees. So they all have the same exact angle. So at minimum, they are similar. And their hypotenuses are the same.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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So we see in all four of these triangles, the three angles are theta, 90 minus theta, and 90 degrees. So they all have the same exact angle. So at minimum, they are similar. And their hypotenuses are the same. So we know that all four of these triangles are completely congruent triangles. So with that assumption, let's just assume that the longer side of these triangles, that these are of length b. So the longer side of these triangles, I'm just going to assume.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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And their hypotenuses are the same. So we know that all four of these triangles are completely congruent triangles. So with that assumption, let's just assume that the longer side of these triangles, that these are of length b. So the longer side of these triangles, I'm just going to assume. So this length right over here, I'll call that lowercase b. And let's assume that the shorter side, so this distance right over here, this distance right over here, this distance right over here, that these are all this distance right over here, that these are of length a. So if I were to say this height right over here, this height is of length a.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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