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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. Now we will do something interesting. Well, first, let's think about the area of the entire square. What's the area of the entire square in terms of c?
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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So if I were to say this height right over here, this height is of length a. Now we will do something interesting. Well, first, let's think about the area of the entire square. What's the area of the entire square in terms of c? Well, that's pretty straightforward. It's a c by c square. So the area here is equal to c squared.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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What's the area of the entire square in terms of c? Well, that's pretty straightforward. It's a c by c square. So the area here is equal to c squared. Now what I'm going to do is rearrange two of these triangles and then come up with the area of that other figure in terms of a's and b's. And hopefully it gets us to the Pythagorean theorem. And to do that, just so we don't lose our starting point, because our starting point is interesting, let me just copy and paste this entire thing.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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So the area here is equal to c squared. Now what I'm going to do is rearrange two of these triangles and then come up with the area of that other figure in terms of a's and b's. And hopefully it gets us to the Pythagorean theorem. And to do that, just so we don't lose our starting point, because our starting point is interesting, let me just copy and paste this entire thing. So I don't want it to clip off. So let me just copy and paste this. Copy and paste.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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And to do that, just so we don't lose our starting point, because our starting point is interesting, let me just copy and paste this entire thing. So I don't want it to clip off. So let me just copy and paste this. Copy and paste. So this is our original diagram. And what I will now do, and actually let me clear that out, clear. I'm now going to shift.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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Copy and paste. So this is our original diagram. And what I will now do, and actually let me clear that out, clear. I'm now going to shift. This is the fun part. I'm going to shift this triangle here in the top left. I'm going to shift it below this triangle on the bottom right.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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I'm now going to shift. This is the fun part. I'm going to shift this triangle here in the top left. I'm going to shift it below this triangle on the bottom right. And I'm going to attempt to do that by copying and pasting. So let's see how much. Well, the way I drew it, it's not that easy.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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I'm going to shift it below this triangle on the bottom right. And I'm going to attempt to do that by copying and pasting. So let's see how much. Well, the way I drew it, it's not that easy. Well, that might do the trick. I want to retain a little bit of the. So let me copy, or let me actually cut it, and then let me paste it.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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Well, the way I drew it, it's not that easy. Well, that might do the trick. I want to retain a little bit of the. So let me copy, or let me actually cut it, and then let me paste it. So that triangle, I'm going to stick right over there. Stick that right over there. And let me draw in the lines that I just erased.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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So let me copy, or let me actually cut it, and then let me paste it. So that triangle, I'm going to stick right over there. Stick that right over there. And let me draw in the lines that I just erased. So just to be clear, we had a line over there. And we also had this right over here. And this was straight up and down.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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And let me draw in the lines that I just erased. So just to be clear, we had a line over there. And we also had this right over here. And this was straight up and down. And these were straight side to side. Now, so I moved this part over down here. So I moved that over down there.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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And this was straight up and down. And these were straight side to side. Now, so I moved this part over down here. So I moved that over down there. And now I'm going to move this top right triangle down to the bottom left. So I'm just rearranging the exact same area. So let me just capture the whole thing as best as I can.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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So I moved that over down there. And now I'm going to move this top right triangle down to the bottom left. So I'm just rearranging the exact same area. So let me just capture the whole thing as best as I can. So let me cut, and then let me paste. And I'm going to move it right over here. And while I went through that process, I kind of lost its floor.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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So let me just capture the whole thing as best as I can. So let me cut, and then let me paste. And I'm going to move it right over here. And while I went through that process, I kind of lost its floor. So let me redraw the floor. So I just moved it right over here. So this thing, this triangle, let me color it in, is now right over there.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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And while I went through that process, I kind of lost its floor. So let me redraw the floor. So I just moved it right over here. So this thing, this triangle, let me color it in, is now right over there. And this triangle is now right over here. That center square, and it is a square, is now right over here. So hopefully, you can appreciate how we rearranged it.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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So this thing, this triangle, let me color it in, is now right over there. And this triangle is now right over here. That center square, and it is a square, is now right over here. So hopefully, you can appreciate how we rearranged it. Now, my question for you is, how can we express the area of this new figure, which has the exact same area as the old figure? I just shifted parts of it around. How can we express this in terms of a's and b's?
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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So hopefully, you can appreciate how we rearranged it. Now, my question for you is, how can we express the area of this new figure, which has the exact same area as the old figure? I just shifted parts of it around. How can we express this in terms of a's and b's? Well, the key insight here is to recognize the length of this bottom side. What's the length of this bottom side right over here? The length of this bottom side, well, this length right over here is b.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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How can we express this in terms of a's and b's? Well, the key insight here is to recognize the length of this bottom side. What's the length of this bottom side right over here? The length of this bottom side, well, this length right over here is b. This length right over here is a. So the length of this entire bottom is a plus b. Well, that by itself is kind of interesting.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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The length of this bottom side, well, this length right over here is b. This length right over here is a. So the length of this entire bottom is a plus b. Well, that by itself is kind of interesting. But what we can realize is that this length right over here, this length right over here, which is the exact same thing as this length over here, was also a. So we can construct an a by a square. So we can construct an a by a square.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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Well, that by itself is kind of interesting. But what we can realize is that this length right over here, this length right over here, which is the exact same thing as this length over here, was also a. So we can construct an a by a square. So we can construct an a by a square. So this square right over here is a by a. And so it has area a squared. Let me do that in a color that you can actually see.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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So we can construct an a by a square. So this square right over here is a by a. And so it has area a squared. Let me do that in a color that you can actually see. So this has area of a squared. And then what's the area of what's left over? Well, if this is length a, then this is length a as well.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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Let me do that in a color that you can actually see. So this has area of a squared. And then what's the area of what's left over? Well, if this is length a, then this is length a as well. If this entire bottom is a plus b, then we know that what's left over after subtracting the a out has to be b. If this whole thing is a plus b, this is a, then this right over here is b. And so the rest of this newly oriented figure, this new figure, everything that I'm shading in over here, this is just a b by b square.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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Well, if this is length a, then this is length a as well. If this entire bottom is a plus b, then we know that what's left over after subtracting the a out has to be b. If this whole thing is a plus b, this is a, then this right over here is b. And so the rest of this newly oriented figure, this new figure, everything that I'm shading in over here, this is just a b by b square. So the area here is b squared. So the entire area of this figure is a squared plus b squared, which, lucky for us, is equal to the area of this expressed in terms of c, because they're the exact same figure, just rearranged. So it's going to be equal to c squared.
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Bhaskara's proof of the Pythagorean theorem Geometry Khan Academy.mp3
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Line segments IN, this is segment IN over here, and TO, this is TO here, are reflected over the line Y is equal to negative X minus two. So this is the line that they're reflected about, this dashed purple line, and it is indeed Y equals negative X minus two. This right over here is in slope intercept form. The slope should be negative one, and we see that the slope of this purple line is indeed negative one. If X changes by a certain amount, Y changes by the negative of that. If X changes by one, Y changes by negative one to get back to that line. If X changes by positive two, Y changes by negative two to get back to another point on that line.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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The slope should be negative one, and we see that the slope of this purple line is indeed negative one. If X changes by a certain amount, Y changes by the negative of that. If X changes by one, Y changes by negative one to get back to that line. If X changes by positive two, Y changes by negative two to get back to another point on that line. And the Y intercept, we see when X is equal to zero, Y should be negative two. When X is equal to zero, Y is indeed negative two. So we validated that.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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If X changes by positive two, Y changes by negative two to get back to another point on that line. And the Y intercept, we see when X is equal to zero, Y should be negative two. When X is equal to zero, Y is indeed negative two. So we validated that. Now they say draw the image of this reflection using the interactive graph. All right, so we can move these lines around, and we wanna reflect these, and I could try to eyeball it, you know, maybe it's something like this. I don't know, this doesn't seem exactly right.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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So we validated that. Now they say draw the image of this reflection using the interactive graph. All right, so we can move these lines around, and we wanna reflect these, and I could try to eyeball it, you know, maybe it's something like this. I don't know, this doesn't seem exactly right. That looks close to the reflection of IN, and for TO, I'd wanna move this down here. TO looks like it would be, I don't know, I'm eyeballing it. This is close, but I can't be close.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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I don't know, this doesn't seem exactly right. That looks close to the reflection of IN, and for TO, I'd wanna move this down here. TO looks like it would be, I don't know, I'm eyeballing it. This is close, but I can't be close. I wanna get exact. So let's, I've copied and pasted the original problem on my scratch pad, so we can find the exact points, and so I don't just have to estimate this. So let's go to the scratch pad.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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This is close, but I can't be close. I wanna get exact. So let's, I've copied and pasted the original problem on my scratch pad, so we can find the exact points, and so I don't just have to estimate this. So let's go to the scratch pad. So exactly what we just saw. And the main realization is, is if we wanna reflect a given point, if we wanna reflect a given point, say point I right over here, what we wanna do is we wanna drop a perpendicular. We wanna find a line that's perpendicular, a line that has the point I on it, and it's perpendicular to this line right over here.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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So let's go to the scratch pad. So exactly what we just saw. And the main realization is, is if we wanna reflect a given point, if we wanna reflect a given point, say point I right over here, what we wanna do is we wanna drop a perpendicular. We wanna find a line that's perpendicular, a line that has the point I on it, and it's perpendicular to this line right over here. Remember, this is the line, let me do this in the purple color. This is the line Y is equal to negative X minus two. Its slope is equal to negative one.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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We wanna find a line that's perpendicular, a line that has the point I on it, and it's perpendicular to this line right over here. Remember, this is the line, let me do this in the purple color. This is the line Y is equal to negative X minus two. Its slope is equal to negative one. So I wanna align that goes through I, point I, that is perpendicular to this line, and I wanna drop it to, I wanna drop it to the line that I'm gonna reflect on, and then I wanna go the same distance onto the other side to find the corresponding point in the image. So how do I do that? Well, if this line, if this purple line has a slope of negative one, a line that is perpendicular to it, a line that is perpendicular to it, so this thing that I'm drawing in purple right over here, its slope is going to be the negative reciprocal of this.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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Its slope is equal to negative one. So I wanna align that goes through I, point I, that is perpendicular to this line, and I wanna drop it to, I wanna drop it to the line that I'm gonna reflect on, and then I wanna go the same distance onto the other side to find the corresponding point in the image. So how do I do that? Well, if this line, if this purple line has a slope of negative one, a line that is perpendicular to it, a line that is perpendicular to it, so this thing that I'm drawing in purple right over here, its slope is going to be the negative reciprocal of this. So the reciprocal of negative one is still just negative one. One over negative one is still negative one, but we want the negative of that. So the slope here needs to be one, and luckily, that's how I drew it.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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Well, if this line, if this purple line has a slope of negative one, a line that is perpendicular to it, a line that is perpendicular to it, so this thing that I'm drawing in purple right over here, its slope is going to be the negative reciprocal of this. So the reciprocal of negative one is still just negative one. One over negative one is still negative one, but we want the negative of that. So the slope here needs to be one, and luckily, that's how I drew it. The slope here needs to be equal to one, which is however much I change in the X direction, I change in the Y direction. We see that. To go from this point to this point right over here, we decrease Y by four, and we decrease X by four.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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So the slope here needs to be one, and luckily, that's how I drew it. The slope here needs to be equal to one, which is however much I change in the X direction, I change in the Y direction. We see that. To go from this point to this point right over here, we decrease Y by four, and we decrease X by four. Now, if we wanna stay on this line to find the reflection, we just do the same thing. We could decrease X by four, so we'll go from negative two to negative six, and decrease Y by four, and we end up at this point right over here. So we end up at the point, this is X equals negative six, Y is equal to negative four.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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To go from this point to this point right over here, we decrease Y by four, and we decrease X by four. Now, if we wanna stay on this line to find the reflection, we just do the same thing. We could decrease X by four, so we'll go from negative two to negative six, and decrease Y by four, and we end up at this point right over here. So we end up at the point, this is X equals negative six, Y is equal to negative four. So this point corresponds to this point right over there. Now, let's do the same thing for point N. For point N, we already know if we drop a perpendicular, if this is perpendicular, it's going to have a slope of one, because this purple line has a slope of negative one, and the negative reciprocal of negative one is positive one. And let's see, to go from this point to this point of intersection, we have to go down one and a half.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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So we end up at the point, this is X equals negative six, Y is equal to negative four. So this point corresponds to this point right over there. Now, let's do the same thing for point N. For point N, we already know if we drop a perpendicular, if this is perpendicular, it's going to have a slope of one, because this purple line has a slope of negative one, and the negative reciprocal of negative one is positive one. And let's see, to go from this point to this point of intersection, we have to go down one and a half. We're going down one and a half, and we're going to the left one and a half. So we wanna do that on the other side. We wanna stay on this perpendicular line, so we wanna go left one and a half, and down one and a half.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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And let's see, to go from this point to this point of intersection, we have to go down one and a half. We're going down one and a half, and we're going to the left one and a half. So we wanna do that on the other side. We wanna stay on this perpendicular line, so we wanna go left one and a half, and down one and a half. And we get to this point right over here, which is the point X equals three, Y is equal to negative eight. And so we are now equidistant. We're on this perpendicular line still, but we're equidistant on the other side.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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We wanna stay on this perpendicular line, so we wanna go left one and a half, and down one and a half. And we get to this point right over here, which is the point X equals three, Y is equal to negative eight. And so we are now equidistant. We're on this perpendicular line still, but we're equidistant on the other side. So the image of IN is gonna go through negative six comma negative four and three comma negative eight. So let me draw that. So let me see if I can remember.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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We're on this perpendicular line still, but we're equidistant on the other side. So the image of IN is gonna go through negative six comma negative four and three comma negative eight. So let me draw that. So let me see if I can remember. Negative six, negative four, three comma negative eight. So I have a bad memory. So negative six, negative four, and three comma negative eight.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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So let me see if I can remember. Negative six, negative four, three comma negative eight. So I have a bad memory. So negative six, negative four, and three comma negative eight. And I was close when I estimated, but I wasn't exactly right. So that's looking pretty good. And then actually we can do the exact same thing with points T and point O.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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So negative six, negative four, and three comma negative eight. And I was close when I estimated, but I wasn't exactly right. So that's looking pretty good. And then actually we can do the exact same thing with points T and point O. Now let me do that. So point T, to get from point T to the line in the shortest distance, once again we drop a perpendicular. This line is gonna have a slope of one because it's perpendicular to the line that has a slope of negative one.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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And then actually we can do the exact same thing with points T and point O. Now let me do that. So point T, to get from point T to the line in the shortest distance, once again we drop a perpendicular. This line is gonna have a slope of one because it's perpendicular to the line that has a slope of negative one. And so to get there, we have to decrease our X by, we have to decrease our X. We're going from X equals five to X equals, looks like half. So X went down by four and a half in the X direction, and Y also needs to go down by four and a half.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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This line is gonna have a slope of one because it's perpendicular to the line that has a slope of negative one. And so to get there, we have to decrease our X by, we have to decrease our X. We're going from X equals five to X equals, looks like half. So X went down by four and a half in the X direction, and Y also needs to go down by four and a half. So if we wanna stay on that line, let's decrease our X by four and a half. So that's half, one, two, three, four. And Y needs to go down by four and a half.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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So X went down by four and a half in the X direction, and Y also needs to go down by four and a half. So if we wanna stay on that line, let's decrease our X by four and a half. So that's half, one, two, three, four. And Y needs to go down by four and a half. So that's half, one, two, three, four. And we get to this point right over here, which is the point X equals negative four, Y is equal to negative seven. Negative four comma negative seven.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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And Y needs to go down by four and a half. So that's half, one, two, three, four. And we get to this point right over here, which is the point X equals negative four, Y is equal to negative seven. Negative four comma negative seven. So this should be at X equals negative four, Y equals negative seven. And there's a couple of things you could do here. You could just say, hey, this is too long, or two units long, not like too long in length somehow.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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Negative four comma negative seven. So this should be at X equals negative four, Y equals negative seven. And there's a couple of things you could do here. You could just say, hey, this is too long, or two units long, not like too long in length somehow. This is two units long, so maybe this is two units long. So this is feeling pretty good. But let's just go through the exercise for point O as well.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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You could just say, hey, this is too long, or two units long, not like too long in length somehow. This is two units long, so maybe this is two units long. So this is feeling pretty good. But let's just go through the exercise for point O as well. So point O, once again, this is going to have, if we drop a perpendicular, it's gonna have a slope of one. So whatever our change in X between this point and this point, we're gonna have the same change in Y. And our change in X to go from seven to one and a half, our change in X is five and a half.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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But let's just go through the exercise for point O as well. So point O, once again, this is going to have, if we drop a perpendicular, it's gonna have a slope of one. So whatever our change in X between this point and this point, we're gonna have the same change in Y. And our change in X to go from seven to one and a half, our change in X is five and a half. So let me do it this way. So the change in X here, so the change in X is equal to negative five and a half. 5.5, if you subtract 5.5 from seven, you get to 1.5.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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And our change in X to go from seven to one and a half, our change in X is five and a half. So let me do it this way. So the change in X here, so the change in X is equal to negative five and a half. 5.5, if you subtract 5.5 from seven, you get to 1.5. And our change in Y, our change in Y, is also negative 5.5. Change in Y is negative, you're not gonna see that, so I'm gonna do it a different way, negative 5.5. And so we need to stay on this line.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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5.5, if you subtract 5.5 from seven, you get to 1.5. And our change in Y, our change in Y, is also negative 5.5. Change in Y is negative, you're not gonna see that, so I'm gonna do it a different way, negative 5.5. And so we need to stay on this line. So we wanna change by those same amounts onto the other side of that line. So if we decrease our X by five and a half, so half, one, two, three, four, five, we get there. And Y by five and a half, half, one, two, three, four, five, three, four, five, we get to the point X equals negative four, Y is equal to negative nine.
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Reflecting segments over line Transformations Geometry Khan Academy.mp3
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What I want to do in this video is I want to prove that segment AC is perpendicular to segment DB based on the information that we have in this diagram over here. This side has the same length as that side. This side has the same length as that side. I'll give you a hint. We're going to use one or more of our congruence postulates. I'll just stick with calling them postulates from now on. The ones that we know, let me draw a little line here.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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I'll give you a hint. We're going to use one or more of our congruence postulates. I'll just stick with calling them postulates from now on. The ones that we know, let me draw a little line here. This is kind of our tool kit. We have the side-side-side postulate. If the three sides are congruent, then the two triangles are congruent.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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The ones that we know, let me draw a little line here. This is kind of our tool kit. We have the side-side-side postulate. If the three sides are congruent, then the two triangles are congruent. We have side-angle-side. If the two sides and the angle in between are congruent, then the two triangles are congruent. We have ASA, two angles with a side in between, and then we have AAS, two angles and then a side.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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If the three sides are congruent, then the two triangles are congruent. We have side-angle-side. If the two sides and the angle in between are congruent, then the two triangles are congruent. We have ASA, two angles with a side in between, and then we have AAS, two angles and then a side. Any of these things we've established, these are our postulates. We're going to assume that they imply congruency. I'm also going to do this as what we call a two-column proof.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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We have ASA, two angles with a side in between, and then we have AAS, two angles and then a side. Any of these things we've established, these are our postulates. We're going to assume that they imply congruency. I'm also going to do this as what we call a two-column proof. You don't have to do something as a two-column proof, but this is what you normally see in an introductory geometry class. I thought I would expose you to it. It's a pretty basic idea.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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I'm also going to do this as what we call a two-column proof. You don't have to do something as a two-column proof, but this is what you normally see in an introductory geometry class. I thought I would expose you to it. It's a pretty basic idea. You make a statement, and you just have to give the reason for your statement, which is what we've been doing with any proof, but we haven't always put it in a very structured way. I'm just going to do it like this. I'll have two columns, write this like that, and I'll have a statement, and then I will give the reason for the statement.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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It's a pretty basic idea. You make a statement, and you just have to give the reason for your statement, which is what we've been doing with any proof, but we haven't always put it in a very structured way. I'm just going to do it like this. I'll have two columns, write this like that, and I'll have a statement, and then I will give the reason for the statement. The strategy that I'm going to try to do is it looks like, right off the bat, it seems like I can prove that triangle CDA is congruent to triangle CBA based on side, side, side. That's a pretty good starting point, because once I can base congruency, then I can start to have angles be the same. The reason why I can do that is because this side is the same as that side, this side is the same as that side, and they both share that side.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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I'll have two columns, write this like that, and I'll have a statement, and then I will give the reason for the statement. The strategy that I'm going to try to do is it looks like, right off the bat, it seems like I can prove that triangle CDA is congruent to triangle CBA based on side, side, side. That's a pretty good starting point, because once I can base congruency, then I can start to have angles be the same. The reason why I can do that is because this side is the same as that side, this side is the same as that side, and they both share that side. I don't want to just do it verbally this time. I want to write it out properly in this two-column proof. We have CD, we have the length of segment CD is equal to the length of CB.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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The reason why I can do that is because this side is the same as that side, this side is the same as that side, and they both share that side. I don't want to just do it verbally this time. I want to write it out properly in this two-column proof. We have CD, we have the length of segment CD is equal to the length of CB. CD is equal to CB, and that is given. These two characters have the same length. We also know that DA, the length of segment DA, is the same as the length of segment BA.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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We have CD, we have the length of segment CD is equal to the length of CB. CD is equal to CB, and that is given. These two characters have the same length. We also know that DA, the length of segment DA, is the same as the length of segment BA. DA is equal to BA, that's also given in the diagram. Then we also know that CA is equal to a CA, I guess we could say. CA is equal to itself, and it's obviously in both triangles.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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We also know that DA, the length of segment DA, is the same as the length of segment BA. DA is equal to BA, that's also given in the diagram. Then we also know that CA is equal to a CA, I guess we could say. CA is equal to itself, and it's obviously in both triangles. This is also given, or it's obvious from the diagram. It's a bit obvious, both triangles share that side. We have two triangles, their corresponding sides have the same length, and so we know that they are congruent.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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CA is equal to itself, and it's obviously in both triangles. This is also given, or it's obvious from the diagram. It's a bit obvious, both triangles share that side. We have two triangles, their corresponding sides have the same length, and so we know that they are congruent. We know that triangle CDA is congruent to triangle CBA. We know that by the side-side-side postulate, and the statements given up here. Let me number our statements so we can refer back to this.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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We have two triangles, their corresponding sides have the same length, and so we know that they are congruent. We know that triangle CDA is congruent to triangle CBA. We know that by the side-side-side postulate, and the statements given up here. Let me number our statements so we can refer back to this. 1, 2, 3, and 4. Side-side-side postulate, and 1, 2, and 3. Statements 1, 2, and 3.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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Let me number our statements so we can refer back to this. 1, 2, 3, and 4. Side-side-side postulate, and 1, 2, and 3. Statements 1, 2, and 3. Statements 1, 2, and 3 in the side-side postulate let us know that these two triangles are congruent. If these are congruent, then we know, for example, that all of their corresponding angles are equivalent. For example, this angle is going to be equal to that angle.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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Statements 1, 2, and 3. Statements 1, 2, and 3 in the side-side postulate let us know that these two triangles are congruent. If these are congruent, then we know, for example, that all of their corresponding angles are equivalent. For example, this angle is going to be equal to that angle. Let's make that statement right over there. This angle DCE, this is going to be statement 5. We know that angle DCE, that's this angle right over here, is going to have the same measure.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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For example, this angle is going to be equal to that angle. Let's make that statement right over there. This angle DCE, this is going to be statement 5. We know that angle DCE, that's this angle right over here, is going to have the same measure. We can even say they're congruent. I'll say the measure of angle DCE is going to be equal to the measure of angle BCE. This comes straight out of statement 4.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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We know that angle DCE, that's this angle right over here, is going to have the same measure. We can even say they're congruent. I'll say the measure of angle DCE is going to be equal to the measure of angle BCE. This comes straight out of statement 4. I can put in parentheses, congruency of those triangles. This implies straight, because they're both part of this larger triangle, they are the corresponding angles, so they are going to have the exact same measure. Now it seems like we can do something pretty interesting with these two smaller triangles, at the top left and the top right of this kite-like figure.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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This comes straight out of statement 4. I can put in parentheses, congruency of those triangles. This implies straight, because they're both part of this larger triangle, they are the corresponding angles, so they are going to have the exact same measure. Now it seems like we can do something pretty interesting with these two smaller triangles, at the top left and the top right of this kite-like figure. Because we have a side, two corresponding sides are congruent, two corresponding angles are congruent, and they have a side in common. They have this side in common right over here. Let's first establish that they have this side in common right over here.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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Now it seems like we can do something pretty interesting with these two smaller triangles, at the top left and the top right of this kite-like figure. Because we have a side, two corresponding sides are congruent, two corresponding angles are congruent, and they have a side in common. They have this side in common right over here. Let's first establish that they have this side in common right over here. I'll just write statement 6. We have CE, the measure or the length of that line is equal to itself. Once again, this is just obvious.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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Let's first establish that they have this side in common right over here. I'll just write statement 6. We have CE, the measure or the length of that line is equal to itself. Once again, this is just obvious. It's the same obvious from diagram. It's the same line, obvious from diagram. But now we can use that information.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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Once again, this is just obvious. It's the same obvious from diagram. It's the same line, obvious from diagram. But now we can use that information. We don't have three sides. We haven't proven to ourselves that this side is the same as this side, that DE has the same length as EB, but we do have a side, an angle between the sides, and then another side. This looks pretty interesting for our side-angle-side postulate.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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But now we can use that information. We don't have three sides. We haven't proven to ourselves that this side is the same as this side, that DE has the same length as EB, but we do have a side, an angle between the sides, and then another side. This looks pretty interesting for our side-angle-side postulate. We can say by the side-angle-side postulate that triangle DCE is congruent to triangle BCE. When I write the labels for the triangles, I'm making sure that I'm putting the corresponding points. I started at D, then went to C, then to E. The corresponding angle or the corresponding point or vertex, I could say, for this triangle right over here is B.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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This looks pretty interesting for our side-angle-side postulate. We can say by the side-angle-side postulate that triangle DCE is congruent to triangle BCE. When I write the labels for the triangles, I'm making sure that I'm putting the corresponding points. I started at D, then went to C, then to E. The corresponding angle or the corresponding point or vertex, I could say, for this triangle right over here is B. If I start with D, I start with B. C in the middle is the corresponding vertex for either of these triangles, so I put it in the middle, and then they both go to E. That's just to make sure that we are specifying what's corresponding to what. We know this is true by side-angle-side, and the information we got from this side is established that these two sides are congruent was from statement 1.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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I started at D, then went to C, then to E. The corresponding angle or the corresponding point or vertex, I could say, for this triangle right over here is B. If I start with D, I start with B. C in the middle is the corresponding vertex for either of these triangles, so I put it in the middle, and then they both go to E. That's just to make sure that we are specifying what's corresponding to what. We know this is true by side-angle-side, and the information we got from this side is established that these two sides are congruent was from statement 1. Then that these angles are congruent is from statement 5, right over here. Then statement 6 gave us the other side, statement 6, just like that. If we know that these triangles are congruent, that means that all of their corresponding angles are congruent.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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We know this is true by side-angle-side, and the information we got from this side is established that these two sides are congruent was from statement 1. Then that these angles are congruent is from statement 5, right over here. Then statement 6 gave us the other side, statement 6, just like that. If we know that these triangles are congruent, that means that all of their corresponding angles are congruent. We know, for example, that this angle right over here is going to be congruent to that angle over there. Let's write that down. We know statement 8, the measure of angle, let's call that DEC, is equal to the measure of angle BEC.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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If we know that these triangles are congruent, that means that all of their corresponding angles are congruent. We know, for example, that this angle right over here is going to be congruent to that angle over there. Let's write that down. We know statement 8, the measure of angle, let's call that DEC, is equal to the measure of angle BEC. This comes straight from statement 7. Once again, they're congruent. Then we also know, we'll make statement 9, that the measure of angle DEC, or maybe we should just write it this way, angle DEC and angle BEC are supplementary.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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We know statement 8, the measure of angle, let's call that DEC, is equal to the measure of angle BEC. This comes straight from statement 7. Once again, they're congruent. Then we also know, we'll make statement 9, that the measure of angle DEC, or maybe we should just write it this way, angle DEC and angle BEC are supplementary. They are supplementary, and you can just look at that from inspection, but I'll write it decently. Supplementary, which means their measures add up to 180 degrees. We know that because they are adjacent, and outer sides form straight angle.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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Then we also know, we'll make statement 9, that the measure of angle DEC, or maybe we should just write it this way, angle DEC and angle BEC are supplementary. They are supplementary, and you can just look at that from inspection, but I'll write it decently. Supplementary, which means their measures add up to 180 degrees. We know that because they are adjacent, and outer sides form straight angle. Then we can essentially, the next step, if we know that these two angles are equal to each other, and if we know that they are complementary, our next step means that we can actually deduce that they must be 90 degrees. 10, measure of angle DEC equals measure of angle BEC, which equals 90 degrees. Then for the reason, it might be a little bit more involved, we could put these two statements together.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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We know that because they are adjacent, and outer sides form straight angle. Then we can essentially, the next step, if we know that these two angles are equal to each other, and if we know that they are complementary, our next step means that we can actually deduce that they must be 90 degrees. 10, measure of angle DEC equals measure of angle BEC, which equals 90 degrees. Then for the reason, it might be a little bit more involved, we could put these two statements together. It would be statements 8 and 9. Then statements 8 and 9 mean that DEC, so I could write this, measure of angle DEC plus measure of angle, of angle, actually let me just, since I don't want to do too many steps all at once, let me just take it little bit by little bit. Let me just do it all like this.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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Then for the reason, it might be a little bit more involved, we could put these two statements together. It would be statements 8 and 9. Then statements 8 and 9 mean that DEC, so I could write this, measure of angle DEC plus measure of angle, of angle, actually let me just, since I don't want to do too many steps all at once, let me just take it little bit by little bit. Let me just do it all like this. Let me say measure of angle DEC plus measure of angle BEC is equal to 180, and this comes straight from.9, that they are supplementary. Then we could say statement, I'm taking up a lot of space now, statement 11, we could say measure of angle DEC plus measure of angle DEC is equal to 180 degrees. We know that from statement 9, we know that from statement 9 and statement 8.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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Let me just do it all like this. Let me say measure of angle DEC plus measure of angle BEC is equal to 180, and this comes straight from.9, that they are supplementary. Then we could say statement, I'm taking up a lot of space now, statement 11, we could say measure of angle DEC plus measure of angle DEC is equal to 180 degrees. We know that from statement 9, we know that from statement 9 and statement 8. We essentially just took statement 9 and substituted that BEC, the measure of BEC is the same as the measure of DEC. Then if we want statement 12, we could say measure of angle DEC is equal to 90, which is equal to the measure of angle BEC. Then this comes once again straight out of point number 11 and 8. You can see I'm taking a little bit more time, going a little bit more granular through the steps.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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We know that from statement 9, we know that from statement 9 and statement 8. We essentially just took statement 9 and substituted that BEC, the measure of BEC is the same as the measure of DEC. Then if we want statement 12, we could say measure of angle DEC is equal to 90, which is equal to the measure of angle BEC. Then this comes once again straight out of point number 11 and 8. You can see I'm taking a little bit more time, going a little bit more granular through the steps. Some of the other proofs I would have just said, obviously this implies this or that. Then we're done, because if these are 90 degrees, let me write the last statement, statement 13, which is what we wanted to prove. We wanted to prove that AC is perpendicular to DB.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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You can see I'm taking a little bit more time, going a little bit more granular through the steps. Some of the other proofs I would have just said, obviously this implies this or that. Then we're done, because if these are 90 degrees, let me write the last statement, statement 13, which is what we wanted to prove. We wanted to prove that AC is perpendicular to DB. AC is perpendicular to segment DB, and it comes straight out of point 12. We're done. We've done a two-column proof, and we have proven that this line segment right over here is perpendicular to that line segment right over there.
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Two column proof showing segments are perpendicular Congruence Geometry Khan Academy.mp3
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I have a pyramid here on the left and I have a cone here on the right. And we know a few things about these two figures. First of all, they have the exact same height. So this length right over here is H and this length right over here, going from the peak to the center of the base here, is H as well. We also know that the area of the bases is the same. So for example, in this left pyramid, the area of the base would be X times, and let's just assume that it is a square. So X times X.
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Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3
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So this length right over here is H and this length right over here, going from the peak to the center of the base here, is H as well. We also know that the area of the bases is the same. So for example, in this left pyramid, the area of the base would be X times, and let's just assume that it is a square. So X times X. So the area here is going to be equal to X squared. And the area of the base, so that's area of this base is equal to X squared. And the area of this base right over here would be equal to, area is equal to pi times R squared.
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Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3
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So X times X. So the area here is going to be equal to X squared. And the area of the base, so that's area of this base is equal to X squared. And the area of this base right over here would be equal to, area is equal to pi times R squared. And I'm saying that these two things are the same. So we also know that X squared is equal to pi R squared. Now, my question to you is, do these two figures have the same volume or is it different?
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Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3
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And the area of this base right over here would be equal to, area is equal to pi times R squared. And I'm saying that these two things are the same. So we also know that X squared is equal to pi R squared. Now, my question to you is, do these two figures have the same volume or is it different? And if they are different, which one has a larger volume? Pause this video and try to think about that. All right, now let's do this together.
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Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3
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Now, my question to you is, do these two figures have the same volume or is it different? And if they are different, which one has a larger volume? Pause this video and try to think about that. All right, now let's do this together. Now, given that we're talking about two figures that have the same height and at least the area of the base is the same, you might be thinking that Cavalieri's principle might be useful. Just a reminder of what that is, Cavalieri's principle tells us that if you have two figures and we're thinking in three-dimensional version of Cavalieri's principle, if you have two figures that have the same height and at any point along that height, the cross-sectional area is the same, then the figures have the same volume. So what we need to do is figure out, is it true that at any point in this height, do these figures have the same cross-sectional area?
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Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3
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All right, now let's do this together. Now, given that we're talking about two figures that have the same height and at least the area of the base is the same, you might be thinking that Cavalieri's principle might be useful. Just a reminder of what that is, Cavalieri's principle tells us that if you have two figures and we're thinking in three-dimensional version of Cavalieri's principle, if you have two figures that have the same height and at any point along that height, the cross-sectional area is the same, then the figures have the same volume. So what we need to do is figure out, is it true that at any point in this height, do these figures have the same cross-sectional area? Well, to think about that, let's pick an arbitrary point along this height. And just for simplicity, let's pick halfway along the height, although we could do this analysis at any point along the height. So halfway along the height there, halfway along the height there.
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Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3
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So what we need to do is figure out, is it true that at any point in this height, do these figures have the same cross-sectional area? Well, to think about that, let's pick an arbitrary point along this height. And just for simplicity, let's pick halfway along the height, although we could do this analysis at any point along the height. So halfway along the height there, halfway along the height there. So this distance right over here, that would be h over two. This distance right over here would be h over two. This whole thing is h. And what we can do is construct what look like similar triangles, and we can even prove it to ourselves that these are similar triangles.
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Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3
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So halfway along the height there, halfway along the height there. So this distance right over here, that would be h over two. This distance right over here would be h over two. This whole thing is h. And what we can do is construct what look like similar triangles, and we can even prove it to ourselves that these are similar triangles. So let me construct them right over here. And the reason why we know they're similar is that this line is going to be parallel to this line, and that this line is parallel to that line, to that radius. And how do we know that?
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Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3
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This whole thing is h. And what we can do is construct what look like similar triangles, and we can even prove it to ourselves that these are similar triangles. So let me construct them right over here. And the reason why we know they're similar is that this line is going to be parallel to this line, and that this line is parallel to that line, to that radius. And how do we know that? Well, we're taking cross-sectional areas that are parallel to the base, that are parallel to the surface on which it sits in this situation. So in either case, these cross-sections are going to be parallel. So these lines which sit in these cross-sections, or sit on the base and sit in the cross-section, have to be parallel as well.
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Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3
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And how do we know that? Well, we're taking cross-sectional areas that are parallel to the base, that are parallel to the surface on which it sits in this situation. So in either case, these cross-sections are going to be parallel. So these lines which sit in these cross-sections, or sit on the base and sit in the cross-section, have to be parallel as well. Well, because these are parallel lines, this angle is congruent to that angle. This angle is congruent to this angle, because these are transversals across parallel lines, and these are just corresponding angles. And of course, they share this angle in common.
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Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3
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So these lines which sit in these cross-sections, or sit on the base and sit in the cross-section, have to be parallel as well. Well, because these are parallel lines, this angle is congruent to that angle. This angle is congruent to this angle, because these are transversals across parallel lines, and these are just corresponding angles. And of course, they share this angle in common. And here you see very clearly, right angle, right angle. This angle is congruent to that angle, and then both triangles share that. And so the smaller triangle in either case is similar to the larger triangle.
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Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3
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And of course, they share this angle in common. And here you see very clearly, right angle, right angle. This angle is congruent to that angle, and then both triangles share that. And so the smaller triangle in either case is similar to the larger triangle. And what that helps us realize is that the ratio between corresponding sides is going to be the same. So if this side is h over two, and the entire height is h, so this is half of the entire height, that tells us that this side is going to be half of r. So this right over here is going to be r over two. And this side over here by the same argument is going to be x over two.
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Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3
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And so the smaller triangle in either case is similar to the larger triangle. And what that helps us realize is that the ratio between corresponding sides is going to be the same. So if this side is h over two, and the entire height is h, so this is half of the entire height, that tells us that this side is going to be half of r. So this right over here is going to be r over two. And this side over here by the same argument is going to be x over two. And so what's the cross-sectional area here? Well, it's going to be x over two squared. So it's going to be x over two squared, which is equal to x squared over four, which is 1 1⁄4 of the base's area, which is equal to 1 1⁄4 of the base's area.
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Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3
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And this side over here by the same argument is going to be x over two. And so what's the cross-sectional area here? Well, it's going to be x over two squared. So it's going to be x over two squared, which is equal to x squared over four, which is 1 1⁄4 of the base's area, which is equal to 1 1⁄4 of the base's area. And what about over here? Well, this cross-sectional area is going to be pi times r over two squared, which is the same thing as pi r squared over four, or we could say that as 1 1⁄4 pi r squared, which is the same thing as 1 1⁄4 of the area of the base. The area of the base is pi r squared.
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Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3
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So it's going to be x over two squared, which is equal to x squared over four, which is 1 1⁄4 of the base's area, which is equal to 1 1⁄4 of the base's area. And what about over here? Well, this cross-sectional area is going to be pi times r over two squared, which is the same thing as pi r squared over four, or we could say that as 1 1⁄4 pi r squared, which is the same thing as 1 1⁄4 of the area of the base. The area of the base is pi r squared. Now we're saying 1 1⁄4 pi r squared. So this is going to be equal to 1 1⁄4 the area. And we already said that these areas are the same.
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Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3
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The area of the base is pi r squared. Now we're saying 1 1⁄4 pi r squared. So this is going to be equal to 1 1⁄4 the area. And we already said that these areas are the same. And so we've just seen that the cross-sectional area at that point of the height of both of these figures is the same. And you could do that 1⁄4 along the height, 3⁄4 along the height, you're going to get the same exact analysis. You're gonna have two similar triangles, and you're going to see that you have the same areas, same cross-sectional areas at that point of the height.
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Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3
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And we already said that these areas are the same. And so we've just seen that the cross-sectional area at that point of the height of both of these figures is the same. And you could do that 1⁄4 along the height, 3⁄4 along the height, you're going to get the same exact analysis. You're gonna have two similar triangles, and you're going to see that you have the same areas, same cross-sectional areas at that point of the height. And so therefore, we see by Cavalieri's principle in three dimensions, that these two figures have the same volume. And what's interesting about that is it allows us to take the formula, which we've proven and gotten the intuition for in other videos for the volume of a pyramid. We've learned that the volume of a pyramid is equal to 1⁄3 times base times height, and say, well, this one must have the exact same volume.
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Volumes of cones intuition Solid geometry High school geometry Khan Academy.mp3
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