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And so this is going to be equal to six plus four, which is 10, over negative two minus negative six. That's the same thing as negative two plus six. So that's going to be over four, which is the same thing as 5 1⁄2. All right, that's interesting. What is the slope of segment BC? The slope of segment BC is going to be ... | Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3 |
All right, that's interesting. What is the slope of segment BC? The slope of segment BC is going to be equal to, once again, change in y over change in x. Our y-coordinates, change in y is two minus six, two minus six, over eight minus negative two. Eight minus negative two, which is equal to negative four, over, and t... | Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3 |
Our y-coordinates, change in y is two minus six, two minus six, over eight minus negative two. Eight minus negative two, which is equal to negative four, over, and then eight minus negative two is the same thing as eight plus two, over 10, which is the same thing as negative 2⁵. Now, in other videos in your algebra cla... | Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3 |
And you can actually see that right over here. These are opposite reciprocals. If you take the reciprocal of this top slope, you'd get 2⁵, and then you take the opposite of it, or in this case, the negative of it, you are going to get negative 2⁵. So these are actually, these are perpendicular lines. So this lets us kn... | Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3 |
So these are actually, these are perpendicular lines. So this lets us know that AB is perpendicular, segment AB is perpendicular to segment BC. So we know that this is the case. And we could keep on to doing that, but in a parallelogram, if one set of segments intersect at a right angle, all of them are going to inters... | Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3 |
And we could keep on to doing that, but in a parallelogram, if one set of segments intersect at a right angle, all of them are going to intersect at a right angle, and we could show that more rigorously in other places, but this is enough evidence for me to know that this is indeed going to be a rectangle. If you want,... | Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3 |
So choice A says yes, and yes would be, it is a rectangle because AB is equal. So the length of segment AB is equal to the length of segment AD, and the length of segment BC is equal to the length of segment CD. So that might be true, I haven't validated it, but just because this is true, and because we do know that AB... | Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3 |
For example, you can have a parallelogram where even all the sides are congruent. So you could have a parallelogram that looks like this. And obviously, if all of the sides are congruent, you're dealing with a rhombus, but a rhombus is still not necessarily going to be a rectangle. And so I would rule this top one out.... | Classifying figures with coordinates Analytic geometry High school geometry Khan Academy.mp3 |
In this first problem over here, we're asked to find out the length of this segment, segment CE. And we have these two parallel lines. AB is parallel to DE. And then we have these two essentially transversals that form these two triangles. So let's see what we can do here. So the first thing that might jump out at you ... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
And then we have these two essentially transversals that form these two triangles. So let's see what we can do here. So the first thing that might jump out at you is that this angle and this angle are vertical angles, so they are going to be congruent. The other thing that might jump out at you is that angle CDE is an ... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
The other thing that might jump out at you is that angle CDE is an alternate interior angle with CBA. So we have this transversal right over here. And these are alternate interior angles. And they are going to be congruent. Or you could say that if you continue this transversal, you would have a corresponding angle wit... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
And they are going to be congruent. Or you could say that if you continue this transversal, you would have a corresponding angle with CDE right up here. And this one's just vertical. Either way, this angle and this angle are going to be congruent. So we've established that we have two triangles. And two of the correspo... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
Either way, this angle and this angle are going to be congruent. So we've established that we have two triangles. And two of the corresponding angles are the same. And that by itself is enough to establish similarity. You can actually, we actually could show that this angle and this angle are also congruent by alternat... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
And that by itself is enough to establish similarity. You can actually, we actually could show that this angle and this angle are also congruent by alternate interior angles. But we don't have to. So we already know that they are similar. Actually, we could just say it just by alternate interior angles, these are also ... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
So we already know that they are similar. Actually, we could just say it just by alternate interior angles, these are also going to be congruent. But we already know enough to say that they are similar even before doing that. So we already know that triangle, I'll try to write it, I'll color code it so that we have the... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
So we already know that triangle, I'll try to write it, I'll color code it so that we have the same corresponding vertices. And that's really important to know what angles and what sides correspond to what side so that you don't mess up your ratios. So that you do know what's corresponding to what. So we know triangle ... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
So we know triangle AB, triangle ABC, is similar to triangle. So A, this vertex A corresponds to vertex E over here, is similar to vertex E. And then vertex B right over here corresponds to vertex D, EDC. Now what does that do for us? Well, that tells us that the ratio of corresponding sides are going to be the same. T... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
Well, that tells us that the ratio of corresponding sides are going to be the same. They're going to be some constant value. So we have corresponding side. So the ratio, for example, the corresponding side for BC is going to be DC. We can see it just the way that we've written down the similarity. This is true. Then BC... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
So the ratio, for example, the corresponding side for BC is going to be DC. We can see it just the way that we've written down the similarity. This is true. Then BC is the corresponding side to DC. So we know that the length of BC over DC is going to be equal to the length of, well, we want to figure out what CE is. Th... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
Then BC is the corresponding side to DC. So we know that the length of BC over DC is going to be equal to the length of, well, we want to figure out what CE is. That's what we care about. And I'm using BC and DC because we know those values. So BC over DC is going to be equal to what's the corresponding side to CE? The... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
And I'm using BC and DC because we know those values. So BC over DC is going to be equal to what's the corresponding side to CE? The corresponding side over here is CA. It's going to be equal to CA over CE. Corresponding sides. This is the last and the first, last and the first. CA over CE. | Similarity example problems Similarity Geometry Khan Academy.mp3 |
It's going to be equal to CA over CE. Corresponding sides. This is the last and the first, last and the first. CA over CE. And we know what BC is. BC right over here is 5. We know what DC is. | Similarity example problems Similarity Geometry Khan Academy.mp3 |
CA over CE. And we know what BC is. BC right over here is 5. We know what DC is. It is 3. We know what CA or AC is right over here. CA is 4. | Similarity example problems Similarity Geometry Khan Academy.mp3 |
We know what DC is. It is 3. We know what CA or AC is right over here. CA is 4. And now we can just solve for CE. So we can, well, there's multiple ways that you could think about this. You could cross multiply, which is really just multiplying both sides by both denominators. | Similarity example problems Similarity Geometry Khan Academy.mp3 |
CA is 4. And now we can just solve for CE. So we can, well, there's multiple ways that you could think about this. You could cross multiply, which is really just multiplying both sides by both denominators. So you get 5 times the length of CE is equal to 3 times 4, which is just going to be equal to 12. And then we get... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
You could cross multiply, which is really just multiplying both sides by both denominators. So you get 5 times the length of CE is equal to 3 times 4, which is just going to be equal to 12. And then we get CE is equal to 12 over 5, which is the same thing as 2 and 2 fifths or 2.4. So this is going to be 2 and 2 fifths.... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
So this is going to be 2 and 2 fifths. And we're done. We were able to use similarity to figure out this side, just knowing that the ratio between the corresponding sides are going to be the same. Now let's do this problem right over here. Let me draw a little line here to show that this is a different problem now. Thi... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
Now let's do this problem right over here. Let me draw a little line here to show that this is a different problem now. This is a different problem. So in this problem, we need to figure out what DE is. And we all, once again, have these two parallel lines like this. And so we know corresponding angles are congruent. S... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
So in this problem, we need to figure out what DE is. And we all, once again, have these two parallel lines like this. And so we know corresponding angles are congruent. So we know that angle is going to be congruent to that angle, because you could view this as a transversal. We also know that this angle right over he... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
So we know that angle is going to be congruent to that angle, because you could view this as a transversal. We also know that this angle right over here is going to be congruent to that angle right over there. Once again, corresponding angles for transversal. And also in both triangles. So I'm looking at triangle CBD a... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
And also in both triangles. So I'm looking at triangle CBD and triangle CAE. They both share this angle up here. Once again, we could have stopped at two angles. But we've actually shown that all three angles of these two triangles, all three of the corresponding angles, are congruent to each other. So we now know, and... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
Once again, we could have stopped at two angles. But we've actually shown that all three angles of these two triangles, all three of the corresponding angles, are congruent to each other. So we now know, and once again, this is an important thing to do, is to make sure that you write it in the right order when you writ... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
We now know that triangle CBD is similar. Not congruent. It is similar to triangle CCAE, which means that the ratio of corresponding sides are going to be constant. So we know, for example, that the ratio of CB to CA, let's write this down. We know that the ratio of CB over CA is going to be equal to the ratio of CD ov... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
So we know, for example, that the ratio of CB to CA, let's write this down. We know that the ratio of CB over CA is going to be equal to the ratio of CD over CE. And we know what CB is. CB over here is 5. We know what CA is. And we have to be careful here. It's not 3. | Similarity example problems Similarity Geometry Khan Academy.mp3 |
CB over here is 5. We know what CA is. And we have to be careful here. It's not 3. CA, this entire side, is going to be 5 plus 3. So this is going to be 8. And we know what CD is. | Similarity example problems Similarity Geometry Khan Academy.mp3 |
It's not 3. CA, this entire side, is going to be 5 plus 3. So this is going to be 8. And we know what CD is. CD is going to be 4. And so once again, we can cross multiply. We have 5 times CE is equal to 8 times 4. | Similarity example problems Similarity Geometry Khan Academy.mp3 |
And we know what CD is. CD is going to be 4. And so once again, we can cross multiply. We have 5 times CE is equal to 8 times 4. 8 times 4 is 32. And so CE is equal to 32 over 5. Or this is another way to think about it. | Similarity example problems Similarity Geometry Khan Academy.mp3 |
We have 5 times CE is equal to 8 times 4. 8 times 4 is 32. And so CE is equal to 32 over 5. Or this is another way to think about it. That's 6 and 2 fifths. Now we're not done, because they didn't ask for what CE is. They're asking for just this part right over here. | Similarity example problems Similarity Geometry Khan Academy.mp3 |
Or this is another way to think about it. That's 6 and 2 fifths. Now we're not done, because they didn't ask for what CE is. They're asking for just this part right over here. They're asking for DE. So we know that this entire length, CE right over here, this is 6 and 2 fifths. And so DE right over here, what we actual... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
They're asking for just this part right over here. They're asking for DE. So we know that this entire length, CE right over here, this is 6 and 2 fifths. And so DE right over here, what we actually have to figure out, it's going to be this entire length, 6 and 2 fifths, minus 4, minus CD right over here. So it's going ... | Similarity example problems Similarity Geometry Khan Academy.mp3 |
So that is our inscribed angle. I'll denote it by psi. I'll use psi for inscribed angle and angles in this video, that psi, the inscribed angle, is going to be exactly 1 half of the central angle that subtends the same arc. So I just used a lot of fancy words, but I think you'll get what I'm saying. So this is psi. It ... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
So I just used a lot of fancy words, but I think you'll get what I'm saying. So this is psi. It is an inscribed angle. It sits, its vertex sits on the circumference. And if you draw out the two rays that come out from this angle, or the two cords that define this angle, it intersects the circle at the other end. And if... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
It sits, its vertex sits on the circumference. And if you draw out the two rays that come out from this angle, or the two cords that define this angle, it intersects the circle at the other end. And if you look at the part of the circumference of the circle that's inside of it, that is the arc that is subtended by psi.... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
So it's all very fancy words, but I think the idea is pretty straightforward. This right here is the arc subtended by psi, where psi is that inscribed angle right over there, the vertex sitting on the circumference. Now, a central angle is an angle where the vertex is sitting at the center of the circle. So let's say t... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
So let's say that this right here, I'll try to eyeball it, that right there is the center of the circle. And so let me draw a central angle that subtends this same arc. So that looks like a central angle subtending that same arc, just like that. Let's call this theta. So this angle is psi. This angle right here is thet... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
Let's call this theta. So this angle is psi. This angle right here is theta. And what I'm going to prove in this video is that psi is always going to be equal to 1 half of theta. So if I were to tell you that psi is equal to, I don't know, 25 degrees, then you would immediately know that theta must be equal to 50 degre... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
And what I'm going to prove in this video is that psi is always going to be equal to 1 half of theta. So if I were to tell you that psi is equal to, I don't know, 25 degrees, then you would immediately know that theta must be equal to 50 degrees. Or if I told you that theta was 80 degrees, then you'd immediately know t... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
So let's actually prove this. So let me clear this. So a good place to start, or the place I'm going to start, is a special case. I'm going to draw a inscribed angle. But one of the chords that define it is going to be the diameter of the circle. So this isn't going to be the general case. This is just going to be a sp... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
I'm going to draw a inscribed angle. But one of the chords that define it is going to be the diameter of the circle. So this isn't going to be the general case. This is just going to be a special case. So let me see. This is the center right here of my circle. Trying to eyeball it. | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
This is just going to be a special case. So let me see. This is the center right here of my circle. Trying to eyeball it. Let me draw it a little bit better. That looks, center looks like that. So let me draw a diameter. | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
Trying to eyeball it. Let me draw it a little bit better. That looks, center looks like that. So let me draw a diameter. So the diameter looks like that. And then let me define my inscribed angle. This diameter is one side of it. | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
So let me draw a diameter. So the diameter looks like that. And then let me define my inscribed angle. This diameter is one side of it. And then the other side maybe is just like that. So let me call this right here psi. If that's psi, this length right here is a radius. | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
This diameter is one side of it. And then the other side maybe is just like that. So let me call this right here psi. If that's psi, this length right here is a radius. That's our radius of our circle. And then this length right here is also going to be the radius of our circle. We're going from the center to the circu... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
If that's psi, this length right here is a radius. That's our radius of our circle. And then this length right here is also going to be the radius of our circle. We're going from the center to the circumference. The circumference is defined by all of the points that are exactly a radius away from the center. So that's ... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
We're going from the center to the circumference. The circumference is defined by all of the points that are exactly a radius away from the center. So that's also a radius. Now, this triangle right here is an isosceles triangle. It has two sides that are equal. Two sides that are definitely equal. And we know that when... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
Now, this triangle right here is an isosceles triangle. It has two sides that are equal. Two sides that are definitely equal. And we know that when we have two sides being equal, their base angles are also equal. So this will also be equal to psi. You might not recognize it because it's tilted up like that. But I think... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
And we know that when we have two sides being equal, their base angles are also equal. So this will also be equal to psi. You might not recognize it because it's tilted up like that. But I think many of us, when we see a triangle that looks like this, if I told you this is r and that is r, that these two sides are equa... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
But I think many of us, when we see a triangle that looks like this, if I told you this is r and that is r, that these two sides are equal, and if this is psi, then you would also know that this angle is also going to be psi. Base angles are equivalent on an isosceles triangle. So if this is psi, that is also psi. Now,... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
Now, let me look at the central angle. This is the central angle subtending the same arc. Let's highlight the arc that they're both subtending. This right here is the arc that they're both going to subtend. So this is my central angle right there, theta. Now, if this angle is theta, what's this angle going to be? This ... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
This right here is the arc that they're both going to subtend. So this is my central angle right there, theta. Now, if this angle is theta, what's this angle going to be? This angle right here. Well, this angle is supplementary to theta. So it's 180 minus theta. When you add these two angles together, you go 180 degree... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
This angle right here. Well, this angle is supplementary to theta. So it's 180 minus theta. When you add these two angles together, you go 180 degrees around, or they kind of form a line. They're supplementary to each other. Now, we also know that these three angles are sitting inside of the same triangle. So they must... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
When you add these two angles together, you go 180 degrees around, or they kind of form a line. They're supplementary to each other. Now, we also know that these three angles are sitting inside of the same triangle. So they must add up to 180 degrees. So we get psi, this psi, plus that psi, plus psi, plus this angle, w... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
So they must add up to 180 degrees. So we get psi, this psi, plus that psi, plus psi, plus this angle, which is 180 minus theta, plus 180 minus theta. These three angles must add up to 180 degrees. They're the three angles of a triangle. Now, we could subtract 180 from both sides. Psi plus psi is 2 psi minus theta is e... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
They're the three angles of a triangle. Now, we could subtract 180 from both sides. Psi plus psi is 2 psi minus theta is equal to 0. Add theta to both sides, you get 2 psi is equal to theta. Multiply both sides by 1 half, or divide both sides by 2. You get psi is equal to 1 half of theta. So we just proved what we set ... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
Add theta to both sides, you get 2 psi is equal to theta. Multiply both sides by 1 half, or divide both sides by 2. You get psi is equal to 1 half of theta. So we just proved what we set out to prove for the special case where our inscribed angle is defined where one of the rays, if you want to view these lines as rays... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
So we just proved what we set out to prove for the special case where our inscribed angle is defined where one of the rays, if you want to view these lines as rays, where one of the rays that defines this inscribed angle is along the diameter, or the diameter forms part of that ray. So this is a special case where one ... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
So now that we know that if this is 50, that this is going to be 100 degrees and likewise, whatever psi is, or whatever theta is, psi is going to be 1 half of that, or whatever psi is, theta is going to be 2 times that. And now this will apply for any time. We can use this notion any time, let me clear this, any time t... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
Let's have an inscribed angle that looks like this. So in this situation, the center, you can kind of view it as it's inside of the angle. That's my inscribed angle. And I want to find a relationship between this inscribed angle and the central angle that's subtending the same arc. So that's my central angle subtending... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
And I want to find a relationship between this inscribed angle and the central angle that's subtending the same arc. So that's my central angle subtending the same arc. Well, you might say, hey Gene, none of these ends or these cords that define this angle, neither of these are diameters, but what we can do is we can d... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
If the center is within these two cords, we can draw a diameter just like that. If we define this angle as psi 1, that angle as psi 2, clearly psi is the sum of those two angles, and we call this angle theta 1 and this angle theta 2, we immediately know that, just using the result I just got, since we have one side of ... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
And so psi, which is psi 1 plus psi 2, so psi 1 plus, let me write the psi a little better, psi 1 plus psi 2 is going to be equal to these two things, 1 half theta 1 plus 1 half theta 2. Psi 1 plus psi 2, this is equal to the first inscribed angle that we want to deal with, just regular psi. That's psi, and this right ... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
What's theta 1 plus theta 2? Well, that's just our original theta that we were dealing with. So now we see that psi is equal to 1 half theta. So now we've proved it for a slightly more general case where our center is inside of the two rays that define that angle. Now, we still haven't addressed a slightly harder situa... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
So now we've proved it for a slightly more general case where our center is inside of the two rays that define that angle. Now, we still haven't addressed a slightly harder situation or a more general situation where if this is the center of our circle, and I have an inscribed angle where the center isn't sitting insid... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
So let's say I have, that's going to be my vertex, and I'll switch colors. So let's say that is one of the chords that defines the angle, just like that. And let's say that is the other chord that defines the angle, just like that. So how do we find the relationship between, let's call this angle right here, let's call... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
So how do we find the relationship between, let's call this angle right here, let's call it psi 1. How do we find the relationship between psi 1 and the central angle that subtends this same arc? So when I talk about the same arc, that's that right there. So the central angle that subtends the same arc will look like t... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
So the central angle that subtends the same arc will look like this. It will look like this. Let's call that theta 1. What we could do is use what we just learned when one side of our inscribed angle is a diameter. So let's construct that. So let me draw a diameter here. The result we want still is that this should be ... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
What we could do is use what we just learned when one side of our inscribed angle is a diameter. So let's construct that. So let me draw a diameter here. The result we want still is that this should be 1 half of this. But let's prove it. Let's draw a diameter. I have to draw a straighter diameter than that. | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
The result we want still is that this should be 1 half of this. But let's prove it. Let's draw a diameter. I have to draw a straighter diameter than that. So let's draw a diameter just like that. And let me call this angle right here, this angle right there. Let me call that psi 2. | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
I have to draw a straighter diameter than that. So let's draw a diameter just like that. And let me call this angle right here, this angle right there. Let me call that psi 2. And it is subtending this arc right there. Let me do that in a darker color. It is subtending this arc right there. | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
Let me call that psi 2. And it is subtending this arc right there. Let me do that in a darker color. It is subtending this arc right there. And so the central angle that subtends that same arc, let me call that theta 2. Now, we know from the last, from really just the earlier part of this video, that psi 2 is going to ... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
It is subtending this arc right there. And so the central angle that subtends that same arc, let me call that theta 2. Now, we know from the last, from really just the earlier part of this video, that psi 2 is going to be equal to 1 half theta 2. They share the diameter is right there. The diameter is one of the chords... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
They share the diameter is right there. The diameter is one of the chords that forms the angle. So psi 2 is going to be equal to 1 half theta 2. This is exactly what we've been doing in the last video, right? This is an inscribed angle. One of the chords that define it is sitting on the diameter. So this is going to be... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
This is exactly what we've been doing in the last video, right? This is an inscribed angle. One of the chords that define it is sitting on the diameter. So this is going to be 1 half of this angle, of the central angle that subtends the same arc. Now, let's look at this larger angle right here. Psi 1 plus psi 2. That l... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
So this is going to be 1 half of this angle, of the central angle that subtends the same arc. Now, let's look at this larger angle right here. Psi 1 plus psi 2. That larger angle is psi 1 plus psi 2. Once again, this subtends this entire arc right here. So it's going to be, and it has a diameter as one of the chords th... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
That larger angle is psi 1 plus psi 2. Once again, this subtends this entire arc right here. So it's going to be, and it has a diameter as one of the chords that defines this huge angle. So this is going to be 1 half of the central angle that subtends the same arc. We're just using what we've already shown in this vide... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
So this is going to be 1 half of the central angle that subtends the same arc. We're just using what we've already shown in this video. So this is going to be equal to 1 half of this huge central angle of theta 1 plus theta 2. So far we've just used everything that we've learned earlier in this video. Now, we already k... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
So far we've just used everything that we've learned earlier in this video. Now, we already know that psi 2 is equal to 1 half theta 2. So let me make that substitution. This is equal to that. So we can say that psi 1 plus, instead of psi 2, I'll write 1 half theta 2 is equal to 1 half theta 1 plus 1 half theta 2. We c... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
This is equal to that. So we can say that psi 1 plus, instead of psi 2, I'll write 1 half theta 2 is equal to 1 half theta 1 plus 1 half theta 2. We can subtract 1 half theta 2 from both sides, and we get our result, psi 1 is equal to 1 half theta 1. And now we're done. We've proven the situation that the inscribed ang... | Inscribed angle theorem proof High School Geometry High School Math Khan Academy.mp3 |
So for example, they say plot the image of point P under a translation by five units to the left and three units up. So let's just do that at first, and then we're gonna think about other ways of describing this. So we wanna go five units to the left. So we start right over here, we're gonna go one, two, three, four, f... | Example translating points.mp3 |
So we start right over here, we're gonna go one, two, three, four, five units to the left, and then we're gonna go three units up. So that's going to be one, two, three. And so the image of point P, I guess, would show up right over here after this translation described this way. Now there are other ways that you could... | Example translating points.mp3 |
Now there are other ways that you could describe this translation. Here we described it just in plain English by five units to the left and three units up, but you could, and this will look fancy, but as we'll see, it's hopefully a pretty intuitive way to describe a translation. You could say, look, I'm gonna take some... | Example translating points.mp3 |
So I would say x minus five comma y, and what do we do to the y coordinate? Well, we're going to increase it by three. We're gonna translate three units up. So y plus three. So all this is saying is whatever x and y coordinates you have, this translation will make you take five from the x. That's what meaning this righ... | Example translating points.mp3 |
So y plus three. So all this is saying is whatever x and y coordinates you have, this translation will make you take five from the x. That's what meaning this right over here is five units to the left, five units to the left, and then this right over here is saying three units up. Increase your y coordinate by three, d... | Example translating points.mp3 |
Increase your y coordinate by three, decrease your x coordinate by five, and so let's just test this out with this particular coordinate, with this particular point. So this point right over, p has the coordinates, its x coordinate is three, and its y coordinate is negative four. So let's see how that works. So I have ... | Example translating points.mp3 |
So I have three comma negative four, and I want to apply this translation. What happens? Well, let me just do my coordinates, and so I started off with three and negative four, and I'm going to subtract five from the three. So subtract five here. We see that right over there, and we're going to add three to the y. So n... | Example translating points.mp3 |
So subtract five here. We see that right over there, and we're going to add three to the y. So notice, instead of an x, now I have a three. Instead of an x, now I have a three. Instead of a y, now I have a negative four. Instead of a y, now I have a negative four. And so another way of writing this, we're going from th... | Example translating points.mp3 |
Instead of an x, now I have a three. Instead of a y, now I have a negative four. Instead of a y, now I have a negative four. And so another way of writing this, we're going from three comma negative four to three minus five is negative two, and negative four plus three is negative one. So what are the coordinates right... | Example translating points.mp3 |
And so another way of writing this, we're going from three comma negative four to three minus five is negative two, and negative four plus three is negative one. So what are the coordinates right over here? Well, the coordinate of this point is indeed negative two comma negative one. So notice how this, as you can see,... | Example translating points.mp3 |
So notice how this, as you can see, this formula, the algebraic formula that shows how we map our coordinates, how it's able to draw the connection between the coordinates. And so you'll see questions where they'll tell you, hey, plot the image and they'll describe it like this. Translate x units to the left or the rig... | Example translating points.mp3 |
And what we have right here is a fairly classic problem. And what I want to do is I want to figure out, just given the information here, so obviously I have a triangle here. I have another triangle over here. We were given some of the angles inside of these triangles. Given the information over here, I want to figure o... | Triangle angle example 1 Angles and intersecting lines Geometry Khan Academy.mp3 |
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