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Inscribed angle theorem proof

Proving that an inscribed angle is half of a central angle that subtends the same arc. Created by Sal Khan.
Video transcript
What I want to do in this video is to prove one of the more useful results in geometry, and that's that an inscribed angle is just an angle whose vertex sits on the circumference of the circle. So that is our inscribed angle. I'll denote it by psi -- I'll use the psi for inscribed angle and angles in this video. That psi, the inscribed angle, is going to be exactly 1/2 of the central angle that subtends the same arc. So I just used a lot a 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 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. 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 that this right here -- I'll try to eyeball it -- that right there is the center of the circle. 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 theta. What I'm going to prove in this video is that psi is always going to be equal to 1/2 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 would immediately know that psi was 40 degrees. So let's actually proved 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 an 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 going to be a special case. So let me see, this is the center right here of my circle. I'm trying to eyeball it. Center looks like that. So let me draw a diameter. So the diameter looks like that. 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 -- that's our radius of our circle. Then this length right here is also going to be the radius of our circle going from the center to the circumference. Your 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. 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 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 this is psi, that is also psi. 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 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 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 equal to 0. Add theta to both sides. You get 2 psi is equal to theta. Multiply both sides by 1/2 or divide both sides by 2. You get psi is equal to 1/2 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, where one of the rays that defines this inscribed angle is along the diameter. The diameter forms part of that ray. So this is a special case where one edge is sitting on the diameter. So already we could generalize this. So now that we know that if this is 50 that this is going to be 100 degrees and likewise, right? Whatever psi is or whatever theta is, psi's going to be 1/2 of that, or whatever psi is, theta is going to be 2 times that. And now this will apply for any time. We could use this notion any time that -- so just using that result we just got, we can now generalize it a little bit, although this won't apply to all inscribed angles. Let's have an inscribed angle that looks like this. So 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 to same arc. So that's my central angle subtending the same arc. Well, you might say, hey, gee, none of these ends or these chords that define this angle, neither of these are diameters, but what we can do is we can draw a diameter. If the center is within these two chords we can draw a diameter. We can draw a diameter just like that. If we 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 you know that, just using the result I just got, since we have one side of our angles in both cases being a diameter now, we know that psi 1 is going to be equal to 1/2 theta 1. And we know that psi 2 is going to be 1/2 theta 2. Psi 2 is going to be 1/2 theta 2. So psi, which is psi 1 plus psi 2, so psi 1 plus psi 2 is going to be equal to these two things. 1/2 theta 1 plus 1/2 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 here, this is equal to 1/2 times theta 1 plus theta 2. 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 1/2 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 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 inside of the two chords. Let me draw that. So 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 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 this. Let's call that theta 1. What we can 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 1/2 of this, but let's prove it. Let's draw a diameter just like that. Let me call this angle right here, 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. So the central angle that subtends that same arc, let me call that theta 2. Now, we know from the earlier part of this video that psi 2 is going to be equal to 1/2 theta 2, right? 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 1/2 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 is sitting on the diameter. So this is going to be 1/2 of this angle, of the central angle that subtends the same arc. Now, let's look at this larger angle. This larger angle right here. Psi 1 plus psi 2. Right, that larger angle is psi 1 plus psi 2. Once again, this subtends this entire arc right here, and it has a diameter as one of the chords that defines this huge angle. So this is going to be 1/2 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 1/2 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 know that psi 2 is equal to 1/2 theta 2. So let me make that substitution. This is equal to that. So we can say that si 1 plus -- instead of si 2 I'll write 1/2 theta 2 is equal to 1/2 theta 1 plus 1/2 theta 2. We can subtract 1/2 theta 2 from both sides, and we get our result. Si 1 is equal to 1/2 theta one. And now we're done. We have proven the situation that the inscribed angle is always 1/2 of the central angle that subtends the same arc, regardless of whether the center of the circle is inside of the angle, outside of the angle, whether we have a diameter on one side. So any other angle can be constructed as a sum of any or all of these that we've already done. So hopefully you found this useful and now we can actually build on this result to do some more interesting geometry proofs.