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## High school geometry

# Inscribed angle theorem proof

CCSS.Math:

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 who's vertex
sits on the circumference of the circle. So that is our inscribed angle. I'll denote it by si -- I'll
use the si for inscribed angle and angles in this video. That si, 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 si. It is an inscribed angle. It sits, it's vertex sits
on the circumference. And if you draw out the to 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 si. It's all very fancy words,
but I think the idea is pretty straightforward. This right here is the arc
subtended by si, where si 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 si, this
angle right here is theta. When I'm going to prove in this
video is that si is always going to be equal
to 1/2 of theta. So if I were to tell you that
si 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
si 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 cords 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 si. If that's si, 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 si. 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
si, then you would also know that this angle is
also going to be si. Base angles are equivalent
on an isosceles triangle. So this is si, that is also si. 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 the kind
of formal 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 si -- this si plus
that si plus si 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. Si plus si is 2 si minus
theta is equal to 0. Add theta to both sides. You get 2 si is equal to theta. Multiply both sides by 1/2
or divide both sides by 2. You get si 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 on 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 si is or whatever
theta is, si's going to be 1/2 of that, or whatever si 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 cords 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 cords 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 si 1, that angle as si 2. Clearly si 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 si 1 is going to be equal to 1/2 theta 1. And we know that si 2 is
going to be 1/2 theta 2. Si 2 is going to
be 1/2 theta 2. So si, which is si 1 plus si 2,
so si 1 plus si 2 is going to be equal to these two things. 1/2 theta 1 plus 1/2 theta 2. Si 1 plus si 2, this is equal
to the first inscribed angle that we want to deal
with, just regular si. That's si. 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 si
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 cords. 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
cords that defines the angle, just like that. And let's say that is the
other cord 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 si 1. How do we find the relationship
between si 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 si 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 si 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
cords that forms the angle. So si 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 cords 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. Si 1 plus si 2. Right, that larger angle
is si 1 plus si 2. Once again, this subtends this
entire arc right here, and it has a diameter as one of the
cords 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 si
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.