0:00:00.690,0:00:03.450
What I want to do in this video[br]is to prove one of the more

0:00:03.450,0:00:08.980
useful results in geometry, and[br]that's that an inscribed angle

0:00:08.980,0:00:14.950
is just an angle whose vertex[br]sits on the circumference

0:00:14.950,0:00:17.080
of the circle.

0:00:17.080,0:00:19.800
So that is our inscribed angle.

0:00:19.800,0:00:24.950
I'll denote it by psi -- I'll[br]use the psi for inscribed angle

0:00:24.950,0:00:27.170
and angles in this video.

0:00:27.170,0:00:33.530
That psi, the inscribed angle,[br]is going to be exactly 1/2 of

0:00:33.530,0:00:37.880
the central angle that[br]subtends the same arc.

0:00:37.880,0:00:40.730
So I just used a lot a fancy[br]words, but I think you'll

0:00:40.730,0:00:41.650
get what I'm saying.

0:00:41.650,0:00:42.820
So this is psi.

0:00:42.820,0:00:44.470
It is an inscribed angle.

0:00:44.470,0:00:48.710
It sits, its vertex sits[br]on the circumference.

0:00:48.710,0:00:52.570
And if you draw out the two rays[br]that come out from this angle

0:00:52.570,0:00:56.040
or the two cords that define[br]this angle, it intersects the

0:00:56.040,0:00:57.340
circle at the other end.

0:00:57.340,0:01:00.390
And if you look at the part of[br]the circumference of the circle

0:01:00.390,0:01:03.730
that's inside of it, that[br]is the arc that is

0:01:03.730,0:01:06.160
subtended by psi.

0:01:06.160,0:01:09.010
It's all very fancy words,[br]but I think the idea is

0:01:09.010,0:01:09.920
pretty straightforward.

0:01:09.920,0:01:28.485
This right here is the arc[br]subtended by psi, where psi is

0:01:28.485,0:01:31.560
that inscribed angle right over[br]there, the vertex sitting

0:01:31.560,0:01:32.400
on the circumference.

0:01:32.400,0:01:37.920
Now, a central angle is an[br]angle where the vertex is

0:01:37.920,0:01:39.460
sitting at the center[br]of the circle.

0:01:39.460,0:01:41.880
So let's say that this right[br]here -- I'll try to eyeball

0:01:41.880,0:01:45.510
it -- that right there is[br]the center of the circle.

0:01:45.510,0:01:51.360
So let me draw a central angle[br]that subtends this same arc.

0:01:51.360,0:01:58.470
So that looks like a central[br]angle subtending that same arc.

0:01:58.470,0:01:59.390
Just like that.

0:01:59.390,0:02:01.440
Let's call this theta.

0:02:01.440,0:02:06.030
So this angle is psi, this[br]angle right here is theta.

0:02:06.030,0:02:10.120
What I'm going to prove in this[br]video is that psi is always

0:02:10.120,0:02:14.050
going to be equal[br]to 1/2 of theta.

0:02:14.050,0:02:18.220
So if I were to tell you that[br]psi is equal to, I don't know,

0:02:18.220,0:02:21.330
25 degrees, then you would[br]immediately know that theta

0:02:21.330,0:02:23.090
must be equal to 50 degrees.

0:02:23.090,0:02:26.080
Or if I told you that theta was[br]80 degrees, then you would

0:02:26.080,0:02:29.300
immediately know that[br]psi was 40 degrees.

0:02:29.300,0:02:31.500
So let's actually proved this.

0:02:31.500,0:02:34.520
So let me clear this.

0:02:34.520,0:02:37.730
So a good place to start,[br]or the place I'm going to

0:02:37.730,0:02:40.460
start, is a special case.

0:02:40.460,0:02:45.250
I'm going to draw an inscribed[br]angle, but one of the chords

0:02:45.250,0:02:47.910
that define it is going to be[br]the diameter of the circle.

0:02:47.910,0:02:50.526
So this isn't going to be the[br]general case, this is going

0:02:50.526,0:02:51.320
to be a special case.

0:02:51.320,0:02:55.325
So let me see, this is the[br]center right here of my circle.

0:02:55.325,0:02:59.030
I'm trying to eyeball it.

0:02:59.030,0:03:00.770
Center looks like that.

0:03:00.770,0:03:04.210
So let me draw a diameter.

0:03:04.210,0:03:06.440
So the diameter[br]looks like that.

0:03:06.440,0:03:09.410
Then let me define[br]my inscribed angle.

0:03:09.410,0:03:11.860
This diameter is[br]one side of it.

0:03:11.860,0:03:15.910
And then the other side[br]maybe is just like that.

0:03:15.910,0:03:20.520
So let me call this[br]right here psi.

0:03:20.520,0:03:27.120
If that's psi, this length right[br]here is a radius -- that's

0:03:27.120,0:03:29.330
our radius of our circle.

0:03:29.330,0:03:33.080
Then this length right here is[br]also going to be the radius of

0:03:33.080,0:03:35.760
our circle going from the[br]center to the circumference.

0:03:35.760,0:03:38.130
Your circumference is defined[br]by all of the points that are

0:03:38.130,0:03:40.340
exactly a radius away[br]from the center.

0:03:40.340,0:03:43.610
So that's also a radius.

0:03:43.610,0:03:47.920
Now, this triangle right here[br]is an isosceles triangle.

0:03:47.920,0:03:49.890
It has two sides[br]that are equal.

0:03:49.890,0:03:51.880
Two sides that are[br]definitely equal.

0:03:51.880,0:03:54.630
We know that when we have two[br]sides being equal, their

0:03:54.630,0:03:57.290
base angles are also equal.

0:03:57.290,0:04:00.640
So this will also[br]be equal to psi.

0:04:00.640,0:04:02.130
You might not recognize[br]it because it's

0:04:02.130,0:04:03.180
tilted up like that.

0:04:03.180,0:04:05.720
But I think many of us when we[br]see a triangle that looks like

0:04:05.720,0:04:10.940
this, if I told you this is r[br]and that is r, that these two

0:04:10.940,0:04:17.860
sides are equal, and if this is[br]psi, then you would also

0:04:17.860,0:04:20.830
know that this angle is[br]also going to be psi.

0:04:20.830,0:04:23.930
Base angles are equivalent[br]on an isosceles triangle.

0:04:23.930,0:04:26.720
So this is psi, that is also psi.

0:04:26.720,0:04:29.770
Now, let me look at[br]the central angle.

0:04:29.770,0:04:32.710
This is the central angle[br]subtending the same arc.

0:04:32.710,0:04:35.920
Let's highlight the arc that[br]they're both subtending.

0:04:35.920,0:04:40.300
This right here is the arc that[br]they're both going to subtend.

0:04:40.300,0:04:44.350
So this is my central[br]angle right there, theta.

0:04:44.350,0:04:49.000
Now if this angle is theta,[br]what's this angle going to be?

0:04:49.000,0:04:50.620
This angle right here.

0:04:50.620,0:04:53.010
Well, this angle is[br]supplementary to theta,

0:04:53.010,0:04:56.640
so it's 180 minus theta.

0:04:56.640,0:04:59.560
When you add these two angles[br]together you go 180 degrees

0:04:59.560,0:05:01.750
around or they kind[br]of form a line.

0:05:01.750,0:05:03.790
They're supplementary[br]to each other.

0:05:03.790,0:05:06.740
Now we also know that these[br]three angles are sitting

0:05:06.740,0:05:08.260
inside of the same triangle.

0:05:08.260,0:05:12.030
So they must add up[br]to 180 degrees.

0:05:12.030,0:05:19.300
So we get psi -- this psi plus[br]that psi plus psi plus this

0:05:19.300,0:05:25.420
angle, which is 180 minus[br]theta plus 180 minus theta.

0:05:25.420,0:05:29.130
These three angles must[br]add up to 180 degrees.

0:05:29.130,0:05:31.740
They're the three[br]angles of a triangle.

0:05:31.740,0:05:34.605
Now we could subtract[br]180 from both sides.

0:05:37.140,0:05:43.260
psi plus psi is 2 psi minus[br]theta is equal to 0.

0:05:43.260,0:05:44.840
Add theta to both sides.

0:05:44.840,0:05:48.770
You get 2 psi is equal to theta.

0:05:48.770,0:05:52.850
Multiply both sides by 1/2[br]or divide both sides by 2.

0:05:52.850,0:05:56.680
You get psi is equal[br]to 1/2 of theta.

0:05:56.680,0:06:00.070
So we just proved what we set[br]out to prove for the special

0:06:00.070,0:06:07.120
case where our inscribed angle[br]is defined, where one of the

0:06:07.120,0:06:11.200
rays, if you want to view these[br]lines as rays, where one of the

0:06:11.200,0:06:15.220
rays that defines this[br]inscribed angle is

0:06:15.220,0:06:17.180
along the diameter.

0:06:17.180,0:06:19.200
The diameter forms[br]part of that ray.

0:06:19.200,0:06:21.720
So this is a special[br]case where one edge is

0:06:21.720,0:06:23.760
sitting on the diameter.

0:06:23.760,0:06:27.660
So already we could[br]generalize this.

0:06:27.660,0:06:30.580
So now that we know that if[br]this is 50 that this is

0:06:30.580,0:06:32.820
going to be 100 degrees[br]and likewise, right?

0:06:32.820,0:06:37.460
Whatever psi is or whatever[br]theta is, psi's going to be 1/2

0:06:37.460,0:06:40.450
of that, or whatever psi is,[br]theta is going to

0:06:40.450,0:06:41.830
be 2 times that.

0:06:41.830,0:06:44.110
And now this will[br]apply for any time.

0:06:44.110,0:06:55.440
We could use this notion any[br]time that -- so just using that

0:06:55.440,0:06:59.460
result we just got, we can now[br]generalize it a little bit,

0:06:59.460,0:07:02.890
although this won't apply[br]to all inscribed angles.

0:07:02.890,0:07:05.090
Let's have an inscribed[br]angle that looks like this.

0:07:10.680,0:07:12.980
So this situation, the center,[br]you can kind of view it as

0:07:12.980,0:07:15.470
it's inside of the angle.

0:07:15.470,0:07:17.150
That's my inscribed angle.

0:07:17.150,0:07:18.890
And I want to find a[br]relationship between this

0:07:18.890,0:07:22.450
inscribed angle and the central[br]angle that's subtending

0:07:22.450,0:07:24.360
to same arc.

0:07:24.360,0:07:29.880
So that's my central angle[br]subtending the same arc.

0:07:29.880,0:07:33.550
Well, you might say, hey, gee,[br]none of these ends or these

0:07:33.550,0:07:37.310
chords that define this angle,[br]neither of these are diameters,

0:07:37.310,0:07:40.400
but what we can do is[br]we can draw a diameter.

0:07:40.400,0:07:43.300
If the center is within[br]these two chords we

0:07:43.300,0:07:46.100
can draw a diameter.

0:07:46.100,0:07:48.920
We can draw a diameter[br]just like that.

0:07:48.920,0:07:51.680
If we draw a diameter just like[br]that, if we define this angle

0:07:51.680,0:07:55.430
as psi 1, that angle as psi 2.

0:07:55.430,0:07:58.320
Clearly psi is the sum[br]of those two angles.

0:07:58.320,0:08:04.350
And we call this angle theta[br]1, and this angle theta 2.

0:08:04.350,0:08:07.240
We immediately you know that,[br]just using the result I just

0:08:07.240,0:08:12.540
got, since we have one side of[br]our angles in both cases being

0:08:12.540,0:08:18.260
a diameter now, we know[br]that psi 1 is going to be

0:08:18.260,0:08:22.010
equal to 1/2 theta 1.

0:08:22.010,0:08:24.870
And we know that psi 2 is[br]going to be 1/2 theta 2.

0:08:24.870,0:08:30.140
Psi 2 is going to[br]be 1/2 theta 2.

0:08:30.140,0:08:39.850
So psi, which is psi 1 plus psi 2,[br]so psi 1 plus psi 2 is going to

0:08:39.850,0:08:41.120
be equal to these two things.

0:08:41.120,0:08:47.580
1/2 theta 1 plus 1/2 theta 2.

0:08:47.580,0:08:51.180
psi 1 plus psi 2, this is equal[br]to the first inscribed

0:08:51.180,0:08:53.850
angle that we want to deal[br]with, just regular psi.

0:08:53.850,0:08:54.980
That's psi.

0:08:54.980,0:08:58.350
And this right here, this[br]is equal to 1/2 times

0:08:58.350,0:09:00.960
theta 1 plus theta 2.

0:09:00.960,0:09:03.960
What's theta 1 plus theta 2?

0:09:03.960,0:09:06.470
Well that's just our[br]original theta that

0:09:06.470,0:09:08.490
we were dealing with.

0:09:08.490,0:09:12.080
So now we see that psi[br]is equal to 1/2 theta.

0:09:12.080,0:09:14.710
So now we've proved it for a[br]slightly more general case

0:09:14.710,0:09:20.020
where our center is inside[br]of the two rays that

0:09:20.020,0:09:21.640
define that angle.

0:09:21.640,0:09:27.100
Now, we still haven't addressed[br]a slightly harder situation or

0:09:27.100,0:09:33.660
a more general situation where[br]if this is the center of our

0:09:33.660,0:09:39.420
circle and I have an inscribed[br]angle where the center isn't

0:09:39.420,0:09:40.990
sitting inside of[br]the two chords.

0:09:40.990,0:09:41.820
Let me draw that.

0:09:41.820,0:09:48.800
So that's going to be my[br]vertex, and I'll switch colors,

0:09:48.800,0:09:51.540
so let's say that is one of the[br]chords that defines the

0:09:51.540,0:09:53.320
angle, just like that.

0:09:53.320,0:09:57.860
And let's say that is the[br]other chord that defines

0:09:57.860,0:09:59.170
the angle just like that.

0:09:59.170,0:10:02.500
So how do we find the[br]relationship between, let's

0:10:02.500,0:10:07.910
call, this angle right[br]here, let's call it psi 1.

0:10:07.910,0:10:13.050
How do we find the relationship[br]between psi 1 and the central

0:10:13.050,0:10:16.160
angle that subtends[br]this same arc?

0:10:16.160,0:10:19.530
So when I talk about the same[br]arc, that's that right there.

0:10:19.530,0:10:22.720
So the central angle that[br]subtends the same arc

0:10:22.720,0:10:23.660
will look like this.

0:10:28.150,0:10:32.910
Let's call that theta 1.

0:10:32.910,0:10:36.770
What we can do is use what we[br]just learned when one side of

0:10:36.770,0:10:39.350
our inscribed angle[br]is a diameter.

0:10:39.350,0:10:41.135
So let's construct that.

0:10:41.135,0:10:44.260
So let me draw a diameter here.

0:10:44.260,0:10:47.010
The result we want still is[br]that this should be 1/2 of

0:10:47.010,0:10:48.180
this, but let's prove it.

0:10:48.180,0:10:57.560
Let's draw a diameter[br]just like that.

0:10:57.560,0:11:09.490
Let me call this angle right[br]here, let me call that psi 2.

0:11:09.490,0:11:14.770
And it is subtending this arc[br]right there -- let me do

0:11:14.770,0:11:16.140
that in a darker color.

0:11:16.140,0:11:19.770
It is subtending this[br]arc right there.

0:11:19.770,0:11:22.360
So the central angle that[br]subtends that same arc,

0:11:22.360,0:11:25.300
let me call that theta 2.

0:11:25.300,0:11:30.890
Now, we know from the earlier[br]part of this video that psi

0:11:30.890,0:11:37.600
2 is going to be equal[br]to 1/2 theta 2, right?

0:11:37.600,0:11:40.760
They share -- the[br]diameter is right there.

0:11:40.760,0:11:44.300
The diameter is one of the[br]chords that forms the angle.

0:11:44.300,0:11:47.500
So psi 2 is going to be[br]equal to 1/2 theta 2.

0:11:50.140,0:11:52.810
This is exactly what we've been[br]doing in the last video, right?

0:11:52.810,0:11:55.430
This is an inscribed angle.

0:11:55.430,0:11:59.550
One of the chords that define[br]is sitting on the diameter.

0:11:59.550,0:12:02.740
So this is going to be 1/2 of[br]this angle, of the central

0:12:02.740,0:12:05.980
angle that subtends[br]the same arc.

0:12:05.980,0:12:09.000
Now, let's look at[br]this larger angle.

0:12:09.000,0:12:11.680
This larger angle right here.

0:12:11.680,0:12:14.240
Psi 1 plus psi 2.

0:12:14.240,0:12:22.720
Right, that larger angle[br]is psi 1 plus psi 2.

0:12:22.720,0:12:28.680
Once again, this subtends this[br]entire arc right here, and it

0:12:28.680,0:12:32.100
has a diameter as one of the[br]chords that defines

0:12:32.100,0:12:34.310
this huge angle.

0:12:34.310,0:12:37.380
So this is going to be 1/2[br]of the central angle that

0:12:37.380,0:12:38.580
subtends the same arc.

0:12:38.580,0:12:42.270
We're just using what we've[br]already shown in this video.

0:12:42.270,0:12:47.390
So this is going to be equal to[br]1/2 of this huge central angle

0:12:47.390,0:12:51.370
of theta 1 plus theta 2.

0:12:54.310,0:12:56.530
So far we've just used[br]everything that we've learned

0:12:56.530,0:12:58.160
earlier in this video.

0:12:58.160,0:13:03.160
Now, we already know that psi[br]2 is equal to 1/2 theta 2.

0:13:03.160,0:13:05.630
So let me make that[br]substitution.

0:13:05.630,0:13:07.030
This is equal to that.

0:13:07.030,0:13:15.330
So we can say that si 1 plus[br]-- instead of si 2 I'll write

0:13:15.330,0:13:26.630
1/2 theta 2 is equal to 1/2[br]theta 1 plus 1/2 theta 2.

0:13:30.340,0:13:34.020
We can subtract 1/2 theta[br]2 from both sides, and

0:13:34.020,0:13:35.740
we get our result.

0:13:35.740,0:13:40.900
Si 1 is equal to 1/2 theta one.

0:13:40.900,0:13:41.970
And now we're done.

0:13:41.970,0:13:44.990
We have proven the situation[br]that the inscribed angle is

0:13:44.990,0:13:50.680
always 1/2 of the central angle[br]that subtends the same arc,

0:13:50.680,0:13:53.980
regardless of whether the[br]center of the circle is inside

0:13:53.980,0:13:58.990
of the angle, outside of the[br]angle, whether we have a

0:13:58.990,0:14:00.950
diameter on one side.

0:14:00.950,0:14:05.860
So any other angle can be[br]constructed as a sum of

0:14:05.860,0:14:08.300
any or all of these that[br]we've already done.

0:14:08.300,0:14:10.190
So hopefully you found this[br]useful and now we can actually

0:14:10.190,0:14:14.630
build on this result to do some[br]more interesting

0:14:14.630,0:14:16.460
geometry proofs.