[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.69,0:00:03.45,Default,,0000,0000,0000,,What I want to do in this video\Nis to prove one of the more
Dialogue: 0,0:00:03.45,0:00:08.98,Default,,0000,0000,0000,,useful results in geometry, and\Nthat's that an inscribed angle
Dialogue: 0,0:00:08.98,0:00:14.95,Default,,0000,0000,0000,,is just an angle whose vertex\Nsits on the circumference
Dialogue: 0,0:00:14.95,0:00:17.08,Default,,0000,0000,0000,,of the circle.
Dialogue: 0,0:00:17.08,0:00:19.80,Default,,0000,0000,0000,,So that is our inscribed angle.
Dialogue: 0,0:00:19.80,0:00:24.95,Default,,0000,0000,0000,,I'll denote it by psi -- I'll\Nuse the psi for inscribed angle
Dialogue: 0,0:00:24.95,0:00:27.17,Default,,0000,0000,0000,,and angles in this video.
Dialogue: 0,0:00:27.17,0:00:33.53,Default,,0000,0000,0000,,That psi, the inscribed angle,\Nis going to be exactly 1/2 of
Dialogue: 0,0:00:33.53,0:00:37.88,Default,,0000,0000,0000,,the central angle that\Nsubtends the same arc.
Dialogue: 0,0:00:37.88,0:00:40.73,Default,,0000,0000,0000,,So I just used a lot a fancy\Nwords, but I think you'll
Dialogue: 0,0:00:40.73,0:00:41.65,Default,,0000,0000,0000,,get what I'm saying.
Dialogue: 0,0:00:41.65,0:00:42.82,Default,,0000,0000,0000,,So this is psi.
Dialogue: 0,0:00:42.82,0:00:44.47,Default,,0000,0000,0000,,It is an inscribed angle.
Dialogue: 0,0:00:44.47,0:00:48.71,Default,,0000,0000,0000,,It sits, its vertex sits\Non the circumference.
Dialogue: 0,0:00:48.71,0:00:52.57,Default,,0000,0000,0000,,And if you draw out the two rays\Nthat come out from this angle
Dialogue: 0,0:00:52.57,0:00:56.04,Default,,0000,0000,0000,,or the two cords that define\Nthis angle, it intersects the
Dialogue: 0,0:00:56.04,0:00:57.34,Default,,0000,0000,0000,,circle at the other end.
Dialogue: 0,0:00:57.34,0:01:00.39,Default,,0000,0000,0000,,And if you look at the part of\Nthe circumference of the circle
Dialogue: 0,0:01:00.39,0:01:03.73,Default,,0000,0000,0000,,that's inside of it, that\Nis the arc that is
Dialogue: 0,0:01:03.73,0:01:06.16,Default,,0000,0000,0000,,subtended by psi.
Dialogue: 0,0:01:06.16,0:01:09.01,Default,,0000,0000,0000,,It's all very fancy words,\Nbut I think the idea is
Dialogue: 0,0:01:09.01,0:01:09.92,Default,,0000,0000,0000,,pretty straightforward.
Dialogue: 0,0:01:09.92,0:01:28.48,Default,,0000,0000,0000,,This right here is the arc\Nsubtended by psi, where psi is
Dialogue: 0,0:01:28.48,0:01:31.56,Default,,0000,0000,0000,,that inscribed angle right over\Nthere, the vertex sitting
Dialogue: 0,0:01:31.56,0:01:32.40,Default,,0000,0000,0000,,on the circumference.
Dialogue: 0,0:01:32.40,0:01:37.92,Default,,0000,0000,0000,,Now, a central angle is an\Nangle where the vertex is
Dialogue: 0,0:01:37.92,0:01:39.46,Default,,0000,0000,0000,,sitting at the center\Nof the circle.
Dialogue: 0,0:01:39.46,0:01:41.88,Default,,0000,0000,0000,,So let's say that this right\Nhere -- I'll try to eyeball
Dialogue: 0,0:01:41.88,0:01:45.51,Default,,0000,0000,0000,,it -- that right there is\Nthe center of the circle.
Dialogue: 0,0:01:45.51,0:01:51.36,Default,,0000,0000,0000,,So let me draw a central angle\Nthat subtends this same arc.
Dialogue: 0,0:01:51.36,0:01:58.47,Default,,0000,0000,0000,,So that looks like a central\Nangle subtending that same arc.
Dialogue: 0,0:01:58.47,0:01:59.39,Default,,0000,0000,0000,,Just like that.
Dialogue: 0,0:01:59.39,0:02:01.44,Default,,0000,0000,0000,,Let's call this theta.
Dialogue: 0,0:02:01.44,0:02:06.03,Default,,0000,0000,0000,,So this angle is psi, this\Nangle right here is theta.
Dialogue: 0,0:02:06.03,0:02:10.12,Default,,0000,0000,0000,,What I'm going to prove in this\Nvideo is that psi is always
Dialogue: 0,0:02:10.12,0:02:14.05,Default,,0000,0000,0000,,going to be equal\Nto 1/2 of theta.
Dialogue: 0,0:02:14.05,0:02:18.22,Default,,0000,0000,0000,,So if I were to tell you that\Npsi is equal to, I don't know,
Dialogue: 0,0:02:18.22,0:02:21.33,Default,,0000,0000,0000,,25 degrees, then you would\Nimmediately know that theta
Dialogue: 0,0:02:21.33,0:02:23.09,Default,,0000,0000,0000,,must be equal to 50 degrees.
Dialogue: 0,0:02:23.09,0:02:26.08,Default,,0000,0000,0000,,Or if I told you that theta was\N80 degrees, then you would
Dialogue: 0,0:02:26.08,0:02:29.30,Default,,0000,0000,0000,,immediately know that\Npsi was 40 degrees.
Dialogue: 0,0:02:29.30,0:02:31.50,Default,,0000,0000,0000,,So let's actually proved this.
Dialogue: 0,0:02:31.50,0:02:34.52,Default,,0000,0000,0000,,So let me clear this.
Dialogue: 0,0:02:34.52,0:02:37.73,Default,,0000,0000,0000,,So a good place to start,\Nor the place I'm going to
Dialogue: 0,0:02:37.73,0:02:40.46,Default,,0000,0000,0000,,start, is a special case.
Dialogue: 0,0:02:40.46,0:02:45.25,Default,,0000,0000,0000,,I'm going to draw an inscribed\Nangle, but one of the chords
Dialogue: 0,0:02:45.25,0:02:47.91,Default,,0000,0000,0000,,that define it is going to be\Nthe diameter of the circle.
Dialogue: 0,0:02:47.91,0:02:50.53,Default,,0000,0000,0000,,So this isn't going to be the\Ngeneral case, this is going
Dialogue: 0,0:02:50.53,0:02:51.32,Default,,0000,0000,0000,,to be a special case.
Dialogue: 0,0:02:51.32,0:02:55.32,Default,,0000,0000,0000,,So let me see, this is the\Ncenter right here of my circle.
Dialogue: 0,0:02:55.32,0:02:59.03,Default,,0000,0000,0000,,I'm trying to eyeball it.
Dialogue: 0,0:02:59.03,0:03:00.77,Default,,0000,0000,0000,,Center looks like that.
Dialogue: 0,0:03:00.77,0:03:04.21,Default,,0000,0000,0000,,So let me draw a diameter.
Dialogue: 0,0:03:04.21,0:03:06.44,Default,,0000,0000,0000,,So the diameter\Nlooks like that.
Dialogue: 0,0:03:06.44,0:03:09.41,Default,,0000,0000,0000,,Then let me define\Nmy inscribed angle.
Dialogue: 0,0:03:09.41,0:03:11.86,Default,,0000,0000,0000,,This diameter is\None side of it.
Dialogue: 0,0:03:11.86,0:03:15.91,Default,,0000,0000,0000,,And then the other side\Nmaybe is just like that.
Dialogue: 0,0:03:15.91,0:03:20.52,Default,,0000,0000,0000,,So let me call this\Nright here psi.
Dialogue: 0,0:03:20.52,0:03:27.12,Default,,0000,0000,0000,,If that's psi, this length right\Nhere is a radius -- that's
Dialogue: 0,0:03:27.12,0:03:29.33,Default,,0000,0000,0000,,our radius of our circle.
Dialogue: 0,0:03:29.33,0:03:33.08,Default,,0000,0000,0000,,Then this length right here is\Nalso going to be the radius of
Dialogue: 0,0:03:33.08,0:03:35.76,Default,,0000,0000,0000,,our circle going from the\Ncenter to the circumference.
Dialogue: 0,0:03:35.76,0:03:38.13,Default,,0000,0000,0000,,Your circumference is defined\Nby all of the points that are
Dialogue: 0,0:03:38.13,0:03:40.34,Default,,0000,0000,0000,,exactly a radius away\Nfrom the center.
Dialogue: 0,0:03:40.34,0:03:43.61,Default,,0000,0000,0000,,So that's also a radius.
Dialogue: 0,0:03:43.61,0:03:47.92,Default,,0000,0000,0000,,Now, this triangle right here\Nis an isosceles triangle.
Dialogue: 0,0:03:47.92,0:03:49.89,Default,,0000,0000,0000,,It has two sides\Nthat are equal.
Dialogue: 0,0:03:49.89,0:03:51.88,Default,,0000,0000,0000,,Two sides that are\Ndefinitely equal.
Dialogue: 0,0:03:51.88,0:03:54.63,Default,,0000,0000,0000,,We know that when we have two\Nsides being equal, their
Dialogue: 0,0:03:54.63,0:03:57.29,Default,,0000,0000,0000,,base angles are also equal.
Dialogue: 0,0:03:57.29,0:04:00.64,Default,,0000,0000,0000,,So this will also\Nbe equal to psi.
Dialogue: 0,0:04:00.64,0:04:02.13,Default,,0000,0000,0000,,You might not recognize\Nit because it's
Dialogue: 0,0:04:02.13,0:04:03.18,Default,,0000,0000,0000,,tilted up like that.
Dialogue: 0,0:04:03.18,0:04:05.72,Default,,0000,0000,0000,,But I think many of us when we\Nsee a triangle that looks like
Dialogue: 0,0:04:05.72,0:04:10.94,Default,,0000,0000,0000,,this, if I told you this is r\Nand that is r, that these two
Dialogue: 0,0:04:10.94,0:04:17.86,Default,,0000,0000,0000,,sides are equal, and if this is\Npsi, then you would also
Dialogue: 0,0:04:17.86,0:04:20.83,Default,,0000,0000,0000,,know that this angle is\Nalso going to be psi.
Dialogue: 0,0:04:20.83,0:04:23.93,Default,,0000,0000,0000,,Base angles are equivalent\Non an isosceles triangle.
Dialogue: 0,0:04:23.93,0:04:26.72,Default,,0000,0000,0000,,So this is psi, that is also psi.
Dialogue: 0,0:04:26.72,0:04:29.77,Default,,0000,0000,0000,,Now, let me look at\Nthe central angle.
Dialogue: 0,0:04:29.77,0:04:32.71,Default,,0000,0000,0000,,This is the central angle\Nsubtending the same arc.
Dialogue: 0,0:04:32.71,0:04:35.92,Default,,0000,0000,0000,,Let's highlight the arc that\Nthey're both subtending.
Dialogue: 0,0:04:35.92,0:04:40.30,Default,,0000,0000,0000,,This right here is the arc that\Nthey're both going to subtend.
Dialogue: 0,0:04:40.30,0:04:44.35,Default,,0000,0000,0000,,So this is my central\Nangle right there, theta.
Dialogue: 0,0:04:44.35,0:04:49.00,Default,,0000,0000,0000,,Now if this angle is theta,\Nwhat's this angle going to be?
Dialogue: 0,0:04:49.00,0:04:50.62,Default,,0000,0000,0000,,This angle right here.
Dialogue: 0,0:04:50.62,0:04:53.01,Default,,0000,0000,0000,,Well, this angle is\Nsupplementary to theta,
Dialogue: 0,0:04:53.01,0:04:56.64,Default,,0000,0000,0000,,so it's 180 minus theta.
Dialogue: 0,0:04:56.64,0:04:59.56,Default,,0000,0000,0000,,When you add these two angles\Ntogether you go 180 degrees
Dialogue: 0,0:04:59.56,0:05:01.75,Default,,0000,0000,0000,,around or they kind\Nof form a line.
Dialogue: 0,0:05:01.75,0:05:03.79,Default,,0000,0000,0000,,They're supplementary\Nto each other.
Dialogue: 0,0:05:03.79,0:05:06.74,Default,,0000,0000,0000,,Now we also know that these\Nthree angles are sitting
Dialogue: 0,0:05:06.74,0:05:08.26,Default,,0000,0000,0000,,inside of the same triangle.
Dialogue: 0,0:05:08.26,0:05:12.03,Default,,0000,0000,0000,,So they must add up\Nto 180 degrees.
Dialogue: 0,0:05:12.03,0:05:19.30,Default,,0000,0000,0000,,So we get psi -- this psi plus\Nthat psi plus psi plus this
Dialogue: 0,0:05:19.30,0:05:25.42,Default,,0000,0000,0000,,angle, which is 180 minus\Ntheta plus 180 minus theta.
Dialogue: 0,0:05:25.42,0:05:29.13,Default,,0000,0000,0000,,These three angles must\Nadd up to 180 degrees.
Dialogue: 0,0:05:29.13,0:05:31.74,Default,,0000,0000,0000,,They're the three\Nangles of a triangle.
Dialogue: 0,0:05:31.74,0:05:34.60,Default,,0000,0000,0000,,Now we could subtract\N180 from both sides.
Dialogue: 0,0:05:37.14,0:05:43.26,Default,,0000,0000,0000,,psi plus psi is 2 psi minus\Ntheta is equal to 0.
Dialogue: 0,0:05:43.26,0:05:44.84,Default,,0000,0000,0000,,Add theta to both sides.
Dialogue: 0,0:05:44.84,0:05:48.77,Default,,0000,0000,0000,,You get 2 psi is equal to theta.
Dialogue: 0,0:05:48.77,0:05:52.85,Default,,0000,0000,0000,,Multiply both sides by 1/2\Nor divide both sides by 2.
Dialogue: 0,0:05:52.85,0:05:56.68,Default,,0000,0000,0000,,You get psi is equal\Nto 1/2 of theta.
Dialogue: 0,0:05:56.68,0:06:00.07,Default,,0000,0000,0000,,So we just proved what we set\Nout to prove for the special
Dialogue: 0,0:06:00.07,0:06:07.12,Default,,0000,0000,0000,,case where our inscribed angle\Nis defined, where one of the
Dialogue: 0,0:06:07.12,0:06:11.20,Default,,0000,0000,0000,,rays, if you want to view these\Nlines as rays, where one of the
Dialogue: 0,0:06:11.20,0:06:15.22,Default,,0000,0000,0000,,rays that defines this\Ninscribed angle is
Dialogue: 0,0:06:15.22,0:06:17.18,Default,,0000,0000,0000,,along the diameter.
Dialogue: 0,0:06:17.18,0:06:19.20,Default,,0000,0000,0000,,The diameter forms\Npart of that ray.
Dialogue: 0,0:06:19.20,0:06:21.72,Default,,0000,0000,0000,,So this is a special\Ncase where one edge is
Dialogue: 0,0:06:21.72,0:06:23.76,Default,,0000,0000,0000,,sitting on the diameter.
Dialogue: 0,0:06:23.76,0:06:27.66,Default,,0000,0000,0000,,So already we could\Ngeneralize this.
Dialogue: 0,0:06:27.66,0:06:30.58,Default,,0000,0000,0000,,So now that we know that if\Nthis is 50 that this is
Dialogue: 0,0:06:30.58,0:06:32.82,Default,,0000,0000,0000,,going to be 100 degrees\Nand likewise, right?
Dialogue: 0,0:06:32.82,0:06:37.46,Default,,0000,0000,0000,,Whatever psi is or whatever\Ntheta is, psi's going to be 1/2
Dialogue: 0,0:06:37.46,0:06:40.45,Default,,0000,0000,0000,,of that, or whatever psi is,\Ntheta is going to
Dialogue: 0,0:06:40.45,0:06:41.83,Default,,0000,0000,0000,,be 2 times that.
Dialogue: 0,0:06:41.83,0:06:44.11,Default,,0000,0000,0000,,And now this will\Napply for any time.
Dialogue: 0,0:06:44.11,0:06:55.44,Default,,0000,0000,0000,,We could use this notion any\Ntime that -- so just using that
Dialogue: 0,0:06:55.44,0:06:59.46,Default,,0000,0000,0000,,result we just got, we can now\Ngeneralize it a little bit,
Dialogue: 0,0:06:59.46,0:07:02.89,Default,,0000,0000,0000,,although this won't apply\Nto all inscribed angles.
Dialogue: 0,0:07:02.89,0:07:05.09,Default,,0000,0000,0000,,Let's have an inscribed\Nangle that looks like this.
Dialogue: 0,0:07:10.68,0:07:12.98,Default,,0000,0000,0000,,So this situation, the center,\Nyou can kind of view it as
Dialogue: 0,0:07:12.98,0:07:15.47,Default,,0000,0000,0000,,it's inside of the angle.
Dialogue: 0,0:07:15.47,0:07:17.15,Default,,0000,0000,0000,,That's my inscribed angle.
Dialogue: 0,0:07:17.15,0:07:18.89,Default,,0000,0000,0000,,And I want to find a\Nrelationship between this
Dialogue: 0,0:07:18.89,0:07:22.45,Default,,0000,0000,0000,,inscribed angle and the central\Nangle that's subtending
Dialogue: 0,0:07:22.45,0:07:24.36,Default,,0000,0000,0000,,to same arc.
Dialogue: 0,0:07:24.36,0:07:29.88,Default,,0000,0000,0000,,So that's my central angle\Nsubtending the same arc.
Dialogue: 0,0:07:29.88,0:07:33.55,Default,,0000,0000,0000,,Well, you might say, hey, gee,\Nnone of these ends or these
Dialogue: 0,0:07:33.55,0:07:37.31,Default,,0000,0000,0000,,chords that define this angle,\Nneither of these are diameters,
Dialogue: 0,0:07:37.31,0:07:40.40,Default,,0000,0000,0000,,but what we can do is\Nwe can draw a diameter.
Dialogue: 0,0:07:40.40,0:07:43.30,Default,,0000,0000,0000,,If the center is within\Nthese two chords we
Dialogue: 0,0:07:43.30,0:07:46.10,Default,,0000,0000,0000,,can draw a diameter.
Dialogue: 0,0:07:46.10,0:07:48.92,Default,,0000,0000,0000,,We can draw a diameter\Njust like that.
Dialogue: 0,0:07:48.92,0:07:51.68,Default,,0000,0000,0000,,If we draw a diameter just like\Nthat, if we define this angle
Dialogue: 0,0:07:51.68,0:07:55.43,Default,,0000,0000,0000,,as psi 1, that angle as psi 2.
Dialogue: 0,0:07:55.43,0:07:58.32,Default,,0000,0000,0000,,Clearly psi is the sum\Nof those two angles.
Dialogue: 0,0:07:58.32,0:08:04.35,Default,,0000,0000,0000,,And we call this angle theta\N1, and this angle theta 2.
Dialogue: 0,0:08:04.35,0:08:07.24,Default,,0000,0000,0000,,We immediately you know that,\Njust using the result I just
Dialogue: 0,0:08:07.24,0:08:12.54,Default,,0000,0000,0000,,got, since we have one side of\Nour angles in both cases being
Dialogue: 0,0:08:12.54,0:08:18.26,Default,,0000,0000,0000,,a diameter now, we know\Nthat psi 1 is going to be
Dialogue: 0,0:08:18.26,0:08:22.01,Default,,0000,0000,0000,,equal to 1/2 theta 1.
Dialogue: 0,0:08:22.01,0:08:24.87,Default,,0000,0000,0000,,And we know that psi 2 is\Ngoing to be 1/2 theta 2.
Dialogue: 0,0:08:24.87,0:08:30.14,Default,,0000,0000,0000,,Psi 2 is going to\Nbe 1/2 theta 2.
Dialogue: 0,0:08:30.14,0:08:39.85,Default,,0000,0000,0000,,So psi, which is psi 1 plus psi 2,\Nso psi 1 plus psi 2 is going to
Dialogue: 0,0:08:39.85,0:08:41.12,Default,,0000,0000,0000,,be equal to these two things.
Dialogue: 0,0:08:41.12,0:08:47.58,Default,,0000,0000,0000,,1/2 theta 1 plus 1/2 theta 2.
Dialogue: 0,0:08:47.58,0:08:51.18,Default,,0000,0000,0000,,psi 1 plus psi 2, this is equal\Nto the first inscribed
Dialogue: 0,0:08:51.18,0:08:53.85,Default,,0000,0000,0000,,angle that we want to deal\Nwith, just regular psi.
Dialogue: 0,0:08:53.85,0:08:54.98,Default,,0000,0000,0000,,That's psi.
Dialogue: 0,0:08:54.98,0:08:58.35,Default,,0000,0000,0000,,And this right here, this\Nis equal to 1/2 times
Dialogue: 0,0:08:58.35,0:09:00.96,Default,,0000,0000,0000,,theta 1 plus theta 2.
Dialogue: 0,0:09:00.96,0:09:03.96,Default,,0000,0000,0000,,What's theta 1 plus theta 2?
Dialogue: 0,0:09:03.96,0:09:06.47,Default,,0000,0000,0000,,Well that's just our\Noriginal theta that
Dialogue: 0,0:09:06.47,0:09:08.49,Default,,0000,0000,0000,,we were dealing with.
Dialogue: 0,0:09:08.49,0:09:12.08,Default,,0000,0000,0000,,So now we see that psi\Nis equal to 1/2 theta.
Dialogue: 0,0:09:12.08,0:09:14.71,Default,,0000,0000,0000,,So now we've proved it for a\Nslightly more general case
Dialogue: 0,0:09:14.71,0:09:20.02,Default,,0000,0000,0000,,where our center is inside\Nof the two rays that
Dialogue: 0,0:09:20.02,0:09:21.64,Default,,0000,0000,0000,,define that angle.
Dialogue: 0,0:09:21.64,0:09:27.10,Default,,0000,0000,0000,,Now, we still haven't addressed\Na slightly harder situation or
Dialogue: 0,0:09:27.10,0:09:33.66,Default,,0000,0000,0000,,a more general situation where\Nif this is the center of our
Dialogue: 0,0:09:33.66,0:09:39.42,Default,,0000,0000,0000,,circle and I have an inscribed\Nangle where the center isn't
Dialogue: 0,0:09:39.42,0:09:40.99,Default,,0000,0000,0000,,sitting inside of\Nthe two chords.
Dialogue: 0,0:09:40.99,0:09:41.82,Default,,0000,0000,0000,,Let me draw that.
Dialogue: 0,0:09:41.82,0:09:48.80,Default,,0000,0000,0000,,So that's going to be my\Nvertex, and I'll switch colors,
Dialogue: 0,0:09:48.80,0:09:51.54,Default,,0000,0000,0000,,so let's say that is one of the\Nchords that defines the
Dialogue: 0,0:09:51.54,0:09:53.32,Default,,0000,0000,0000,,angle, just like that.
Dialogue: 0,0:09:53.32,0:09:57.86,Default,,0000,0000,0000,,And let's say that is the\Nother chord that defines
Dialogue: 0,0:09:57.86,0:09:59.17,Default,,0000,0000,0000,,the angle just like that.
Dialogue: 0,0:09:59.17,0:10:02.50,Default,,0000,0000,0000,,So how do we find the\Nrelationship between, let's
Dialogue: 0,0:10:02.50,0:10:07.91,Default,,0000,0000,0000,,call, this angle right\Nhere, let's call it psi 1.
Dialogue: 0,0:10:07.91,0:10:13.05,Default,,0000,0000,0000,,How do we find the relationship\Nbetween psi 1 and the central
Dialogue: 0,0:10:13.05,0:10:16.16,Default,,0000,0000,0000,,angle that subtends\Nthis same arc?
Dialogue: 0,0:10:16.16,0:10:19.53,Default,,0000,0000,0000,,So when I talk about the same\Narc, that's that right there.
Dialogue: 0,0:10:19.53,0:10:22.72,Default,,0000,0000,0000,,So the central angle that\Nsubtends the same arc
Dialogue: 0,0:10:22.72,0:10:23.66,Default,,0000,0000,0000,,will look like this.
Dialogue: 0,0:10:28.15,0:10:32.91,Default,,0000,0000,0000,,Let's call that theta 1.
Dialogue: 0,0:10:32.91,0:10:36.77,Default,,0000,0000,0000,,What we can do is use what we\Njust learned when one side of
Dialogue: 0,0:10:36.77,0:10:39.35,Default,,0000,0000,0000,,our inscribed angle\Nis a diameter.
Dialogue: 0,0:10:39.35,0:10:41.14,Default,,0000,0000,0000,,So let's construct that.
Dialogue: 0,0:10:41.14,0:10:44.26,Default,,0000,0000,0000,,So let me draw a diameter here.
Dialogue: 0,0:10:44.26,0:10:47.01,Default,,0000,0000,0000,,The result we want still is\Nthat this should be 1/2 of
Dialogue: 0,0:10:47.01,0:10:48.18,Default,,0000,0000,0000,,this, but let's prove it.
Dialogue: 0,0:10:48.18,0:10:57.56,Default,,0000,0000,0000,,Let's draw a diameter\Njust like that.
Dialogue: 0,0:10:57.56,0:11:09.49,Default,,0000,0000,0000,,Let me call this angle right\Nhere, let me call that psi 2.
Dialogue: 0,0:11:09.49,0:11:14.77,Default,,0000,0000,0000,,And it is subtending this arc\Nright there -- let me do
Dialogue: 0,0:11:14.77,0:11:16.14,Default,,0000,0000,0000,,that in a darker color.
Dialogue: 0,0:11:16.14,0:11:19.77,Default,,0000,0000,0000,,It is subtending this\Narc right there.
Dialogue: 0,0:11:19.77,0:11:22.36,Default,,0000,0000,0000,,So the central angle that\Nsubtends that same arc,
Dialogue: 0,0:11:22.36,0:11:25.30,Default,,0000,0000,0000,,let me call that theta 2.
Dialogue: 0,0:11:25.30,0:11:30.89,Default,,0000,0000,0000,,Now, we know from the earlier\Npart of this video that psi
Dialogue: 0,0:11:30.89,0:11:37.60,Default,,0000,0000,0000,,2 is going to be equal\Nto 1/2 theta 2, right?
Dialogue: 0,0:11:37.60,0:11:40.76,Default,,0000,0000,0000,,They share -- the\Ndiameter is right there.
Dialogue: 0,0:11:40.76,0:11:44.30,Default,,0000,0000,0000,,The diameter is one of the\Nchords that forms the angle.
Dialogue: 0,0:11:44.30,0:11:47.50,Default,,0000,0000,0000,,So psi 2 is going to be\Nequal to 1/2 theta 2.
Dialogue: 0,0:11:50.14,0:11:52.81,Default,,0000,0000,0000,,This is exactly what we've been\Ndoing in the last video, right?
Dialogue: 0,0:11:52.81,0:11:55.43,Default,,0000,0000,0000,,This is an inscribed angle.
Dialogue: 0,0:11:55.43,0:11:59.55,Default,,0000,0000,0000,,One of the chords that define\Nis sitting on the diameter.
Dialogue: 0,0:11:59.55,0:12:02.74,Default,,0000,0000,0000,,So this is going to be 1/2 of\Nthis angle, of the central
Dialogue: 0,0:12:02.74,0:12:05.98,Default,,0000,0000,0000,,angle that subtends\Nthe same arc.
Dialogue: 0,0:12:05.98,0:12:09.00,Default,,0000,0000,0000,,Now, let's look at\Nthis larger angle.
Dialogue: 0,0:12:09.00,0:12:11.68,Default,,0000,0000,0000,,This larger angle right here.
Dialogue: 0,0:12:11.68,0:12:14.24,Default,,0000,0000,0000,,Psi 1 plus psi 2.
Dialogue: 0,0:12:14.24,0:12:22.72,Default,,0000,0000,0000,,Right, that larger angle\Nis psi 1 plus psi 2.
Dialogue: 0,0:12:22.72,0:12:28.68,Default,,0000,0000,0000,,Once again, this subtends this\Nentire arc right here, and it
Dialogue: 0,0:12:28.68,0:12:32.10,Default,,0000,0000,0000,,has a diameter as one of the\Nchords that defines
Dialogue: 0,0:12:32.10,0:12:34.31,Default,,0000,0000,0000,,this huge angle.
Dialogue: 0,0:12:34.31,0:12:37.38,Default,,0000,0000,0000,,So this is going to be 1/2\Nof the central angle that
Dialogue: 0,0:12:37.38,0:12:38.58,Default,,0000,0000,0000,,subtends the same arc.
Dialogue: 0,0:12:38.58,0:12:42.27,Default,,0000,0000,0000,,We're just using what we've\Nalready shown in this video.
Dialogue: 0,0:12:42.27,0:12:47.39,Default,,0000,0000,0000,,So this is going to be equal to\N1/2 of this huge central angle
Dialogue: 0,0:12:47.39,0:12:51.37,Default,,0000,0000,0000,,of theta 1 plus theta 2.
Dialogue: 0,0:12:54.31,0:12:56.53,Default,,0000,0000,0000,,So far we've just used\Neverything that we've learned
Dialogue: 0,0:12:56.53,0:12:58.16,Default,,0000,0000,0000,,earlier in this video.
Dialogue: 0,0:12:58.16,0:13:03.16,Default,,0000,0000,0000,,Now, we already know that psi\N2 is equal to 1/2 theta 2.
Dialogue: 0,0:13:03.16,0:13:05.63,Default,,0000,0000,0000,,So let me make that\Nsubstitution.
Dialogue: 0,0:13:05.63,0:13:07.03,Default,,0000,0000,0000,,This is equal to that.
Dialogue: 0,0:13:07.03,0:13:15.33,Default,,0000,0000,0000,,So we can say that si 1 plus\N-- instead of si 2 I'll write
Dialogue: 0,0:13:15.33,0:13:26.63,Default,,0000,0000,0000,,1/2 theta 2 is equal to 1/2\Ntheta 1 plus 1/2 theta 2.
Dialogue: 0,0:13:30.34,0:13:34.02,Default,,0000,0000,0000,,We can subtract 1/2 theta\N2 from both sides, and
Dialogue: 0,0:13:34.02,0:13:35.74,Default,,0000,0000,0000,,we get our result.
Dialogue: 0,0:13:35.74,0:13:40.90,Default,,0000,0000,0000,,Si 1 is equal to 1/2 theta one.
Dialogue: 0,0:13:40.90,0:13:41.97,Default,,0000,0000,0000,,And now we're done.
Dialogue: 0,0:13:41.97,0:13:44.99,Default,,0000,0000,0000,,We have proven the situation\Nthat the inscribed angle is
Dialogue: 0,0:13:44.99,0:13:50.68,Default,,0000,0000,0000,,always 1/2 of the central angle\Nthat subtends the same arc,
Dialogue: 0,0:13:50.68,0:13:53.98,Default,,0000,0000,0000,,regardless of whether the\Ncenter of the circle is inside
Dialogue: 0,0:13:53.98,0:13:58.99,Default,,0000,0000,0000,,of the angle, outside of the\Nangle, whether we have a
Dialogue: 0,0:13:58.99,0:14:00.95,Default,,0000,0000,0000,,diameter on one side.
Dialogue: 0,0:14:00.95,0:14:05.86,Default,,0000,0000,0000,,So any other angle can be\Nconstructed as a sum of
Dialogue: 0,0:14:05.86,0:14:08.30,Default,,0000,0000,0000,,any or all of these that\Nwe've already done.
Dialogue: 0,0:14:08.30,0:14:10.19,Default,,0000,0000,0000,,So hopefully you found this\Nuseful and now we can actually
Dialogue: 0,0:14:10.19,0:14:14.63,Default,,0000,0000,0000,,build on this result to do some\Nmore interesting
Dialogue: 0,0:14:14.63,0:14:16.46,Default,,0000,0000,0000,,geometry proofs.