[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.00,0:00:00.59,Default,,0000,0000,0000,,
Dialogue: 0,0:00:00.59,0:00:03.64,Default,,0000,0000,0000,,Let's say we have a circle,\Nand then we have a
Dialogue: 0,0:00:03.64,0:00:05.28,Default,,0000,0000,0000,,diameter of the circle.
Dialogue: 0,0:00:05.28,0:00:09.08,Default,,0000,0000,0000,,Let me draw my best diameter.
Dialogue: 0,0:00:09.08,0:00:09.76,Default,,0000,0000,0000,,That's pretty good.
Dialogue: 0,0:00:09.76,0:00:12.58,Default,,0000,0000,0000,,This right here is the diameter\Nof the circle or it's a
Dialogue: 0,0:00:12.58,0:00:14.70,Default,,0000,0000,0000,,diameter of the circle.
Dialogue: 0,0:00:14.70,0:00:16.11,Default,,0000,0000,0000,,That's a diameter.
Dialogue: 0,0:00:16.11,0:00:19.22,Default,,0000,0000,0000,,Let's say I have a triangle\Nwhere the diameter is one side
Dialogue: 0,0:00:19.22,0:00:26.04,Default,,0000,0000,0000,,of the triangle, and the angle\Nopposite that side, it's
Dialogue: 0,0:00:26.04,0:00:28.96,Default,,0000,0000,0000,,vertex, sits some place\Non the circumference.
Dialogue: 0,0:00:28.96,0:00:34.20,Default,,0000,0000,0000,,So, let's say, the angle or the\Nangle opposite of this diameter
Dialogue: 0,0:00:34.20,0:00:35.26,Default,,0000,0000,0000,,sits on that circumference.
Dialogue: 0,0:00:35.26,0:00:38.02,Default,,0000,0000,0000,,So the triangle\Nlooks like this.
Dialogue: 0,0:00:38.02,0:00:44.16,Default,,0000,0000,0000,,The triangle looks like that.
Dialogue: 0,0:00:44.16,0:00:47.17,Default,,0000,0000,0000,,What I'm going to show you\Nin this video is that
Dialogue: 0,0:00:47.17,0:00:50.70,Default,,0000,0000,0000,,this triangle is going\Nto be a right triangle.
Dialogue: 0,0:00:50.70,0:00:54.29,Default,,0000,0000,0000,,
Dialogue: 0,0:00:54.29,0:00:57.04,Default,,0000,0000,0000,,The 90 degree side is going\Nto be the side that is
Dialogue: 0,0:00:57.04,0:00:58.55,Default,,0000,0000,0000,,opposite this diameter.
Dialogue: 0,0:00:58.55,0:01:00.34,Default,,0000,0000,0000,,I don't want to label it\Njust yet because that would
Dialogue: 0,0:01:00.34,0:01:02.14,Default,,0000,0000,0000,,ruin the fun of the proof.
Dialogue: 0,0:01:02.14,0:01:05.07,Default,,0000,0000,0000,,Now let's see what we\Ncan do to show this.
Dialogue: 0,0:01:05.07,0:01:08.91,Default,,0000,0000,0000,,Well, we have in our tool kit\Nthe notion of an inscribed
Dialogue: 0,0:01:08.91,0:01:12.97,Default,,0000,0000,0000,,angle, it's relation to\Na central angle that
Dialogue: 0,0:01:12.97,0:01:14.83,Default,,0000,0000,0000,,subtends the same arc.
Dialogue: 0,0:01:14.83,0:01:15.72,Default,,0000,0000,0000,,So let's look at that.
Dialogue: 0,0:01:15.72,0:01:18.95,Default,,0000,0000,0000,,So let's say that this is an\Ninscribed angle right here.
Dialogue: 0,0:01:18.95,0:01:22.76,Default,,0000,0000,0000,,Let's call this theta.
Dialogue: 0,0:01:22.76,0:01:25.07,Default,,0000,0000,0000,,Now let's say that\Nthat's the center of
Dialogue: 0,0:01:25.07,0:01:27.37,Default,,0000,0000,0000,,my circle right there.
Dialogue: 0,0:01:27.37,0:01:30.19,Default,,0000,0000,0000,,Then this angle right here\Nwould be a central angle.
Dialogue: 0,0:01:30.19,0:01:32.62,Default,,0000,0000,0000,,Let me draw another triangle\Nright here, another
Dialogue: 0,0:01:32.62,0:01:33.46,Default,,0000,0000,0000,,line right there.
Dialogue: 0,0:01:33.46,0:01:35.13,Default,,0000,0000,0000,,This is a central\Nangle right here.
Dialogue: 0,0:01:35.13,0:01:38.19,Default,,0000,0000,0000,,This is a radius.
Dialogue: 0,0:01:38.19,0:01:40.07,Default,,0000,0000,0000,,This is the same radius\N-- actually this
Dialogue: 0,0:01:40.07,0:01:41.23,Default,,0000,0000,0000,,distance is the same.
Dialogue: 0,0:01:41.23,0:01:44.48,Default,,0000,0000,0000,,But we've learned several\Nvideos ago that look, this
Dialogue: 0,0:01:44.48,0:01:48.71,Default,,0000,0000,0000,,angle, this inscribed angle,\Nit subtends this arc up here.
Dialogue: 0,0:01:48.71,0:01:52.42,Default,,0000,0000,0000,,
Dialogue: 0,0:01:52.42,0:01:55.85,Default,,0000,0000,0000,,The central angle that subtends\Nthat same arc is going
Dialogue: 0,0:01:55.85,0:01:57.40,Default,,0000,0000,0000,,to be twice this angle.
Dialogue: 0,0:01:57.40,0:01:59.04,Default,,0000,0000,0000,,We proved that\Nseveral videos ago.
Dialogue: 0,0:01:59.04,0:02:02.15,Default,,0000,0000,0000,,So this is going to be 2theta.
Dialogue: 0,0:02:02.15,0:02:05.26,Default,,0000,0000,0000,,It's the central angle\Nsubtending the same arc.
Dialogue: 0,0:02:05.26,0:02:10.12,Default,,0000,0000,0000,,Now, this triangle right here,\Nthis one right here, this
Dialogue: 0,0:02:10.12,0:02:11.62,Default,,0000,0000,0000,,is an isosceles triangle.
Dialogue: 0,0:02:11.62,0:02:13.80,Default,,0000,0000,0000,,I could rotate it and\Ndraw it like this.
Dialogue: 0,0:02:13.80,0:02:16.48,Default,,0000,0000,0000,,
Dialogue: 0,0:02:16.48,0:02:22.16,Default,,0000,0000,0000,,If I flipped it over it would\Nlook like that, that, and then
Dialogue: 0,0:02:22.16,0:02:25.00,Default,,0000,0000,0000,,the green side would\Nbe down like that.
Dialogue: 0,0:02:25.00,0:02:28.98,Default,,0000,0000,0000,,And both of these sides\Nare of length r.
Dialogue: 0,0:02:28.98,0:02:31.16,Default,,0000,0000,0000,,This top angle is 2theta.
Dialogue: 0,0:02:31.16,0:02:33.53,Default,,0000,0000,0000,,So all I did is I took it\Nand I rotated it around to
Dialogue: 0,0:02:33.53,0:02:35.06,Default,,0000,0000,0000,,draw it for you this way.
Dialogue: 0,0:02:35.06,0:02:37.05,Default,,0000,0000,0000,,This side is that\Nside right there.
Dialogue: 0,0:02:37.05,0:02:41.66,Default,,0000,0000,0000,,Since its two sides are equal,\Nthis is isosceles, so these to
Dialogue: 0,0:02:41.66,0:02:43.98,Default,,0000,0000,0000,,base angles must be the same.
Dialogue: 0,0:02:43.98,0:02:47.58,Default,,0000,0000,0000,,
Dialogue: 0,0:02:47.58,0:02:49.82,Default,,0000,0000,0000,,That and that must be the same,\Nor if I were to draw it up
Dialogue: 0,0:02:49.82,0:02:55.15,Default,,0000,0000,0000,,here, that and that must be\Nthe exact same base angle.
Dialogue: 0,0:02:55.15,0:02:58.15,Default,,0000,0000,0000,,Now let me see, I already\Nused theta, maybe I'll
Dialogue: 0,0:02:58.15,0:02:59.80,Default,,0000,0000,0000,,use x for these angles.
Dialogue: 0,0:02:59.80,0:03:05.23,Default,,0000,0000,0000,,So this has to be x,\Nand that has to be x.
Dialogue: 0,0:03:05.23,0:03:08.00,Default,,0000,0000,0000,,So what is x going\Nto be equal to?
Dialogue: 0,0:03:08.00,0:03:12.12,Default,,0000,0000,0000,,Well, x plus x plus 2theta\Nhave to equal 180 degrees.
Dialogue: 0,0:03:12.12,0:03:13.97,Default,,0000,0000,0000,,They're all in the\Nsame triangle.
Dialogue: 0,0:03:13.97,0:03:15.77,Default,,0000,0000,0000,,So let me write that down.
Dialogue: 0,0:03:15.77,0:03:23.01,Default,,0000,0000,0000,,We get x plus x plus 2theta,\Nall have to be equal to 180
Dialogue: 0,0:03:23.01,0:03:30.88,Default,,0000,0000,0000,,degrees, or we get 2x plus\N2theta is equal to 180 degrees,
Dialogue: 0,0:03:30.88,0:03:35.97,Default,,0000,0000,0000,,or we get 2x is equal\Nto 180 minus 2theta.
Dialogue: 0,0:03:35.97,0:03:42.98,Default,,0000,0000,0000,,Divide both sides by 2, you get\Nx is equal to 90 minus theta.
Dialogue: 0,0:03:42.98,0:03:50.59,Default,,0000,0000,0000,,So x is equal to\N90 minus theta.
Dialogue: 0,0:03:50.59,0:03:52.89,Default,,0000,0000,0000,,Now let's see what else\Nwe could do with this.
Dialogue: 0,0:03:52.89,0:03:55.13,Default,,0000,0000,0000,,Well we could look at this\Ntriangle right here.
Dialogue: 0,0:03:55.13,0:03:59.16,Default,,0000,0000,0000,,This triangle, this side over\Nhere also has this distance
Dialogue: 0,0:03:59.16,0:04:01.93,Default,,0000,0000,0000,,right here is also a\Nradius of the circle.
Dialogue: 0,0:04:01.93,0:04:04.08,Default,,0000,0000,0000,,This distance over here we've\Nalready labeled it, is
Dialogue: 0,0:04:04.08,0:04:05.06,Default,,0000,0000,0000,,a radius of a circle.
Dialogue: 0,0:04:05.06,0:04:08.87,Default,,0000,0000,0000,,So once again, this is also\Nan isosceles triangle.
Dialogue: 0,0:04:08.87,0:04:12.77,Default,,0000,0000,0000,,These two sides are equal,\Nso these two base angles
Dialogue: 0,0:04:12.77,0:04:13.50,Default,,0000,0000,0000,,have to be equal.
Dialogue: 0,0:04:13.50,0:04:17.16,Default,,0000,0000,0000,,So if this is theta,\Nthis is also going to
Dialogue: 0,0:04:17.16,0:04:17.90,Default,,0000,0000,0000,,be equal to theta.
Dialogue: 0,0:04:17.90,0:04:20.77,Default,,0000,0000,0000,,And actually, we use that\Ninformation, we use to actually
Dialogue: 0,0:04:20.77,0:04:25.10,Default,,0000,0000,0000,,show that first result about\Ninscribed angles and the
Dialogue: 0,0:04:25.10,0:04:27.34,Default,,0000,0000,0000,,relation between them and\Ncentral angles subtending
Dialogue: 0,0:04:27.34,0:04:27.98,Default,,0000,0000,0000,,the same arc.
Dialogue: 0,0:04:27.98,0:04:29.67,Default,,0000,0000,0000,,So if this is theta, that's\Ntheta because this is
Dialogue: 0,0:04:29.67,0:04:31.12,Default,,0000,0000,0000,,an isosceles triangle.
Dialogue: 0,0:04:31.12,0:04:36.15,Default,,0000,0000,0000,,So what is this whole\Nangle over here?
Dialogue: 0,0:04:36.15,0:04:39.85,Default,,0000,0000,0000,,Well it's going to be theta\Nplus 90 minus theta.
Dialogue: 0,0:04:39.85,0:04:41.65,Default,,0000,0000,0000,,That angle right there's\Ngoing to be theta
Dialogue: 0,0:04:41.65,0:04:44.69,Default,,0000,0000,0000,,plus 90 minus theta.
Dialogue: 0,0:04:44.69,0:04:46.27,Default,,0000,0000,0000,,Well, the thetas cancel out.
Dialogue: 0,0:04:46.27,0:04:49.69,Default,,0000,0000,0000,,So no matter what, as long as\None side of my triangle is the
Dialogue: 0,0:04:49.69,0:04:53.07,Default,,0000,0000,0000,,diameter, and then the angle or\Nthe vertex of the angle
Dialogue: 0,0:04:53.07,0:04:56.62,Default,,0000,0000,0000,,opposite sits opposite of\Nthat side, sits on the
Dialogue: 0,0:04:56.62,0:05:01.78,Default,,0000,0000,0000,,circumference, then this angle\Nright here is going to be a
Dialogue: 0,0:05:01.78,0:05:08.75,Default,,0000,0000,0000,,right angle, and this is going\Nto be a right triangle.
Dialogue: 0,0:05:08.75,0:05:11.64,Default,,0000,0000,0000,,So if I just were to draw\Nsomething random like this --
Dialogue: 0,0:05:11.64,0:05:16.01,Default,,0000,0000,0000,,if I were to just take a point\Nright there, like that, and
Dialogue: 0,0:05:16.01,0:05:19.75,Default,,0000,0000,0000,,draw it just like that,\Nthis is a right angle.
Dialogue: 0,0:05:19.75,0:05:23.22,Default,,0000,0000,0000,,If I were to draw something\Nlike that and go out like
Dialogue: 0,0:05:23.22,0:05:25.24,Default,,0000,0000,0000,,that, this is a right angle.
Dialogue: 0,0:05:25.24,0:05:27.86,Default,,0000,0000,0000,,For any of these I could\Ndo this exact same proof.
Dialogue: 0,0:05:27.86,0:05:30.09,Default,,0000,0000,0000,,And in fact, the way I drew it\Nright here, I kept it very
Dialogue: 0,0:05:30.09,0:05:33.81,Default,,0000,0000,0000,,general so it would apply\Nto any of these triangles.
Dialogue: 0,0:05:33.81,0:05:34.13,Default,,0000,0000,0000,,