0:00:00.000,0:00:00.590 0:00:00.590,0:00:03.640 Let's say we have a circle,[br]and then we have a 0:00:03.640,0:00:05.280 diameter of the circle. 0:00:05.280,0:00:09.080 Let me draw my best diameter. 0:00:09.080,0:00:09.760 That's pretty good. 0:00:09.760,0:00:12.580 This right here is the diameter[br]of the circle or it's a 0:00:12.580,0:00:14.700 diameter of the circle. 0:00:14.700,0:00:16.110 That's a diameter. 0:00:16.110,0:00:19.220 Let's say I have a triangle[br]where the diameter is one side 0:00:19.220,0:00:26.040 of the triangle, and the angle[br]opposite that side, it's 0:00:26.040,0:00:28.960 vertex, sits some place[br]on the circumference. 0:00:28.960,0:00:34.200 So, let's say, the angle or the[br]angle opposite of this diameter 0:00:34.200,0:00:35.260 sits on that circumference. 0:00:35.260,0:00:38.020 So the triangle[br]looks like this. 0:00:38.020,0:00:44.160 The triangle looks like that. 0:00:44.160,0:00:47.170 What I'm going to show you[br]in this video is that 0:00:47.170,0:00:50.700 this triangle is going[br]to be a right triangle. 0:00:50.700,0:00:54.290 0:00:54.290,0:00:57.040 The 90 degree side is going[br]to be the side that is 0:00:57.040,0:00:58.550 opposite this diameter. 0:00:58.550,0:01:00.340 I don't want to label it[br]just yet because that would 0:01:00.340,0:01:02.140 ruin the fun of the proof. 0:01:02.140,0:01:05.070 Now let's see what we[br]can do to show this. 0:01:05.070,0:01:08.910 Well, we have in our tool kit[br]the notion of an inscribed 0:01:08.910,0:01:12.970 angle, it's relation to[br]a central angle that 0:01:12.970,0:01:14.830 subtends the same arc. 0:01:14.830,0:01:15.720 So let's look at that. 0:01:15.720,0:01:18.950 So let's say that this is an[br]inscribed angle right here. 0:01:18.950,0:01:22.760 Let's call this theta. 0:01:22.760,0:01:25.070 Now let's say that[br]that's the center of 0:01:25.070,0:01:27.370 my circle right there. 0:01:27.370,0:01:30.190 Then this angle right here[br]would be a central angle. 0:01:30.190,0:01:32.620 Let me draw another triangle[br]right here, another 0:01:32.620,0:01:33.460 line right there. 0:01:33.460,0:01:35.130 This is a central[br]angle right here. 0:01:35.130,0:01:38.190 This is a radius. 0:01:38.190,0:01:40.070 This is the same radius[br]-- actually this 0:01:40.070,0:01:41.230 distance is the same. 0:01:41.230,0:01:44.480 But we've learned several[br]videos ago that look, this 0:01:44.480,0:01:48.710 angle, this inscribed angle,[br]it subtends this arc up here. 0:01:48.710,0:01:52.420 0:01:52.420,0:01:55.850 The central angle that subtends[br]that same arc is going 0:01:55.850,0:01:57.400 to be twice this angle. 0:01:57.400,0:01:59.040 We proved that[br]several videos ago. 0:01:59.040,0:02:02.150 So this is going to be 2theta. 0:02:02.150,0:02:05.260 It's the central angle[br]subtending the same arc. 0:02:05.260,0:02:10.120 Now, this triangle right here,[br]this one right here, this 0:02:10.120,0:02:11.620 is an isosceles triangle. 0:02:11.620,0:02:13.800 I could rotate it and[br]draw it like this. 0:02:13.800,0:02:16.480 0:02:16.480,0:02:22.163 If I flipped it over it would[br]look like that, that, and then 0:02:22.163,0:02:25.000 the green side would[br]be down like that. 0:02:25.000,0:02:28.980 And both of these sides[br]are of length r. 0:02:28.980,0:02:31.160 This top angle is 2theta. 0:02:31.160,0:02:33.530 So all I did is I took it[br]and I rotated it around to 0:02:33.530,0:02:35.060 draw it for you this way. 0:02:35.060,0:02:37.050 This side is that[br]side right there. 0:02:37.050,0:02:41.660 Since its two sides are equal,[br]this is isosceles, so these to 0:02:41.660,0:02:43.980 base angles must be the same. 0:02:43.980,0:02:47.580 0:02:47.580,0:02:49.820 That and that must be the same,[br]or if I were to draw it up 0:02:49.820,0:02:55.150 here, that and that must be[br]the exact same base angle. 0:02:55.150,0:02:58.150 Now let me see, I already[br]used theta, maybe I'll 0:02:58.150,0:02:59.800 use x for these angles. 0:02:59.800,0:03:05.230 So this has to be x,[br]and that has to be x. 0:03:05.230,0:03:08.000 So what is x going[br]to be equal to? 0:03:08.000,0:03:12.120 Well, x plus x plus 2theta[br]have to equal 180 degrees. 0:03:12.120,0:03:13.970 They're all in the[br]same triangle. 0:03:13.970,0:03:15.770 So let me write that down. 0:03:15.770,0:03:23.010 We get x plus x plus 2theta,[br]all have to be equal to 180 0:03:23.010,0:03:30.880 degrees, or we get 2x plus[br]2theta is equal to 180 degrees, 0:03:30.880,0:03:35.970 or we get 2x is equal[br]to 180 minus 2theta. 0:03:35.970,0:03:42.980 Divide both sides by 2, you get[br]x is equal to 90 minus theta. 0:03:42.980,0:03:50.590 So x is equal to[br]90 minus theta. 0:03:50.590,0:03:52.890 Now let's see what else[br]we could do with this. 0:03:52.890,0:03:55.130 Well we could look at this[br]triangle right here. 0:03:55.130,0:03:59.160 This triangle, this side over[br]here also has this distance 0:03:59.160,0:04:01.930 right here is also a[br]radius of the circle. 0:04:01.930,0:04:04.080 This distance over here we've[br]already labeled it, is 0:04:04.080,0:04:05.060 a radius of a circle. 0:04:05.060,0:04:08.870 So once again, this is also[br]an isosceles triangle. 0:04:08.870,0:04:12.770 These two sides are equal,[br]so these two base angles 0:04:12.770,0:04:13.500 have to be equal. 0:04:13.500,0:04:17.160 So if this is theta,[br]this is also going to 0:04:17.160,0:04:17.895 be equal to theta. 0:04:17.895,0:04:20.770 And actually, we use that[br]information, we use to actually 0:04:20.770,0:04:25.100 show that first result about[br]inscribed angles and the 0:04:25.100,0:04:27.340 relation between them and[br]central angles subtending 0:04:27.340,0:04:27.980 the same arc. 0:04:27.980,0:04:29.670 So if this is theta, that's[br]theta because this is 0:04:29.670,0:04:31.120 an isosceles triangle. 0:04:31.120,0:04:36.150 So what is this whole[br]angle over here? 0:04:36.150,0:04:39.850 Well it's going to be theta[br]plus 90 minus theta. 0:04:39.850,0:04:41.650 That angle right there's[br]going to be theta 0:04:41.650,0:04:44.690 plus 90 minus theta. 0:04:44.690,0:04:46.270 Well, the thetas cancel out. 0:04:46.270,0:04:49.690 So no matter what, as long as[br]one side of my triangle is the 0:04:49.690,0:04:53.070 diameter, and then the angle or[br]the vertex of the angle 0:04:53.070,0:04:56.620 opposite sits opposite of[br]that side, sits on the 0:04:56.620,0:05:01.780 circumference, then this angle[br]right here is going to be a 0:05:01.780,0:05:08.750 right angle, and this is going[br]to be a right triangle. 0:05:08.750,0:05:11.640 So if I just were to draw[br]something random like this -- 0:05:11.640,0:05:16.010 if I were to just take a point[br]right there, like that, and 0:05:16.010,0:05:19.750 draw it just like that,[br]this is a right angle. 0:05:19.750,0:05:23.220 If I were to draw something[br]like that and go out like 0:05:23.220,0:05:25.240 that, this is a right angle. 0:05:25.240,0:05:27.860 For any of these I could[br]do this exact same proof. 0:05:27.860,0:05:30.090 And in fact, the way I drew it[br]right here, I kept it very 0:05:30.090,0:05:33.810 general so it would apply[br]to any of these triangles. 0:05:33.810,0:05:34.132