WEBVTT 1 00:00:00.426 --> 00:00:01.946 Euclidean Geometry 2 00:00:02.367 --> 00:00:03.547 Logic and Problem Solving 3 00:00:04.670 --> 00:00:09.710 Eucildean Logical Thinking Skills for Digital Content Creator Minimal knowledge for logical thinking 2 4 00:00:30.359 --> 00:00:32.199 Hello this is Jonghwa Park 5 00:00:32.199 --> 00:00:38.279 In this lecture we will learn about the minimal knowledge that Euclidean geometric requires 6 00:00:38.739 --> 00:00:42.599 We often emphasize idea as more important than knowledge 7 00:00:42.919 --> 00:00:47.239 You will also feel this as digital content creators 8 00:00:47.499 --> 00:00:52.339 But can we come up with ideas without any knowledge? 9 00:00:52.599 --> 00:00:53.939 We can't 10 00:00:53.939 --> 00:00:58.799 There is no idea made with no knowledge at all 11 00:00:58.799 --> 00:01:02.279 Basic knowledge is needed to create an idea 12 00:01:02.379 --> 00:01:03.960 What we must be cautious of is 13 00:01:03.960 --> 00:01:09.362 being tangled up in the former knowledge and doing only what was done before without anything new 14 00:01:09.680 --> 00:01:15.220 In this lecture, we will learn about the minimal knowledge that Euclidean geometric requires 15 00:01:15.680 --> 00:01:20.780 If you know this, you won't need anything more to make ideas 16 00:01:21.180 --> 00:01:23.240 I hope you can create more ideas more confidently 17 00:01:23.241 --> 00:01:27.088 with the knowledge from this lecture in the future 18 00:01:27.749 --> 00:01:31.246 Euclid's axiom 19 00:01:31.726 --> 00:01:34.740 Euclidean geometry starts from very basic things 20 00:01:34.740 --> 00:01:39.740 and creates mathematical theorems with error-less deductive logic 21 00:01:40.240 --> 00:01:44.040 The most basic starting point of Euclidean geometry is 22 00:01:44.041 --> 00:01:46.181 corresponding angles are identical 23 00:01:46.181 --> 00:01:49.980 This explained in diagram is as followed 24 00:01:49.980 --> 00:01:55.300 If a straight line crosses two parallel lines 25 00:01:55.301 --> 00:01:58.440 angle alpha and beta are corresponding angles 26 00:01:58.440 --> 00:02:01.940 The size of angle alpha and beta are identical 27 00:02:01.940 --> 00:02:06.900 These two angles facing each other are called vertical angles 28 00:02:06.900 --> 00:02:09.960 Their sizes are also identical 29 00:02:10.200 --> 00:02:13.100 This can be proved like following 30 00:02:13.300 --> 00:02:19.940 With the red line, the dotted angle is 180-beta 31 00:02:20.280 --> 00:02:27.800 and with the blue line, the dotted angle and angle alpha's sum would be 180 32 00:02:27.800 --> 00:02:32.800 So 180-beta+alpha is 180 33 00:02:33.480 --> 00:02:37.520 So we can conclude that alpha=beta 34 00:02:37.960 --> 00:02:44.860 Angle alpha's verticle angle and angle beta are alternate interior angles 35 00:02:44.860 --> 00:02:48.800 Angle alpha and beta are identical as corresponding angles 36 00:02:48.800 --> 00:02:53.800 and angle alpha and alpha prime are identical as vertical angles 37 00:02:53.800 --> 00:03:00.280 which makes alternate interior angles angle alpha prime and beta identical as well 38 00:03:00.972 --> 00:03:06.172 Euclidean geometry starts from the fact that corresponding angles 39 00:03:06.173 --> 00:03:08.293 vertical angles, and alternate interior angles are identical 40 00:03:08.704 --> 00:03:13.200 Remembering this and applying it to many situations to solve problems 41 00:03:13.201 --> 00:03:15.281 is Euclidean geometry 42 00:03:15.281 --> 00:03:19.420 We know that the sum of the interior angles of a triangle is 180 43 00:03:19.880 --> 00:03:22.640 Then how can we prove this 44 00:03:23.440 --> 00:03:27.240 When triangle ABC is given as followed 45 00:03:27.240 --> 00:03:33.140 when we draw a straight line L that is parallel to the base BC and crosses point A 46 00:03:33.140 --> 00:03:35.240 the interior angles of the triangle 47 00:03:35.240 --> 00:03:40.120 are identical as alternate interior angles 48 00:03:40.560 --> 00:03:46.180 Like this, we can see that alpha, beta, and gamma are on top of a straight line 49 00:03:46.180 --> 00:03:51.100 Therefore alpha+beta+gamma is 180 50 00:03:51.440 --> 00:03:54.983 Like this, based on what we already know 51 00:03:54.984 --> 00:03:58.405 logically calculating what we don't know 52 00:03:58.406 --> 00:04:00.013 is what math is 53 00:04:00.632 --> 00:04:04.220 From the fact that the sum of interior angles of a triangle is 180 54 00:04:04.221 --> 00:04:07.021 we can think of the following relationship 55 00:04:07.801 --> 00:04:11.880 This relationship appears very often in problem solving 56 00:04:11.881 --> 00:04:14.521 which is also called the external angle theorem 57 00:04:14.661 --> 00:04:19.899 It would be good to remember this and apply it in problem solving 58 00:04:20.799 --> 00:04:26.919 One vertical angle's external angle is identical to the sum of other two vertical angles' internal angles 59 00:04:27.699 --> 00:04:31.119 It would be good to understand this with problems 60 00:04:31.599 --> 00:04:35.859 Problem 1, two straight lines are parallel 61 00:04:35.859 --> 00:04:39.259 What is the size of angle X? 62 00:04:39.779 --> 00:04:42.699 We can solve this with what we've just learned 63 00:04:42.699 --> 00:04:43.779 Let's see 64 00:04:44.419 --> 00:04:48.650 Just with the fact that corresponding angles and alternate interior angles are identical in parallel lines 65 00:04:48.651 --> 00:04:53.496 we must understand the relationship between the straight lines and angles to solve this problem 66 00:04:54.159 --> 00:05:00.919 Let's think of the given angles divided 67 00:05:01.119 --> 00:05:03.559 For other angles as well, we can draw a parallel line 68 00:05:03.560 --> 00:05:06.420 and divide the angles 69 00:05:06.679 --> 00:05:10.379 We can calculate the divided angles 70 00:05:10.380 --> 00:05:13.643 and eventually get the size of X 71 00:05:13.959 --> 00:05:19.599 As you can see, the size of angle X is 44 72 00:05:20.059 --> 00:05:28.579 Problem 2, there is quadrilateral ABCD between two parallel straight lines E and F 73 00:05:28.579 --> 00:05:32.499 The angles of same color are of identical sizes 74 00:05:32.499 --> 00:05:35.179 What is the size of angle D? 75 00:05:35.319 --> 00:05:39.419 Let's say the angle between A and C is angle AB 76 00:05:39.799 --> 00:05:42.999 A line parallel with straight lines E and F 77 00:05:43.000 --> 00:05:46.180 and that passes point B is drawn 78 00:05:46.799 --> 00:05:49.578 From the fact that two alternate interior angles are identical 79 00:05:49.579 --> 00:05:54.740 we can conclude that a+b is 150 80 00:05:55.339 --> 00:05:58.659 The sum of the internal angles of a quadrilateral is 360 81 00:05:58.659 --> 00:06:05.059 Therefore a+b+150+angle D is 360 82 00:06:05.339 --> 00:06:08.499 Therefore angle D is 60 83 00:06:09.259 --> 00:06:12.759 The sum of internal angles of a quadrilateral is 360 84 00:06:12.819 --> 00:06:17.179 Because all quadrilaterals can be divided into two triangles 85 00:06:17.179 --> 00:06:20.819 and since the sum of internal angles of a triangle is 180 86 00:06:20.819 --> 00:06:26.779 the sum of internal angles of a quadrilateral is 180X2, 360 87 00:06:27.859 --> 00:06:34.299 Problem 3, what is the sum of all five colored angles? 88 00:06:34.519 --> 00:06:39.879 You might try solving this problem yourself first and then look at the solution 89 00:06:46.219 --> 00:06:49.219 The sum of internal angles of a triangle is 180 90 00:06:49.219 --> 00:06:51.899 And the external angle of an angle 91 00:06:51.900 --> 00:06:55.849 is identical to the sum of two other angles' internal angles 92 00:06:55.849 --> 00:06:59.079 This was the external angle theorem 93 00:06:59.359 --> 00:07:03.219 The given shape can be divided into a few triangles 94 00:07:03.379 --> 00:07:06.239 As you can see 95 00:07:06.239 --> 00:07:11.000 you can use the external angle theorem with two triangles 96 00:07:11.359 --> 00:07:15.799 Like this, the five angles 97 00:07:15.799 --> 00:07:21.939 are the same as on triangle's internal angle 98 00:07:22.599 --> 00:07:25.099 Since the sum of internal angles of a triangle is 180 99 00:07:25.100 --> 00:07:29.040 the sum of the five angles are also 180 100 00:07:29.360 --> 00:07:31.540 How were these problems 101 00:07:31.540 --> 00:07:32.960 Did you solve them well? 102 00:07:33.081 --> 00:07:35.158 Or were they harder than you expected? 103 00:07:35.399 --> 00:07:39.959 There are very basic necessary knowledge 104 00:07:40.399 --> 00:07:44.019 and using them many things are solved 105 00:07:44.439 --> 00:07:50.319 But without the basic knowledge, everything just feels difficult 106 00:07:50.319 --> 00:07:52.199 Content creation is also the same 107 00:07:52.499 --> 00:07:56.899 It is creation of new things but there are basic necessary things 108 00:07:57.239 --> 00:07:59.999 I hope you look for what you need 109 00:08:00.000 --> 00:08:02.720 like the Euclidean axioms, what the 110 00:08:02.720 --> 00:08:05.879 basic knowledge needed are 111 00:08:06.455 --> 00:08:09.995 Proportion 112 00:08:10.375 --> 00:08:15.599 The beginning of math starts from Thales who lived 600 years BC 113 00:08:15.899 --> 00:08:19.199 The reason why Thales is considered the first mathematician 114 00:08:19.199 --> 00:08:24.799 is because he gathered mathematical knowledge and taught people 115 00:08:25.179 --> 00:08:27.579 Thales' idea applied in math 116 00:08:27.579 --> 00:08:31.680 is that shapes that are similar keep the same proportion 117 00:08:32.099 --> 00:08:35.259 Even if the sizes are different, when there is a small triangle 118 00:08:35.260 --> 00:08:39.446 if we expand that into a larger one keeping the proportions 119 00:08:39.446 --> 00:08:44.679 the angles and the ratio of the length of the sides remain the same 120 00:08:45.019 --> 00:08:49.439 Thales was who solved many problems with this simple idea 121 00:08:49.759 --> 00:08:55.219 For example, if this rectangle was expanded keeping the proportion 122 00:08:55.219 --> 00:08:58.559 the rectangle that had the width of 6 and height of 2 123 00:08:58.560 --> 00:09:02.422 will become 12 wide and 4 tall 124 00:09:02.422 --> 00:09:04.799 It increased in a certain ratio 125 00:09:05.319 --> 00:09:07.579 This is applied to all shapes 126 00:09:07.579 --> 00:09:11.477 and it goes the same for the simplest shape, triangle 127 00:09:11.779 --> 00:09:15.899 Two similar triangles have maintain a certain ratio for the length of their sides 128 00:09:16.359 --> 00:09:22.359 Remember what you see in the screen 129 00:09:22.799 --> 00:09:25.059 It's very obvious in a way 130 00:09:25.059 --> 00:09:28.839 Similar triangles maintain a certain ratio 131 00:09:29.059 --> 00:09:31.639 This was Thales' idea 132 00:09:31.639 --> 00:09:36.399 and this became the key in Euclidean geometric problem solving 133 00:09:36.399 --> 00:09:39.899 There is a key idea in problem solving 134 00:09:39.899 --> 00:09:42.139 It would also be good for you 135 00:09:42.139 --> 00:09:45.559 digital content creators, to create and find 136 00:09:45.559 --> 00:09:49.555 a key idea for your creation 137 00:09:50.099 --> 00:09:54.179 When content consumers experience content 138 00:09:54.179 --> 00:09:58.339 around that idea, they will understand and enjoy the content easier 139 00:09:58.599 --> 00:10:03.699 We'll solve problems to see how useful Thales' idea is 140 00:10:03.879 --> 00:10:09.959 Problem 4, find the area of the colored rectangle 141 00:10:10.359 --> 00:10:11.879 At first sight you might be confused 142 00:10:11.879 --> 00:10:15.780 and think we only have information about what we don't need while there is no information on what do need 143 00:10:15.781 --> 00:10:16.880 How can I solve this problem? 144 00:10:16.880 --> 00:10:18.679 You might think this way 145 00:10:19.119 --> 00:10:22.319 But people born in 600 BC 146 00:10:22.320 --> 00:10:24.320 learned math from Thales 147 00:10:24.600 --> 00:10:28.699 and will have thought 'this is an easy one' 148 00:10:28.699 --> 00:10:30.699 You can try as well 149 00:10:35.819 --> 00:10:41.159 In the problem, you can see that for triangles A and B 150 00:10:41.159 --> 00:10:46.039 sizes of all three angles are identical 151 00:10:46.479 --> 00:10:49.819 because they are corresponding angles 152 00:10:50.139 --> 00:10:55.039 We say that A and B are similar 153 00:10:55.379 --> 00:10:58.699 Using the fact that A and B are similar 154 00:10:58.699 --> 00:11:02.099 we can think of the rate of similar 155 00:11:02.099 --> 00:11:04.959 7:a = b:12 156 00:11:05.439 --> 00:11:08.479 Therefore by using the rate of similar 157 00:11:08.479 --> 00:11:11.681 ab=7 X 12 158 00:11:11.682 --> 00:11:13.399 84 159 00:11:13.759 --> 00:11:17.239 The area of the rectangle we are looking for is a, b 160 00:11:17.240 --> 00:11:19.240 so the answer is 84 161 00:11:19.419 --> 00:11:22.559 To add on to the calculation 162 00:11:22.560 --> 00:11:26.259 from a:b=c:d 163 00:11:26.260 --> 00:11:29.790 to b x c = a x d 164 00:11:30.030 --> 00:11:35.499 In the following proportionality, the product of the inner two are the same as the outer two 165 00:11:35.499 --> 00:11:37.299 This is how it was done 166 00:11:37.739 --> 00:11:42.639 This is a calculation of the following fraction expression of proportions 167 00:11:42.879 --> 00:11:45.139 a / b = d / c 168 00:11:45.140 --> 00:11:49.132 This is calculated as 169 00:11:49.312 --> 00:11:50.559 b x c = a x d 170 00:11:50.919 --> 00:11:58.139 Problem 5, what is the area of the colored square? 171 00:11:58.139 --> 00:12:03.599 In this problem as well, finding similar triangles is needed 172 00:12:04.059 --> 00:12:06.999 Let's find similar triangles 173 00:12:07.539 --> 00:12:13.319 We can see that large triangle ABC and 174 00:12:13.319 --> 00:12:15.016 small triangle DEC are similar 175 00:12:15.299 --> 00:12:20.859 If we say the length of a side of the colored square is a 176 00:12:20.859 --> 00:12:24.979 Triangle DEC's height is 10-a 177 00:12:25.027 --> 00:12:27.787 and the base is a 178 00:12:27.867 --> 00:12:31.999 Since triangles ABC and DEC are similar 179 00:12:32.000 --> 00:12:34.563 we can make the following expression 180 00:12:34.899 --> 00:12:39.179 10:15 = 10 - a:a 181 00:12:39.659 --> 00:12:42.399 and we can calculate this 182 00:12:42.959 --> 00:12:47.079 There for the length of a side of the colored square is 6 183 00:12:47.080 --> 00:12:49.897 and the area of the square is 6 squared 184 00:12:49.898 --> 00:12:51.419 36 185 00:12:51.419 --> 00:12:55.179 One most basic thing about triangles is about the area 186 00:12:55.379 --> 00:12:58.479 The area of a triangle is as follows 187 00:12:58.479 --> 00:13:01.199 Base multiplied by height multiplied by 1/2 188 00:13:01.339 --> 00:13:03.979 This is what we saw at lecture 1 189 00:13:04.259 --> 00:13:08.599 This applies the same for not only acute triangles 190 00:13:08.600 --> 00:13:11.850 but also for obtuse triangles with one angle larger than 90 191 00:13:11.850 --> 00:13:17.639 It's good to remember that two triangles with the same base length and height have the same area 192 00:13:17.859 --> 00:13:23.279 That is because we often need to remember this when solving problems 193 00:13:23.279 --> 00:13:25.599 Let's take a look at some more problems 194 00:13:26.319 --> 00:13:31.379 Problem 6, find the area of x 195 00:13:32.699 --> 00:13:38.239 We can think of two similar triangles in the problem 196 00:13:38.239 --> 00:13:45.119 First is saying that we draw a foot of perpendicular from point E to AD and name that H 197 00:13:45.119 --> 00:13:52.519 triangle ADE's area is 1/2 x EH x 8 198 00:13:52.520 --> 00:13:53.940 8 199 00:13:53.940 --> 00:13:58.839 Therefore the length of EH is 2 200 00:13:58.839 --> 00:14:04.579 The second triangle is, drawing foot of perpendicular from point C to AB and calling it J 201 00:14:04.579 --> 00:14:10.699 Triangles AEH and ACJ are similar 202 00:14:10.699 --> 00:14:16.259 Therefore 3:2=11:CJ 203 00:14:16.759 --> 00:14:22.080 The area of triangle ABC is 1/2 x 12 x 22/3 204 00:14:22.080 --> 00:14:23.540 44 205 00:14:23.839 --> 00:14:27.139 Therefore x is 44-8 206 00:14:27.140 --> 00:14:28.736 36 is the answer 207 00:14:29.474 --> 00:14:32.755 Similarity and congruence 208 00:14:32.979 --> 00:14:37.139 Thinking that A and B are similar is an abstract thought 209 00:14:37.439 --> 00:14:42.819 People who lived 600 BC also thought abstractly 210 00:14:42.819 --> 00:14:46.259 and used this to solve specific problems 211 00:14:46.739 --> 00:14:50.039 In many problems we face, there are similarities 212 00:14:50.040 --> 00:14:53.431 and there are also ones that are 213 00:14:53.431 --> 00:14:55.499 of exact same sizes 214 00:14:55.739 --> 00:14:57.959 This is called congruence 215 00:14:57.959 --> 00:15:03.439 We can use the relationship of similarity and congruence very usefully in problem solving 216 00:15:03.439 --> 00:15:05.939 Let's see in problems 217 00:15:06.539 --> 00:15:11.899 Problem 7, find the area of the colored area 218 00:15:11.899 --> 00:15:16.499 The key of this problem is that the two triangles facing each other are similar 219 00:15:16.499 --> 00:15:21.399 and the rate of similar is 2:1 from the base length 220 00:15:21.399 --> 00:15:24.398 Therefore the height is also 2:1 221 00:15:24.758 --> 00:15:26.798 and it's four grids horizontally 222 00:15:26.799 --> 00:15:32.634 so 2:1=4x2/3:4x1/3 223 00:15:32.799 --> 00:15:36.899 Therefore the colored area's area can be calculated as followed 224 00:15:36.899 --> 00:15:44.839 1/2x2x8/3+1/2x1x4/3 225 00:15:45.179 --> 00:15:49.239 This comes out of 10/3 226 00:15:49.439 --> 00:15:53.679 Problem 8, there is trapezoid ABCD 227 00:15:53.679 --> 00:15:58.139 When the areas of the two parts of the four parts are given as followed 228 00:15:58.139 --> 00:16:01.919 find the area of trapezoid ABCD 229 00:16:02.079 --> 00:16:07.379 The key to solving this problem is again finding two similar triangles 230 00:16:07.659 --> 00:16:10.059 Which two triangles are similar? 231 00:16:10.439 --> 00:16:16.439 Triangles AED and BEC are 232 00:16:16.959 --> 00:16:21.959 The two similar triangles' area's rate is given as 1:4 233 00:16:22.179 --> 00:16:26.319 Therefore the ratio of the two triangles' base and height 234 00:16:26.319 --> 00:16:28.919 is all 1:2 235 00:16:29.179 --> 00:16:33.000 Since the area of triangle AED is 1 236 00:16:33.000 --> 00:16:35.900 a x h is 2 237 00:16:35.900 --> 00:16:39.739 The area of trapezoid ABCD can be calculated as followed 238 00:16:39.939 --> 00:16:44.959 1/2 x 3a x 3h = 9 239 00:16:45.359 --> 00:16:51.819 Also, areas of parallelograms and trapezoids can be calculated as shown on the screen 240 00:16:52.099 --> 00:16:55.059 The proof for these formula 241 00:16:55.060 --> 00:16:58.040 are applications of what we've learned 242 00:16:58.040 --> 00:17:00.819 so you could think about it 243 00:17:01.219 --> 00:17:07.440 Try thinking of the triangles inside the parallelogram and trapezoid 244 00:17:07.600 --> 00:17:10.660 Euclidean geometry, an ancient Greek math, began 245 00:17:10.660 --> 00:17:14.442 from understanding and using similarity and congruence 246 00:17:14.800 --> 00:17:18.559 The first mathematician Thales learned a lot of 247 00:17:18.560 --> 00:17:24.041 useful math knowledge going back and forth to Egypt during his trade business 248 00:17:24.360 --> 00:17:28.000 In Egypt, there are giant Pyramids 249 00:17:28.000 --> 00:17:31.440 To the people of that time, Pyramids were more than great architectures 250 00:17:31.441 --> 00:17:33.901 but also a subject of respect 251 00:17:34.360 --> 00:17:38.520 The height of a Pyramid is known to be about 150m 252 00:17:38.540 --> 00:17:42.740 and the length of one side of the base plane is over 200m 253 00:17:43.100 --> 00:17:46.600 To people of Thales' time, 2600 years ago 254 00:17:46.600 --> 00:17:50.940 Pyramids would have been considered an incredibly huge and great architecture 255 00:17:51.160 --> 00:17:56.720 It is said that Egyptians of that time also couldn't calculate the exact height of the Pyramids 256 00:17:57.160 --> 00:18:01.740 They could measure the length of the base plane with a tape measure, but they couldn't for the height 257 00:18:02.080 --> 00:18:06.360 But Thales accurately calculated the height of the Pyramids 258 00:18:06.361 --> 00:18:09.801 using similarity and ratio 259 00:18:10.480 --> 00:18:14.200 What's more important than what you can see is the invisible key idea 260 00:18:14.200 --> 00:18:16.460 I hope you content creators 261 00:18:16.460 --> 00:18:20.653 also find out the invisible key idea needed for you 262 00:18:20.653 --> 00:18:22.273 and use that 263 00:18:22.730 --> 00:18:24.590 Euclid's Axiom The fundamental starting point of Euclidean geometry is the principle that "corresponding angles are equal" In parallel lines, corresponding angles, vertically opposite angles, and alternate angles all have the same measure 264 00:18:24.590 --> 00:18:26.751 Exterior Angle Theorem: An exterior angle equals the sum of the two opposite interior angles To solve problems, one must understand the basic knowledge and apply it to various situations 265 00:18:26.751 --> 00:18:28.852 Proportion Thales' idea applied in math is that "similar objects maintain a certain proportion" 266 00:18:28.852 --> 00:18:29.935 When a shape is expanded, if the angles are maintained the length of bases are also maintained Using similarity and proportion, the length and area can be calculated 267 00:18:29.935 --> 00:18:32.756 Similarity and Congruence Euclidean geometry starts from understanding and applying similarity and congruence 268 00:18:32.756 --> 00:18:36.837 Similarity is two shapes with same shape and when they are shrunk or expanded they become congruent Two identical figures in shape and size, fitting perfectly when overlaid, are called congruent