 Hi, I'm Zor. Welcome to Unisor Education. Today we will solve a couple of geometrical problems. Now this problem presented as part of the course called Mass Plus and Problems. It's part of the Unisor.com website. On the same website there is a prerequisite course called Mass Fatines. There are physics fatines, there is a relativity for all courses, etc. So everything is totally free. There are no advertisement, no strings attached. You don't even have to sign in unless you are studying under somebody's supervision. And then we just needed to establish the sign-in to establish connection between you and the supervisor, teacher, parent or somebody. But otherwise if you just do it yourself, you don't have to sign in. Now, this course, Mass Plus and Problems, it's dedicated to the development of your creativity, your analytical thinking, just to force your brain to start working outside of the box. So that's the purpose. The prerequisite course called Mass Fatines is a classic course of mathematics for high school, maybe a little bit above that, but generally speaking it's for high school. And it presents the theoretical material. There are some problems there as well, just to illustrate the theory, the theorems, the proofs, etc. Now, these problems are totally different. They are not related to any particular part of the theoretical course, just to illustrate it. No, you have to really come up with something first, something creative, and then it will actually lead you to using some kind of theorems or definitions or whatever you have learned from the theoretical course. So this is to think about, basically. And every problem which I present, I have it as a lecture, it's on the website, and on the same website, together with the lecture on the same screen, you have notes, and notes are like a textbook. It's exactly the same thing, sometimes with the proof, sometimes without proof, if I have the proof during the lecture. But it's very important for you, instead of just listening to whatever I'm saying or reading, whatever the proof is in the notes with the lecture, it's very important for you just to read the conditions, whatever the problem is about, and try to solve it yourself. This is the most important part, and I repeat it in every lecture which I have. Try first to solve the problem yourself, and then watch the lecture or read the notes if there is a proof or solution or whatever. Okay, now, being as of me, let's start with the problem. So the first problem, I do remember this problem given to me when I was probably in high school, very, very long time ago. And I don't remember I was ever solving this problem, I just know that it existed. Now, for this lecture, I just decided to solve it, and well, it was not easy, basically. That's all I'm saying. But nevertheless, whenever you have already come up with a solution, it's kind of easy to explain. It's not easy to really make something like a first step towards the solution. Okay, so here it is. I need a very large triangle. Okay, something like this. Okay. And, okay. So this is ABC triangle. And I saw this. So sides are equal, and this is 20 degree. So that's what's given about this triangle. Now, this line is at 60 degree. This line at 60 degree to this. This is at 50 degree. And what we do know to find out is this angle. So again, I saw this 20 degree, 60 from this line to base, and 50 degree from this line to base. Well, they intercept the opposite sides at points D and E, and you have to find the angle A, D, E. Okay. And again, to tell you the truth, I did not know how to approach it. And, well, first what I did, okay, let's start from calculating. Okay, if this is 20, how much this total angle, this one? Well, we subtract 20 from 180 degrees, which is sum of all angles. That's 160. Now, these angles are equal. So this one is 80, and this one is 80. Right? Okay. That's what I just started with. That's normal, right? Okay. Now, if this is done, then I put some other angles as well, whatever is needed. For example, how can I calculate this angle, for example? A, D, C. Well, I have 60 here and 80 here. That's what, 140. So this one would be 40 degree. I don't know if I need it or not, but I just put it in. But anyway, so let's just continue. What is this angle? So this one is 40. What is this one? So this is 80. This is 50. So it's 130. From 180 remains to be 50. Oops. Now, this is very important. Whenever I hit this particular angle, and I'm just talking about how I was thinking, it immediately obvious that since this is 50 and this is 50, D, C, A is 50, and C, A, C, E, A is 50. Which means that these two sides are equal to each other. Okay. That's fine. Then I decided to draw a parallel line here and connect it with this. So this would be 60 as well. Let's call this point O. So these are two symmetrical lines at 60 degrees. Now, why is it 60? This is parallel and this is connected and this is 60. So this is obviously, I don't want to spend time to prove that if this is parallel line, then this is 60, then this is 60 as well because these are a social triangle, et cetera. 80 degree on both sides. That's obvious. So we'll skip the easy part. Now, let's just think a little bit more. If this is 60 and this is 60, O, O, O, C, A is 60 and O, AC is 60 degrees. Then this is an equilateral triangle. A, A, O, C is equilateral. Which means that this one is equal to O, C and equal to O, A. But now look at this. EA was equal to AC because these are two 50 degrees angles. Now, AO is equal to AC because this is equilateral triangle. Which means these two, this one and this one are equal to each other. Which means that these angles are also equal to each other. Now this one is what? If this is 80 minus 60, this is 20, right? And since these are equal to each other because AE and AO are of the same lengths and equal to AC, then this is supposed to be what? If this is 20 and this is again a social triangle, that would be 80 and 80. So this one would be 80 and this one would be 80. Alright, that's good. Now we know that this one angle which is basically vertical to this one is 60. So now we can determine this one if we want to, right? 80 and 60, that's 40. 60 and 80 is 140 minus the whole angle 180. It will remain 40. Okay, that's great. Now what is this one? This is 60, this is 80, this is 60, this is 80, 140, this is also 40. So these two lines, this one EO and EF are also equal to each other because these angles are 40. And now you see what actually prompted me is that this EFDO looks like a kite. You see, these are perpendicular to each other, at least on my drawing. That actually prompted me to consider this thing. So I just consider this particular shape and look at this. FD and DO and FO actually are all the same because they're all sides of equilateral triangle. Remember, 60 and 60, so this is 60 and 60, every single 60 degrees, so this is equilateral triangle. So this is equal to this. Now ED is a common line between these two triangles, this one EFD and EOD. So side is equal to side, this side is common, and now these are 40 degrees, 40 degrees, so these two are the same. So we have these two triangles EFD and EOD equal to each other by three sides. Which means these are two equal angles. Now the total angle is 60, half of this is 30. And that's what's necessary to determine. AD is 30 degrees. Okay, so what's important here? First important is you draw this parallel line and connect it with this C. So you will have this equilateral triangle and this equilateral triangle. That's easy. How I came up with this, I don't know. Now after that, when I draw all this thing, what prompted me is if you have an exact drawing that I was trying to do as exact as possible. It actually was kind of suspicious that it looks like a perpendicular and it looks like a kite. And then I started to prove it. And basically I put all the angles whatever I could. And then again all of a sudden I had this 50 and this 50 and then this 40 and this 40. So these are kind of steps which I took by proving this type of thing. Sometimes you can just have a good guess what exactly you have to draw in addition to whatever it's given to make some reason and some way to prove whatever is necessary. Sometimes it takes time and that's very important. You have to take time if you have a problem and you don't know how to solve it. And I'm talking about problems you don't know how to solve. You have to think yourself, you have to come up with a solution. Because again more I would say typical way somebody in school will just give you some theoretical material and even the problems they're saying, okay this is a recipe, how to solve a square, the quadratic equation for example. Okay these are the steps, this is the formula, use it. So you know how to solve the problem and you're just using already written for you algorithms. You are using the recipe which has already been given to you and it does not require the brain power. This requires the brain power and that's exactly the purpose why I'm putting all these problems for you. Just to force you to think about outside of the box. Nobody gives you the solution, you have to come up with a solution yourself. With algorithm, with recipe, with step-by-step instruction how can I do it. Okay, second problem. Second problem is as follows. So you have two triangles, two triangles. Now we have to prove their equality. Well by the way whenever I'm saying equal in geometry it's kind of more customary right now to say congruent. And basically it's the same thing and if I'm just using one instead of another, I mean the same. It means basically that two different figures if they are congruent they can be actually somehow move in such a way that they will completely be the same figure or whatever it is. Sometimes it's just movement, sometimes you have to turn it around and then you have to put them together one on the top of another. Sometimes you have to turn it around or like reflect it and then they can actually be the same. So in all these cases sometimes we are saying that they are congruent, sometimes I used to say equal but that's basically the same thing. It assumes certain process after which they coincide completely. For example if you have this triangle and this triangle they need reflection first to coincide but they are equal because all three sides are equal. So that's just different kind of movement which you have to different transformations and all these transformations are supposed to be legitimate. So you have to like move it, you have to turn it and you have to reflect it and if under these circumstances you can make them to coincide with each other they are called congruent or equal. Okay so what's known about these things? Well known that the base of one is equal to the base of another, means congruent or whatever. Also we have an angle on the top. So it's angle A1, C1, B1 equals to angle A2, C2, B2. So the base and the angle on the top. And bisectors, C1, X1 equals to C2, X2 where X1 and X2 are intersections between this is bisector, this is X1, A1, B1, C1. Okay so bases are the same, angles on the top are the same and the top angle bisectors are the same. We have to prove that these are equal, congruent triangles. Now, so what I will do, I will assume that C, I will just put A1 and A2 at the same point, B1 and B2 at the same point and C1 and C2, let's assume they are different. Let's say this is something like this. This will be my second triangle and it has certain bisector, that will be X2, C2. So here are my two triangles which are not coinciding right now, but that's how they will probably be positioned on the plane if I will put the bases on the top of each other. So the bases are supposed to be the same because they are by definition of the same lengths. But let's assume that the vertex C1 is not the same as C2 and that's why we have different intersections of two different bisectors with the base. So this will be A1 and 2 and this is B1 and B2. Okay, now let me put it here, X2. Now, what do I do next? Okay, here is again some kind of additional construction which I did, which helped basically to do whatever I am going to do. My first point was if these angles are the same, then if I will inscribe one of those triangles into a circle, since these angles are the same, this one and this one, point C2 is supposed to belong to a circle circumscribed around the first triangle. Why? Well, because I hope you remember the theorem, I will remind it right now, because I will use it. So if you have a circle, then angle which is inscribed, which is two quarts, which one side coincides. This angle is measured half of the central angle, which is subtended by the same arc. So this is twice as much as this. This is half of the central angle. Or sometimes I can say that this angle is measured by half of this arc, means actually half of the central angle which is subtended by this arc. Now, if you have two quarts which are intersect, how about this angle? Well, this angle is measured by half of the sum of this and this. Well, more precisely if you have two central angles, this one and this angle, so some of these central angles divided by two. That's what I mean. It's measured by half of the sum of these two arcs. And finally, if you have an outside angle, it's measured by half of the difference. Half of the difference between these two. So half of the difference between these two central angles. So this is a known fact and I presented in the theoretical course, Mass for Teens, which is a prerequisite for this one. And I'm going to use it. So at this point, C2 doesn't belong to a circle. It should be either inside or outside. If it's inside, it would be, let's say, here. It would be half of this arc plus half of something else, right? Now, this one is only half of this because it's inscribed into a circle. But near is supposed to be equal. So we cannot add anything. Or if it's outside, we cannot subtract anything from half of this central angle. That's why it's supposed to belong. Now, what's interesting to use, let's put it here. Now, this is bisector. So this is supposed to divide this in half. This is bisector. Cx, this is Cx1, not Cx. So Cx1 is a bisector. So whenever it goes to this point, it's supposed to divide it by 2, right? Since it's a bisector, it's supposed to divide the arc into halves. That's why they are intersecting at the same point, P. Now, we consider this triangle. Pc1, C2. Okay. First, let's take a look at this angle. C2, C1, P. Now, it's inscribed. So it's supposed to be half of this arc. Well, it means half of the central angle which is subtended by this arc, right? So half of this arc. So this angle is equal to one-half of C2B2P. C2BPP, right? Now, how about this angle? And how about this angle? First, let's consider this one. This is C2x2B2. It's equal to half of this one. Half of CB2 plus half of A1P. Remember, I was just saying half of the arcs which are supported by this angle. Okay. Now, how about C1C2P? This angle. Well, it's this arc, right? So it's equal to one-half of C1A1P. C1A1P, right? And how about which one? This angle, which is C1x1A1. This angle is half of this arc, C1A1 plus this one. C1PB2. Okay. So basically we have expressions. Now, this arc and this arc, these are equal to each other, right? So consider this angle, this little triangle. This angle was equal to this, and it's a vertical to this one. Now, this angle was compared to this one. It was equal to this one, but it's vertical to this. So it looks like this angle in the big triangle is equal to this one equal to this and this equal to that. So basically what I'm saying is C1, angle C1C2P was equal to this, X2X1P. And angle C2C1P was equal to opposite, X1X2P. Which means what? Which means that these two triangles are similar. We have two angles of this equals two angles of this. Now, it's not parallel. I'm not saying this parallel because it's basically crossover from this angle to this and from this angle to that. So this angle was equal to this and this to this. That's why it's a cross. Now, if they are similar to each other, and we know that these pieces C1X1 and C2X2 are equal to each other, from this very easily follows that these two pieces are also supposed to be equal to each other. Because we have the proportionality of the sides. Again, that's very easy to prove. Let's call this little thing X1, this little thing X2. And these two pieces, let's say D and D, because the bisectors are supposed to be the same. So what is proportionality? Well, D plus X1 divided by X1 is supposed to be divided by X2. Proportional. Supposed to be equal to D plus X2 divided by X1, right? That's proportionality. Because X1 is supposed to be proportional to this as this is supposed to be proportional to this. Because the angles are crossing each other, right? From which follows DX1 plus X1 squared is equal to DX2 plus X2 squared. From this we do DX1 minus X2. We put this here and this there is equal to X2 squared minus X1 squared, which is X1 minus X1, X2 minus X1, X2 plus X1. Now, obviously X1 and X2 are supposed to be equal to each other. Because if they are not, if X2, one second, X1 minus X2 and X2 minus X1, because if they are not equal to each other, you see the opposite signs. That means D will be equal to minus X1 plus X2. If they are not equal to each other, I mean if they are not equal to zero, if X1 not equal to X2, which means their difference not equal to zero, we can basically cancel it with reverse signs would be minus and that's impossible. So they are supposed to be equal to each other. So if these two are equal and these are proportional, then these two must be equal as well. And this is equilateral triangle and this is equilateral triangle. Now, from the fact that this is equilateral triangle, this PC1C2 follows one. Well, the fact that this is equilateral triangle means that these two triangles are basically symmetrical relative to this diameter line. And that's very easy to prove. I actually don't even want to prove it because it's kind of obvious. Since these two angles are equal and these two are equal as well, you basically have the whole angles are equal to each other, which means that these sides are equal to each other. I mean it's really kind of easy to prove. I don't want to spend time on trivial proof, which basically means that these are exactly the same. Since these angles are equal, the arcs are equal, these two are equal. Okay, I have already proved it. And that means that these two arcs are equal, which means that the sides are equal, the quarts basically are equal. That's it. Again, it requires certain, I would say, creativity. The first kind of creative piece is whenever you circumscribe the triangle with a circle, and basically kind of an easy proof that C2 belongs to the same circle as well. Because it cannot go either outside or inside the circle because that would be a change of the angle. The angles are the same, and they are the same by definition. See, the top angles are the same by definition, so it must be on the circle. Then the second point was that if you will continue these two bisectors, they will go into the same point. Again, you have to actually kind of remember this to the future because basically if you have any circle and you have equal angles, let's say these two angles, bisectors will always go into the same point because they have to divide this arc into equal parts. Any other angle which is supported by the same arc would have bisectors. The bisectors will go into the same point. It's a fact which we have just proven during the proof of this problem, and everything is fine. Okay, after that, there was really another kind of revelation that these two triangles C1, C2 and P and X1, X2 and P are similar to each other because I was using this theorem about the angle between two quarts intersecting inside that is supposed to be half of the sum. Well, after that everything else was really kind of easy. Well, again, the more problems you solve, the more different approaches you will accumulate in your brain and it will be easier to do it next time. Okay, that's it for today. I do suggest you to read notes for this lecture. They are more accurate and the drawings are maybe a little better. And basically that's it. Thank you very much and good luck.