 Hi, I'm Zor. Welcome to the new Zor education. I will continue talking about certain problems and theorems actually related to similarity and those which are using similarity maybe to get some interesting facts. Now, this particular set of problems is something which I like a little bit more, maybe, than the simple stuff which I was talking before. This is also simple, but it has this creative spirit, if you wish, which I am kind of insisting that Unisor.com, if you really go through all these theorems and problems, whatever is offered, and try to do everything yourself before, after the lectures, etc., it will develop your very creative thinking. Okay, so this lecture is a little bit with more of this creative spirit in it, if you wish. Alright, so let's prove a few theorems. Given the triangle ABC, besides A, B, B, C, and place A, C, so it goes something like this, A, B, C. B, P is an angle bisector of angle ABC. This is angle bisector, okay. Prove the segments A, P, and P, C are the base cut by an angle bisector B, P are proportional to sides A, B, and B, C. So what we have to do is we have to prove that A, P over P, C is equal to A, B over B, C, where this is an angle bisector. Well, as you see, this is an interesting property of the triangle. I mean, there are certain things which are intuitively obvious. This is not. It's not obvious that the bisector divides the opposite side proportionally to sides of the triangle. So this is one of those theorems which are not obvious, and to prove it requires certain, well, creativity, I should say. Now, what kind of creativity? We can't just look at this picture, prove it. I mean, there is something missing here. It's certain additional constructions, maybe, which would be kind of helpful in proving whatever we want to prove. And to do this type of thing, to do this type of thing, I will have to just, you know, come up with this new, interesting construction, whatever can be. Now, you know what it's too big for my additional construction. So let me just write it slightly differently. I'll do it this way. A, B, C, and this is P. So what's my additional construction which I'm talking about? I will, these are equal ones. Okay. So I will draw a line from C parallel to BP, and it will cross somewhere at point D. So BP and CD are parallel to each other, from which we can conclude that this is also the same angle because these are too parallel, transversal, and these are interior, alternate interior angles, right? At the same time, this angle, ABP, is equal to this one, as these two parallel, this is transversal, and these are corresponding angles. But since these two angles are equal between themselves, then these two will be equal between themselves, because this is equal to this, this is equal to that. These are equal, so these are equal. So B, G, C is equilateral triangle with these two sides equal to each other. Now, let's consider something which we have already learned from the previous lecture. Angle, G, A, C, two parallel lines, this is parallel to this by construction, and they are cutting proportional segments. So the AB to AP would be the same as BG to PC. What can we do from here? Well, actually it's already basically proven. From the BG, we go to another BC, which is, as I was saying, equal in lengths. So it's AB to AP equals to BC to PC. Now this and this, if you compare them, they are exactly the same thing, because this means that AP times BC equals to PC times AB, and this means AB times PC equals to BC times AP. PC, AP, BC, AP, they just commute and PC, AB and PC, AB. Same thing. So this is the same as this. So that's how we can prove in it. Now what's important lesson to learn from this particular problem? Sometimes it's not obvious how to solve the problem, how to construct something or prove the theorem. You need some additional construction, additional drawing, additional element to be introduced into the system. Now this parallel line, it's not trivial, actually, to just come up with this idea. Okay, I need this parallel line and consider this triangle and it will be equilateral. I know it's not trivial, but again, the more problems like this you will solve, the more trivial this will be actually for you. Let's put it this way. So you just do the quantity in this particular case, quantity after quantity of different problems like this, and you will develop this field. Okay, what do I need to help me to prove this theorem? Okay, I need this additional line or this additional circle. That's basically the whole purpose of all these problems, the problem solving. Let me just give you an analogy again, another digression. You would like to build a space shuttle or something like this. Now, if you remember when you were watching this space shuttle launch, you have the whole construction around it, which holds it in place and helps to fill it up with some kind of whatever they are using for energy source, service it, enter it after all when the astronauts are entering this. So all these constructions around the spaceship are necessary to build it and to launch it. This is exactly the same case. You have to build something around your problem to basically address it, to solve it, if you wish. Without these constructions around it, without the anal cranes, for instance, which are helping to hoist certain items when you're building the building, without all those extra equipment around it, you will not be able to build the contemporary building. Well, that's the purpose of these problems. So you have to understand what additional do I need to solve the problem, what kind of environment I have to build. Okay, next. Given triangle ABC with size rk, more or less the same thing, ABC equals 10, BC equals 15, and AC equals 20. BP is an angle bisector, the same thing as before. What are the lengths of segments AP and PC? If this is x, this is 20 minus x, correct? So we will use basically algebraic approach to solve this problem. Now, we know from the previous theorem that AP to PC is related as AD to BC, if this is an angle bisector. So let's just write this particular equation. This is an algebraic problem right now. So we have 10 over x is equal to 15 over 20 minus x. Forget about our drawing, we don't need it anymore. And just solve the algebraic equation like this. Now, we know that none of these is equal to zero, so I'm not going to go into the details what if this is equal to zero, etc. So we'll just multiply it. 10 times 20 minus x is equal to 15x. 200, open the parenthesis, minus 10x equals 15x. 200 equals to add 10x to both sides of the equation. This disappears, this will be 25x. From this, x is equal to 8. That's the answer. 20 minus x is equal to 12. So these two segments are 8 and 12. Now, again, lesson from this particular problem. Well, you can use algebra in geometry. Using certain properties of geometry, you can put together some kind of an equation. And using the algebra, you can solve the equation. And that's how you solve geometric problems in this case. Okay, next. Given the right triangle ABC, it was right angle ABC. Okay, ABC. And this is the right angle. BH is altitude, perpendicular. Okay. Prove that AH divided by BH, is the same as BH divided by HC. How can we prove that? Now, I'm talking about lengths of these segments, obviously. We can divide segments, we can divide their lengths. None of them is equal to zero, again. By the way, it's almost automatic for me to, whenever I see the fraction, think about what if the denominator is equal to zero. And that's why I mentioned this. It should be automatic for you as well. But since we're talking about geometry, we definitely know these are segment lengths which are not equal to zero. So it's okay. So this side is to this as this to this. Now, this thing is called mean proportional. Or this is also a geometrical mean, if you wish. You know the geometrical mean of two numbers is square root of their product. Geometrical mean of three numbers is cubic root of their product, et cetera. If you have N, it will be N multiplied by N. And different components multiplied by each other and the root will be of N's degree. That's a geometrical mean. In this case, it's a geometrical mean of the second degree between, because BH times BH is BH squared. It's equal to AH times HC from which BH is equal to square root of AH times HC. So that's the geometrical mean. Or in this form, it's called mean proportional, proportional mean to the other. All right, so how can we prove that? Well, I could spend so much talking about what to do and what this means, but the proof itself is trivial. Think about this. We have three different triangles, small, medium and large. They are all similar to each other. Why? Because the angles are the same. Now, if you, for instance, compare small with a large, they have a common acute angle. And since they're both right triangles, obviously the second angle is 90 degrees. Same thing with this guy, the medium one. It's also similar to the big triangle because they have common acute angle. And another angle is 90 degrees. So basically they're all similar to each other because they have the same angles. And let me now do it this way. Now it's quite obvious. This angle is equal to this. This is equal to this. Why? Well, obviously because this one is 90 degree minus this one if you consider this triangle. This triangle is 90 degree minus this one if you consider this right triangle. So they are equal to each other. Same thing with these guys. And these are two right angles. So these are similar triangles. And if you will take a look at this, this is nothing but proportionality of sides between small and medium. Small catatoules of the small triangle to a large catatoules of a small triangle which is BH is the same as small catatoules in medium triangle which is BH relative to the bigger catatoules of this medium triangle. Now I'm using small catatoules and bigger catatoules. At the same time I can say catatoules which are lying across the corresponding angles. So what's the catatoules which is lying corresponding to this angle in a small triangle? It's this one. And what's the catatoules which is lying against the same angle in the medium triangle? This one, BH. Now, what's the catatoules which is lying against this angle in the small triangle? This one, BH. And what's the catatoules which is lying against the same angle in the medium triangle? It's this one, which is HC. So, this is just the proportionality of the similar triangles. Nothing to it. Next, given the right triangle, I'll use the same picture, actually, with the right A, B, C, and hypotenuse A, C. Denote the lengths of the calculus. Okay. We'll use letters. Let's call this side opposite to the vertex C. We will use C as its lengths. And the calculus, which is opposite to the vertex A, I will use A. And the hypotenuse, which is opposite to vertex B, would be B. And I will put this as a B1 and this as a B2. So B is divided into B1 and B2. Alright, so what I have to do is I have to prove the Pythagorean theorem, which I think everybody knows about. Now, what's the theorem in this case? A squared plus C squared is equal to B squared. So, square of one calculus plus square of another calculus is equal to square of hypotenuse. You know what? I think people are more used to this. It's kind of, it looks better, right? So, let's just change the letters. This would be C. This would be C1 and C2. This would be A, oops, lowercase A. And this would be B, which means this would be A, this would be B, and this would be C. By the way, it is traditional in mathematics to use names of the sides in a triangle. The same lowercase letter as the name of the vertex opposite to the side. That's why I used against vertex B, I have a lowercase B side and against vertex A, I have lowercase A side. So, we have to prove the Pythagorean theorem, which again everybody heard about. Many people know how to prove it, but I'm going to suggest a very, very easy proof. And the proof is based on whatever we were actually talking just a second ago, that if you draw a perpendicular and altitude from the vertex of the right angle to hypotenuse, then all three triangles, this small one medium and the big one, are proportional to each other. So, well, let's do it. For instance, triangle ACH is proportional to the big one, ABC, right? So, how can I express the proportionality of the sides? Well, again, very easily. Let's consider the small triangle opposite to this angle. It's C1. What's in the big triangle opposite to the same angle? It's B. Okay. Now, let's consider this angle in the small one and the... Well, let's take hypotenuse then. Hypotenuse is B on the small one and hypotenuse of the big one is C. By the way, instead of C, I can put C1 plus C2. That would be easier. Now, let's consider the medium triangle and do exactly the same thing. First, we will consider this angle and the side opposite to this angle in the medium triangle, which is C2. Opposite to the same angle in the bigger triangle, opposite to this angle is A. So, these are two casualties and now two hypotenuses. In the medium triangle, hypotenuse is A and the big one is a C. A, C, which I will write as C1 plus C2. Okay. Yeah. Now, I have these two equations, right? So, let's forget about geometry and think about algebra. So, multiply this cross by 1 is equal to B square and this cross multiplication C2 times C1 plus C2 is equal to A square. I will add them together. What do I have? C1 plus C2 is the same here and here. I will put it outside of the parenthesis and inside would be C1 plus C2. A square plus B square. Well, C1 plus C2 is C. C and C, C square is equal to A square plus B square. As you see, it's a very, very easy proof of the Pythagorean theorem using similarity. Now, there are dozens of different proofs of the Pythagorean theorem. This is just one of them, one of the easy ones. Probably there are some even easier, but this is easy enough. Let's put it this way. And if you want, you can actually search the web for many other proofs of the same theorem. It's a famous theorem. It has a name. And well, actually, I think everybody knows the Pythagorean theorem. But if they are asked how to prove it, well, they might actually think of it before presenting some kind of a proof. Okay, this is an easy proof, so you can just take it under consideration. One more interesting thing, actually, which is related to whatever I said before. If I will ask somebody, okay, this is a triangle. It's a right triangle. This is right angle. Prove that A square plus B square is equal to C square. Let's say the person doesn't know anything about just proof, which I could just offer. Just think about how can it be proved. It's not easy. Well, it's easy if you know what kind of additional construction you have to do and what to use. This is additional construction. The altitude and considering this triangle as similar to Big One and this triangle as similar to Big One. After that, everything is really easy. What's most important and most difficult in this particular problem is to come up with idea that the altitude is actually needed. So it's not just, you know, once you know what kind of a tool you need and you know how to use this tool, everything is easy. Try to build a chair if you don't have proper tools. You won't be able to. So that's why it's very important to understand what kind of tools and in this case that's the tool. Similarity and altitude and similarity of the triangles is the tool. Okay, let's move on. Given a circle with a diameter AB, point P anywhere on the circle. Okay, so we have a circle, we have a diameter AB and point anywhere P on the circle. pH is perpendicular. Okay, prove that. AH to pH is equal to pH to HB. It looks very much like the previous problem. Now, what should you do to basically completely reduce this problem to the previous ones? Well, very simple. Connect these guys. Now, this is the right angle. Why is the right angle? This is the diameter. So this is 180 degree. This is inscribed angle, which means it's measured half of the arc supporting this angle. So 180 divided by 2 is 90. This is 90 degree. And now, all of a sudden, the picture becomes familiar. This is exactly the problem which we have just solved. Right triangle and altitude towards the hypotenuse. And these are two segments. The hypotenuse is divided by an altitude. And this is exactly the proportionality we have just proven. That's it. End of story. All you have to do again is to realize that forget about the circle, forget about the core, the diameter. These are all language related to the circles. Now, if you will transfer it into the language related to the right triangle, which is very easy to do by just connecting the dots, you will have the problem which you have just solved. Again, just a little push in the right direction will give you the solution. Okay. And the last problem is given two segments. Construct a new segment with the length equal to geometric means of the lengths of two given segments. So, there is a segment A and a segment B. And you need to build X in such a way that it's geometrical mean. Or, which is the same thing. Or, which is the same thing. Right? That's obvious. From this follows this, from this follows this. X square equals A times A. This is the same thing. So, you have to find X, which would satisfy this particular equation. Again, let's just remember what we have just learned. If you have a circle, and these are, let's say, A and B, then this would be X. Right? Because A over X would be X over D. So, how to build this? Well, very simply. Just attach one segment to another. One segment to another. Well, in this case, A is longer. A would be approximately like this. And B is a shorter one. In my picture, it's B. You attach one into another. Then, using the center of the combined segment, draw a circle. And at this particular point, where A ends and B begins, you just draw a perpendicular. And that would be the segment which you're looking for. Okay. I promised that this lecture would be a little bit more creative. And I hope it was. At least it was for me. I would like to encourage you more in solving the problems, whatever is presented on Unisor.com and maybe some other things. By the way, the internet is just filled with the problems of this type. Some of them are maybe a little bit more difficult than the others. Now, I will try to present as much problems myself. But in any case, the more problems you solve, the richer your toolbox is, and more developed your mind will be to solve any practical problems. Like, for instance, how to build the space shuttle. Right? All right. So, I do not pretend I'm building the space shuttle, but what I'm doing is hopefully it would help you to do this. Thanks very much. That's it for today.