 Hi, I'm Zor. Welcome to Unisore Education. I will continue talking about certain theorems related to similarity. And basically they are, well, some of them actually, just a little bit more difficult, but a couple of really simple ones. I would like actually to encourage you to try to solve all these problems and prove all these theorems just by yourself. And then listen to this lecture. The nodes contain all these theorems. And please try to do it every time, actually, with all the problems and all the theorems. Try to do it yourself first. Then listen to whatever the lectures are and then try to do it again just to make sure that you remember everything. In any case, all the material is on Unisore.com website. Please refer to the nodes to this lecture with all the conditions, all the theorems. And there are actual hints in the nodes which might help you to basically do it yourself. All right. Without further ado, let me just go straight to these theorems. Given two circles, radiuses are one and arc two with centers, q1 and q2, externally tangential to each other. So this is one circle. This is another circle. q1, q2. Radiuses are one and arc two. All right. Straight line tangential to both of these circles. It points m and n such that circles are on one side of the line. Proof that half of the segment mn is a mean proportional between radiuses of these circles. Basically, we have to prove that half of mn is mean proportional between the radiuses. That's what it means. Half of this is square root basically of the product of radiuses. Okay. How can it be proven? All right. Let's draw another tangent here. Let's call it x. I would like to consider triangle q1, x, q2. Think about this. For obvious reasons, from x, you have two tangents which makes triangles mx, q1. And, well, let's put this point t. And x, t, q1 basically are congruent to each other, which means these angles are congruent as well. Now, why? Well, for obvious reasons, because these are right triangles, common hypotenuse, and the same radiuses on both sides. So, we have a hypotenuse and the candidate, the same thing. So, triangles are equals or angles are equal. Same thing on this side. If I will connect this to this, I will also have these two angles equal to each other, which means that the angle q1, x, q2 is actually half of the 180 degrees angle mxm, because this is the same as this. This is the same as this. So, some of these two is exactly half of the whole thing, which means it's 90 degrees. So, it's a right triangle. q1, x, q2 is right triangle. Okay. Since it's a right triangle, we can use a theorem. I hope you remember that in the right triangle like this, if you have an altitude, then altitude is new proportional between pieces of the hypotenuse it cuts through. I'm talking about an altitude from the vertex of the right angle. Now, this theorem was actually proven a couple of times before in one of my lectures, and I'll just use it. Now, in this particular case, xt, which is this particular altitude, is exactly equal to xm, and it's exactly equal to xn. So, xt is actually half of the mm. xt is half of mm. And using this theorem, I know that xt squared is equal to r1 times r2, because this is mean proportional between these two, which basically proves our theorem. So, what's the necessary thing? What's the lesson of this? Well, a couple of additional constructions like connecting q1 and q2 to this point x, which is the midpoint, is all that is required, and then all you have to do is just to see one of the theorems which has already been proven in the past, the theorem about this right triangle and its altitude. All right, next. I just want to mention one more time, and I know I mentioned it many, many times, that mathematics, especially these problems related to some theorems proving construction, they do develop your creativity, because you have to basically kind of invent something new which would help you to solve the problem. That's the creativity. Next, given two circles externally tangential to each other, same as before, basically, and the common tangent which is going through this point. So, these are all tangent to each other at one point. All right, choose any point x and draw two seconds, any seconds, a, b, m, c, d. Prove that these four points, a, b, c, and g, are lying in the same circle. Well, obviously, we know that we can draw a circle around three points which are not lying in the same line, but these are four lines, four points. So, that's not necessarily the truth for any four points. These four points do have this property of lying in the same circle. Now, if you remember, and again I'm referring to one of the prior lectures, if you have a convex quadrilateral inscribed into a circle, then the characteristic property of this inscribed into a circle quadrilateral or quadrangle is that some of the opposite angles, this plus this, is equal to 180 degree as well as this plus this. For obvious reasons, because every inscribed angle can be measured by half of the arc supporting it, and so the arc which supports this angle is this one, and the arc which supports this angle is this angle, this arc. So, some of these arcs is 360 degrees, and that's why some of these angles is half of this, which is 180. And this is a characteristic property, which means if a quadrangle has the property of some of the opposite angles equal to 180, then it can be inscribed into a circle. And if it can be inscribed into a circle, then it has this property. So, I'm referring to this and now I'm using this particular property which I have already proven before to prove that this quadrangle has this property of having opposite angles which means this plus this equal to 180 degrees. Alright, how can we do it? Let's think about two triangles, XAC, the small one, and XBD, the bigger one. They are actually similar to each other. I'm going to prove it right now. Now, if I do, if I prove that this is the right, I'm just angolizing, then the corresponding angles will be to this, will be this, and to this angle will be this. Now, let's think about it. If these two angles are the same, because triangles are similar, and this angle plus this obviously are equal to 180 degrees, that means that this plus this combined together will give me 180 degrees. Similar to the same, if these two angles, XBD and XCA are equal to each other, these two in sum gives 180. That means these two in sum equal to 180. So some of the opposite angles is 180 degrees, which means I can inscribe this quadrangle into a circle. So all I have to do is similarity of these two triangles, AXC and BXD. Now, how can I prove that? Let's think about it. They do have a common angle, right? So if two triangles have a common angle, what do I have to prove to prove their similarity? Well, the proportionality of the sides which form this angle, right? So these sides, to this should be equal to this to that. Okay, how can I prove that? Now, let's recall another theorem. Another theorem is that if I have a tangent and a second, then XA times XB is equal to XT squared. So the square of the tangent from a point is equal to a product of the whole second by its external piece. Again, this is the theorem which I have already proven in one of the prior lectures. Now, I will use it here right now. Let's call this point T. For this purpose, XT is a tangent, XB is a second, which means XA times XB equals XT squared. Now, from the same X, the same tangent, and let's consider this tangent of this circle, I will have XC times XD is equal to XT squared, correct? Well, obviously they are equal to each other, so XA times XB is equal to XC times XD, from which we conclude that XA divided by XD is equal to XC divided by XB. And that is exactly the proportionality of the sides which we wanted to prove. That's it. From proportionality, we infer the similarity and with similarity, we infer that the opposite angles are summed up to 180 degrees. Now, next. Next is an inverse model. Okay, if I have two circles tangential to each other and have another circle which intersects these, then these points of intersection between seconds from these intersect points with the common tangent, they are all intersecting in the same point. Okay, how can I prove that? So now, again, I have four points, A, B, C, and D. But now it's given that they are on the same circle. Now I have to prove that seconds A, B, and C, D continue to intersect with the tangent will intersect at the same point. Well, but let's do it this way. First, we will draw A, B, and get the point X. Now, what I will do, I will connect that point X with point C and continue. My question is, will it hit the point D? Well, obviously yes, because consider this. I know that the point which intersects, the further point of intersection of this second lies on the same circle with A, B, and C. Because I will put it D prime. I know that A, B, C, and D prime lie on the same circle. That's from the previous theory which I can just prove it. Now, so point D, D prime, is supposed to belong to a circle which is intersecting at points A, B, and C. But there is only one circle which you can draw around three points A, B, and C. And it's already given. Which means that the point D prime should really coincide with the point D. There is no other way. It's the same circle and the D prime is actually intersection of the same circle with the same circle and it's only this one. That's easy. By the way, I use the fact that I can draw only one line through two points. Point A and C, I draw the line. What if I can draw another line through these two points which will go to point D and not D prime? No, that's impossible if you remember from one of the postulates. We have only one line which can be drawn through two points. Okay, next problem. Given two circles with centers P and Q, consider points A and B. I know. Two points on circles P and Q. And what's the name of these points? Okay, it's A and B. So we choose A and B in such a way that P A is parallel to Q B. And now we connect these two points. You get that? So the theory is that regardless of how we choose these two radiuses, as long as they are parallel to each other, the line which connects the ends of this will hit the center line at exactly the same point, regardless of the position of the circle. So if I will have some other, let's say here, and parallel to this, I will hit exactly the same point. Okay, how can it be P? Well, let's think about it. Well, it's obvious that these two triangles, A, P, X, and B, Q, X, are similar to each other. These are parallel, which means these angles are the same, and this is the common angle. So we have two angles congruent to each other. So triangles are congruent, which means, I mean, sorry, not congruent, single, which means that the sides are proportional. All right, let's put the equation. X, Q from this small triangle across this angle relates to X, P, as in the small triangle across this angle is the radius. Let's say this is radius R1, and this is radius R2. So it's R2 to R1. X, Q over X, P is a fixed ratio. I don't care about R1 or R2, but since our circles are given, this ratio is fixed. Now, what's also fixed is P, Q. And instead of X, P, I can write X, Q plus Q, P. X, Q plus Q, P. So this E is constant. Now, for convenience purpose, if this is constant, it's more convenient for me to reverse it. It would be X, Q plus X, sorry, plus X, P over X, Q. This is also constant. But if I will divide this, this is 1 plus X, P over X, Q. I'm sorry, not X, Q, P. Sorry, I made a mistake. So it's Q, P over X, Q. Now, this is also constant, regardless of the position of A and B, right? So this is 1, this is constant, this is X, Q, which we don't know, and this is constant. So basically it's like an equation if you wish. 1 plus, let's say Q, P is A over unknown X, Q, which is X is equal to B. I mean, obviously this E is an equation which has only one solution. X over A is equal to B minus 1, sorry, 1 over B minus 1. So X is equal to A divided by B minus 1. So if B is not equal to 1, now B cannot be equal to 1 because it's basically the length of this piece. So if B is not equal to 1, there is one and only one solution to this equation. So this particular X, Q, this is X, Q, this is X. So this length is constant. It's related to some kind of equation from known variables. The variable A, which is Q, P, which is this, and the variable B, which is this ratio, which is basically this ratio. So everything seems to be working in case this B not equal to 1. Now B is equal to 1 only when the radiuses are equal to each other, but if they are equal to each other, then the line Ax will be parallel and it will not intersect this line. You see, if two circles are the same and you have these two radiuses, it will be parallel and there is no intersection. So if there is an intersection, radiuses are different. So basically this equation has only one solution, which is a fixed length. So X, Q, which is X in my equation, is fixed, which means X position is also fixed. It's very noisy outside today. Hope it doesn't disturb your train of thoughts. It does actually disturb mine a little bit. Next. Given two concentric circles, choose point X on one of them and diameter Mn on another. Prove that sum of squares of segment Xm and Xm, Xm squared plus Xm squared is constant, regardless of the position of the point X. Okay, here is how we can do it. This is the center. Let me turn the whole picture by 180 degree around the center. Now X will be turning into Y, let's say. Now Xm will be Yn because point M will be transformed into M and N into M. So Xm will be transformed into Yn and Xn will be transformed into My. And for obvious reasons, this is parallelogram. Now, why is it parallelogram? Well, because I'm turning the segment Xm, let's say, by 180 degrees, which means it turns into a parallel to itself. Okay, now since this is a parallelogram, I can recall a theorem. And again, this theorem was proven when I was talking about parallelograms. That sum of squares on the sides of the parallelogram is equal to sum from the squares of the diagonals. Now, but what is the sum of the squares of all the sides? Well, that's Xm squared plus Xm squared plus Yn squared, which is the same as this one, plus My squared, which is the same as this one. So basically, it's two times this sum. And sum of the squares of diagonals is basically one diameter squared times another diameter squared, right? So it's g1 squared plus g2 squared, g2 and g1, which is constant, obviously, because the circles are given to us. So all I have to do is basically think about, okay, how to convert this into something more familiar, and how is turn the whole thing by 180 degrees and consider parallelogram and recall this property of the parallelograms. Well, that's why actually I'm asking you to, after you finish your own self-study and listening to the lecture, go again through all these problems or theorems which I have proven during the lecture just by yourself. So it will kind of inculcate into your mind. All right. Proving theorems and solving problems is basically like finding your way from point A to point E if you don't know the way. So it's very, very useful in real life. I mean, if you know how to get from here to there and nobody actually told you how, well, mass helps. All right. One more problem I think I have, yes. Given the circle, it's diameter AB and point C on the continuation of this diameter outside. This is C. This is AB. And then we draw a perpendicular here. Take point M and connect this. This is A prime. Prove that A M times A A prime is constant independent of the position of the point M. Basically you understand that if M goes up, then A M would be lengthening, but A A prime would be shrinking. So the product is actually the same. Okay. How can I prove it? Well, obviously, I mean, here I think it's just drawing this particular segment, this chord just asks to be drawn. And now we have two triangles. Obviously this is the right triangle because it's perpendicular by construction. This is the right triangle because this is an inscribed angle which is supported by half a circle. AB is a diameter, remember that, right? So both are right triangles. They share one particular acute angle, so they are similar. Okay, let's just write the similarity equation. Okay. The bigger catheters from the smaller triangle, which is A A prime, which is across this angle, relates to the catheters of the bigger triangle across the same angle, which is AC. As a hypotenuse of the smaller, which is AB, relates to hypotenuse of the bigger, which is A M. From which we can prove the A A prime times A M is equal to AB times AC. AB and AC, this is constant, which means that this is constant as well. And the proof. Okay. Thanks very much for listening to me. Don't forget what I said before. Go through all these problems, use the notes, which are on this website for this lecture. And there are some hints, which is making your life much easier and try to prove it again just by yourself. Parents are encouraged very much to take charge controlling the educational environment of their students by enrolling them and asking them to take exams. All exams are available on the website for registered students. That's it. Thank you very much.