 Hi, I'm Zor. Welcome to Unisor Education. During the last few lectures, we were talking about similarity. And the last lecture which I have recorded was about basically problems related to similarity and how similarity can be used to solve some problems. So, I will continue this concentration on problems. But let me digress a little bit about the problems in general. I'm trying to convey that point quite often, I'm afraid, but anyway, repetition is always helpful. So, why do we need to solve these problems? In your real practical life, I almost guarantee you will never have to solve any similarity-based problems or any other problems we were talking about primarily. So, why do we do it? Well, you see, solving problems is essential for development of analytical thinking and logic and creativity, if you wish, because certain problems require real creativity. So, let's just consider that the best purpose of solving these problems is not to teach you particular skill, but to develop the creativity and open-mindedness and logic and some analytical thinking. So, just bear it in mind as this development kind of a tool, rather than certain amount of knowledge which I would like you to learn by heart and utilize in your practical life, that will never happen. Alright, let's finish with this preaching and get to the problems. So, again, these are problems related to similarity and the more problems you solve yourself, the better will you be equipped with certain very practical problems which you will meet. And that's why I insist on the methodology of first trying to solve the problems yourself, then listening to the lecture, and then try again to solve the same problems after you have finished the lectures. So, if you will go through this routine and basically will be able to solve by yourself all the problems which are presented on the website, I'm pretty sure you'll be well equipped to life in general. Alright, prove that similar triangles have radiuses of inscribed circles proportional to their sides. Alright, it's kind of obvious, I guess, everything even obvious things need to be proved. So, you have two triangles which are similar to each other. Now, you have inscribed circles. Now, if you remember, inscribed circle is on the crossing of angle by sectors. Here, you have inscribed circle. Same thing here. Alright, let's call it ABC. And this would be A prime, B prime, and C prime. So, let this be P and P prime. Now, you know that similar triangles have congruent angles, which means this angle is congruent to this, this to this, and this to this. Now, halves of these angles will also be congruent to the same purpose, obviously. So, half of this angle, this is the bisector, right? Angle bisector. And this is half of this angle. They will be congruent. Same are these. So, two triangles, ABP and A prime, B prime, P prime, have two angles congruent to each other, which means they are similar. Which means that in these two similar triangles, ABP and A prime, B prime, A prime, B prime, P prime, certain elements which we have already learned about will also be proportional. Now, we have learned that angle bisectors and regions and altitudes in similar triangles are proportional to their sides. But now, what is the radius of the inscribed circle? This is an altitude of this particular triangle, right? The altitude is the radius of the inscribed circle. They are particular from P to the side AB or from P prime to side A prime, B prime. And we know that since these two triangles are similar, the altitudes are proportional, which means that the radiuses of inscribed circles are proportional. That was it. Now, how about circumscribed circles? Obviously, they are also proportional. This is the next problem. And here is how we will address this. You have one triangle and you have another one, which is similar to this one. Now, you have a point, let's say here maybe, which is the center of circumscribed circle, which means that the distance from Q to vertices A and B and C are the same. Same thing in this case. Distances from Q prime to the vertices A prime, B prime and C prime are the same because this is the center of circumscribed circle. Okay, so these distances are the same. This is the center of circumscribed circle. Okay, now we know that ABC triangle is congruent to A prime, B prime, C prime. Now, let's forget about the congruent component of the transformation of one into another and let's concentrate on scaling only. Now, how can we forget about it? Well, just move the whole thing, let's say, to this particular place and make A prime coincide with A prime, A prime, B prime going along the AB and A prime, C prime along C. Doesn't really matter. So, from A, we can scale one triangle into another. Okay, now, scaling transforms basically everything on the plane, right? So, scaling from point, you know what, let me actually draw this thing. So, let's consider this would be my B prime and C prime and this would be my Q prime. Alright, so I just congruently moved it here. Now, let's talk about scaling from point A as a center of scaling for a factor of scaling equal to the ratio between AB and AB prime. Obviously, C prime would be scaled into C and now what's very important is if Q prime is a center of circumscribed circle around AB prime and C prime, then as a result of the scaling if AB prime, C prime is transformed into ABC then Q prime must be transformed into Q. Why? For a very simple reason. The distances from Q prime to A and B prime and C prime are exactly the same. Now, all segments during the scaling are transformed in such a way that their lengths is multiplied by the same factor. So, these three segments were equal in lengths. They're not equal in my picture, but let's consider they are. Then, after the transformation, they will also be equal in lengths. So, wherever Q prime will be converted, whatever the point is, the distance of that new point, the image of Q prime to the images of B prime, which is B and the images of C prime, which is C and the images of A, which stands in place, will be the same. So, the center of circumscribed circle is transformed into a center of circumscribed circle since the distances to the vertices will be the same. But there is one and only one center for a circumscribed circle. We have already proved that during our triangle lectures, whatever. So, there is no other point but Q, which is the center of circumscribed circle around ABC where Q prime must be actually transformed during this transformation. So, Q prime is transformed into Q, which means that the radius, which is AQ, is exactly proportional for similar triangle because if AQ prime is transformed into AQ and the factors of transformation is exactly the same for sides and for everything else, so the proportionality is preserved because the factor is exactly the same. So, the basis for this, again, is equal radiuses are transformed into equal radiuses. So, center of the circumscribed circle is transformed into a center of circumscribed circle and there is the proportionality. Okay. So, we can construct a triangle by two interior angles and the radius of the inscribed circle. Actually, during the last lecture, we did circumscribed circle with the inscribed circle is exactly the same. Two angles. So, this is inscribed circle. Now, two angles are sufficient to build any triangle, which is similar to the one which we would like to build. So, let's just build any triangle with these two angles. Inscribed circle in it. We got that. Now, since we know that the circle is supposed to have this radius, but we have this radius, well, let's just preserve the center, make this circle, it's a bigger one and what we have to do is just to put tangents in these points and the new triangle will be the one which we need. And these points are actually perpendicular because this is an inscribed circle. So, to do the tangent, you just do the perpendicular at the point where the radius is intersecting the bigger circle. That's very simple. Now, the way how we use the similarity is we have two angles and some linear dimension, which is the radius of inscribed circle. So, we forget for linear dimension for a second because we know the linear dimension is scaling and we just use the angles and two angles are sufficient to build a triangle similar to the one which we need. And then we just shrink or expand it using the difference in radiuses. Okay, prove that parallel lines intersecting both legs of an angle cut proportional segments onto legs. All right, so you have an angle and you have parallel lines. Now, let's say this is A, B and C, A prime, B prime and C prime. Well, what you have to really prove is quite frankly, we don't even need to see. Let's just concentrate on A and B that's sufficient. Let's just prove that A divided by A prime is equal to B divided by B prime. So, these are lengths of these segments. Now, in this particular case, let's put, let's say, this is K, L, P, Q. Now, obviously triangles M, L, Q and M, Q, P the bigger and smaller are similar because K, P is parallel to L, Q. Right? Just two triangles and the bases are parallel, we know they are similar. Now, from similarity of these triangles follows the proportionality of the sides. So, what we can say is that A is related to A prime as A plus B related to A prime plus B prime, right? Because the side of the small one is A and this one is A prime. The big one is A plus B and this is A prime plus B prime. Now, or it's exactly the same. You can say that the corresponding sides are related in the same proportion, which is A to A plus B would be equal to A prime divided by A prime plus B prime. Obviously, these two equations are completely equivalent so we can use either one of those to prove this one. Now, how can we prove this relationship if we have this or this? It doesn't really matter which one we will use. Let's use the second one. Well, obviously from this proportion we can conclude what? A times A prime plus B prime equals to A prime times A plus B, right? Open the parenthesis. A A prime plus A B prime equals A prime A plus A prime B. Now, A A prime subtract both sides of the equation by the same A A prime. A A prime and A prime A are the same because multiplication is commutative and what we have is this. Now, let me just put it again. A B prime equals A prime B. And what is this? This is exactly the same. A times B prime equals to A prime B. So these two are equivalent. Well, when I'm saying equivalent just don't forget the term talking about lengths of the segments which means none of them is equal to zero because otherwise I cannot really say that algebraically for any A prime B prime this is, these two are equivalent. If one of them is equal to zero the top doesn't make any sense and the bottom does make sense. So we are talking about geometry so none of them is equal to zero so these two are completely equivalent to each other. So basically from this which is the similarity we can prove this. So not only small to the bigger is the same as small to the bigger but also each individual is proportional to each individual. Now, if I have another line let's say X, Y and we will call it C and C prime so what can we say here? How can we say that C to C prime is also in exactly the same proportion? I'll just use exactly the same methodology. For obvious reason we understand that MLQ is similar to MXY because these two are a parallel side. So let's do again exactly the same principle of similarity. What is the side ML? It's A plus B it relates to MX and this is A plus B plus C and this proportional to A prime plus B prime over A plus B plus C. Now what we have to prove is again that A over A prime is equal to C over C prime. The fact that it's equal to B over B prime we have already proven. So how can we prove this? In exactly the same fashion. Let's just multiply from this we will get A A prime plus A B prime plus A C prime plus B A prime plus B B prime plus B C prime equals this and this. A prime A plus A prime B plus A prime C plus B prime A plus B prime B plus B prime C. Now let's reduce this thing. A A prime A A prime B B prime B B prime What do we need? A C. A B prime yes A B prime A B prime B A prime B C prime A prime B A prime. So what do we have? We have A plus B C prime equals A plus B C prime equals to C equals to A plus B time C. So basically what we have come up with is that C relates to C prime as C relates to C prime as A plus B relates to A prime plus B prime. But from the previous theorem we have already proven that this is exactly the same as A over A prime and B over B prime. That's the one which we have just proved before. So that actually proves what we need to do. So each pair is kind of corresponding to each other. And this relative to this is equal to this relative to this and this relative to this. And no matter how many lines we will put it will be exactly the same thing. They're all proportional to each other. Okay next. Prove that a set of straight lines are originated at the same point and intersecting two parallel lines proportional segments on these lines. Okay so basically what we have to do is A B C A prime B prime C prime. But we have to prove that these are proportionate to each other. Or if you wish this to this is the same as this to this. So let's call this particular A and this is A prime B and B prime. So what we have to do is A over A prime is equal to B over B prime. How can we prove that? Well that's actually easy. Let's think about it. Triangle Q A B is similar to Q A prime B prime. Which means A over A prime is the same as Q B over Q B prime. So A over A prime is equal to Q B over Q B prime. This to this as this to this. Since these two triangles are similar. Now same thing these triangles are similar. Which means B to B prime B to B prime is like Q B to Q B prime. This is the same. So that's why we have this. And no matter how many other lines originated from this withdraw you will still have exactly the same situation because let's say we have the new one C and C prime. Again C relates to C prime as Q C relates to Q C prime. Which in turn is the same as Q B to Q B prime etc. They're all equal to each other because these triangles are all similar to each other. Pair by pair. That's why this ratio is retained. Divide a segment in three parts in proportion 4 to 13 to 7. How can we divide this particular segment using something like this? How to find these two points which would divide it into sub-segments lengths of which would be proportional to these numbers? Well, easy actually. Let's have an angle and let's have some unit lengths. Whatever that unit is. One centimeter, one inch, one whatever. So four. Let's just count. One, two, three, four. Okay. Then 13. One, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen. Thirteen. And seven. One, two, three, four, five, six, seven. Okay. Now connect this to this. The end of the segment with end of this chain of units. And from this draw a parallel since we know that parallel lines cut segments on one side of the angle proportional to another side of the angle and these are in proportion 4 to 13, 4 to 13, to 7. That means these also would be in proportion 4 to 13 to 7. That's just using the previous way. That's just using the previously proven theorem. Okay. Given segments a, b, and c. a, b, and c. Construct the four segment x in such a way that a divided by b is equal to c divided by x. Well, this is, if you wish, more or less exactly the same story. You can have an angle and two parallel lines. If this is a, this is b, this is c, and this is x, then as we have just learned from one of the previous theorems, a relates to b and c relates to x. Right? So we have to build this picture basically. Now, how can we do it? Well, very simply. How can we construct it? Just draw two parallel lines, any two parallel lines, half a and then c and then anywhere b and then draw lines this and this. We get this point and we draw another line and that will be our x. So using a and b and c, we constructed this picture. Now, when would not be, when this would not be possible to construct? Obviously, if a is equal to b, then these two lines will be parallel to each other and they will not have any interception. So it's a special case. If a is equal to b, then we just check. If a is equal to b, then we choose x equals to c and that would be the end of it because then a over b would be one and c over x would be one as well. But if a not equal to b, then something like this would help. Well, that's it. This is the series of problems I would like to address. I will have some others. Don't forget my very, very strong recommendation. Go to the website first. Go to the notes of the lecture. Try to solve the problems yourself before the lecture and after the lecture independently. That would be a great exercise for you and it will help you to not only just to remember, I'm not really sure whether you will remember or not, these particular proofs. However, what you will probably retain the more problems you will solve the more you will have it. You will retain something like a tool. Similarity is a tool. My goal is for you to accumulate a toolbox of all the different kinds of methodologies which you can use to solve problems. The more you will accumulate, the easier it will be to go into something which you just don't know how to solve. That's the purpose. I'm solving these problems not to teach you how to solve these problems, but how to solve problems which I did not address at all. The new ones. That's why it's very, very helpful in the real life, because in the real life you always solve the problems which you just don't know how to do. That's the key to everything. There are a lot of tools to do, but that's not fun. That's it for today.