 Hi, I'm Zor. Welcome to your new Zor education. I have completed basically talking about some theory behind similarity, and I would like to spend some time solving problems. First of all, as usual, let me just give a general kind of a guidelines about the problems on this website, Unizor.com. I think it is imperative that you first try to solve the problems offered on that website yourself. I'm not talking about exams. I'm talking about lectures like this one, where I explain the solutions to these problems. But before listening to me, read the notes where all the problems are stated. Try to solve them yourselves. Whether you're successful or unsuccessful doesn't matter. Then you listen to the lecture. I might suggest some new way to solve the problem which you have already solved differently, which is fine, or maybe you didn't find the solution that listened to me. After the lecture is completed, do again your solving problems exercise. Again, going through the same notes, try to solve the problems yourself. Most likely you will remember what I said during the lecture. If not, and you still cannot solve it yourself, listen to the lecture again. Basically, this is some kind of an approach which would enable you to remember certain ways and means to solve problems. Now, the next thing is, I'm usually putting the problems after certain theoretical materials covered, like in this particular case, the problems which I'm talking about right now would be about similarity. This is a hint already. So, any problem which I'm presenting in this particular lecture is about similarity and should be solved about similarity. It would be much more difficult if this problem is just given to you without referring to, okay, this is about similarity, kind of a hint. So, consider yourself in an easier situation than in, I would say, something like during the exam or whatever, where it's not obvious which tool to use. And speaking about tools, solving problems actually is supposed to develop a toolbox for you, so to speak. Every tool is basically the method you can use to solve certain problems. For instance, you can solve using algebraic equations, or you can solve using locus, the geometrical set of points which have the same property, or similarity, or I don't know, something else, whatever. So, this is a toolbox which I wish you developed by the end of the whole course, and it would be at your disposal at any time. All right, now, the problems. Problems are relatively easy, and as I was saying before, they're all about similarity. Construct a triangle by angle, a ratio between sides forming this angle, and an altitude from this angle onto the opposite side. Okay, so let me start the game. You have to construct a triangle knowing the angle, and then ratio between the sides and altitude to opposite side. Am I right? Angle, ratio between sides forming this angle, and an altitude from this angle onto the opposite side. All right, now, as far as the similarity is concerned, you basically understand that if I have a ratio of these sides, it means I can build a triangle similar to this one. How? Well, very easily. I have an angle, right? Now, on one side of the angle, I will put the point a prime, which is on a distance, I'll take the point b prime, I'm sorry, it's c prime, so that bc prime is equal to 1, some unit, whatever it is. Then, since I know this ratio, it's a numerical value, I will choose a prime b equals to a over b. Since I have explicit lengths of these two segments, this would be equal to 1, this would be equal to a over b, I can build a triangle a prime bc prime. And since the ratio between these two sides is proportional to these, an angle was constructed as a given, I can say that these two are similar to each other. Since they are similar, what I can do is I can just draw a perpendicular and continue it in such a way that I will have this bh equal to the altitude which I need. After that, now this is the right angle. After that, I draw a perpendicular and call these two points a and, sorry, c. So, now since these two triangles are similar to each other, these lines are parallel and that's enough for similarity. Since ratio between these sides is equal to a over b, ratio between these sides also will be equal to a over b, as was basically asked in the problem. And the altitude is also given, that's how I basically constructed the point h on the distance h from b. My problem is solved, the triangle a b c is the one which I would like to build. It has the angle, it has the ratio between the sides and it has an altitude. So, basically I would like you to just emphasize what I did. I was not able to build the triangle straight away whatever was required, but I was able to build a similar triangle using the ratio between the sides a over b and the angle. And then I just stretched or shrink whatever my triangle which I have built to the proper size so my altitude would be given. Basically all the problems which I'm talking about right now are that type. We inscribe a circle into a given angle such that this circle contains given point inside an angle. So, you have an angle, you have a point inside that angle p and you have to construct this particular circle. So, it will intersect the point and it will contain the point and it will be inscribed into the angle. How to do that? Well, okay, we cannot build this circle right now. I cannot construct it because I don't know the center. However, I can construct any other circle which would not go through the point p but would still be inscribed into the angle, like this one. How do I do it? You know all these inscribed into the angle circles are lying on an angle bisector. So, I pick any point on an angle bisector and draw these perpendicular and this is my radius. So, this is an inscribed circle. But what's the problem with this circle? It's not the one which I would like to construct because it does not go through the point p. So, what do I do? Let me connect with p and that's where it intersects my small circle, point m. Now, as you see, everything is scalable along that angle. If I will use b as a center of scaling and I will scale point m into p, as I told many times before, this center x would be transformed into a center of the bigger circle y. The reason is that when I am scaling bm to a factor which will convert point m into point b, my ratio bm over bp, I will use the same ratio. Now, obviously, since bx is proportional to by by the same ratio, that's how I can determine the by. So, all I do is, as far as construction is concerned, if I don't know this circle, I construct any circle which is inscribed into an angle, draw the line, take the intersection, and all I have to do is to draw py parallel to mx. That's basically enough because as soon as I draw it parallel, my angles are the same. Now, the triangle bmx is similar to triangle dpy because there are two angles. One is common and these two are corresponding with parallel mx and py and transversal dp. So, triangles bmx is similar to dpy, which means that bm relates to bp like bx related to by equals to mx related to py. Obviously, if I draw a perpendicular here, let's say these are c and d. I can see that my triangles bcx is similar to bdy from which radius is cx and a segment cd are in ratio of bx to dy, which is the same thing as this one, right? But now, since cx and mx are both radiuses, cx and mx are both radiuses, so equality between this bx over dy and this is obvious, so I can equate these two. And since mx and cx are equal, cd and py are equal as well. Not cd, wait a minute. cx to dy, I'm sorry. cx to dy is dx. So, which means py and dy are also equal to each other. And that means that this is a circle. Since this is the perpendicular, it will be tangent and the radiuses are equal to each other. That's why this particular circle will be inscribed and passing through the point p. Yeah, that's it. So again, the construction itself, after I did all the analysis, is just construct a small circle, connect BP and intersection m, draw line, the radius actually, from x to m, and py would be parallel to mx. Whenever it crosses the y-sector, that's the center. Everything else is just proof that this is correct. So first I did some analysis, I realized how to do it. Then I basically constructed and then I proved that the construction is what it requires. This is basically a typical way of solving the problems. You analyze it, you construct whatever is necessary, and then you prove that whatever you have constructed is right. Inscribe a rhombus with a given acute angle into a given triangle such that one side of the rhombus goes along the side of the triangle and two endpoints on opposite side of rhombus lie on two other sides. So basically this is the rhombus which you would like to build. a, b, c, c would be here, and this would be k, l, m, n. So what I know, I know that k, l, m, n is a rhombus and I also know that I am given this angle, the acute angle of the rhombus. And obviously, what am I saying? This is c, I don't know why I put d. So I have a triangle a, b, c as given and I have this angle and I also know that it's supposed to be a rhombus which means all sides are supposed to be equal. Okay, obviously it's not easy to create it immediately but however, if I have any other rhombus which, let me just do it this way, I pick any point here, let's say k' and I make the line here to n' in such a way that this is a given angle. So now I have the lengths k' and n' and I am using it to basically find my m' which is equal in length since it's a rhombus and I am building the rhombus this way and this would be l'. So as you see, k', l', m' and n' is a rhombus which is similar to klmn which we have to construct. Now, I can obviously construct k', l' and n' because I just picked the point k' and then used the angle. Now, construction is actually obvious right now after I have constructed any rhombus with this acute angle with one side lying on AC and one particular vertex lying on AB. Now, if I can stretch it along this line so I'm connecting a to l' and continue until it intersects with BC and that would be the point l. Now, this is the construction. So that's how I came up with this particular construction methodology. Pick the point, draw the line with this angle which is given and then using the fact that this is a rhombus I found the point m' and then l' which completes this particular rhombus. Then I connect a with l' until it intersects with BC. Now, all I have to do is that the resulting point l from which I can obviously put one line parallel to k' and n' and another parallel to AC that that particular klmn would be the rhombus which I need. Well, and it obviously follows from the fact that all the components here are similar to each other. A triangle let's say ak'l' is similar to akl. Why? Well, primarily because they have a common angle and since this is parallel to this one this angle is also equal. So ak'l' is similar to akl. Now, with the same similarity I can say that al'm' is similar to alm because again these are parallel so angles are the same and there is one common angle. So all these triangles are proportional to the ratio of al' to al which means if they used to be whatever the elements used to be equal will remain equal. Like for instance this line, this particular kl related to k'l' as al' to al'. But same thing can be said about lm to l'm' it's also in the same ratio and since these two are equal, these two are equal. So this is rhombus. Well, that's basically enough. I mean we can do exactly the same with two other sides. So I don't want to go into the trivial details but I basically understand that it's enough to build a small rhombus of required proportions and then stretch it such that the fourth vertex will coincide with bc. Okay, construct triangle by two angles and reduce of circumscribed circle. Okay, two angles and reduce of circumscribed circle. If I have two angles I can always construct a triangle which is similar to the given one, right? So I pick any base and construct a triangle similar to the one which is required. I'm actually constructing right now without analysis because it's kind of obvious. Then I find the center and circumscribed a circle around this triangle. So is it the one which is required a triangle? No, it's not. It's similar to the one which is required and the difference is that it has this radius. Now, there is a radius which is given to me. Construct triangle by two angles and a radius of circumscribed circle. So this is a circle circumscribed over a triangle which is similar to ours. It's not ours. Our radius is different. It's not this one. It's a different one. Whatever the radius is. So I take this radius. I use the same center and just continue each radius so that the new points which I have are all on this distance from the same center. So basically what I'm doing is I'm stretching my picture from this radius to this radius and these will be the new points. This is my triangle which I really need. Why? Well, obviously the fact that I'm stretching by the same factor triangles, let's say a prime, b prime, o is stretched into a, b, o and these two lines used to be radiuses of whatever the triangle I built and this is the one which I need so the ratio is exactly the same and there is an angle which means a prime, b prime, o is similar to a, b, o which means all elements are stretched by the same factor including a prime, b prime is stretched to this one. Similarly in this case b prime, o, c prime is similar to b, o, c and a prime, o, c prime is similar to a, o, c they are all similar to each other. Lines are parallel obviously because when I'm scaling segment a prime, b prime is parallel to a, b which means all angles are preserved all angles are preserved since all lines are shifted parallel to themselves and the distance from o, a to o, b and o, c exactly the ones which I need so o is still a center of circumscribing so that's basically the construction as you see every problem in this series I cannot construct directly the figure which I need I construct something else which is similar and then stretching or scaling rather the picture to the proper size construct a triangle by angle ratio between altitude from this angle and the opposite side this altitude falls on and a median to another side wow that's too many so again there is a triangle we have an angle ratio between altitude b, h, o, c and a median construct a triangle by an angle ratio between altitude from this angle and opposite side this altitude falls on and median to another side okay it's not this median it's different that's my telephone but it's just waiting to finish okay now angle, angle, altitude okay so let's think about this okay angle, ratio okay here's how we can do let's speak ac as any base whatever we want so since we have ratio we will put ac a prime, c prime equals to 1 now that's number 1 number 2 we know this angle which means I can build a circle around it which is a locus of points from which this particular segment would be let me just draw a better picture let's start with a circle okay so this is I started with this a, a prime, c prime which is equal to 1 then I draw a circle with which is a locus of points from which a prime, c prime is viewed at this particular angle now that's actually easy just in case you don't remember this is an inscribed angle so it's half the size of a central angle so if I have this angle I basically have this one which is twice as big and this is an isosceles triangle so I know these angles as well so I can build o, c a, a prime, o, c prime by this particular base which is equal to 1 and these two angles which are half of 180 minus this one half of 180 minus double this angle so I build the center and then I build the circle around the center radius o, a prime or f, o, c prime now I know this ratio and I know that a prime, c prime is equal to 1 which means b prime, b prime h prime b prime, h prime is equal to this ratio ratio which is given to me so how do I do it? from a, c, a prime, c prime and the distance equal to b to this ratio actually I draw a line and this is my real b prime so this is an angle it's the same angle as this one because this is the circle it's the locus of all points where this angle is the same but now my h prime would be here now I know that this triangle a prime, b prime, c prime is exactly what I need except one particular case it has the proper ratio between b prime, h prime and a prime, c prime it has a proper angle at the top but it does not have the median whatever we need now what should I do? well basically I should construct the median here that would be my median a prime, m prime basically stretch a prime, m prime to this particular length and after I stretch it let's say everything else should be just drawn parallel and I will get what I want now let me just summarize how I constructed before we go into the details so I take any segment a prime, c prime of the length not any at the length one then I'm building a circle around it so that this is a locus of points from which angle would be the same now how can I find the point of a prime? I'm taking this ratio and have the line on this side on this distance this would be my b prime so bh prime over ac prime is equal to my ratio angle is given and all I have to do is draw this median and to m prime and then expand it to whatever am is given and then I build parallel lines here and these are just continuation now obviously since this this is b and this is c since I built bc parallel to b prime, c prime the triangles abc and ab prime, c prime are similar the angle because it's parallel is the same and the ratio between the altitude and this is exactly the same between the ratio as the ratio between the big altitude and the big base because of similarity of triangles so that's why my abc is the one which we need to construct and again we have built a smaller or a different rather triangle which is similar to the one which I need and then we scale it using the only linear dimension we have linear dimension is the median so I ignore the linear dimension of this triangle which I have to construct and just use the rest which is angle and the ratio between altitude and the base to construct a similar triangle and then just scale ok so that's the same basically as all other problems and they have one left given an angle and point inside it find a point on one of its side equidistant from this point and another side so this is an angle which is given this is the point which is given we have to find an equidistant from side ab along the perpendicular of course the distance between the point and the line and mx so basically we have to find m such that mp is equal to mx well exactly the same thing let's just have any other point pick any one of them draw the perpendicular this will be p' connect since mx and mp are the same size this is a circle with a center m which is tangent to this line and intersects this one so here from m' we also make a tangent circle and that's where my ax is intersecting this is x' intersecting this circle so basically now it's obvious that ax' m' triangle is similar to axm so this is analysis now how to construct well we don't know point m so all we know is let me start from the beginning so now the construction itself we have a point x and what I'm doing is I pick any point m' construct a circle around it in such a way that it's tangent to this line connect x with a and now just do a parallel one xm is parallel to x' m' since it's parallel then these angles are corresponding with two parallels and ax as a transfer so this is the common angle these triangles ax' m' and axm so triangles are similar ax' m' and axm and that means that the ratio between xm and x' m' is the same as xa to x' a now what we can do is basically say that if I will draw this one and make this circle obviously mp would be similar mpa would be similar to m' p' and mp would be in ratio with m' p' as am to am' and these in turn because of those triangles similarity are exactly the same as ratio between this and this and that's basically enough mp relates to m' p' as am to am' that's from similarity of these triangles now and this is equal to mx over m' x' from similarity of these triangles mx' is equal to m' x' x' is equal to m' p' because these are ranges from which we have that mx and mp are equal to each other mx and mp are equal to each other and that proves that we have constructed correctly I hope it wasn't really very difficult you just have to understand that there is always one underlying principle in all these problems you just try to construct something similar to whatever you have to construct and then use the proper scaling to get to the object of your choice alright that's it for today now unizor.com as I was saying before contains all the information parents and teachers are encouraged to use this site to individually control the educational process of their students and thank you very much for your attention that's it for today