 Hi, I'm Zor. Welcome to Unisor Education. I will continue talking about problems related to similarity. These lectures about construction problems, there will be other construction related problems which are using similarity. I would like to remind a couple of very, very simple techniques which I will be using probably almost like every construction problem which is using similarity. I did discuss these techniques before. I'm just reminding them because it's very relevant to this particular lecture. All right. These are fundamental constructions, if you wish, which you can employ in all these problems. One is if you have, for instance, three different segments, a, b, and c, and you would like to find the fourth segment, the lengths of which is in this particular relationship. How to construct x if you know a, b, and c? Well, it's simple. Just have an angle. This is your a, this is your b, this is your c. Draw a line parallel to this one, and then parallel to this one, and this will be your x. Now, I'm not going to prove it, it's obvious because of similarity of the triangles. Now, very analogous problem. What if you have some numerical ratio, and you have to find x in this particular case? How to do this? So, you need to find a second segment which would be related to the a, as far as the length is concerned, in this way. Well, again, simple, a. Now, obviously, if you have segments and the length is involved, we have also assumed that you have a unit length segment. So, we will use it. In this case, we actually can use any other segment, but unit is just more convenient. So, use m times this particular segment, one, two, three, four, five, m, and then m times exactly the same segment. So, you connect this to the boundary between these two groups, and from the end, you draw a parallel line. And obviously, this is your x. Now, finally, another, the third, very, very simple task is what if you have a segment a which you would like to divide in ratio m over n, which means you have to find segment x and y, which are together to give you a, but between themselves, they are related as this. How to do this? Again, same technique. This is your a. This is your m unit segments. This is your m unit segments. Connect the end of it, and now, if you draw a parallel line from the border between groups, then this would actually divide in x and y. Obviously, x and y are satisfying this because they are actually part of the same segment, and this because of similarity of the triangles. So, this is an introduction to this and some other construction problems, and it shows the basic technique which is used as basically the final point of any solution which I will present. So, whenever I will reduce my problem to this particular task, one of these three particular tasks, I basically say, okay, the problem is solved. We know how to do that. All right, so let's go one by one with the problems. Problem number one, given an angle, a qv, qv, and the point p in between. Now, p actually can be either inside this angle or outside of the angle. Angle can be acute. It can be obtuse. Doesn't really matter. I will consider only this particular case when it's in the acute angle and the p is inside. All other situations are exactly similar and you can do it yourself. So, the problem is construct a line through p, x, y, in such a way that xp relates to py as known ratio m over m. Well, in this case, we don't know either of these two. So, what can we do? Well, let me reduce this problem to something which we know better, which is the following. Let's draw perpendicular from here, perpendicular from here, and parallel line to qv from here. So, this is c, d, e. So, xd is perpendicular to qv, p is perpendicular, and this line, pc, is parallel to qv. Now, what can we say? Obviously, these two triangles are similar. They are both right triangles. They have the same angles. So, they are similar, which means that these two segments, xp and py, are relating to each other as these two. The hypotenuses are in the same ratio as these ketatins. So, xc over pe is also m over e. Now, have I made my problem easy? Sorry, m, m, m. Have I made my problem easier? Oh, absolutely, because pe, we know. And m, m, m over m, we also know. So now, using one of the basic techniques, I can construct fc. So, all I need to do is, from this line, which I know, it goes through pe parallel qv. I have to basically draw another line, which is distance on this xc from this parallel line. And wherever it intersects my leg aq, you basically have to choose the point x. So that's what it is. OK, that's the solution. Now, after point x is found, obviously, you just draw through x and pe line, and that would be the solution to the problem. Next problem is, find the point inside the given triangle, such as its distances from three sides of a triangle have a given ratio. All right. So, you have to find a point inside the triangle, let's say, which, if you draw perpendicular to all three sides, then these perpendicular as three known numbers. How to do this? This is t-y, by the way. It's my handwriting, which is, OK. Now, how to find this point? Let me refer you back to the previous lecture, where I was talking about locuses. There is a problem which I have solved during that lecture. If you have an angle, what's the locus of points whose distances from both legs of the angle are in certain ratio? Now, I have solved this problem, and I'm again referring you to the previous lecture where I just presented. By the way, the notes for this lecture contain some hints, including this one, with exact reference where you should go on unizord.com. Now, if I know how to construct, how to do this particular locus of points, which is the line from the vertex of this angle, because obviously all other points on this line have exactly the same ratio between these distances. So if I know this, I can solve this problem by, first, considering this angle, dhc. And px over pz would be m over p. And I will construct the locus of points where the ratio of distances from these two sides of this angle is m over p. Then I consider this angle, abc, and construct another locus of points with the ratio which is equal to px over py, which is m over n. On the crossing of these two locuses, I have my point, and obviously the distance from these two sides would be m over p, which is very easily seen from this equation. So I'm just using this locus construction twice, and I'm getting the solution. All right, problem solved. Next one, construct a triangle by its angle, one of its two sides that form this angle, and ratio of this side to the third side. OK, that seems to be an interesting. So you have an angle. You have a side, vc, and you have a ratio. vc over ac is equal to m over n. OK, how to construct this triangle? Well, this is easy, because now in this particular equation you have this, and you have this, so you can find ac. So now, basically, you construct by start with an angle, then use your bc here, and then find ac using one of the techniques which we were talking about in the beginning. So you basically have an ac. So with a c as a center, an ac as a radius, you draw a circle, and it can either intersect in no points. This side, or in one point, if it's tangent, or two points, or if it's larger than this, it will be again one point of intersection. So depending on the number of points of intersection, you have number of solutions. It can be 0, 1, or 2 solutions. That's it. Instruct a triangle by its angle, an altitude originated with the vertex of this angle, and the ratio of the segments when opposite side, this angle divides. All right, this seems to be a little bit more complex. So let me just put it in perspective. So what you have here is you have an angle, you have an altitude, bh, and you have a ratio, ah over hc. OK, now, this problem belongs to another category. Category of problems where you cannot really construct in one step, so to speak, this particular triangle. But you can use two steps. Number one, you disregard some linear dimension which you have, bh is linear. And you build a triangle which is similar to the one which we need to build with basically, with no restriction on the altitude. So the ratio should be preserved, and the angle should be preserved, but no restriction to the altitude. Now, if I will be able to construct this particular triangle, then all I have to do is just squeeze it or stretch it to a proper size of the altitude. So let's forget about the altitude for a while, how to construct a triangle using an angle and this particular ratio. OK, very easy. Just let's have one particular segment. I don't care which one, any lengths. It will be my future base of a triangle. So number one, I will divide it in this ratio. I know how to do it. So this will be my h prime. So a h prime, a prime h prime relates to h prime, c prime, as m over n, the same ratio. Now, how can I get the third point, the third vertex of this triangle? Now, the third vertex should actually confirm to two different properties. Number one, the angle should be given. Number two, it obviously should lie on this perpendicular. So how can I get that point b? So let me just wipe out this. How can I get that point b? Well, easy. Obviously, we can draw a perpendicular here. And number two, if I am given this particular angle, what is the locus of all points from which given segment is viewed at a given angle? Again, go back to a circles lecture about inscribed angles. These angles are all on a circle where a prime, c prime is a quart. Now, how to construct this particular circle? Well, I'm absolutely sure that we discussed it in one of the previous lecture. But if you really want to know right now, without referencing back, look, I mean, if you will have midpoint of a prime, c prime, which is this, it goes through the center of this circle, right? So we know this thing. Now, how to find this point? Well, since this is a given angle, then this is half of this angle, right? Because this is a perpendicular through a midpoint of a segment. So you can build a prime, let's call it d prime, a prime, b prime, what else, d prime. This is the midpoint. You can build triangle a prime, b prime, d prime, because it's the right triangle. You know the casualties. And you know the cute angle which flies in the opposite side. So you build a triangle. You have the point d prime. And now, using three points, a prime, b prime, and c prime, you construct a circle which circumscribes this particular triangle. So you've got this circle. And now, the intersection of this circle and this perpendicular through the point which divides in the proper ratio is my point b. Now, triangle a prime, b, c prime, is the one which satisfies this condition, because these are all inscribed angles which are supported by the same arc. It satisfies this condition, because this is divided. Altitude is dividing the opposite side in the proper ratio. Now, all it doesn't have is a proper altitude. But that's the easiest now. Now we just have this particular segment start from b, cut it to the proper length of the altitude, and draw a line perpendicular to it, or parallel to this. And that would be my real points a and c. Obviously, since these lines are parallel, they are perpendicular to the same line. Triangles a, b, c, and a prime, b, c prime are similar. So division between a, h, and hc is exactly the same as between a prime, h prime, and h prime, c prime, which is m over n. So all conditions are satisfied. Now, what's the lesson to be learned from this? If you cannot build the triangle which you have to build using some particular linear dimension, build a similar one if you can. And then stretch it or squeeze it to a proper linear dimension. All right, so this is done. What's next? OK, construct a triangle by its two angles, and some of the side common for these angles with an altitude. It's very similar actually. So you have a triangle about which you know two angles, which immediately triggers you, OK, I know how to construct a similar triangle, since I know two angles, and all triangles with the same two angles are similar to each other. So you have two angles, b, a, c, and b, c, a. And you also have a plus h, a plus h, the side plus altitude. Now, how to build this? Same story. Forget about the linear restriction on our triangle. Let's construct any triangle which has these particular angles, OK? So let's say I constructed this triangle. So a prime, b, c prime is similar to the one which I need. All I have to do is I have to squeeze it somehow. But now let's think about this. All linear dimensions in similar triangles are proportional to each other. And it's related not only to sides, but also to altitudes and medians and bisectors, et cetera. I probably proved it before if I didn't. You can do it yourself. That's trivial stuff. So in this case, if I have built a prime, b, c prime, I can construct actually my new sum, a prime, c prime, plus b, h prime. So this is my new sum of the side and an altitude. Since I have constructed it, it's the known thing. Now, I also know that it's related to a plus h, which is this plus this, as a known, as a ratio of, let's say, sides, a prime, b over a b. So similar triangles have everything similar, including sides are related to each other as some of the corresponding sides and corresponding altitudes. So now, what exactly I have to really find here? a prime, c prime, plus b, h prime is a known segment because it's a sum of these two things which we have constructed. This is also given. So this is the ratio which has a prime b, which is, again, known thing, and only one, a b, which is unknown. So I have to build the fourth. You remember in the beginning, if you have three known elements and one unknown, how to construct it? So this is exactly this case. You know this, you know this, and you know this. This one, a b, you don't know. So that's how you determine it from this particular equation. And since you know a b, you know the point a, and then parallel to a prime, c prime, gives you a triangle. That's it. Same technique, by the way. I have disregarded the linear dimension, which is a plus h in this case, and built triangle just based on two angles, which is similar to the one which I need. And then just squeeze it or stretch it. All right, construct an isosceles triangle by its angle between equal sides and difference between the base and the null features. Same thing, basically. So if you have, let's say, isosceles triangle, a b c a b is congruent to a c, and you have an angle b a c. What you also have is b c minus a h. Well, let me do exactly the same situation. Since I know this angle, I can construct an isosceles triangle, which is similar to the one which I need, because these two angles are equal to each other, and some of them complements this angle to a 180 degrees. So basically, I know all angles in the triangle, and I can construct the one which is similar. Let's call it b a prime and c prime. Now, having the difference between, well, let me just use letters which correspond actually. This would be a, and this would be b prime. That's easy. So now, considering I can basically find what's the difference between b prime, c prime, and a h prime, b prime, c prime minus a h prime, considering this, and considering I know b c minus a h, and considering I know that, let's say, this is the same ratio as, let's say, a b prime over a b, knowing this from this angle which I have constructed, and this which is given, and this which is, again, from this constructed triangle, I can determine a b. So I find my point b, draw a parallel line, and that's the triangle we change. That's it. That's the end of this particular lecture. As usually, I do encourage you to go through the same problems again just by yourself. Go to theunisor.com and just inculcate it into your mind, if you wish. There will be other problems related to similarity. So I'm definitely encouraging you to continue learning this stuff. And also, you know that there is a way to register on theunisor.com, in which case, as a registered student, you will be able to take exams. And your supervisors or parents will be able to, who also have to register, obviously, will be able to control the educational process by basically enrolling you in certain topics and marking them as complete if the exam was successful passed. That's it for today. Thank you very much.