 Hi, I'm Zor. Welcome to Unizor education. Well, by now you probably realize that construction problems in geometry is kind of my favorite, just because it really develops your creativity and analytical thinking, et cetera. So, it's very important for you to listen to this lecture prepared, which means you have to really go to the website and look through the notes to this lecture where all these problems are presented to you. Think about them and try to solve them yourselves. They're not really difficult at all. Basically, solving them is the main purpose. What I'm doing right now is just suggesting my personal solution to these problems, whatever I come up with, which might actually be the same as yours, might be different. But in any case, again, the purpose of the whole exercise is for you to think about these problems. Even if you didn't really solve the problems, listen to the lectures, and when it's finished, go again through the problems and try to remember at least my proof or solution and repeat it again, at least once. So, it will inculcate in your mind that there are certain approaches which we are using and it will help you in the future for the next problems. Again, creativity and analytical mind. That's the development which we're looking for. Okay, now thanks for listening to this and let me go to the problems. Construct a quadrangle by two diagonals, two sides with the common vertex and an angle formed by two other sides. Okay, so we have it. Let's just analyze. Quadrangle. So, two diagonals. We have AC, we have BD, two sides with the common vertex. Well, let's say these two sides, AB and BC and an angle formed by two other sides and this angle, which is angle ADC. Okay, these are given. We have to construct the quadrangle. Well, let's just think about if both sides and this diagonal are given, we can definitely construct the triangle ADC. Now, that's basically half the job, right? Now we have to find where is the point D, which would satisfy both. The BD should be equal to whatever BD is given and the angle ADC would be again given. But let's just do it this way. Obviously, the locus of all points D with the distance from B equals to whatever is given is obviously a circle with the center at point B and the radius equal to BD. So, that's one locus and then another locus are all the points from which the segment AC is visible at certain angle, which is also given, which is kind of an arc, if you remember. From the previous lecture, it's an arc. So, by combining these two locuses, whatever the intersection between them is, that's the solution, that's where point D is supposed to be located. So, this is part of basically analysis, finished. Now, let's do the construction. By three sides, AB, BC and AC, AC, AB and BC, we build this triangle, this is ABC. And then, using the center at B and the radius BD, we draw one circle, this is the center and the radius is equal to BD. And using AC segment, we build a locus of points from which, this is an arc, from which AC is visible at this angle. So, all points on this arc have the same property. They are all congruent to this angle. And we know how to do it. If you don't go to the previous lecture, that's where I explain it. So, intersection of these two circles, in this case, I have two intersections. Basically, both satisfy our quadrangle. So, it's either ABCD or ABCD prime. So, both quadrangle have the same angle and the same second diagonal. That's it. That was easy. Now, if you didn't catch it for whatever reason, if I'm too fast, send me an email. I'll try to be more specific in the return email if you want to. Okay. Given three points on the plane, construct a line that contains the first point, such that the distance between the perpendicular to this line from two other points equal to a given segment. Okay. So, we have to find a line, three points, A, B and C are given. We have to construct the line in such a way that these two perpendicular DE equals some kind of a given segment. So, DE is given. That seems to be easy. Let's draw a line, let's analyze the game. Consider the line is there. Obviously, this, if AF is parallel to DE, and these are two perpendicular, then obviously DEFA is a rectangle. Now, let's think about AFB triangle. What do we know about it? AB is given because points A and B are given on the plane. And AF is obviously equal to DE because this is a rectangle. So, in this right triangle, we know the casualties, AF and hipogenous, which means we can build it. Now, that's basically it. Analysis is finished. The construction itself is the following. So, if you have two points A and B, and this segment, so you built a right triangle by, well, it's not exactly right. That probably would be better. AF. You built the right triangle with AF casualties equal to our segment, and AB is a hipogenous. Now, we know, now this is point C. Now, we know that the line is supposed to be parallel. So, we just draw a parallel line to AF. And that's the line which we need. Because if you will go with perpendicular, obviously we will get this DE. The distance between perpendicular is equal to AF, which is equal to DE. So, what was the constructive element in this particular problem? Draw a line AF parallel to the line which we need and construct this particular triangle because you know all the elements of it. By the way, all construction problems are like that. You cannot really build exactly what is required immediately. You have to reduce this problem to some other problem, which is so little easy, and then go back to whatever you need. Right now, we have reduced the problem of drawing the line to the problem of drawing a triangle, right triangle, for which we know everything, and then the line will be parallel to the casualties. Inscribe a triangle with two given angles into a given circle. So, well, this is actually something which I probably have to talk about a little bit. If you know the angle, let's say you know angle A and angle B, and you know the circle into which this triangle is inscribed. Basically, what you know is you know the side opposite to this angle because no matter how you draw this angle inscribed this angle into a circle, let's say this way and this is the same angle B prime. No matter how you inscribe it, you will always have exactly the same lengths of the cord, which is connecting the end points of an arc supporting this. Why? Well, obviously, because this angle is always inscribed angle, is always half of the central angle, and central angle, therefore, is the same. In this case and in this case. So, if angles are the same, if inscribed angles are the same, then the cords which connect the arc supporting this angle are always the same. Now, that means that if angle is given, then the cord is given. So, if you have, let me just draw it again, so if you have two angles, let's say ABC angle BCA, these two angles. If these two angles are given, then it's very easy to get the cords which are opposite to these angles. How? Just draw any angle in this circle and inscribe any angle. Any cord here, and using this angle, put another cord, and if this is an angle equal to ABC, then this is a cord which is equal to this one. And same thing with this angle. Again, take any other point, let's say this point, and inscribe the angle which is congruent to BCA. Let's say it's this angle. Then the cord would be exactly the same as this cord, because all cords in the same circle which support congruent angles are congruent among themselves. So, the construction is very simple, that's why. You have a circle and you have a couple of angles. From the angles you derive two segments, one side and another side. One would be AC and another would be AB. Now, knowing these two sides, the situation is very simple. Just have any point, use one side and use another side. And that's your triangle. So again, knowing an inscribed angle in this circle is equivalent to knowing a cord which supports this particular angle. Because all inscribed angles in the same circle which are supported by the same arc are congruent, and if they are congruent, then supporting arcs are congruent. Arcs and angles are always going together. I mean, arcs and inscribed angles supported by these arcs. Angles are equal, arcs are equal or congruent. All right, circumscribe a triangle with two given angles around a given circle. Okay, now instead of inscribing, we have to circumscribe a triangle. So we have a circle and we have a couple of angles. So we have angle BAC and angle ABC. So you know these two angles. How to construct a triangle? Knowing the circle of this. Well, as you know, if this circle is inscribed into an angle, then the center is supposed to be located on the bisector of this angle. Same thing here, right? Now, what does it mean? Well, it basically means that you can start from, let's say, construction. You can start from, let's say, choosing tangent to this particular circle. And then from the center, you draw a segment under the angle equal to half of this one from ABC. And to this side, we draw a segment under this angle, which is half of BAC. Since we know these angles, we know they're halves, right? And that basically gives us two points. Now, from these two points, we just draw two tangents to a circle and that gives us the triangle. Now, I don't stop on something like how to draw a tangent from a given point outside of a circle because we have already covered that many times, so I'm not going to stop. If you need the clarification of this, go to the previous lectures and you can find it over there. These are trivial problems which I have solved once and use it basically as given. Construct a triangle by a radius of its circumscribing circle. Okay, triangle. You have a radius of a circumscribing circle, an angle and an altitude from the vertex of this angle. Okay, so you have the radius of a circle, which means you have a circle. You have an angle. And since you have an angle, as I was saying before, you can construct a chord actually. So you know the length of it. Well, not that you know, but you can derive actually the chord itself. And that's why what you can do, as far as the construction is concerned, you do the following. First, you determine the chord. Now, having the chord, you basically use it here. Here. Anywhere. Now, since you know an altitude, you just draw a line parallel to this one at this distance and here you have two solutions. One and two. Both triangles have the same height and the same angle which is given because they are supported by the same arc and they are inscribed into the same circle. Well, obviously if height is too high, if altitude is too big, then this parallel line goes outside and there is no solution. Or if it's a tangent, there is one solution. Inscribed into a given circle, triangle was given some of its two sides and an angle opposite to one of these. So, if you have triangle inscribed into a circle, let's say you have AB plus BC. Inscribed into a triangle with a given sum of its two sides and an angle opposite to one of these. And let's say you have this angle. Opposite to AB. All right. So you have this basically. Let's put it C prime. So BC prime is congruent to BC. So you have this segment AC prime and you have this angle. But look, again, again and again, if you have an angle and you have a circle where this angle is inscribed, it's equivalent to having a chord which supports this angle. So you can derive the value of AB. Now, having AB, you obviously having this sum can derive BC. So the construction is first using an angle derived with AB by basically putting an angle somewhere, as I was saying, doesn't really matter in the circle. And as soon as you inscribe this angle, you get the chord. So once you get the chord, all you have to do is subtract this length from the given length of the sum and you get the length of another side. So that will be A, that will be B, and that will be C. That's it. So I'm trying to be a little bit more specific what's the analysis and what's the construction. So maybe it will be just a little bit clearer that these are two independent stages, if you wish, of solving construction problems. First you analyze what and how you can construct and then you construct. Okay. Inscribe into a given circle a quadrangle with known side and two angles opposite to this side. That's the same thing basically. If you have a quadrangle, you have a side, let's say you have AB, and you have two angles opposite to each, which is B, C, D, and C, D, A. These two angles. What should we do about this? Well, knowing an angle again, knowing an angle again, you know the chord which supports this angle, which means diagonal. So you have one side and you have a diagonal. So that actually means that you can have in your circle the construction can actually be have this particular AB side anywhere. Then since you have this diagonal, you just use a compass. So if this is, let's say, B and this is A. Then from the B, using the radius of B, D, you just draw a circle and you've got point D. Now, similarly, since the D angle is given, C, D, A, you have this diagonal, AC. So again, using a compass A as a center and AC as a radius, you draw another and you have your C. And this is your quadrangle. So from the angles, we go to diagonals and having diagonals, one side and two diagonals, we just using these radiuses find these two points C and D. Well, if you have noticed almost all the problems in this particular series are related to this relationship between the inscribed angle and the quad. If you know the inscribed angle within a circle, you know the quad. I mean, you can derive the quad, you can construct the quad. And vice-versa, actually, because if you have a quad, just draw any angle inscribed which is supported by this quad and you have an angle. They're always together. Okay. Inscribe a circle into a given rhombus. All right. So if you have a rhombus, now you have to inscribe a circle. Now, if you remember, this circle is always when it's touching these both sides, these are perpendicular to the sides. So the center of the circle is at the crossing of the diagonals of the rhombus. That's obvious. Now, the radius is, well, you have to drop the perpendicular from the intersection of the diagonals onto any side, and that would be your radius. That's very simple, actually. Since it's rhombus, all these radiuses are the same. Inscribe three circles in a given equilateral triangle, such that each circle is tangent to two other circles and two sides. So if you have equilateral triangle, now you have to inscribe three circles in something like this fashion. So each of them, each of these circles is touching two other circles and two sides. Well, again, let's start from bisectors, angle bisectors, and that's where the centers will be located, right? Now, at the same time, you can obviously derive that these three bisectors will also pass through the tangential points of tangency between the circles. I don't want to interprove it. It's kind of obvious, and if you want, you can spend some time to prove this type of thing. It's very easy. Now, since this is true, then obviously all you need to do is to find the center and how to find the center. Well, consider this, for instance, angle. This circle is inscribed into this angle, which means it's supposed to be on its bisectors. And same thing with this circle. Not only it is on this bisector, but it's also on this bisector. And same thing here. So, by constructing three-angle bisectors, or medians or altitudes, they're all the same in the equilateral triangle. And then, from the midpoints, bisectors of these angles, you get the centers. Now, once you get the centers, the radius would be just drop of perpendicular to any side and you would get the radius. Well, that's it. This completes this particular lecture. As you saw, problems are not really very difficult. And I do encourage you to go through the problems again and try completely by yourself to construct whatever is necessary. Use the notes from Unizord.com. It's construction problems number 8, four circles. Don't forget that you can actually go through a much more well-controlled educational process by enrolling into a specific course. Then you can start basically taking exams. Parents or teachers are encouraged actually to do it with their students and check the results of the exam and make sure that your students are actually going through all the exams getting the maximum score available, which means actually they have mastered the material. Well, that's it for today. Thank you very much. Good luck.