 I'm Zor. Welcome to Unizor Education. Today is one of my favorite types of lectures. It's basically solving problems, continuation of construction problems in geometry. Without any further ado, I'll just go through my list of problems and obviously you have to try yourself first. They're all in the notes for this lecture and when you are done, just listen to the lecture and see if you're right or am right or wrong. Anyway, so let's go one by one. Construct a right triangle by its two legs. Right triangles by two legs. These are legs or cacti as they are called in Latin sometimes. So construction is very simple. You start with the right angle somewhere. Now how to start with the right angle? Well obviously start with a line and choose some kind of a point outside and drop a perpendicular. Now on one side of this angle, you basically, using the compass, you mark one particular side and on another angle side, you mark another point and just connect them. So that's easy and obviously we're using the fact that this is the right triangle. That's why this angle is known. Basically it's an equivalent to constructing any triangle by two legs and angle between these legs. In this case, angle is 90 degrees so it's known perfectly. Simple. Construct the right triangle by one of its legs, cacti and a half point. All right, so now we have other two elements. Usually it is sufficient to have two elements in the right triangle, three elements in the regular triangle. So we have one cacti, one leg and hypogenous. Okay, how to construct this particular triangle using these three elements? Well very simple. Again, you start with the right angle. You use the compass to measure this particular side and mark the point here. Then using compass set to this particular side, to hypotenuse, you basically mark the circle with this at the center and wherever it crosses this line, that's our triangle. Now all these problems are extremely simple and what I'm basically doing is accumulating certain sets of simple problems so in the future when something more complex comes along, I won't really go into the details how to construct, for instance, the right triangle by two cacti. I'll just say, okay, we know these two x of this right triangle, that's why we can build it over there, construct it over there without any details. So I'm continuing with these little things, small, simple. Construct the right triangle by a cacti and an acute angle. It forms with hypotenuse. All right, so now we have these cacti and this angle. Okay, how can we do this? Well, as usually, if we know some kind of an angle, we should start with this angle. So let me start with this angle. I'll build an angle which is given. On this leg, I will mark a segment which is congruent to a given leg and from this point, I'll just construct a perpendicular from the point to this line and wherever it crosses the other side of the angle, that's my third point of the triangle. Construct an isosceles triangle by altitude to a base and one of two congruent sides. Okay, so we have an isosceles triangle, we have altitude to the base and one of the sides. Now, when I'm saying that these are given elements, basically it means that somewhere on the plane, there are two segments, one and two. And what's given is, okay, this element is my altitude, this element is my side of the isosceles triangle and using these two elements, I have to reconstruct or construct my triangle. So this picture is very helpful to analyze the problem. So what if, for instance, I have this particular triangle, so how does it look? Okay, this is my side and this is my altitude. But now, what do we see? We cannot construct immediately the whole triangle, but we can construct half of this triangle, this one, a, b, d, a, b, d. Since in the isosceles triangle, the altitude to the base and median and angle bisector and perpendicular segment bisector, they're all the same, basically. Then I can just construct half of this triangle and then construct another half by symmetry. So in this particular triangle, the half of the, say, isosceles triangle, it's the right triangle because it's altitude. So I can construct it by the casualties, the leg, and the hypotenuse. So this is hypotenuse, this is the casualties. Now, I have already done, before this particular problem, a couple of other construction problems related to right triangles, I've done that already. So I'll just use, basically, what I have already done. I know that I can construct right triangle by hypotenuse and the casualties. So I'm saying, okay, let's consider a bd. We have two elements of this right triangle known, which is hypotenuse and the casualties. All right, so I'll build it. I'm not going into the details about how to build it. I refer to the prior construction problems which have been already solved. So I can do this one. Now, I continue this line and build the symmetrical point here. So these two segments are equal to each other. And that's my full triangle. Now, why can I do it? Well, because the altitude is also a medium, so it breaks the whole base into two halves. So I'm analyzing the problem first by drawing the picture of the ready-to-be, ready-to-use triangle and see what elements actually constitute this particular triangle. And I know which elements I can construct. Actually, half of the triangle I can construct. Then I continue with another half. And again, notice that I have already used one of my construction problems which I have done earlier about right triangles in subsequent construction problems when I had to construct an exhaustive triangle. All right, what's next? Construct the exhaustive triangle by altitude to a base and an angle opposite to a base. Okay, at the vertex. So again, let's consider we have this already constructed. So what's known? Known this angle and altitude bg. Now, again, we know that altitude to the base of the exhaustive triangle is also a bisector of the angle and bisector of the base. So since it's bisector, consider again the right triangle and altitude. The right triangle abg. We don't know what this right triangle is. We know one catheter, the leg, because that's the altitude of the exhaustive triangle. And we actually know this angle because it's half of the angle which is given. Now, angle abc is given. So half of this angle I can consider as basically almost given. I can construct it. So how do I proceed now? I can construct abg knowing the leg and the angle. So if I have an angle and altitude, so what I do first I bisect the angle using this half and in this segment I build the right triangle. And to complete the picture I just reflect this point to this side so it's the same length because it's an edging as well. This one. Okay, let's go. Construct the exhaustive triangle by a base and altitude towards one of two congruent sides. That's slightly different. So we have an exhaustive triangle. So we have a base, AC. These two sides are equal, congruent to each other. So I have this and I have altitude to one of congruent sides given. Now, well, obviously as you understand, I can consider the triangle abc. It's a right triangle since this is an altitude. And I know the partners and the calculators of this right triangle so I can build it. So if my segments, this is AC and this is AD, this is altitude and this is the base. So using these elements I construct right triangle. Okay, that's the beginning. Now what should we do? How can I get this third point of the triangle? Well, actually it's quite simple. What you can do is you can continue this line. And since this point B lies on the perpendicular bisector of AC, I can build perpendicular bisector AC. And that's where they cross. That's my third line. All right. And again, I'm not going into the details about how to construct the perpendicular bisector or how to bisect an angle or how to drop a perpendicular to a line, etc. All these elementary problems have been discussed in the prior lectures and they should be at your disposal at any time. Construct right triangle by a hypotenuse and the cute angle it forms with one of the cavities. Okay, so you have a hypotenuse and this angle. Well, I think I did solve this problem before. Well, in any case, it's easy. You start with an angle which is given. You draw a hypotenuse. Oh, I think before I was solving when it was one of the cavities and an angle. Now it's a hypotenuse and an angle. All right, so you have an angle, then you have this segment. So you mark it here. So this is your point here. And what you should do is draw a perpendicular to this line and you got your triangle. Given an angle and a point inside it. Okay, angle and point inside it. Construct a straight line crossing this point and cutting congruent segments from the sides of a given angle. So I have to construct the line in such a way that these two segments are congruent to each other. Well, I remember one of the previous theorems or construction problems, whatever we had was the following. If you have an angle bisector and draw a perpendicular anywhere to this angle bisector, then this perpendicular would cut congruent segments from both sides. Why? Well, it's obvious since this is an angle bisector, these angles are congruent. And these are right angles since we draw a perpendicular to bisector. And this category is common for these two right triangles. That's why hypotenuses are equal to each other. So how can I apply it to this particular problem? Well, first of all, it's good that I remember that particular theorem, now I can use it. Now, how can I use it? Well, obviously I have to draw an angle bisector first and from M I have to draw perpendicular to this particular bisector. Now, this perpendicular obviously is crossing the point M which is required and since it's perpendicular to the bisector, it cuts both sides of the angle in equal lengths of the segments. So that's it. That's a very simple one. But you do have to remember, as you see, you do have to remember certain things. And that's typical in mathematics. You always have to do something new based on something which you have already learned and basically know. Okay, construct two segments where they're given sum and differences. Uh-huh. It's similar to the angles actually. We already covered this with angles. So if you have two segments X and Y, now you are not given these two segments. What you are given is sum of these two segments called A and difference of two segments which is B. So the A is equal to X plus B and D is equal to X minus. All right? Well, actually let's draw a little smaller. Now, what should we do now? Well, obviously we have to solve this particular equation for X and Y and that's very easy. You just sum them up. It will be A plus B equal to 2X and if you subtract one from another, it will be A minus B equals to 2Y. And as a result, you have X equals A plus B by half and Y equals A minus B by half. All right? We make it pretty longer. So the difference is B and sum is A. Okay. So now, knowing A and B, we algebraically solved for X and Y. So how can I construct these two segments X and Y if I know A and B? Well, elementary. I just add A plus B and divided by 2. How to add two segments you have to basically know. You just draw one of them equal to A and then continue with another and then how to divide it by two? Well, you draw a perpendicular bisector. So half of this is A plus B divided by two and that's exactly the X which we need. Now, similarly, how to construct A minus B divided by two? Well, you construct A again. Now, you construct B to the opposite direction towards another end of the first segment and this one, whatever is left of it, A minus B, you draw a perpendicular bisector and this part is your white. Okay. And the last, but not least, construction problem for this particular lecture is how to divide a segment into four, eight, 16, etc. parts. Well, again, it's very similar to whatever we did with angles and basically the solution is exactly the same. Since you know how to divide a segment by two, you just draw a perpendicular bisector, that means that you can continue this process of dividing by two. So if you divide this particular part, which is half, again into two, you will get a quarter, then you will get one eighths, one sixteenths, etc. Now, I actually addressed a very interesting problem when discussing the angles about how to trisect an angle and I told that it is impossible to do using the standard tools of geometry compass and the ruler. With segments, it's completely different. You can divide a segment into any number of parts, three, five, 27, whatever you want. Well, there is a difference. I mean, there is a similarity between dividing segments and dividing angles, but this is the difference. So segments can be divided into basically into any number. All right, that's it for this lecture. Don't forget that there is a website Unisor.com, which contains all the different educational material and it will definitely help parents and students supervisors to control the study, the educational process by enrolling the students into certain subjects, controlling their results of their taking the exams, score you can examine and basically you can either check that this particular subject is passed or failed and if it's failed then the student must actually go again through theoretical material and exam questions until he gets reasonable results. All right, good luck and thanks very much.