 Hi, I'm Zor. Welcome to Unisor Education. I am continuing solving problems with the quadrangles, triangles, parallel lines, etc. So all these problems are spread in eight sets. This is the set or series number seven out of eight. Okay. So I have these problems in piece of paper. They are in the lectures, in the notes of the lectures, and I'm sure you have already spent some time just looking at them and thinking about how to solve these problems. So if you came up with a different solution to whatever I'm suggesting right now, great. Now you will have just yet another solution. But it's very important for you to try to solve it with yourself. All right, construct a parallelogram by one side and one diagonal. So let's think about it. You have parallelogram. You have a side which is equal to this one. And you have two diagonals. It's direct by one side and two diagonals. Well, this is actually simple because diagonals are divided by their intersection point in two halves, which means if you have two diagonals, you can actually take two halves of these diagonals. So AP will be one half, PD would be another half of another diagonal, and AD is a given side. So APD can be built by three sides. So after you build APD, you just extend one diagonal by the same length, another diagonal, and you will get a parallelogram. So as for analysis, what you see is you basically see that APD at triangle you can construct using the side which is given and halves of diagonals. You don't even need any additional construction, additional lines or whatever. Next, construct a parallelogram by two diagonals and angle the form with each other. Well, that's basically the same thing. Parallelogram, two diagonals. So you have the lengths of two diagonals and angle. Now, if you have two diagonals, you have two halves of these diagonals. And again, this triangle can be built by side, angle and side. And then you extend the triangle to both sides to get vertices of the parallelogram. Construct a parallelogram by one side, one diagonal, one side, one diagonal, and altitude between a given side and opposite side parallel to it. And this altitude. So you have A, B, C, D, E. So you have A, D, A, C, and C, E. Well, since you have A, C, and C, E, which is actually two sides, hypotenuse and the characters of right triangle A, C, E, you can build it. So you can build the triangle A, C, E by hypotenuse and one of the characters. Fine, now this is done. Now you have A, D, so you just measure it here. Here you go. And now you have half of your parallelogram. So all you do now is draw a parallel from C parallel to A, D, and from A you draw a parallel to C, D. And here is your point. Well, these are all relatively easy exercises just to know that you have techniques, basically, that you understand that the parallelogram has certain properties and right triangles has certain properties. They're quite obvious. So this is just some kind of your, if you want, a toolbox for really complex problems which might be in the future. So just consider these as different tools in your toolbox. Construct a rectangle by a diagonal and an angle between two diagonals. So you have a rectangle, you have one diagonal, and you have an angle between diagonals. But in the rectangle two diagonals are congruent, which means you have both, actually. And we have already solved this for parallelograms when you have two diagonals and an angle between them. If you remember, we're using just halves of diagonals to do it. Probably there is a better and easier way in case you know this is a rectangle. But for me, it's easy just to refer to the previous problem which we have already solved. So without wasting any more time, I can just refer to the previous problem. Construct a rhombus by a side and one diagonal. Okay, if I know that this is a rhombus, which means parallelogram with all sides congruent to each other. And if I have a diagonal, that means that I have three sides of this triangle because these two sides are congruent. So this is given, this is given as well. And diagonals, so I built the triangle and then the couple of parallel lines gives me the rhombus. Construct a rhombus by two diagonals. Now, we know that two diagonals in rhombus are not only divided by the point of intersection into congruent parts, but also they are perpendicular to each other. So having two diagonals, you can actually build any of these right triangles. They are all congruent to each other. And basically, you know, two cacti of this right triangle, half of one diagonal and half of another diagonal. So you start from the triangle and then you extend it both ways to get the rhombus. Construct a rhombus by diagonal and angle this diagonal begins at. So you have a rhombus, again. You have an angle and you have a diagonal. Now, remember the diagonal in the rhombus is a bisector of the angle it goes from, right? So if you have an angle, the way how to construct this particular rhombus is to make your angle bisected and whatever the lengths of diagonal just cut it here and then draw a couple of parallel lines, parallel to this and parallel to this. Here is your rhombus. So you just use the property of the rhombus diagonal to bisect an angle. Construct a rhombus by diagonal and angle opposite to it. Well, this is not much difference, I guess. So you have a diagonal, but instead of this angle, you have this angle. Now, first of all, the easiest way basically is just think about if you have this angle which is equal to alpha, let's say, then what is the size of this angle? Obviously, it's 180 minus alpha divided by 2. Why? Because the sum of these three angles is 180. So subtract alpha, you get some of these two angles and basically divided by 2, you have one angle. So having this angle is equivalent to having this angle. So again, this is angle which I could just calculate it. I have the size of diagonal, I put it here. Another angle of the same size here because the diagonal that is bisector and couple of parallel lines. Construct a rhombus by a sum of two diagonals and an angle one diagonal forms with a rhombus side. Okay, this might be a little bit more interesting. So let me draw a rhombus like this. So again, construct a rhombus by sum of two diagonals. So sum of these two diagonals and angle one diagonal forms with a rhombus side. Let's say this angle. Okay, but let's consider a triangle C, D, P. It's a right triangle because diagonals of the rhombus are perpendicular to each other. Now, you know sum of D, D plus H, C. Now each diagonal is cut in half by this point P, which means having sum of these two it's the same as having sum of P, D plus P, C. These are just halves. So basically what you have is how to build a right triangle by one acute angle and sum of two triangle by sum of two legs and an end an acute angle. Basically that's enough to build the whole rhombus because we just have one triangle and then extend with the parallel lines, etc. Okay, so how to build a triangle knowing sum of two cartotons and an angle. Okay, let's think about this. Let's extend this guy to C prime. Now, what do we know about B, C, C prime? We know B, C prime because this is the sum of these two. A, C prime and A, C are congruent by construction. So you know that B, C prime is the sum of your two legs, okay? You know this angle and actually you know this one which is 45 degrees. So you know a side and two angles of this triangle. So you can do it. So build B, C prime, C and then from the C you draw a perpendicular to this line to get A. That's how you build a B, C triangle. Now having this triangle, you just put it here, extend your diagonals by the same size in both directions and you will get the rhombus. So I have reduced my a little bit more complex problem about rhombus to a simpler problem about triangle. I'm not sure actually maybe we have already solved this problem about triangles some time ago. I just don't remember. Anyway, construct a square by its diagonal. So if you have a square, let's analyze this problem. You have a diagonal of the square. How to construct a square knowing only its diagonal? Well, the simplest way is if you have A, C, cut it in half, draw a perpendicular bisector of A, C, cut these lengths on both sides and you have two other vertices of the square. Okay, construct a square by a sum of its side and diagonal. Okay, so you have a sum. This plus this. Okay, so what do we do now? Well, as usually, if you have sum of two sides, you usually extend, let's say, which side to extend? Well, let's extend hypotenuse to D prime. So A D prime and A D are the same. Now, what do you know about triangle C D D prime? Well, you know this side. Obviously, it's C D prime, right? It's given to you. It's a sum of diagonal and a side, which is, these are congruent to each other. So you know this. You know this angle is 45 degrees and we know how to build this 45-degrees angle, all right? And also, what we know is that this 45 is also, and this is exterior angle to A D D prime, which is a socialist, which means each one of them is equal to 22.5 degrees, half of 45, right? So you know this angle as well. So you take 45 degrees angle, no matter how you got it. Let's say you, you know, built a triangle with two congruent legs, doesn't matter. You got 45 degrees, you drew a bisector and that's how you get 22.5 degrees. So you have basically this angle, this angle, and you have C D prime. So you build this triangle and from this point D, you, well, since you do have this point D, so you can do it, you can do it perpendicular to this line to get the B to the same distance. And from B, you get parallel to C D to get the point A. So start from C D D prime by having the side and two angles near it. And then from the point D, you get point B, which is symmetrical relative to this. And then from B parallel to C D and you get the point A. So that's your square. I'm sure there is maybe a better way or different for sure. There are many different ways how you can solve the problems like this. But anyway, that's it for this particular lecture. And so this is lecture number seven. There is an eighth. That will be the next one. Meanwhile, I do encourage you to take a look at the Unisor.com. And if you have a parent or a supervisor, or if you are a parent or a supervisor, interested in a deeper and more controlled environment of learning math, Unisor is basically an excellent choice for you. You can enroll the student into a specific program of your choice. And you can watch the exams, which are part of this Unisor.com website. Check the scores and pass or fail, basically, based on the scores relative to the maximum score available for this particular exam. That's it for today. Thank you very much and good luck.