 I'm Zor. Welcome to user education. Today I will continue solving different problems about triangles, quadrangles, parallel lines, etc. This is a series of questions number six out of eight. And we have 10 different problems to solve today. Okay, so let me just start. Construct a triangle by two sides and an altitude onto one of them. So we have a triangle. We have two sides and altitude towards one of these sides. Well, let's do it this way. Since we know this side and an altitude, first let's put this side somewhere on the plane. And now let's build a locus of all the points which can serve as the third vertex of the triangle with this base and give an altitude. Now, as we know from one of the previous problems, which is obviously without the problems, all these vertices are located on the line parallel to our base and on the distance equal to an altitude. Now, out of these lines, we have to pick up one which has a particular distance from A. So we just take this distance in the compass, use A as a center and just do this, basically. So that's the point C which we are looking for. Now, obviously, you can have more than one solution if this circle intersects the parallel line in another point. So this is another C prime, a triangle A, B, C prime, which has exactly the same base, A, B, exactly the same side, which is A, B, and an altitude which is drawn towards the base. The same altitude. This one is equal to this one. So this particular problem might have two solutions, might have one solution if the circle is just touching the parallel line, or no solutions if the segment which is given to us as the side A, D is too short. Okay, number one. Number two, construct a triangle by a side, an angle it forms with another side, and an altitude onto it. And an altitude. Okay, it's more or less the same as before, we just have an angle instead of a side. By the way, I would like to mention something. Whenever you're talking about triangles, well, you know that there are three different most important properties of triangle. They are congruent if either two sides and an angle in between, or two angles and a side in between, or three sides are congruent. So in all cases, we have three different elements, either three sides or side and an angle, two angles, whatever. Now, in these cases, all of these cases, which we are, which I'm going through, there are also three elements. So it's not coincidental. Triangle is something which can be defined by three different elements. By the way, it cannot be defined by three angles, because they are dependent on each other. If you remember, the third angle is always 180 degrees minus two other angles. So getting three angles is actually the same as getting two angles. But let's say two angles and some kind of side, or altitude, or anything actually, three elements independent of each other really gives you more or less unique identification, more or less because sometimes, as you saw from the previous problem, you can have two solutions or no solutions depending on their mutual dimensions and relationship. But in general, triangle requires three elements. And you can actually go back to algebra. I'm not sure if you know this property, but in general, if you have n different unknowns, you do need n different equations to find these unknowns. With linear equations, it's kind of obvious. If you have linear equations, then number of equations should be equal to the number of unknowns to have a solution. Okay, so let's go back. So we have a side, we have an angle and we have an altitude. Well, we start the same way as before. We put the base somewhere and then draw a locus of all the vertices which have the same altitude as the altitude given to us. So basically A, B, C, D, A, C. This is given. Now we have to find a point B. Now, how can we find a point B on this line? Well, in the previous problem, we knew the side A, B. In this case, we know the angle, which means from this point A, we have to just build an angle equal to the given one. And that would be the point B. And obviously, the altitude will be correct because the distance between these parallel lines is equal to the altitude which is given to us. Simple. Construct a triangle by an angle and two altitudes onto sides that form the angle. Okay, so again, three elements, angle and two altitudes to this and to this. Okay, so how can we build this particular triangle? Let's think. Let's think about it this way. Consider a triangle A and C. We know, first of all, that this is a right triangle. We know the acute angle. And we know the cataclysm, which means we can build it. I don't want to go into the details about how to build a right triangle by a cataclysm and an opposite angle. We already considered that before. So we can always construct this triangle. So I'll just construct it, basically, using the same. So that's how we do it. A and C. Well, if you are very much interested, okay, I'll just talk about this a little bit. You have the right angle first. You know these cataclysm. So you basically know the point C. And now you have to either subtract from 90 degree this angle to get this one, and you draw the line. Or you draw somewhere else, somewhere else, the line with this angle and draw a parallel through C. So anyway, we can do it. So we have this. And we have this. Now, now what should we do? Well, now we have two different ways. First of all, since we know this altitude, we can draw a line parallel to a C on the distance equal to this altitude. Continue this, and you will get the point B. Because the distance between these parallel lines is equal to our altitude. Alternatively, by the way, you can consider a triangle ABM. ABM. So instead of drawing parallel line here and continuing ABM towards intersection, you can just build ABM by exactly the same two things. You have the cataclysm and the angle. And we know how to do that. So either or we've built the whole triangle starting from the right triangle, which is made by an altitude. Okay, number four, construct a triangle by site, sum of two other sides, and an altitude onto one of those sides. Again, construct a triangle by a site, which is a C, let's say, sum of two other sides. So sum of AB plus BC and altitude onto one of them. You know what it appears to be right triangle. Let's try to do it more general. Okay, and now the altitude would go something like this. Okay, so this is point B, and this is B. All right, so we know the site, we know sum of these two sides, and we know the altitude. So the way how to do it is the following. Let me sum of these two sides this way. I will continue lying AB towards C prime on the on the lengths equal to BC. So BC prime is congruent to BC. What will it give me? Since I have one side and sum of other two sides, so basically AC prime is sum of AB plus BC because BC prime is equal to BC. All right, now let's think about what we have, AC prime C. It's a triangle where we know one side, let's call this side the base, and the altitude, and another side, which is exactly what we have already solved before. So using whatever the technology we had, we can always build AC prime C by a side, by an altitude towards that side, and another side. So let's consider we do it. So we built AC prime C. Now, how to find the point B? Well, very easy. Since BC and BC prime are congruent, BC prime C is a socialist triangle and median and bisector and altitude from the vertex B towards C prime always fall into the middle. So we have the middle, we built the perpendicular, and that will be the point B. And ABC, this is not prime, so ABC is the triangle which we need. Now, the altitude is exactly the way how we had it. AC prime is sum of these two sides. Yep, everything seems to work fine. Notice that in many cases I'm just referring to something which has already been covered and one of the previous problems. That's how one problem helps to solve another problem. That's basically the whole building of mathematics is built from the foundation of. Construct a triangle by its perimeter, an angle, and altitude onto one of two sides forming that angle. Okay, so we have an angle and altitude, let me repeat. Construct a triangle by its perimeter which is sum of three sides, an angle, this is an angle, and altitude onto one of two sides forming an angle. Okay, fine. Now, I think that since we have a perimeter, we have to really stretch this triangle the way similar to the previous problem. So, we will continue this line to C prime, so these are equal to each other, and continue this line to know what prime, double prime, it's too many primes. Let's just continue with E and F. So, I basically opened up our triangle. From BC, I turn it to BE and from AC turn it to AF. That's what usually is done when you have perimeter or sum of two sides, etc. Now, what's interesting actually is that you see the triangle ECF, ECF is not really easy to build using exactly the same technique as before because we do have the side, we do have an altitude towards that side, but we don't have an angle, as in the previous problem. However, what's interesting here is that you see FAC is an isosceles triangle and this is exterior angle which is equal to sum of two interior angles, not supplemental to it. But since it's isosceles, these guys are equal to each other, which means the angle FAC is half of the angle which is given to us, which means we basically have it. We just take the angle which is given to us and divide it in half. So, we take the angle, divide it in half, and we get this angle. Now, using this angle and this side and this altitude, now we can build this triangle. So, this is FEC because we now know this angle which is equal to half of the one which is given to us. Now, how to get points A and B? Well, the same way as before, since FEC can be divided in half and this perpendicular would hit the A since this is isosceles triangle, so I do this and that's how I get A perpendicular through the middle of this segment. And same thing here, perpendicular bisector of this will give me B. So, I have ABC. This is equal to this. This is equal to this. Angle is twice the size of this angle and the altitude is also the one which we need. Okay, let's continue. That was actually not a bad problem, quite frankly. Draw a line in the triangle parallel to its base such that segment between the points of intersection of this line with other two sides is equal to some of segments this line and a parallel to its base cut from this side. So, if you have a triangle you have to draw a line such that mn is equal to am plus nc. Well, it seems to be kind of a difficult problem. However, again, I remember we already actually addressed this issue before. There was another problem. If you have two bisectors, angle bisectors and on their crossing you draw a line parallel to the base, then this line exactly will be the one which possesses this quality. Why? Because this angle is equal to this angle as alternate interior. Now, this is equal to this because it's a bisector, which means these two angles are congruent to each other, which means these two segments are congruent to each other. This is isosceles triangle, MP8. Same thing here. These two angles are congruent because two parallel and transversal alternate interior angles. These two angles, this and this, are congruent because Cp is an angle bisector, which makes these two angles mpc and ncp congruent to each other, which means these two segments congruent to each other, which means that this entire segment mn is equal to sum of this and this because this is equal to this and this is equal to this. So, what's interesting here is it's easy when you remember that the point of intersection of bisectors has this property. If you don't remember that, well, you might actually spend some time basically thinking about how to solve this problem. Well, that's why it's important not only to solve the problems, but also to remember what exactly the problems were, and how we solved it. Okay, construct a polygon congruent to a given one. Okay, let's say you have some kind of a polygon. How to construct a polygon equal to this one, congruent to this one? Well, basically there's a very simple answer. You know how to construct a triangle congruent to a given one. Let's say by three sides. That's the easiest, right? So, what we do is we just break it into triangles and copy every triangle separately. Build a triangle congruent to this one. Okay, by three sides. Done. Now, build a triangle congruent to this one by three sides. We have this as a diagonal, this one. So, you have another two sides. And continue, continue, continue until you finish up. Basically, the whole process of constructing a polygon is reduced to many smaller and simpler problems to construct congruent triangles. Because otherwise, you will have to think about angles and how to copy the angle. You still have to break it into triangles no matter what. Construct a quadrangle by three angles and two sides that form the fourth angle. All right. So, you have two sides and three angles. So, again, three angles and two sides that form the fourth angle. All right. So, what can we do? You know what? Here is the easiest, I think, method. You know that the sum of angles of a quadrangle is 360 degrees. So, knowing these three angles, we just calculate, we just build, construct, whatever, draw an angle which is 360 degrees minus these three. So, consider this as well given to us. Now, and if it's given the way how we construct this particular quadrangle is simple. First, you do this. You construct this angle, which is also given as we just stated. Then, knowing these sides, you basically know this point and this point. Right? Now, knowing this angle, you build from this point an angle which is congruent to this one and from this point an angle congruent to this one. On the crossing, you have your quadrangle. So, what's the important thing here? Well, understand that the fourth angle actually is given to you basically with a very simple construction problem to subtract from 360 degrees three given angles, which we did many times before. Construct a quadrangle by three sides and two diagonals. Again, quadrangle, we have both diagonals and three sides. Well, let's see this, this and this. Well, but think about it. If we have both diagonals and three sides, we have a, b, c, g. We have a triangle a, b, g where we know all three sides. So, we can start basically from having this triangle by three sides. So, we have already these points. Now, how to position my point c using a, c, d? We have this a, d thing, right? Now, using a, c, and c, d, we do two parts centered in a and d corresponding way and that's where the point c is. And that's our triangle, the quadrangle. So, okay. The last problem, construct a parallelogram by two non-parallel sides and one diagonals. Well, that's basically the same thing. If you have a parallelogram, you have two non-parallel sides given to you and a diagonal. Well, if it's this diagonal, which is given to you, then you just construct a triangle and do a couple of parallel lines. If it's this side, which is a diagonal, which is given to you, then you notice that this side and this side are congruent. So, you can still build this triangle. So, in any case, you build triangle using three elements which are given to you, two sides and a diagonal. And then, basically, you can reconstruct the parallelogram from the triangle. So, if you have a triangle, let's say this one. How to build a parallelogram? These are sides and this is diagonal. Well, you just do the parallel here and parallel here. And here you have your parallelogram. That's it. That was the last problem. I hope I wasn't too fast. In any case, please continue solving the problems. Pay attention to Unisor.com. For parents, it's the perfect tool for homeschooling. It has a lot of exams. You can enroll your student or, basically, you can check the score on the exam. You can pass or fail the student. So, basically, it's a tool to do the homeschooling for responsible parents who want their children to know math a little bit deeper than is traditionally started at school. That's it. Thank you very much.