 Hi, I'm Zor. Welcome to Unizor Education. I continue solving different problems about parallel lines, triangles and quadrangles. This is series of the problems number 5. And it has, let me see, 10. It's my favorite number, 10 different problems. Hopefully they're not very difficult, because I will have to basically solve it on the fly with you. Alright, so let's just go on. Position a given segment between the rays that form a given angle perpendicular to one of these rays. Okay, so I have a segment and I have an angle. I have to position this segment into this angle so that it touches basically both sides of the angle and perpendicular to one of these. Well, here's how we do it. We take this and on any perpendicular we measure it. So this is congruent to this. And then I draw a parallel line to this leg of the angle. Now, this is another perpendicular. Now, since these are two perpendicular to the same line, they're parallel. And these are also parallel because I have constructed it this way. So it's parallelogram, which means this is congruent to this. And that's why to this, and it's perpendicular. Problem solved. So position a given segment between the rays that form a given angle parallel to a given straight line that intersects both legs. So again, you have a segment, you have an angle, and you have some kind of a straight line. So you have to position this segment again between these two legs of the angle, but it should be parallel to this gap. Well, how can we do it? Very easily. We just take this and measure it here. And then draw a parallel line. And then draw a parallel line to this one. So again, this is parallelogram by construction. Now, this side is congruent to our segment again because I have constructed it this way, which means this one as well. So it's parallel to this line and equal in length to our initial segment. Problem solved. Easy. Position a given segment between the rays that form a given angle so that the segment between an angle's vertex and each of two ends of this segment are congruent. So again, you have a segment, you have an angle, and we have to construct it in such a way that these two segments are congruent to each other, and this is still congruent to our segment. How to do it? Well, obviously you remember that if you draw a bisector here, it will be perpendicular to any segment which cuts equal segments from both legs. Why? Because obviously these triangles A, M, N and M, B, P are congruent by side, angle, angle, this is the bisector, right? So these angles are congruent and common side. So since they are congruent to each other and this is a hundred and eighty degree, it means each one of them is ninety. So these are right angles between the bisector and our segment which we need. So basically what I would like to say is that I have the direction, basically, I have the direction into which I have to draw our segment. This direction is perpendicular to the bisector. So what I do, I just take any line perpendicular to it, basically I'm reducing the problem to the previous one. So now I have to build, I have to construct this particular segment which is congruent to this one parallel to this one. And how do I do it? The same way as I did before. I just cut this length, this length and draw a parallel line. And the point where I get it, that's the one which I need. So this is exactly the same. Now another parallel will actually cut another line and since it's a perpendicular to a bisector, these two are congruent to each other. Construct a right triangle by one acute angle and opposite catatris. So you have a triangle and you have one particular catatris and you have an angle which is opposite to it. Well, how to construct a triangle? Well, basically very easy. If you have this angle, let's say, you take any point, you build an angle equal to it and from here you draw a line parallel to this one. Since these are parallel lines, these angles will be congruent as corresponding. This is the catatris of a given length. So basically that's exactly the triangle which we need. Construct a triangle by two angles and decide opposite to one of them. Okay, so let's analyze it. So you have two angles, let's say this one and this one. And decide opposite to one. Let's say this side is given. How to build a triangle in this case? Well, let's start from what we know. We just take our segment, whatever is given to us, that's given. Then we can have this angle. That would be nice. Now, well, what we can do actually, we can do two things. Number one, knowing two angles, we can always build the third angle of a triangle. The previous lecture actually was dealing with it. Well, that's one of the ways. And knowing that third angle, I can actually build it from here. And that would be enough. Now, another way to do it is the following. Somewhere on this line, pick a point, any point, and have this angle from this point. Now what do we do? You have this line and you draw parallel to this. And these angles will be congruent to each other because they are corresponding with these parallel and the transverse. Again, two different ways to solve this problem. Okay, what's next? Construct an isosceless triangle by a base and an angle opposite to it. A isosceless triangle by base and angle at the top. Well, let's think about this way. You can always draw a median and bisector and altitude in this particular isosceless triangle. Now, if we know the angle at the top, we know it's half. And if we know this segment, we know it's half. So we can always build a triangle ABD, which is the right triangle by knowing the characters in the opposite acute angle. We just did it before. So basically that's enough. You build this triangle and then you just continue this line to the same distance and you get the third point of the triangle. So basically we reduce the isosceless triangle to two right triangles. Construct an isosceless triangle by a leg and an altitude dropped onto it. So again, you have an isosceless triangle. You have one leg, let's say this one, and an altitude dropped onto it and this one. Alright, so what can we do about it? Well, since we know the altitude, we know the locus of all the points of a triangle which have the same side and the same altitude. So what we do is we build ABD and on the distance equal to CD, our altitude, we just drop another one. Now, where on this line is point C? Well, since this is a isosceless triangle, we can take this radius AB and basically mark it here. So with the center at B. So from B to A, the distance is exactly the same as from B to C. So that makes our triangle isosceless. And obviously since point C belongs to the line parallel to AB, on this distance, the altitude is equal to exactly what we need. These are all relatively simple problems. The more difficult one I will leave for exams. If you are signing in as a student and enrolled in a particular course. Alright, that with that. Okay, construct equilateral triangle by its altitude. So we have an equilateral triangle and altitude in it. So we know only the altitude. We don't know anything else. Well, here is how we can do it. We can take any segment as the side of an equilateral triangle. Let's say a big one. And you build some equilateral triangle with this side, which you have picked by three sides. You draw its median and measure the altitude which we have on this median. And then you draw a parallel line to this one. So since this is parallel to this one, this is also equilateral triangle. Well, obviously because all the angles are the same, they are corresponding. All of them are 60 degrees, etc. In this case, since these are two parallel, then the altitude is common for both of them. So if we measure the altitude equal to ours, so that would be actually our equilateral triangle we are looking for. So we build a bigger one and then cut it to size using an altitude which is given to us. Construct an angle of 30, 45, 60 and 75 degrees. Okay, 30 degrees. Well, look at this picture. It's already obvious how to construct an angle at 30 degrees. You just construct an equilateral triangle and draw a median. Now, you have any equilateral triangle using any segment as a side. And then you just divide one of the segments in half, let's say this one, and connect it with the opposite vertex and that would be 30 degrees because these are 60, 60 and 30 again. So we covered 30 and we covered 60. Now, how to do 45 degrees? Okay, you remember that in right isosceles triangle, right isosceles, they have two legs congruent to each other. Since this is 90 degrees, some of these is 90, but since this is isosceles, they are supposed to be equal to 45. So your task is to build a isosceles right triangle. How to do it? Have the right angle perpendicular to each other lines. Have any segment and cut it in both sides exactly the same distance. You connect the points and here is your 45 degrees. And next is 75. Well, just not to think about it too much. If you have 30, you can divide it in half using just bisect it and you'll get 15. Take 15 at 60, you'll get 75. That's easy. Alright, that's it for this series of problems. Don't forget Unisor, that's your site for homeschooling, for advanced level of mathematics where you're looking for rigorous problems and stuff like this. And if you are a parent, a responsible parent who really cares about education of a student, a child, just use this site to enroll your child into some advanced math courses or anything else and let him go through exams, check the score in the exams and you decide whether your children, your child passed or failed that particular exam. And if he or she fails, just let him do it again. No problem. Alright, good luck. Thank you very much.