 Hi, I'm Zor. Welcome to Unisorification. I will continue solving different problems related to parallel lines, triangles, and quadrangles. This is the series of problem number four, and I have them here on this paper, so I will go just one by one. I'm sure you have already read all these problems and tried to solve them yourselves. If you didn't, they did successfully great. This is just yet another opinion about these problems. If you don't, well, then I'll just suggest my solution. All right, locus. Okay, so a couple of problems about locus. What is the locus? Locus is in geometry, at least, whatever I'm using it for. It's a set of points which satisfy a certain criteria. So if I'm asking what is the locus of points that, and then specify the criteria, it means that you have to find some kind of a line or a curve or area on the plane, all points of which satisfy this criteria. What is the locus of points that are dividing all segments connecting given point with all points in a given line into two congruent parts? So you have a point, and you have a line, and you have all the different segments which connect the point to this line. Now, my question is, where are the midpoints of each of these segments lie? Well, you obviously know that they lie on a straight line, which is parallel to this one, and it's equidistant, so I will draw a perpendicular here, so equidistant from point and line. So if you draw a perpendicular line, and through its middle, if you draw a parallel line to the given line, that would be the locus of all points of all the midpoints of all the segments which connect this point with the segment. Now, how to prove it? Well, very easy. So let's start with constructing this line. So let's say I have already drawn this line through a midpoint of a perpendicular from the given point to a line. Now, let's take any other segment which connects the point to a given line. Now, since these lines are parallel, if you consider a triangle ABC, so this point M is a midpoint of AB because that's how I constructed this line, and the line is parallel to the given line because, again, that's how I constructed it. So MN is a line drawn parallel to the base of the triangle ABC through the middle of one of its sides, and the theory which was proven before was that this line cuts the other side in half as well. So all these segments will be cut by this particular line in half, and that's the proof that all the mid-segments of all these segments which I can draw from a given point to a given line are on this line. Okay, next. What are the locus of points equidistant from two parallel lines? Well, obviously, you know the answer. If you have two parallel lines, then the point which is equidistant from it is supposed to be on a perpendicular and in the middle of this perpendicular because how to measure the point, the distance from a point to a line? Well, the distance from this point to this line is the length of the perpendicular, and from this to this is the length of this perpendicular. Now, since these are perpendicular to parallel lines, they basically coincide. These are because they are containing this one line. So these are two perpendicular, actually, which are coinciding with each other. So this is a straight line, and that's why the length of this piece is equal to the length of that piece, which means this is the midpoint of this particular segment. Now, if you will take any other point which is equidistant from both lines, again, it lies on the neutral perpendicular, cutting it in half. So if you will connect all these points together, well, let's connect only these two points. Now, what do we have about this figure? Now, why is it parallelogram? Well, because the A, B, C, D is a rectangle, since these are perpendicular, and opposite sides of the rectangle are congruent to each other, which means caps of these opposite sides are also congruent and parallel to each other, which makes M, N, C, G also, well, at least parallelogram. It's actually a rectangle. And so is A, B, N, M. Now, if that's true, it means that M, N is parallel to C, G. And then if you take any other point somewhere, let's call it T, and draw the perpendicular, it will, again, be parallel to any one of those guys, which means it will be parallelogram. And if it divides in half, again, this will be equal to this. But since these are two equal to each other, P is supposed to lie exactly on the line M, N. Because line M, N is dividing every neutral perpendicular in two halves, two equal halves. So the line in between, that's the locus. The locus of vertices of all triangles with the given base and altitude. All right, so you have a base and you have certain altitude. So the triangle in this case will be like this. But so will be any other triangle which has these vertices on the line parallel to the base in the same distance as this. So this is the same altitude. This is the same altitude as this one. And this one is also the same altitude. Because altitude is always at lengths of the perpendicular dropped from the point to the base line. So since these lines are parallel, we already actually proved that the distance between the two parallel lines is constant and it doesn't really depend on the point. So every point here will actually have the same lengths of the altitude. So all the vertices of all these triangles lie on the line parallel to the base of this triangle. By the way, it would be actually more rigorous I would say to prove that any other point not lying on this particular parallel line has a different altitude. And it's kind of obvious. So if you have two parallel lines which are on a certain distance from each other and you have a point outside, then this distance would be greater than this distance. Why? Because these pieces are equal and this is extra. So that's why any other line not lying on a parallel line is not making the same altitude if you draw a triangle. So that actually is, you know, it's a more rigorous proof because not only you have to prove that whatever your locus is set of points which satisfy this criteria you also have to prove that this is all the points which satisfy this criteria which means any other point does not satisfy the criteria. So that's what I did. So basically to proving that a set is a locus it's kind of two theorems. All points within whatever you point as a set, whatever you decide is a set. They do satisfy the criteria and whatever outside of this set points do not need to cover all the different points which satisfy the criteria. That would be a more robust proof. Construct the third angle of a triangle if two other angles are given. Well, look, this is just a simple thing. So if you have a triangle, you know that three angles are summed up give the hundred and eighty degrees, right? If you know two angles what you have to do, you have to subtract from a hundred and eighty degrees angle, you have to subtract these ones. Now how to do it? Well, actually it's very simple. So first this is a hundred and eighty degrees angle, right? The whole thing. Now how to subtract an angle, let's say this angle. Well, you just build a triangle using this side here and this side and this side giving this point. So you get a triangle by three points, right? By three sides. So this is an angle which is congruent to this one. Okay, so we basically subtract it. So from the entire hundred and eighty degree we subtract it only this part and whatever is left is the difference. Now we have to subtract from this angle this one. Okay, how to do it? Well, let's just forget about this piece. We have already subtracted this angle and now this one. Well, we do it very similarly. You take this radius, put it here, take this radius, this radius make this and this make it this. So again, you have a triangle this, right? This is equal to this and this is equal to this and this is equal to this, right? So now you got this angle. So from this angle, whatever is a remainder after the first subtraction, you can subtract the second end and that's what's left. But basically this angle is congruent to this. So basically it's an exercise for subtraction of angles. That's easy. Construct an acute angle of a right triangle if another acute angle is given. Well, same thing. If you have a right triangle then you know that two acute angles are summed up as 90 degrees. Right angle, right? So sum of these two angles is 90 degrees. So what you have to do is if you have one of those angles, you again have a right angle, that's 90 degrees. And now you have to subtract this one which you have to basically build a triangle like this, right? So how do you build it? You take this side and then using typogenous and another leg, you just have two different semicircles. You get these points, you got the triangle. So this angle is congruent to this one. So whatever is left is supposed to be this angle. I'm using any triangle basically here. It doesn't really matter. I don't want you to have an impression that I have a triangle. I don't have a triangle. I have just an angle basically. But if I have an angle, I can always build any right triangle with this angle as one of the acute angles. So basically whatever is left, actually probably the easier way, not actually the easier way, but just another way what you can do is if you have an angle, you can just take any point and drive perpendicular until it crosses another side of this angle and whatever you have here is basically the second acute angle. But to draw a perpendicular is probably not easier than to subtract the angles the way I did. So just two different solutions in this case. Right, construct a straight line parallel to a given straight line and position it from it on a given distance. So you have a straight line and you have some kind of a segment which constitutes a distance which you have to draw another line parallel to this one on this distance. Well, how can you do it? Well, first of all, you take any point, you draw a perpendicular, then you have this length cut here, and then you draw a perpendicular through this line again. How to draw a perpendicular through a line I'm not really discussing. You're supposed to know from previous lectures. So that's easy. By sect an angle if its vertex is not reachable. Oh, that's an interesting point. So consider you have an angle which you have to bisect. Well, basically you know how to bisect the angle. You can just choose for instance any radius, have two marks with the same radius here. You can have another radius a little bigger then you can draw something like this and it will be a bisector and it's very easy to prove basically because this side is equal to this side, this side is equal to this side, and this one is common. So you have a couple of triangles basically equal. That's actually besides the point. This is an old problem which we have solved before. You have to know the vertex in this case. But what if you don't know the vertex? What if you have from this angle it somehow cut and you don't see this piece? Well, what should you do in this case? Well, you can't really have the same construction. You can't have these equal segments from the vertex because we don't have the vertices. So what should we do? But if you remember we had a problem before, the problem like this. Let me go back to my original drawing. If you have an angle and you have two parallel lines on the same distance from the angle, from the angle sides. So this side is on the same distance as this one. Then the bisector is the same for both angles. So you can basically go back to the lectures where the proof actually is. But you feel that this is right actually, right? So if you don't have this vertex, if it's not reachable, it's easier if I would just erase everything and draw it again. So if you do not have, so you have one side and another side of an angle. So if you will be able to come up with a way to draw one parallel line on some distance, doesn't matter on which distance, what matter is that it's supposed to be the same for this side and for this side. So for this side and for this side. So if they cross somewhere on this part of the plane which is accessible for you, then you can just have a bisector of this angle and that would be a bisector of this as well using that theorem which I was just referring before. So all you have to know is basically what kind of distance to choose. Well, this is actually easier. What you can do is the problem basically to make sure that you hit the right distance. Because you see, if you use a small distance, let's say this one and this one. Again, these lines will not intersect in the known part of the plane. So you have to actually have a distance big enough so they cross. So how to choose it? Well, here is something which I can suggest. I'm sure there are many different ways to do it. Well, just connect two points. It doesn't really matter which ones. From this point, you draw a parallel line to this one. Then you measure this distance and using this distance, draw a perpendicular here and draw another one. So this distance is equal to this distance. These lines will definitely cross in the known part because you are actually going away from it. And having the crossing point, you have an angle which you can bisect and you know it will be a bisector of the whole angle here. So what's the most important part? To find basically the distance large enough so these two parallel lines will cross in the known part of the universe. I hope I'm not too fast. If I am sending an email and I repeat something. Given a straight line, point not on it and an angle. Straight line, point not on it and an angle. So there is a straight line, point and an angle. Okay? Construct a new line that contains a given point and forms a given angle with a given line. So we have to construct basically the line which has the same angle as this line. How to do it? Well actually it's very easy. First of all, instead of angle, let's just make it a triangle and do the triangle here. Have this piece, then radius and another radius have another point. Do the triangle by three sides. So this angle is what you need. Now from this line, you can always draw a line parallel to this one and since they are parallel, these angles are corresponding with this transversal. So this line parallel to this one will be the one which we need. Given an angle and a point inside it. Okay? An angle and a point inside it. Construct a line intersecting both lengths of the angle and containing a given point such that the segment of this line between its points, the intersection with the left, is divided into two congruent parts. Okay, so it's something like this. So we have to find the line which goes through this point in such a way that these two are equal to each other. Alright, so what can we do about it? Let me think. What if you draw a line parallel to this one and parallel to this one? Will that help? I don't know, let's think about it. These two triangles are obviously the same. So these are equal sides. So what can we build based on this? So you have an angle m x and we have p. This is not m, but we have to find out which. These lines are known. This is something like m and this is r. Okay. So basically what we know is we have the angle and we can construct two parallel lines so we have the parallelogram. Okay. What else do we know? That's not the way to do it. Yes, this is the proper way. Now, that I understand. You see, since triangle, let's consider this particular triangle, call it what, AB. AB x. In this triangle, AB is divided in half by point p. The line p r is parallel to line A x, which means it divides this, A xB into two congruent parts. So if you know the point r, you can always take this distance and make exactly the same segment here and the end of the segment would be the point p. So how to construct it? You don't really need this line actually. You don't need this line. So you draw the line parallel to this side and have this particular distance. This is an equal distance and that would be the point p. Notice that I did not really prepare myself for these problems. I'm just trying to solve them on the way and I'm thinking basically together with you if you're following me. All right. Okay, this is the last one. Given a segment, two parallel lines and point in between. Two parallel lines, point in between and some kind of segment. Construct a traversal that intersects both parallel lines contains given point in such a way that it's segment between the intersection with parallel lines is congruent to a given segment. Well, in this particular picture you belong to segment I guess. Something like this. Okay, so we have to construct some kind of a line which is intersecting both parallel lines contains our point and the length of this segment to the length of this segment. Well, it's actually easy. You pick any point, any other point on one of the lines using this as a radius mark a point which makes this particular segment congruent to this one and now all you have to do is from this point draw a line parallel to this and since these are parallel it's a parallelogram so they are congruent to each other and congruent to this particular segment. Well, that's it. That was the last problem in this series number four. I still have other four series. I think I have eight all together. All right, so don't forget that Unisor.com is an excellent source of information for advanced level mathematics and a certain number of relatively rigorously proven theorems and properties and parents and supervisors will have the ability to control the educational process of their children and students by basically signing in as a parent and enrolling your child or your student into this or that particular series of lectures and exams which you can check how your student actually did with the exams with the score basically and decide whether to repeat the same course again if the score is not high enough or just pass it and go forward. For home schooling it's excellent. All right, thanks very much and good luck.