 Hi, I'm Zor. Welcome to Unizor education. Today, we will continue talking about symmetry in three-dimensional space, primarily symmetry relative to a plane called the plane of reflection. This is part of the Advanced Mathematics course presented on Unizor.com. I do suggest you to watch this lecture from that website because it contains notes. Notes are like a textbook, basically, and whatever I'm presenting during the lecture is explained in written format in those notes. All right, so the previous lecture was about symmetry relative to a point, a central symmetry. Now, we have proven that the image of the line during the central symmetry is a line, image of the plane is a plane. Now, we will talk about different kinds of symmetry. Symmetry relative to a plane of reflection or reflection plane. So, let me first explain how this process of symmetry is organized and how to construct a symmetrical image. And then we will prove exactly the same two theorems that the image of a line is a line, image of a plane is a plane. Okay, so how it is constructed? Well, imagine that you have a plane which we will call a plane of reflection or reflection plane or plane of symmetry, whatever you want. And for instance, it's vertical like this one, something like this. Okay, this is the plane. Now, if I would like to construct an image relative to this plane of this point A, I have to first drop a perpendicular to the plane and then continue it on the same distance to get A prime. So, perpendicular to the plane and extended by the same lengths. Let's call this point B. So, my first statement is, it's quite obvious from this construction that this reflection is actually symmetrical in both ways. So, if A prime is symmetrical relative to this plane of reflection to the point A, then point A is symmetrical to A prime relative to the same plane. Because what do I have to do if I would like to build to construct a point which is symmetrical to this one? Drop a perpendicular, which is exactly the same line, right? Because this is the perpendicular extended. So, it's exactly the same perpendicular and extended by the same lengths and they are equal to anyway by construction, right? So, basically from A I get A prime, from A prime I get A. So it's the symmetry of reflection is reflexive. That's the word reflexing and reflection are, they have the same root, but it's in different meaning right now. Reflection is one thing and reflexive is basically being the same in both directions. All right. Anyway, my first theorem, as I was saying, to prove that image of a line is a line. And I will do exactly the same thing as I did for central symmetry. So, if I have some kind of a line, I will take two points A and B, I will take image of both, connect the images. And now I'm stating that any other point on the original line, let's call it D, will find its reflection exactly on the line which connects A prime and B prime. So, this is the point Q. Now, my first statement right now is that I would like actually you to recall that all particulars to the same plane are parallel to each other, which means AP is parallel to BQ or A A prime, actually, is parallel to B B prime. And actually, it's also parallel to M M prime. Wherever this M prime is, I mean, I can assume it's not falling on the line A prime B prime and then come to some kind of a contradiction. All right, now, since A A prime and B B prime are parallel to each other, it means I can basically draw a plane through them because two parallel lines are lying in the same plane. So, let's call this plane beta. So, beta is the plane where A A prime B and B prime belongs to these four points. Now, not only these four points belong to this plane, also the whole line D, because if two points belong to the plane, then the whole line which connects them belong to the same plane. Now, the way how I constructed D prime is just I connected A prime and B prime, which means D prime is also falling in the same plane. Now, point M as belonging to the line D also belong to this plane. Now, what follows is that the entire M M prime should also belong there because it's very, actually, it's very simple. Because if you will consider this plane beta, which is basically the plane of this board, right? Now, you know that these three points are in that plane and these two also are in this plane and these are parallel to each other. And this one is also parallel to this and this one. Now, if I will draw a line from M within my plane beta parallel to A A prime, that will be a line parallel to A A prime drawn from the point M. Now, if I assume that M M prime doesn't belong to beta to my plane, it will mean that I would have two different lines M M prime and another one which I can construct from point M within the plane beta. So, I will have two different lines parallel to A A 1 drawn from the same point M, which is impossible in Euclidean geometry. So, it looks like M M prime also belongs to this plane beta. So, what I'm doing right now is very simple. I'll just forget about three-dimensional space and I will consider everything within the plane beta, which is the plane of my board. Because A A prime belongs to this plane, B B prime belongs to this plane, M M prime also belongs to this plane. Now, what's given is that if I will connect P and Q, that would be an intersection between the beta plane and my plane of reflection, right? Now, the intersection is a straight line. Obviously, point P belongs to this line, point Q belongs, so PQ is a straight line, right? So, that's an intersection. And now, I will reduce completely this problem to a two-dimensional case. What do I have? I have A A prime B prime B, which is a trapezoid within the plane beta. These are parallel to each other. I also know that A P is equal to P A prime, so P is the midpoint of the top base of this trapezoid. Q, similarly, is a midpoint of this one. So, what I'm doing right now, I am basically from point M, I draw a line parallel to this one, and I know that this particular distance, let's call it R, and I know that M R is equal to R M prime by definition of the symmetry, right? I drop a perpendicular to the reflection point plane and then extend it by the same length. So, I know that. So, I have to actually prove that M prime falls on this side of the trapezoid. Well, again, that's very easy to do. Let's call this point M. Let's consider, for instance, N and M prime are not on the same place. Now, obviously, among all the different statements which I made about this trapezoid, what's obvious was that M R and R N are equal to each other. Well, it's just plain problem of the plane geometry. How can I prove it? Well, there are many different ways to prove it. For instance, if I draw a perpendicular here and here, I can very easily prove that these two triangles are congruent to each other. Why? Well, let me just think why. These are parallel lines. A, B, and A prime, B prime are symmetrical to each other relative to this middle line. So, that means that since they are symmetrical, these angles must be the same. Because symmetry within the plane, basically, you can just fold it along this. And this gadget is equal to this gadget is because these lines are parallel and this is distance between them. So, the angles are the same, the cadets are the same, which means that this piece is equal to this piece and this is obviously a rectangle, so this is equal to this piece. So, obviously, M R is equal to R N. I mean, probably there are many different ways to prove it in a two-dimensional case. But R M prime is also supposed to be equal to M R. So, that necessitates that M and M prime are one and the same point. That cannot be different points because the difference, because the same distance from R to N is equal to M R and from R to M prime is equal to M R. So, basically, that's it. So, it's a very easy kind of a plane geometry problem to prove this. And I'm sure you can come up with many different other proofs. I mean, I just... something which came to my mind the first thing. But it's definitely very easy theorem to prove. Well, which basically proves that any point M, because I don't have any restriction on M, I just took any point within this segment AB, any point will be reflected relative to this reflection plane into a point on this line, which means every other point also will be there. Now, if this is outside of this trapezoid, there is absolutely no difference in the logic side. I'm not going to spend any time on this. So, wherever M is here inside or outside of this segment, doesn't really matter. Its image will fall into the line, which connects A prime and B prime. Okay, that's my first theorem. So, line is reflected into line. By the way, there are some trivial cases. What if this line is within the plane of reflection? Well, obviously, the distance, the perpendicular from any point to the plane will be zero, right? So, every point will be reflected to itself. So, if you are talking about plane of reflection, then any point on that plane will be reflected into itself, obviously, right? Alright, fine. Next, theorem is about the plane. So, how can we prove that the plane would be reflected into the plane in three-dimensional space relative to some plane of reflection? And again, I will employ exactly the same technique as I did for central symmetry. So, let's say this is my plane of reflection. And I have certain plane here. Now, my point is that the reflection would be the plane here. Something like this. Now, why? Well, let's do this. Let's pick three points here, A, B, C. Now, these three points I can pick not lying on the same line. They define this plane uniquely. Now, they have images. B prime, A prime and C prime. And I draw a plane through these three points. Now, I will take any other point M here. Well, let me start not from any point inside. Let me just start from the triangle A, B, C. Now, I know from the previous theorem that if point A goes to A prime, point B goes to B prime, then the whole line which connects A and B will be reflected to this line. And this line obviously belongs to the plane which I construct from these three lines, these three points. Same thing with any other. So, all three sides of my triangle are inside that plane which I constructed from three images of three points here. Now, let's take any other point inside. Let's call it F, for instance. Now, what I can do is I can draw a line through F which intersects two sides of my triangle. Now, this point, this end of this segment is somehow reflected into the point on A prime, B prime. So, let's call it M and N. So, that will be M prime. And N is within A, C somewhere here, M prime. So, M goes to M prime on the A prime, B prime side. And N, as part of the A, C, will go somewhere within A prime, C prime. Which means that the whole line which connects M and N also will be part of this triangle. Part of this, and it belongs to this plane. And therefore, each point on it, including this one, will also be on this plane. Since two points belong to the plane and the whole line belongs to the plane. So, basically, I was using the first theorem and using, and by using this theorem, I have reduced the whole plane into triangle and a couple of segments. Well, that's basically it. I wanted to stop here as far as the properties of symmetrical figures are concerned. So, lines go to lines, planes go to planes. And it's for both central symmetry, which is the previous lecture, and symmetry relative to a plane of reflection, which is this lecture. Now, next lecture will also be dedicated to properties of symmetrical figure. But now we will go deeper and not only I will prove that, let's say, line goes to line, I will also prove that two figures which are symmetrical to each other, whether it's centrally symmetrical or reflexively symmetrical, they are usually congruent. So, that's what we will be doing the next time. That's all for today about symmetry. Thanks very much and good luck.