 Hi, I'm Zor. Welcome to a new Zor education. We were talking about symmetry in three-dimensional space, basically two kinds of symmetry. One is central symmetry, which is symmetry relative to a point, and another was a reflection, which is symmetry relative to a plane. Now, what's important about symmetrical figures just in general? Well, they look alike. So, this lecture, which is a very, very short one, and actually trivial, is about what exactly is look-alike in the symmetrical figures. Well, obviously, we all know that symmetrical figures are almost equal. I mean, they are either completely equal in the terms of you can just put one on the top of another and they will coincide. Or they are, so to speak, symmetrical, like two different hands when you have to really, like, turn something around and then they will coincide. So, basically, that's what I'm going to prove right now, that in any kind of symmetry which we were talking about, we have this property of likeness. And in mathematical terms, it's called congruence. Well, in my young days, it was called basically quality. Nowadays, it's more customary to use the term congruence. So, two figures are congruent rather than equal, primarily because it's like two hands. They're not really equal, but they are congruent because you can do something like this. Alright, so a few mini theorems about congruence and symmetry. So, I have a few different kinds of proofs in this particular case. They're all about symmetrical objects and they're all very, very simple. So, let's just start with something simple. This simplest, actually. You have a central symmetry relative to a point and you have a segment. So, you reflect this segment and we know that the reflection, well, I should say symmetry relative to a point of a straight line is a straight line. So, a segment is transformed into a segment. So, these are equal and these two are equal. These angles are obviously vertical. Now, the thing is going on in three-dimensional space. However, if we are talking about a segment and a point, it's three points actually, two intersecting lines. So, everything can be considered as happening in the plane which contains a, b, a prime, b prime, and the center of symmetry. That's called p. So, in plane geometry, it's absolutely obvious that these two triangles are congruent because it's side-angle side, side-angle side. So, that's the end of the proof. Very simple, right? So, segments are, symmetrical segments are equal in lengths, basically. And as far as their position in the central symmetrical case, they are also parallel. So, that's my first theory. So, the central symmetry preserves the segment lengths. Okay, great. How about a symmetry relatively to a plane? So, let's say we have a reflection relative to a plane. And let me draw a plane, something like this. This is my plane of reflection. So, I drop a perpendicular from a onto this plane and extend it by the same lengths. Drop a perpendicular to b, from b to the plane. And this is my, and this is my image. And this is the result. Now, these are two perpendicular to the same plane. They are parallel, which means they align the same plane. So, again, I can consider the whole thing as happening in the plane, which contains a, b, b prime, and a prime. And incidentally, two bases of perpendicular, if you wish. And now, what we are talking about is a plane geometry theorem where I can just wipe out my plane of reflection, as if I don't have it. And I can consider the plane problem, which basically states that, by the way, these are perpendicular, because this is perpendicular to the whole plane, right? So, these are perpendicular, and these are equal to each other by construction. These are equal to each other by construction. And all I have to prove that a, b is equal to a prime, b prime, which is, again, a trivial task. I mean, how can we do it? Well, for instance, we can drop perpendicular here and here, which means now this is a rectangle. So, this is equal to this, and this is a difference between this and this. And it's exactly the same thing here. And these are also equal to each other, because these are equal to this one. So, we have two right triangles with two categories correspondingly equal to each other. So, obviously, the symmetry relative to a plane, the reflection relative to a plane, also preserves the lengths of the segment. Alright, two down. Next, we will have a little bit more complicated, but not by much case. Instead of a segment, we will use this transformation of symmetry with triangles. So, let's say we have a triangle, a, b, c, and let's say central symmetry p. So, we go with b to here, with c to here, and with a to here, something like this. Too much, I guess. No, that's not too much. Yeah, I think it's okay. This is a prime, b prime, and c prime. So, a, b, c goes into a prime, b prime, c prime. Now, these are equal, this is equal to this, and a, p is equal to p, a prime. Now, why are these triangles congruent? Well, because every triangle has three sides, every side is a segment, and we know that this segment is equal to this segment, this to this, and this to this, because of the previous theorem about segments. So, these two triangles are congruent by three sides. That's it. Now, the next theorem would be in exactly the same fashion proved about reflection of the triangle. So, if you have a triangle, a, b, c, and the plane of reflection, and we reflect it here, we will also have a congruent triangle because of three sides. Each side is a segment. So, side is equal to side, and we have congruent triangles. Next, so we can forth down. Next, too, will be about angles. Now, what I'm staging right now is that angles are also preserved by the symmetry. So, if you have an angle, which is actually like two rates, and you have a center of symmetry. Now, this ray is reflected into this one, and this is the vertex, and this ray is reflected into this one. So, if this is a, this is a prime, now this is line b, this is line c, this is line b prime, this is c prime. So, why angles are the same? Well, there are a couple of ways I can prove it. Well, first of all, we have already proven that lines are parallel in the central symmetry. So, in this particular case, we have two different angles with corresponding with parallel sides. So, that's one thing. The second thing is, I can just choose a couple of points here and here, have their symmetrical counterparts, and consider a triangle. Now, since triangles are preserved, the angles are also preserved, so this angle is equal to this angle. Now, this last approach with the triangle, I will use for reflection for symmetry relative to a plane. So, instead of this picture, I have again an angle and some kind of a plane, and I have a reflection of this angle, which is probably something like this. Now, it's all in three-dimensional space, right? So, again, what I will do, I will choose two points and make it a triangle, and this is my image triangle. And since triangles are congruent, the corresponding angles are congruent as well. Slightly more difficult is the last couple, which I would like to prove. This is more three-dimensional case. So, let's consider we have a dihedral angle, which means an angle between two half planes. So, we have some kind of a border. This is one, and how can I put it? Let me just do it slightly differently. I will use this. So, this is one plane, and this is another plane. This is my dihedral angle. Now, somewhere I have a plane relative to which I reflect this. Oh, no, first is a central symmetry. So, I have a point. Now, let's just think. We know that image of a plane is a plane. Image of a line is a line. The plane will be parallel, by the way, and the line will be parallel. So, in this particular case, this line would be reflected into this, this plane would be reflected into, I guess, this one, into this, and this one into something like this. Right? So, this is alpha, this is beta, this is beta, this is alpha prime. And this point is in between, center of symmetry. Okay. Now, how can I prove that these dihedral angles are the same? Well, here is what I can do in this particular case. Well, number one, obviously, I can refer to the parallelism. So, if you have two planes, two dihedral angles with correspondingly parallel planes, I can actually derive from this the quality of the dihedral angles. But probably I would choose something else, because I would like to use the same proof for reflection, where I will not have this parallelism. So, here is what I would do. So, here is the point, and obviously it's in here. Now, in one plane, I will do the perpendicular to the edge and in another. So, these lines are perpendicular to the edge, this line was in beta and this line was in alpha. Now, the flat angle between these two lines is actually a measure of the dihedral angle, right? Now, same thing here, perpendicular to this, perpendicular to this. Now, we know that plane angles are always preserved. So, the image of this right angle is this right angle, because it must belong to the plane, right? So, the angles must be equal, since it's just flat angles. So, this angle and this angle must be the same, since they are image of each other. Same thing with this one, also should be the right angle, right? So, they are also right angles here and there. Now, what I can do, I can use the triangle thing if you want to. Something like put a couple of points here, find their images here. So, triangles are preserved. So, all in all, what I'm saying is that all these elements, all these three points will be symmetrically transformed into these three points and everything about these three points supposed to be the same, including the angle between these two, which is a measure of the dihedral angle, right? So, let's do it again. So, I have this edge, I have the point here and this image, a couple of perpendicular here and there, here and there. And since angles are preserved, I can say that this angle between this line and this is the same as this one and this. And then, just again, two points in half triangles and equal to the triangles. Now, I can use exactly the same approach to have symmetry relative to a plane. So, if I have some kind of a plane, which is symmetrical, and now I have this, something like this. So, that would be my A, that would be my I alpha and beta. Same thing. So, first, so this alpha goes to alpha prime. So, this perpendicular should go to this perpendicular, this perpendicular to this. And now, I can choose a couple of points and the equality of the triangles will give me equality of the dihedral angles, exactly the same thing. So, I was trying to prove that basically all the elementary objects of 3D geometry, like segments, angles, dihedral angles, they are preserved. Now, as a consequence, you probably can think about that any complicated geometrical object, if symmetrically transformed, whether it's centrally symmetrical or reflection relative to the plane, it should produce something similar, well, congruent, actually, to the original. Now, why? The reason is very simple. Any complicated geometric object can be subdivided into more elementary pieces. So, all elementary pieces we have basically covered, like for instance, we have covered triangle, but we did not cover, let's say, pentagon. But, hey, pentagon can be divided into triangles, right? And therefore, the corresponding pieces of the corresponding pentagon will also be congruent piece by piece, and therefore the whole pentagon will be congruent. Now, if I have proven that dihedral angles, for instance, are the same, and all faces are basically the same, congruent in the symmetrical figure, then that actually means that the whole geometric figure, however complex it is, would be properly reflected, properly means congruently reflected into its image, whether it's a central symmetry or plane symmetry. Well, that was relatively short and relatively trivial subject of this lecture. I would suggest you to read all these proofs in the notes on theunisor.com. I think it's very helpful. Well, other than that, that's it. Thank you very much and good luck.