 Hi, I'm Zor. Welcome to a new Zor education. I would like to talk about symmetry in three-dimensional space. Well, first of all, let me just remind very briefly about symmetry in two-dimensional space. So, if this is the plane, there is a central symmetry, which means symmetry relative to a point. So, if you have some kind of point P, which is the center of symmetry, and then you have a point somewhere else, then what you do, you connect and extend by the same length. So, if this is A, this is A' and the AP is equal to PA'. Now, another way to present it on the plane is you can just turn the whole plane by 180 degrees. But let's not talk about this particular because it's not really applicable to three-dimensional space. But the first one, the first way to construct it is definitely applicable. So, you connect your point with the center and then extend it by the same length. So, that's one thing. Another is a reflection relatively to some kind of an axis. So, if you have a point here, you drop a perpendicular and extend it by the same length. Now, this we will not discuss today in three-dimensional space. But what we will discuss is this, the center of symmetry. And exactly the same way it's defined on the plane, it is defined in three-dimensional space. So, assume that the point A and point B are two completely different points in the space. This one is called the center of symmetry. And this is the point which we want to reflect relatively to the center of symmetry. So, what we do, well, these are two points in space, right? We can always connect the line, connect them with the line and extend the segment by the same length. Exactly the same rules. And we get the image. So, if this is the prototype, this is the image. This is a regional point and this is a reflection using the center of symmetry relative to a point P, which is called a center of central symmetry. Okay, this is all fine. By the way, let's just mention a very trivial fact that if you built a point A prime, central is symmetrical to A, relatively to point P. Obviously, if you start from the point A prime, you will get the point A, because you have to draw the line, which is exactly the same line. And it's exactly the same length, so you will definitely hit the point A, which means that if A is symmetrical to A prime, A prime is symmetrical to A. This is a symmetrical relationship between a regional prototype and its image. Okay, now, what's important about central symmetry is that it preserves the general shape of the geometrical figures, geometrical objects. Now, what I mean is the following. Well, obviously, point is symmetrical to a point. Now, how about a line? So what if you have a line? Will the image of this line, which means a set of all points symmetrical to all points on this line, will be the line? I mean, intuitively, everybody understands that if you have some kind of a geometric figure, object in three-dimensional space, solid object, and you symmetrically reflect it relatively to a line, which means every point goes to an opposite beyond this line. It should be, well, almost the same. I mean, it's a reflection, which means it's not exactly the same, like left becomes right, but the general shape is preserved, line to line, triangle to triangle, hexagon to hexagon, etc. Parallel lines to parallel lines. I mean, everything should be preserved as far as these properties are concerned. Now, obvious things are sometimes very difficult to prove, by the way. So let's just prove, for instance, this particular theorem, that the image of a line is a line which is parallel to this one, and also these two lines are equidistant from the center of symmetry P, which again seems to be obvious, but still needs to be proven. Okay, so how do we do this? Well, here's what I suggest. Line is defined by two points. So let's just take another point, B. Now, we don't have this line yet. All we have is image of the A and image of the B. Now, what I'm going to do is I will connect them with a line, and now I will prove that this entire line is actually image of this entire line, which means that every other point C would be reflected to a point C prime on this line. Okay, that's how I'm going to prove it. Let's consider these two lines, A A prime and B B prime. Both are going through the point P, which means they intersect, and if you have two intersecting lines, there is always a plane, which contains these two intersecting lines. Well, that's one of the main things which we were talking about, planes, lines, etc. If you have two intersecting lines, there is one and only one plane which they define, which contains these two lines, which means points A and B and B prime and A prime are all in that line, in that plane, sorry. So let's call this plane gamma. In this case, it's actually a plane of this board. Right? Alright, so A B, A prime, B prime belong to the same plane, which means I can actually consider the whole picture as two dimensional. Now let's consider triangle A B P. Now, it's quite obvious it's congruent to A prime B, P B prime. Why? Because these two are equal. These two are also equal by construction, because that's how we constructed the symmetrical point A prime. We just connect it and extend it by the same length, right? Same thing here. And these angles are vertical, because these are straight lines, right? So triangles A B P and A prime P B prime are congruent. You have side, angle, and side, side, angle, and side. Now, what we have right now is that these two angles are also equal to each other, because of the equality of the triangles, right? Triangles and congruent, therefore the angles opposite to equal sides are supposed to be equal. Which means, by the way, that this line is parallel to this line, because this is two lines, this is a transversal, and these are alternate interior angles. So incidentally, this line A prime B prime is parallel to A B. Now let's talk about why the point C, I chose in between A and B just for convenience, why the image of the point C should fall on C prime, which belongs to this particular line A prime B prime. Now, here is why. Let's consider triangle, let's forget about this point here, let's put it a little bit further, just for illustration purposes, and connect it to these two. Now, let's consider triangle A P C and A prime P C prime. C prime is here, not here, C here. Now, these two triangles are obviously congruent for the same exact reason why A P B and A prime P B prime are congruent, because C P is equal to P C prime by construction, angles are vertical, this one and this one, and A P is equal to P A prime, which means that B C and, sorry, not B C, A C. A C is equal to A prime C prime, must be, right? Similarly, if I consider B P C and B prime P C prime, same thing, triangles are congruent because you have vertical angles, you have sides equal, and you have this side equal, so B C also is equal, B prime C prime. Well, let's add them up together. A C plus B C is A B, right? Because this is a straight line. And here we have A prime C prime plus B prime C prime. But from the other thing, from the other hand, we already proven that A P B and A prime P B prime are equal, so A B is equal to, is equal to A prime B prime. So I have A prime C prime plus B prime C prime equals A prime B prime. So this plus this should be equal to this. Well, obviously C must be on this segment because we know that the straight line between A prime and B prime is shorter than any other line wherever C prime might be if it's outside. So this equality is possible only if C prime is lying exactly on the straight line between A prime and B prime. Now, if a point is outside here or here, point C, the logic is exactly similar, so I'm not going to go into this. So what we have proven is that the image of any point on this line is on this line. So image of the whole straight line is this straight line. We just take two points, make these images go with the straight line, and then we have proven that this straight line is the place where all other points will actually be reflected to. Images of all other points on this line are lined here. Alright, so the image of the straight line is straight line. It's parallel because I was just talking about alternate interior angles. And why these two lines are equidistant from P? Well, again, the same thing. What is the distance from P? It's basically an altitude of triangle A P B and altitude of triangle A prime P B prime. That's distance from this. So this is distance and this is distance. Obviously, if you have congruent triangles, their altitudes are equal. So they are equidistant. Well, that's basically the full theorem about two lines which are symmetrical to each other relatively to a center of symmetry. The second part of this lecture, I would like to talk about planes. So image of the line is a line. Image of the plane is a plane. Parallel to the original plane and equidistant, these two planes are equidistant from the central point. Exactly the same thing as with the lines. Okay, so let's go to the planes. Obviously, I will be using the results of this theorem when I will be talking about the planes. Okay, so now we have a plane and we have a point which is the center of symmetry. I would like to reflect every point on this plane to other points. But I'm going to prove that the result will be, the image will be the plane parallel to this one and on the same distance from the P as the original. Alright, so what we do is basically, logically equivalent to whatever we did with lines. Line is defined by two points, right? So I picked two points A and B and then draw the line from images of these two points. I will do very similar with the three points A, B and C. Three lines not lying on the same line define this particular plane. So I just pick any three lines on this plane which are not on the same line. So we have some kind of a triangle here. Now what I'm going to do right now is I'm going to reflect these points. This is B, this is C and this is A. And I have a triangle A prime, B prime and C prime. So picked three points, have these images and draw a plane through these three points. And what I'm going to do is I will take any point, let's take it inside the triangle. And I'm staging right now that the image of this point will be on this plane somewhere within this triangle. Let's call this point F, that would be F prime. How can I prove it? Okay. Here is how. Let's just draw any line on this plane through point F which intersects our triangle ABC. Let's say this line. And it intersects the points M and M. Obviously, wherever the point F is on the plane and there is a triangle, I can always find a line which intersects two sides of this triangle. I need two sides. Why? Because, well, you have three different sides of the triangle, line can be parallel only to one of them. So the other two are definitely not parallel, and if it's not parallel, then I can always draw a line. I mean, there are many different ways to do it. All right, so I have a line which goes through F and intersects M and N with the sides. But now let's think about this way. A and C belong to line, some kind of line, AC. So A prime and C prime also define the line on which images of every point on this line is reflected to this. That's the previous theorem, right? Line is reflected to line. So if A prime is image of A, C prime is image of C, then A prime, C prime is image of AC. That's what we have just proven before in the first theorem. So every point on AC is reflected into some point on A prime, C prime. So point M is definitely M prime. Definitely lies here. Similarly, point N lies on BC, B prime, C prime. This is N prime. So N prime definitely lies on this line. And by the way, if it lies on this line, it lies within this plane because B prime and C prime belong to a plane, so the line in between belongs to the same plane. All right, so if M prime and N prime belong to this plane, then the line between them, again, which is a straight line, has image on the plane, which is a line from M prime to N prime, which means every point on line MN, including our F, is reflected somewhere within this segment M prime, N prime. So basically we have proven that any point wherever it is on the plane, well in this case it's inside but doesn't really matter, it can be outside as well, every point is reflected on the point lying on this plane. So the plane is reflected into plane. That's number one. Okay, now we have two other points, parallelism between these planes and equidistance from P. Here is how I suggest to do it. Let's consider one of these lines, let's say A A prime. Well, probably it's better if I will draw another picture of this. I need a perpendicular. So this is one, this is another, and my point A, now this is P. So I draw a perpendicular from P to this original plane and that's where I will choose point A. And then this is a reflection, this is A prime. Now this is B and this is C. This is B and this is C prime. No, that's wrong, it's central symmetry. So it goes this way. This is B prime and this is C prime. Okay, that's how it is. Now P A is a perpendicular to this plane. Now we have made an image of the A which is A prime, which means we just continue this by the same distance and we have already proven the triangles, let's say BAP and B prime, A prime, P, P are congruent, right? Because A B, B prime, A prime belong to the same plane, A B is parallel to A prime, B prime and everything is equal, right? That's basically what we did in the previous theorem. Which means that since angle BAP, the right angle, and it is the right angle because it dropped the perpendicular, right? It dropped the perpendicular to any line. This angle P A prime B is also right angle. Same thing with two triangles ACP and A prime, C prime, P prime. This angle is 90 degrees since AP perpendicular to the plane and therefore to each line and therefore this angle is also right angle. So it looks like P A prime is perpendicular to two lines, which means it's perpendicular to the whole plane. So that's perpendicularity. And the equidistance means because, I mean it's obvious because this is equal to this from the equality of the triangles ADP and A prime B prime AP. Well, so as you see, central symmetry is a very nice feature. It reflects point to point, line to line, plane to plane and the resulting features are exactly the same. They are parallel to each other, plane to plane, line parallel to line and they are equidistant from the center of symmetry. So these are the properties which I wanted to discuss today about central symmetry in the three dimensional space. I would recommend you to read the description of whatever I have just was talking about on theunisor.com that probably would bring some more logic. And again, things which are obvious and this is an obvious thing is very difficult sometimes to logically prove and you can observe it in any arguments, by the way. Well, one person is considered something to be absolutely obvious which doesn't really require the proof. Another person doesn't really consider it obvious. So we have to really be able to prove. And that's everywhere. Once actually I observed the trial and the lawyers were trying to explain something and it was very obvious for me how poor their logic sometimes is. I mean we expect lawyers to be able to prove that somebody is guilty or not guilty, but that's not easy thing to do. So even the obvious things require a significant effort sometimes to prove. So in this case there are many axioms which we have referred to before like for instance two lines intersecting to each other define the plane and some others like two points belong to a plane and therefore the line between them belongs to a plane. So some of them are axioms, some of them are previously proven theorems but anyway you have to really employ this sequence of logical statements one immediately following from the previous one to prove anything. And that's a very good exercise in this logic. Thanks very much and good luck.