 Hi, I'm Zor. Welcome to a new Zor education. We are continuing talking about solid geometry elements and today's lecture is about parallel planes. Well, obviously we start with definition. The parallel planes are two planes in three-dimensional space which do not have any common points. Basically, it's as simple as this. Now speaking about properties, I have four theorems, actually, which I would like to prove right now. They're all presented on unizor.com. I do suggest you to go to this website and watch this lecture from that website, but even before you watch the lecture, try to prove these theorems yourselves. Using whatever the axioms, you know, we have already went through a certain number of very simple axioms and a couple of theorems, which we have proven in the previous couple of lectures, which are dedicated to parallel lines and line parallel to the plane. So this is just a continuation, plane parallel to plane. So without further ado, I'll just go through these elementary theorems. The whole theory is really very simple. Parallel planes, we all kind of understand what basically this is. In Euclidean geometry, and let's just go for these theorems one by one. Theorem number one. Actually, it can serve as a sufficient condition for two planes to be parallel. And here it is. Let's say you have two planes. I usually use gamma. Okay, gamma and another plane delta. Now, let's assume that we have two straight lines on this plane. One is called A and I will index this with gamma, which means it belongs to gamma and B also indexed with gamma. And you have two lines here. Also A indexed with delta and B indexed with delta. Now, let's assume that you know that these lines are correspondingly parallel to each other. So A gamma parallel to A delta and B gamma parallel to B delta. Now, I didn't really say it before, but I do assume that these two lines are intersecting. Symbolically, it sounds like this. Intersection between A gamma and B gamma is not empty and correspondingly here. Delta. Not empty. Okay. So basically I would like to prove based on these two conditions, the parallelism of the corresponding lines and the lines are intersecting, of course, I would like to prove that the planes are parallel. Basically, it means that gamma and delta should have no common points. Okay, fine. As in many different cases, we assume the opposite and come to some kind of a contradiction, right? So let's assume that gamma and delta do intersect somewhere. Well, one of the axioms says that if two planes intersect at some point, they intersect through a whole line, straight line, which contains this point. So let's assume that we have some kind of a straight line C, which is intersection of these two planes. Now let's see what happens. Line C, which is an intersection between gamma and and delta, has a very interesting property, which we have already learned before. Look at it this way. Since this line, a gamma is parallel to a delta, you know from the previous theory that if line is parallel to another line on a plane, then it's parallel to an entire plane. That was a sufficient condition for this line to be parallel to the plane. And so is bq, b gamma. B gamma is parallel to b delta, and if one line is parallel to another inside the plane, then it's parallel to an entire plane. So basically what we know about this is that both a gamma and b gamma parallel to the whole plane. Now let's consider another piece, which we have already learned. If you have a plane, in this case it's a delta, and a line outside, which is parallel to the plane, in this case it's a gamma, and you have some kind of a plane which intersects delta and goes through the line, which is parallel to delta. Then the intersection of this plane is parallel to the line a gamma. This is one of the theorems in the previous lecture. So if there is a line outside of a plane, which is parallel to a plane, and there is a plane going through this line and intersecting this plane, then the intersection will be parallel to this line. Now in this particular case, this plane which intersects delta is gamma. We have assumed that gamma and delta intersect somewhere, and this is c. This is an intersection. So it looks like c should be parallel to a gamma within this plane, which is plane gamma. Now, similarly, we have another line here, b gamma, and the same line, the same plane, which is gamma, intersects the same plane delta along the same line c. So it looks like c should be parallel to b gamma as well. So within this plane gamma, we have some kind of a line c, which is an intersection of this and this, which is supposed to be parallel to both these lines, but these lines are not parallel to themselves. So that's the contradiction. We cannot have one line parallel to two intersecting lines in the plane. That's basically the plane geometry. So that's the contradiction we're looking for. So our initial assumption that the planes gamma and delta have some kind of a common point, and therefore common line c is just invalid assumption. It cannot be. So they are parallel. They do not have any common points, and that's why they are parallel. That's my first theorem. And it's completely derived from the previous couple of theorems, which we have already learned from the parallelism between lines and planes. Now, the theorem number two. Okay, so we have two planes gamma and delta, and in this case we already have a condition preposition that they are parallel. So that's given. Okay. Okay, then we have some kind of a plane which intersects them both. This is a plane which intersects them both, and this is the intersection. So this is plane alpha. Now this is intersection between plane alpha and gamma. I call it a gamma, and this is a delta. So a gamma is intersection between alpha and gamma, and a delta is intersection between alpha and delta. Now I have to prove that these two intersecting intersections are parallel to each other. That's supposed to be proven. Well, this is a very easy theorem. Well, when two lines are parallel to each other, well, we have two conditions in three-dimensional space. One, they should belong to the same plane, and two, they should have no common points. Now, obviously they belong to the same plane because this is the plane alpha. They are intersections of alpha with correspondingly gamma and delta. So they are in the same plane. Now, can they have common points? Of course not, because if these two have common points, it means the two planes they belong to gamma and delta have common points, and we in the very beginning assumed that they are parallel. So there are no common points. So that's why a gamma and a delta cannot have common points. So they're in the same plane, alpha, and have no common points. That's why they are parallel. You see, all these theorems are really very, very simple. You just have to, you know, think logically and basically that's it. I spent more time just writing down all these prepositions of the theorem than I spent to really prove the theorem. All right, now the third theorem. Okay. So again, we have two different planes parallel to each other. Gamma, delta, and we have two lines parallel to each other, A and B, and they are parallel to each other. So gamma is a plane parallel to delta, and A is a line parallel to line B. Well, which means they line the same plane and do not have any common points. Now let's call this point A gamma and this point A delta. This is B gamma and B delta. So we have two parallel planes, and they're intersected by two parallel lines. Well, basically the theorem is that these two segments should be, well, basically that the whole thing is supposed to be parallelogram. So we have to prove that A gamma, B gamma, B delta, A delta, so this is parallelogram. So that's what we have to prove. Okay, that's also very easy. Think about this way. Consider a plane which contains A and B because they're parallel. They're supposed to belong to some plane, right? So this plane intersects two planes parallel between themselves, gamma and delta. And we have already learned from the previous problem that if this is the case, then the lines of intersection are parallel to each other. So A gamma, B gamma is parallel to A delta, B delta. So these two lines are parallel because of the previous theorem. When we have a plane, Calpha, which intersects two parallel plane, gamma and delta. So the lines of intersection parallel to themselves. All right, fine. Now, above and beyond that, we know that A gamma, A and B parallel to each other, right? So A gamma, B, sorry, A delta parallel to B gamma, B delta. So these lines are also parallel. Now, go back to plane geometry. If you have a quadrilateral or quadrangle, whatever you call it, with opposite lines parallel to each other, if you remember, we did prove that this is parallelogram. Basically, that's it. The proof is finished. So since this is parallel to this and this is parallel to this, this is a parallelogram. With all consequences, for instance, one of the consequences is that these two segments are equal in lengths and these two segments are equal in lengths. These are just consequences from the fact that this is parallelogram. Okay, fine. And the last but not least theorem, which I wanted to prove today was, as follows. Let's say you have two planes. Now we will have to prove the parallelism of the planes. It's kind of expansion to the first theorem. You remember, the first theorem was, if I have two lines which are intersecting to each other and they're correspondingly parallel to the two lines on another plane, then the whole planes are parallel to each other, okay? Now I will do just one step further. Let's say you have, again, two lines which are intersecting. Actually, I took only rays, half lines. It's more convenient for me. And again, they are parallel to each other and directed the same way. That's why I'm just using rays instead of the whole lines. So the fact that they are in the parallel planes, we have already proven. So let's call this point B gamma and C gamma, B delta and C delta. So I know that A gamma, B gamma is parallel to A delta, B delta. A gamma, C gamma is parallel to A delta, C delta. And I have to prove not only the parallelism of the planes. Actually, this is already proven. But that angles B gamma, A gamma, C gamma equals B delta, A delta, C delta. So I have to prove that these angles are also parallel. Well, you remember that, again, in the plane geometry, there was a theorem. If you have two angles and their sides are correspondingly parallel, then the angles are directed the same way. The angles are equal, congruent, to be more exact. Well, the same thing basically is in three-dimensional space. That's all I'm just saying. All right. So how can we prove this in the three-dimensional space? So these two angles are not in the same plane. So we cannot really resort to that old theorem. So they are in space, but they are on parallel planes, as we have already proven. So what we will do is the following. We will choose points B gamma, C gamma, B delta, and C delta in such a way, and this is just our decision, our construction. I would like to choose them as this. So this is equal to this. And correspondingly, this to this, A gamma, C gamma equals to A delta, C delta. So basically, I would like to include these two angles into triangles and prove the congruence of the triangles. All right. How can I do it? Here is the way. Let's just connect A to A. B to B and C to C. Yeah, something like this. Not very preaching, but that's okay. All right. So what can we say now? Based on the previous theorem, I can say the following. A, C and A, C. I know they are parallel. And I know that they are of equal lengths because that's how I constructed them. Same thing, AB and AB. They're also parallel. That's the preposition of the theorem. And by construction, I made them equal. So what does it mean? It means that A gamma, B gamma, B delta, A delta is parallelogram, AB, BA parallelogram. Again, why? Because you have parallel and equal opposite sides. There was a theorem in the plane geometry which basically proved that this is the parallelogram. Now, same thing with A, C. A gamma, C gamma, C delta, A delta is also parallelogram. Now, since these are parallelograms, then from the first parallelogram follows that A, A, Q, A gamma, A delta parallel and equals to the BB parallel and equals. That's how I will use it. B gamma, B delta. In the parallelogram, opposite sides parallel and equal. Same thing here. A, Q, A delta parallel and equal to C gamma, C delta. So what follows from this? That B gamma, B delta is equal and parallel to C gamma, C delta. So this is also parallelogram. This one. And this is, since this is a parallelogram, BC is equal to BC. B gamma, C gamma equals to B delta, C delta. And this is the third side of this triangle. So now we have triangle, ABC and ABC and all three sides are correspondingly equal to each other. So that's why they're congruent. That's it. And since they're congruent, well, not that's it. Since they're congruent, then the corresponding angles are equal and that's exactly what we have to prove. Okay, so these are these four very, very simple theorems. Each one of them is just one or two statements, basically, to prove them. But nevertheless, it's a good kind of a logical schooling, if you wish. So you can very accurately deduce something new from whatever you have already known before or proven before. So this is how the whole building of mathematics is built. You assume something simple, then the simple theorems, more complicated, more complicated, etc. And every new theorem is based on whatever you have already proven before in the axioms. And that's the pure geometry. That's what it is. Okay, so that's it. Thank you very much and good luck.