 Hi, I'm Zor. Welcome to a new Zor education. I would like to present a couple of exercises or mini theorems, if you want, related to concepts of similarity and Cavalier principle. The Cavalier principle is something which was discussed in the previous lecture, and obviously these mini theorems will be very closely related to this principle. Also, it will serve as a preliminary introduction into the volume of the pyramids, which is part of the next topic. So, I have two-dimensional and three-dimensional cases of Cavalier principles, and I will start with a two-dimensional just as an illustration of the approach, and then I will use it for a three-dimensional case, basically to present something which I will really be using in the volume of the pyramids topic. Okay, so, first of all, I'm sure you remember the previous lecture about Cavalier principle, but I'll just remind that in the two-dimensional case if you have two different flat figures on the plane and there is a base line such that each line which is parallel to the base intersects both figures along segments of the same lengths. If that is true, so, every line has these two segments of the same lengths. If this is true, then the area of these two figures is the same. Okay, now, here is the problem which I would like to present right now. A concrete case of two geometric objects on the plane, and I would like to prove that the Cavalier principle for these two geometric objects is really held. So, here is what I would like to do. I have a line and I have two equal segments on it, and I have any point outside. Now, let's consider two triangles, this triangle and this triangle. Now, for those of you who remember, and I'm sure it's everybody, that the area of a triangle is a half product of lengths of the base times altitude. And altitude, obviously, in both cases is the same because it's a distance from this point to the line. So, this is our altitude and, obviously, these two triangles have the same altitude. And, as I said from the very beginning, KB is equal to CD, these two segments are equal as given. Then, obviously, the area is the same. So, why bother? Well, let's consider you don't know this formula, and I would like to prove using the Cavalier principle that the areas are the same. Now, what do I have to do for this? I have to prove that any line which is parallel to some baseline cuts across these two triangles and the sections are of the same lengths. That's what I have to prove. Well, what is the baseline? Well, obviously, the baseline should be this line where both bases of the triangles are aligned. So, now, let's just draw one line, any line, actually, which crosses both of those guys. And let's prove A prime, B prime, C prime and D prime. And let's prove that A prime, B prime is equal to C prime, D prime. Now, if I prove that, that actually is a premise of the Cavalier principle that every line parallel to the base, and this is our base, is cutting segments of the same lengths. Then the Cavalier principle will be held, and if I accept the Cavalier principles, which I do, I can conclude that the area of these two triangles is the same. All right. So, how can I prove that these two segments are equal to each other? Well, it's actually very simple. Now, since these lines are parallel, then obviously, these angles are equal to each other. Right? Two parallel lines and one which transcends them. Now, therefore, I can use the similarity. Now, the similarity of these triangles, S A prime, B prime and S AB, follows from the equality of the pairs of angles. That's basically the known theorem similarity. Now, if they are similar, then all the linear dimensions are proportional. So, in particular, I can say that A prime, B prime relates to AB as, let's say, S A prime relates to S A. Or from absolutely similar, let's call this H and this is H prime. From absolutely analogous similarity between S H prime, A prime and S HA, follows that S A prime relates to S A as S H prime relates to S H. Now, let's switch to this triangle. Now, in this triangle, C prime, G prime relates to C G because these, again, the same angles. Right? So, these triangles, S C prime, D prime is similar to S C D. So, C prime, D prime relates to C D as, let's say, S C prime relates to S C. And, again, from similarity of S H prime, C prime and S H C follows that this ratio is equal to this ratio. Now, these are equal, which means these are equal. So, A prime, B prime divided by AB is equal to C prime, D prime equals divided by C D. Now, AB and C G are the same. That's why these two guys should be the same. So, these are equal. And that proves that any line, because we didn't really fix a particular height, so it can be on any height. So, we have proven that the Cavalieri principle is held for these particular two triangles, which means that their areas are the same. Now, that's my first theorem. Now, the second theorem is analogous, but it's three-dimensional. Now, in a three-dimensional case, let's assume that on this plane I have two triangles of the same area. Now, let's assume I have a point. Yes. And let's assume that we have constructed two triangular pyramids. One with base ABC, this is one triangular pyramid, and another triangular pyramid with base GEF, two triangular pyramids. So, again, I know that the area of ABC triangle ABC equals to the area of triangle DF. So, remember, in two-dimensional case, I had two segments lying along the same line of the same length. Here I have two triangles having the same area lying on the same plane. So, we're just switching by one dimension up from a segment to a triangle, from length to the area, and from line to the plane. Everything is one dimension up. Now, my question is, for these two triangular pyramids, can we say that the Cavalier principle is held? So, obviously, we have to choose the base plane, which obviously is this one. And now I have to prove that if I will cut any other plane parallel to this one, which cuts both pyramids, I will have, in the section, I have triangles of the same area. If I will prove that, that actually means that my Cavalier principle is held. So, let's cut with some parallel plane. And these are the sections, something like this and something like this. These two triangles. So, this is the plane which cuts these two pyramids. It's parallel to this plane. Now, remember what the similarity is in the three-dimensional space. We have to have a point, remember, and we should really introduce some kind of a scaling factor. Basically, it's the same as the two-dimensional case, obviously. So, my point is that triangles A' B' and C' triangle A' B' C' is similar to ABC with S as a center of scaling and the same factor, obviously, for all of these. Now, why is that the case? Well, let's just think about it. These are two parallel planes and SAB is another plane which cuts these two parallel planes, which means that A' B' is parallel to AB, right? If we have two planes parallel to each other and the one which cuts them both transcends, then the intersections will be parallel. We did learn this in one of the previous lectures. Similarly, all other sides would be parallel. So, all sides of this triangle are parallel to this one. Now, if you consider triangle SAB and SAB, they are similar to each other, obviously. Well, because they are in the same plane and these two lines are parallel, AB and A' B' which means angles are equal, etc. So, that's very easy. Which means that A' B' relates to AB as SA' relates to SA. Now, absolutely the same, we can say about B' C' and A' C' and they are all related as far as the ratio is concerned to SA' to SA. Because SB' to SB has also the same ratio and then SC' to SC has exactly the same ratio from different triangles, right? So, they are all proportional which means that all these three sides are proportional to these three sides. Which means that these are similar triangles. The ratio, the factor of scaling we know, this is basically some factor which depends actually on the distance between S and these two planes. Because if I will draw a perpendicular, obviously it will be the same ratio between SH' and SH, right? So, these are all trivial stuff. And, if you remember, whenever we have a scaling in the three-dimensional world, if you have a flat figure and then it's scaled to another, similar to this, the factor should be squared if we want to compare the areas. So, the linear dimensions are changing by a factor, let's say F, this is F factor. But the aerial dimensions are scaled by the factor of F square. So, I can say that the area of triangle A' B' C' is equal to, well, multiplied by F by the factor. Whatever that factor is, square equals to the area of ABC. Okay? Now, in absolutely similar fashion, we can prove that the area of triangle of D' D' E' F' D' E' F' F' times the same F square. Because, again, it's all related to the ratio between SH' and SH. So, the same ratio between this multiplied by F gives you this. Equals to the area of triangle D' F. Now, but these areas are the same by condition, right? Which means that since it's the same factor, these areas are the same. So, what we get is that the areas of the section which are two different triangles are exactly the same, no matter where we put this plane. So, that's my three-dimensional equivalent. We have proven that if you have two triangles of the same area and common apex of pyramid, then we will have the same volumes. Very similar to the two-dimensional case. So, that's the end of my second theorem. And the third one is basically a trivial consequence of the one which we have just proven. And here it is. And this is actually the theorem which I will be using when talking about volume of the pyramid. So, let's consider you have a plane again. And you have some kind of a parallelogram on this plane. And you have a point somewhere. Now, let's connect and get the pyramid. This is my pyramid. This is my pyramid. Now, let's consider two pyramids. Instead of one quadrilateral pyramid, we will consider two triangle pyramids. SABG and SBCD. Now, these two triangles are two triangles of the parallelogram if we divide it by diagonal. And they are obviously congruent and they have the same area. So, basically we are just using the previous theorem which I have proven that these two pyramids, since they have congruent bases in the same common apex, they must have the same volume. So, what I am saying is that if you consider the quadrilateral pyramid and the plane SBD, which basically cuts it into different triangular pyramids, it cuts the volume in half because the volumes are the same. So, that's a very important point. Don't forget it and whenever we will go into the volume of the pyramid, we will use it. Well, that concludes today's lecture. I do suggest you to read the notes on Unisor.com website which accompanies this lecture. Well, that's it for today. Thank you very much and good luck.