 Hi, I'm Zor. Welcome to Unizor Education. I would like actually to present you another problem related to pyramids. The problem is presented on Unizor.com as part of the advanced course of mathematics for high school students. I do recommend you to try to solve this problem just by yourself. It's problem number six under pyramids. And after you've done that, you can listen to this lecture and I'm going to present the solution to this problem myself. Okay, so what's the problem? The problem is as follows. We have a triangular pyramid. S is the apex, A, B, C is the base. Now, I know lengths of the side edges. S A has a length, lengths A, S B has a length B, and S C has lengths C. Next, I choose any point P on the base within the triangle ABC. No restrictions, whatever the point is. Now, from this point P, I draw lines parallel to side edges. So, P A prime would be parallel to S A. And it will hit the plane S, B, C somewhere at point A prime, let's say. And the lengths from P to A prime would be lowercase A prime. Same thing for B. So, this is S B H, right? And I'm constructing the line parallel from point B on the base from P to B prime, which hits the opposite side, opposite face of the pyramid at some point B prime. And the length is lowercase B prime. And finally, the same thing with C. So, S C is here, so parallel would be here, and it will hit my front face at point C prime, with the lengths from P to C prime being lowercase C prime. I have to prove this. So, the lengths of these three segments are related to corresponding lines from which they are parallel to, and the sum is equal to one, sum of this ratio. So, A prime divided by A, but A plus B prime divided by B plus C prime divided by C should be equal to one. That's an interesting property, because apparently it doesn't depend on the point B. So, whatever point you choose. Alright, so, how can we prove that? Here is what I suggest. Now, S A and P A prime are parallel lines by construction, right? So, I can draw a plane, which would, through these two parallel lines, which would cut my A, B, C for instance along this line, let's call it A1, and it will cut the opposite face, S, B, C, sorry, through again some kind of line. So, this is a plane, which goes through parallel lines P A prime and C A, and it cuts through the pyramid two faces and goes into point A1 on the edge from B to C. Now, since these two are parallel and A A1S is plane, right? Obviously, we have similarity between triangles A A1S and P A1A prime. Now, from this obvious similarity, I can say that A prime is related to A the same way as A1P relates to A1A, right? Let's just consider again. So, this is triangle S A A1. Now, this is all in one plane, right? And this is A prime. And this line is parallel. That's P. This is what happens in the plane A A1S. Now, these are parallel. So, relationship between A prime and A, this is A prime and this is A. A prime divided by A is exactly the same as A1P divided by A1A. A1P divided by A1A, right? Okay. So, let's remember this. And we will do exactly the same with other two. If I will connect, let's say, B to P to get to B1. So, this is B1 point on AC edge. I can say that B prime to B is equal to B1P divided by B1B, my writing B1B, B1P. And the same thing for the C. C prime divided by C is equal to... Okay. C, C, C. Here is C prime. So, that would be on the front page, on the front face. This would be C1. So, C prime divided by C would be equal to C1P divided by C1C. Now, this is lowercase C, by the way. This is uppercase C, because these are lengths and these are letters. All right. Now, why did they do it? Is it better right now that instead of this, I have to prove a similar equation with these guys? It's immensely better. Because everything happens right now within the plane ABC, within the base. So, all these lines A1P, A1A, B1P, B1P, C1P and C1C, they are all pieces of this triangle. Everything is here. So, let me just wipe out my pyramid and replace it with flat plane triangle ABC. So, I have ABC triangle. I have any point on it, P. And I have three lines through this point. One, two and three. To A1, B1 and C1, the point is P. And I have to prove that, well, A1P divided by A1A plus B1P divided by B1B plus C1P divided by C1C is equal to one. So, it's a plane geometry problem, which has its own interest, by the way. And I would love to put this problem, for instance, in one of my plane geometry lectures. I didn't, so it came out for solid geometry case. Alright, so how can we prove that? Here is a very easy, actually, way to prove it. Let me delineate three triangles. This is one triangle, B, P, C. This is another triangle, A, P, C. And the third triangle, A, P, B. Now, obviously, my total triangle is divided into three smaller triangles. And the area of the total triangle ABC is a sum of these three areas, right? Now, let's just think about it. What is A1P divided by A1A, for instance? Well, or similarly, B1P divided by B1B. Let's say, in case of B, it's, or C, it's a little bit more visible. So, if I will draw an altitude from B, is this an altitude? No, it's not an altitude. Let me draw it better. Altitude is this. Okay, let's call it an altitude from A and altitude from C. Too many lines, right? Okay, point P is somewhere in the middle. Now, the altitude should have letters, right? So, from ABC we have D, E and F. So, AG is perpendicular to the BC. BE is perpendicular to AC. And CF is perpendicular to AB. Now, let's consider triangle, let's say, B, E, B1. And think about what is B1P divided by B1B. Well, if you drop a perpendicular from P, you would see that this is exactly the ratio. So, this relative to this is exactly the same thing as this relative to this. So, this is the ratio of altitudes of triangle A, P, C. Altitude is P, how can I call it? F, G, H, okay, let's say. So, what I'm saying is that the ratio between, let's say, B1P and B1B is the same as ratio between GP and EB. Because these triangles are similar, because these are two perpendiculars. But this is the altitude of this triangle. So, the PG is altitude of this triangle. And BF is altitude of the bigger triangle, ABC, with the same base, AC. So, the ratio of these altitudes is exactly the same as the ratio of the areas, because they have the same base. So, AC is the base, and PG is the altitude of APC. For ABC, correspondingly, AC is the same base, and BE is altitude. So, the ratio between B1 and P1, B1 and P is two B1Bs, the same as ratio of altitudes. And since they have the same bases, it will be the same as ratio of the areas, sorry. So, if I will call the area S1, S2 and S3, I have these three triangles. So, an area of the entire triangle would be S. So, I can say that B1P divided by P1 by B1. B, B1P, B1P is exactly as S1 to S. And similarly, A1P divided by A1P divided by A1A would be the ratio between altitude of this triangle divided by altitude of the bigger triangle. And again, they have the same base. So, it will be S2 divided by S, where S2 is area of this small triangle. And finally, the third triangle would be this one, and the ratio between C1P and C1C is S3 divided by S. And what happens if you add them together? This plus this plus this. Well, you will have S1 divided by S plus S2 divided by S plus S3 divided by S, which is equal to sum of S1, S2 and S3 is actually the area of the entire triangle. So, it's S divided by S, which is 1. And that's exactly what is necessary to prove. Now, what did I use many times in this particular solution? The similarity. Similarity of the three-dimensional elements to corresponding similarities in the plane geometry I basically reduced using this. So, these are three-dimensional lines in all different directions. These are, everything is in the base, in the plane, which is the base triangle ABC. So, first of all, these three ratios, I have replaced with three ratios which relate only the elements within the triangle. And then, within the triangle, I also kind of replaced ratio between this and this and this, with a ratio between altitude this and altitude that. And since they have the same basis, these triangles, small one and the big one, the ratio of the altitude is the same as the ratio of areas, right? So, playing on this similarity many times, I basically reduced it to a very simple and obvious formula that the area of the triangle is equal to sum of the area of three pieces, whatever point you choose in the middle of this triangle doesn't really matter. Well, that's it for today. I suggest you to go through this proof yourself. It would be great if you can write it down. This equals to this, or this is similar to that, etc. And it's very useful if you can write it with an explanation. For instance, this is similar to that because such and such property of the similarity. All right, that's it for today. Thank you very much and good luck.