 Hi, I'm Zor. Welcome to your new Zor education. I would like to talk about Pythagorean theorem today. I'm sure I mentioned it before in one of those lectures, but I would like actually to dedicate the whole lecture today to this particular theorem for a few reasons. Well, number one, although this is probably the most famous formula in all the development of the science, mathematics, whatever in history of the world, this is probably the most memorable one. So whoever remember anything about mathematics or geometry, they probably remember this formula, which is basically the sum of two squares of lengths or quantity of right triangle is equal to a square of hypotenuse. So that's number one consideration. This is a memorable formula and obviously, it deserves certain special attention. Now, another more, I would say, pedagogical consideration is, I'm going to present more than one proof of this particular theorem. As a matter of fact, four different proofs. It's like an illustration that the same kind of statement, a truthful statement can be proven in many different ways. You start from certain axioms and then you do some logical conclusions and inferences and ultimately, you arrive to the statement which you would like to prove and there are many different ways from A to B. There are many ways to go and it's extremely important to understand that there are many different ways to get to the same destination point. Well, obviously, people who drive the cars know about this, but it's true in life or any profession. There are many solutions to the same problems and somehow we have to deal with this. We have to judge which solution is better. Maybe there is some price attached to a solution which we would like to minimize and stuff like this. So, it's just an illustration that in mathematics we are dealing with the same type of very real life problems. Without further ado, let me get to all these proofs and here they are one by one. Well, first of all, yes, I did mention the formula, but what I actually mean is this. So, you have a right triangle, A, B, and C, or whatever letters you use are two-catchity, two-legs, and the hypotenuse and the sum of squares. These two is equal to the square of hypotenuse. All right, now. One more consideration. This is part of the more general topic about lengths and areas and why? Because certainly when we are talking about A square, this is an area of this particular square, which is having this leg as a side. Similarly, B square is the area of this particular square and C square is the area of that square. So, basically what we are talking is, we are talking about the sum of these two areas is equal to the area of the square built on hypotenuse. All right. So, it's about areas. Not only you can approach it algebraically, like really like a formula, A square plus B square is equal to C square, but also from a pure geometrical standpoint where area is actually at play. Okay. Number one. Here is what I am suggesting following the original truth which belongs to Pythagoras. Now, this is our original right triangle. Now, what Pythagoras suggested is the following. Let's continue this by the length A, this by the length B. So, this is A plus B and this is A plus B, and this is the right angle. Then we complete a square. Then we put B here and A and B and A here. Now, we consider this. Now, it's very easy to prove that if I have a square of A plus B side and then I choose these points so that this is B and this is A, this is B and this is A, this is B and this is A, and this is B and this is A. Then obviously all these four triangles at the corners are congruent to each other by two categories, for instance, and so they all have exactly the same area. It is also obvious that not only these four sides of this square triangle in the middle are equal but also these are right angles. Why? Because again, this angle and this angle and this angle and this angle are congruent to each other. So are these and we know that the sum of these two is equal to 90 degrees. So the sum of these two is equal 90 degrees. So this is 90 degrees in the middle. So this is a square and actually what the theorem says that this area is actually C square. All right. Now, what to do next? To do next is to perform the following rearrangement of the squares and triangles actually. So let's consider this triangle, for instance, top right. And let's move it parallel to itself towards bottom left. What happens? Well, this will be moved to this position and this will be in this position. Now, these lines obviously are parallel. So that's why this is a rectangle for obvious reasons. This is the parallelogram. So all sides are equal, et cetera, et cetera. So to get about this particular triangle, we moved it here. Now, let's talk about top left triangle. We move it to the right. So it will take position. This will be moved to this and it will be in this position. So from this point to this to this. So top left will be this triangle. I don't want to put letters at the vertices, just not to overbound the picture. And again, we are just moving parallel shift to the right. So this will take this position. This vertex will go to this. So this is the resulting triangle. And finally, the bottom right triangle, we move upwards. So this will take this position. So this is the same as this. This is the same as this, et cetera. Now, what have we done? We have rearranged triangles, but the components are basically the same. If before, the area of this square was equal to a plus b square minus 4 areas of triangle, a plus b square minus 4 triangles. That was the area inside, which is basically a c square, right? Now, when we have rearranged the whole thing, we did not really change anything. We just moved these triangles around. So now, what are the areas of these 4 triangles? It's the same as it used to be. But now, you really have to see that this is two triangles out of those fours. And these are two triangles out of those four. Now, this is b times a. And this is b times a, the area of these two triangles. So now, I can say that minus 2a times b. So now, it's obvious that the area of these four triangles is actually some of the areas of these two rectangles. Each one of them is a times b. Well, which is a square plus 2ab plus b square minus 2ab a square plus b square. So that's the Pythagorean proof. Well, that's it, not with you. Now, what's a little bit more complicated, I would say, in my personal view is that all these movements of triangles, quite frankly, I don't think we really needed to do this. We could have calculated the area of triangles just by themselves. But that would be a different proof, right? Anyway, what we will do right now is another proof, which, well, basically, does just that. So we have exactly the same picture. So this is b, this is a, this is b, a, b, a, b, a, c, c, c. And now, instead of moving triangles around like Pythagorean proof, Pythagoras was a smart guy. But it's not really necessary to move things around if you can do it without this type of moving. So again, let's do it a little bit differently. The whole area is equal to a plus b square, right? And it's comprised from the area of the inside square and four areas of triangle. But we know what the area of triangle is. We already learned this. It's a times b over 2. Why? Well, obviously, the general formula of the area of triangle is base times altitude. Now, if this is the base, then this is an altitude, right? In the right triangle, the area is just multiplication of two categories divided by two, which is equal to exactly the same as before. And if you will compare these things, obviously, a plus b square is a square plus 2ab plus b square is equal to c square plus 2ab. And obviously, we can do this and we have the same formula. I think it's a little bit easier just because we don't have to move things around and et cetera. But this proof doesn't have any name attached to it. All right. Now we will proceed to Euclid. OK, Euclid is the famous guy in geometry who proved a lot of different theorems. And for whatever reason, he has a more, I would say, I think it's the most complicated proof of this theorem. So I put a little drawing here, but I will do it right now in a more bigger environment. So this doesn't look like a square, right? Let's reduce the size. OK, that would be better. So now we will do it more genetically, so to speak. Because probably the purpose, Euclid's purpose was to do it without algebra, which we kind of used, right? All right, so now he would like to geometrically prove that the sum of these two areas is equal to this area. All right, how can it be done? Here is what he has suggested. Let's draw an altitude to where it's a part of this. Now what Euclid suggested is the following. This area is equal to this piece, and this area is equal to this piece. So if he proved that, and if we will actually prove after him this, that would actually mean that the sum of these two is the same sum of these two, which is the entire square built on top of this. All right, now, instead of proving that the whole square is equal to this, the area of this square is equal to area of this rectangle, I suggest to use the following. I will use half of this square, which is this triangle. And I will prove that its area is equal to half of this rectangle, which is this triangle. And to do that, I will do the following. Let me use another color. Let's connect this to this, and this to this. Complicated, right? It is kind of complicated. Now consider a triangle. Actually, I do need letters now. Let's have this C, A, B, C, OK, A, B, C. Now, B, E, F, G, H, I, what else? M, right? Now, let's consider two triangles, congruence of which I'm going to use. A, B, F, you see this one? A, B, F, and E, B, C, this one. They look the same, right? And actually, it's very easy to prove that they are congruent. Look at this. If you will turn this triangle by 90 degrees from 90 degrees into this position, right? What happens? Point A will go to E, because this is 90 degrees. This is square. And A, B, and B, E are of equal lengths, right? So that's why A goes to E. And if I turn this triangle by 90 degrees. Now, since this is perpendicular and equal to this, then the F will go to C. So when I turn around the point B, the A, B, F would take position E, B, C. So these triangles are congruent. We basically kind of decided this not even based on theorems, which we can definitely do based on theorems. But we also did it by direct transformation. One of the transformations is rotation, right? Obviously, it can be proven by this side is equal to this one. This is equal to this. And the angle is 90 plus this degree. And this is also 90. We have this remodeling behind the door. Sorry about that. All right, so let's disregard that. So these triangles are congruent. That's obvious, right? Now, well, they are congruent. And that's why their area is obviously the same. But now let's think about this way. Half of this square, which is B, F, C, B, F, C. This triangle has exactly the same area as A, B, F, Y, because they have the same base, right? And the same altitude, which is G, F. It's a perpendicular to this. And these are parallel lines, right? So from A to this B, F, the perpendicular will have exactly the same length as F, G. So the altitude of B, F, C, which is B, C, basically is exactly the same as the altitude of A, B, F, which is the distance from A to this line. So the same base, this triangle, and this triangle, the same base, and the same altitude, which is distances between these two parallel lines. So the areas that are the same. Now let's talk on this side. This triangle, which is half of this rectangle, this triangle, has exactly the same area as this big C, C, E, B. Why? Because again, they share the same base. And the altitude is the distance between these two parallel lines. C, E, B, and M, E, B have the same base and the same altitude, which is distance between these parallel lines. So they have the same area. So what happens is triangle area of triangle B, C, F is equal to A, B, F, area of triangle A, B, F. Now A, B, F is congruent to C, E, B. That's why the area is equal to triangle E, C, B, which is in turn equal to the area of triangle M, E, B. So we have proven purely geometrically that half of this square area is equal to half of this rectangle area. So the whole square, obviously, is equal to this particular rectangle. Now I'm not going to do it, but I do encourage you to prove in exactly the same fashion that this particular square has exactly the same area as this rectangle. It's exactly analogous. You just have to switch lines around. But it's exactly analogous. And that concludes the proof by Euclid that some of these two areas is equal to the area built on hypotenuse. Some of two areas of squares built on quantity is equal to an area of a square built on hypotenuse. Yeah. Anyway, that's the Euclid. He insisted on purely geometrical proof of this particular problem. Well, and he did it. OK, so let me wipe it out. And we still have one last proof, which, again, has no name, but much simpler than anything else, quite frankly. So this Euclid's proof is, I would say, it's rather complicated. Although I agree that it has absolutely no algebra in it. It's pure geometry, which makes sense. Now, let's not afraid to use some algebraic methods. And I'm going to introduce something which is, I would say, one of the most simple cases. This is our right triangle. This is the right angle. And so let's basically consider a very simple fact that the big triangle and both smaller triangles, this is an altitude, by the way, which is dividing the hypotenuse into segments x and y. So these are all similar triangles. Now, similarity is obvious because angles are the same. So this angle is equal to this one, and this one, and, sorry, just the cleaner. OK, so we have proven many times that this angle is congruent to this one. Well, primarily because these two lines are corresponding with perpendicular to these lines. This perpendicular to this, this perpendicular to this. So angles with mutually perpendicular legs are congruent. So these angles are correspondingly equal to each other, and that's why all these triangles, the big one and two small ones, have their all right triangles, and they all have exactly the same angles. That's why they're all congruent. So let's just write this congruency as the proportionality of the sides. Now, the small one is congruent to, let's say, the big one. So x is a smaller calculus of the smaller triangle relates to a hypotenuse as a smaller calculus of the big triangle, which is a, which lies against the same angle. x against this double r angle, and in the small triangle, and a in the big triangle is also against the same angle, is related to hypotenuse, right? Now, similarly, this triangle is also similar to the big one. So its angle, which lies across the single arc angle, which is y, is related to its hypotenuse, b, as the angle as the calculus, which lies in the big triangle against the same angle, which is b, towards its hypotenuse, right? So x squared, sorry. So x is equal to a squared over c. y is equal to b squared over c, right? That's what we can conclude from these guess, from the proportionality. x times c is equal to a squared, so x is equal to a squared over c. y times c is equal to b squared, so y is equal to b squared over c. And we know also this, right? x plus y is equal to c. So that's why a squared over c plus b squared over c is equal to c. Well, multiply by c everything, and you will see a squared plus b squared is equal to c squared. Same equation. So again, the purpose of my lecture was to introduce many different proofs. Not everywhere. Well, you know what, there are probably dozens, if not hundreds, of different proofs of the same Pythagorean theorem. My purpose was to present to some historical important, like the one which belongs to Pythagoras and the one which belongs to Euclid. And also, what's important was that some of them are algebraic, and some of them are geometry-based, and some of them are mixture of both of those guys. So different proofs, they always exist, no matter what kind of a problem you decide. I'm sure there are many different solutions. So be open to this. The whole course of mathematics, which I'm trying to present to you, is about creativity and developing of your creativity. Now, if you will be able to come up with yet another proof, yeah, not come up by looking at the Google or anything like that, but just think about it yourself. If you can, send it to me. I'll mention it in one of the lectures. All right, so that's it for today. I do recommend you to go to notes to this lecture on Unisor.com and try to just go through these proofs again. I think it's very educational. And obviously, as usually, I encourage parents and supervisors to become registered parents and supervisors and enroll your students into classes, into courses, so they can take exam and basically be a little bit more active in the whole educational process. It's supposed to be like a big course with these exams, et cetera, to bring up your analytical thinking, your creativity, et cetera. Thanks very much. That's it for today.