 Hi, I'm Zor. Welcome to Unizor Education. You know, my most favorite type of lessons are those related to problems, and I will continue talking about problems related to similarity. This is lecture number four with different problems about this. And if you didn't look at the notes to this lecture and did not try to solve the problems yourself, please stop listening to this lecture and just go to the website Unizor.com, go to similarity problems lecture number four, and attempt to solve all these problems yourself first. Only after that, whether you will be successful or not, only after that listen to these lectures, and when I finish, try to do exactly the same thing again by yourself. That's the best way to approach all these problems, and that's exactly how I would recommend you do it. So I will continue with problems for similarity, and these problems are related to, well in some ways related to Pythagorean theorem, they are a little bit more general if you wish, not only for right angle triangles, but also for triangles with acute or two angles. So, well, let me remind first that if you have the right triangle, then notice that I'm using the lowercase letters for the side, the same as uppercase for an opposite vertex. So the Pythagorean theorem is a squared is equal to b squared plus c squared, and obviously you understand that a, b, and c are the sides, but when I'm writing something like this, what I actually mean is the lengths of this corresponding side. So if you take the lengths of the k-potonous a and square it, it will be equal to sum of squares of two-pattergy. Proof was actually in one of the prior lectures, so I'm just referring you to one of them. It's very easy actually, and it's using the similarity of three triangles. This is in the altitude, so small one is similar to the bigger one, and the similar to the biggest one. So from this similarity you can very easily get this. Now, what if my angle is not right angle? What if it's, well, let me just draw it again. What if my angle is acute? So again, a, b, c, and b, a, c. Well, as you understand, this angle is smaller, angle a I'm talking about, is smaller than the right angle. So if you will increase its measure to the right angle, then this will be the sum of squares of these two. But if you make it smaller, you kind of feel intuitively that the square of this side should be less than the sum of these squares. So I'm going to quantify this particular relationship between these two. Now, before attempting to do that, what I will do is with some kind of auxiliary thing. Let me draw a perpendicular from the vertex b. Now the, let's look at the h here. Now, there is a concept of projection in geometry. Projection of the point, for instance, onto the line is just the base of the perpendicular to this line. So if you have a line and a point, you draw the perpendicular and this point is a projection of that point on the line. If you have a segment, then the projection is basically a segment of a line between two projections of the points. So knowing that, I can say that a h is a projection of a b. Well, this is just beside the point, just the terminology. But now what I'm going to do is the following. If you have this triangle, and you have, I'm using the different letters here in my notes. If you have bm as an altitude from the vertex b to side ac, then if you have a similar altitude towards side ab from the c. So the point which I would like to make right now, equation basically is that ac, which is this. And I'm talking about the lengths obviously, not about segments. Segments are not multiplied, it's the lengths which is multiplied. But by the lengths am of projection, if you wish, of another side onto this side is the same as if I would take ab other side and projection of the first side on it, which is am. Now, why do I actually need this? Well, for the future, for the next problem actually, I will be using this type of projection of one side onto another. And I don't want to be asymmetrical, so to speak. I would like to say that this multiplication of the lengths of the side by the lengths of the projection of another side on it is actually the same if you take this side or that side around this angle. Now, how can I prove it? Well, actually it's extremely simple, because if you will notice abm is the right triangle, because this is an altitude, right? And so is acm, also right triangle. Now, these right triangles have the common acute angle, which means that the other angle is also congruent to each other, right? So, two right triangles, one angle is common, it means other angles are congruent to each other, which means they are similar. You remember we are talking about problems related to similarity. So, the similarity of these triangles actually mean, well, that their sides are proportional, right? You remember that. So, let's say hypotenuse of one side, which is ac, over hypotenuse, now this is the hypotenuse of anc, right? Now, over to hypotenuse of abm, which is ab, is the same as the catatose am opposite to this angle, to the catatose opposite to the same angle, opposite in another triangle, which is am. And this, as you see, is exactly the same as this, because these are proportions, which means ac times am is equal to ab times am, which was supposed to be proven. So, this is a very simple thing, and this is not really the problem which is kind of a central problem for this particular lecture. For this particular lecture, I would like actually to express the square of this side, which is a square, as some kind of a formula related to b and c. Well, I can't really relate it to b and c. I also need this particular piece, am, and let me call it, since it's a projection of c, I will call it c prime. Now, what is the square of this particular side, which is opposite to acute angle a? Well, let's think about it. Considering bmc is a right triangle, I can use the Pythagorean theorem, right? Which means this hypotenuse square is equal to the sum of square of this plus square of this, right? Let's call it h. h square plus... Now, what is this particular mc? mc is b minus c prime, equals to... Now, what is h square? Well, h square, let's consider now abm, right, triangle. hc, ah square is basically the square of the catatose vm, which is equal, again, Pythagorean theorem. Square of hypotenuse, which is c square, minus square of another catatose. Since this square plus this square is equal to this, so this square is equal to c square minus c prime square. That's what h square is. And we still have this sum. Let me go here. Well, let's open the parenthesis. c square minus c prime square plus b square minus 2bc prime plus c prime prime square. c prime square plus and minus if 0, and the resulting formula is... I will use b square first, plus c square, b square plus c square minus 2bc square. c prime, sorry. So, as we see, the square of one particular side of a triangle, which is opposite to acute angle, I would like to emphasize the word acute here, because I will definitely consider another case when the angle is obtuse and the formula will be slightly different. So, opposite to acute angle is equal to sum of squares of two other sides. But then you have this minus something, minus double lengths of one of the sides multiplied by projection of another on it. Now, and this is exactly something which is like asymmetry. Why do we use this particular b and the projection of c on to b as some kind of, you know, special case or whatever? Well, it's not a special case, because based on the problem which I have just proven before that, it's exactly the same as if I will take instead of this. So, b square plus c square still in place, but instead of multiplication, b times projection of the c on b. I can take c and projection b prime. And b prime is this. So, they are symmetrical. These two formulas are symmetrical because these members are exactly the same. So, I don't want you, you see, symmetry sometimes can be, you know, some kind of a criteria because if you are starting from something similar and obtain asymmetrical formula, well, that's always a suspicion that something might be actually wrong in your deriving, in your derivation of this formula. But in this particular case, although the formula looks asymmetrical, this asymmetry is not the real one because if I will use a symmetrical variation of this formula, I will get exactly the same thing. Okay, so basically what I can say is that the square of the side which is opposite to the q-tangle is equal to sum of squares of other two sides minus, minus double product of one of the sides onto projection of another side on it. Regardless of which side you take, this b or this c. Which means it's less as we see. By this particular member of this formula, it's less than sum of two squares as is the case with right triangle. By the way, how does this formula actually look like if I will start increasing the angle b, a, c to the right angle to 90 degrees? Obviously, as I increase this angle, this altitude will come closer and closer and by the time it's right triangle, a, b, c, my altitude actually is coinciding with the catatose which means that this particular length, c' becomes zero and if c' becomes zero, obviously this member disappears and the formula now is exactly the Pythagorean formula which basically is not a proof obviously of the Pythagorean theorem because we were using the Pythagorean theorem to prove this one but it's kind of an indication that there is nothing terribly wrong with our formula of this. So we have two actually intuitive checkings which we have made. Number one, that this is symmetrical and number two, that it actually transforms into the known Pythagorean formula in case the angle is increasing to 90 degrees. Okay, so this is two different problems. One was the equality of these products. Another is the formula for a cute angle and now the third one would be the formula for a choose angle. I'll do exactly the same thing. The drawing will be slightly different. First of all, let's think about what happens with our formula what would happen with our formula in case of choose angle the multiplication of the side by the projection of another side on it. So let's take this altitude and that was our point M and let's take this altitude. This is our point M. Now this is a choose angle which we were discussing. Basically do exactly the same thing. Now triangle ABM is the right triangle and triangle ACM is also the right triangle. Now BM is perpendicular to AC and this is a continuation of AC. CN is perpendicular to AB and this is a continuation of AB. Now since these are continuation of the sides these two angles are vertical and therefore congruent to each other and that's why these particular angles are also congruent to each other. Since these are right triangles ACM and ABM are right triangles pair of acute angles are congruent to each other so the other pairs as well. Now we will just write the expression of similarity based on the fact that these two triangles are indeed similar. Now proportionality between hypotenuses so AC to AB is equal to proportionality between contratus lying opposite to this angle in this triangle which is AM to a contratus which is lying opposite to the same angle in another triangle to AM. And we again see exactly the same equality. AC times AM equals to AB times AM. So the result of the product of a side by projection of another side on it which is AM is equal to again result of the product of another side AB by projection of the first side on it which is AM. So this is exactly the same symmetry is preserved and it's the same for acute angle or for obtuse angle. But now let's talk about our square of one side. So this was A, this is C, this is B. Now the projection of C onto B we mark as C prime and projection of B on C we mark as B prime. Now everything is just a couple of theorems, a couple of Pythagorean equations. Now MBC is the right triangle so A square equals to H square plus in this case it's B plus C prime square. So this catatose of this triangle is B plus C prime. Now if you remember in case of acute angle it was minus C prime because the point was here and you have to subtract it. So that's basically the difference this sum which is equal to instead of H square I can always write considering this right triangle H is the catatose which means it's square is equal to square of hypotenuse minus square of another. So it's C square minus C prime square equals open the parenthesis here C minus C square minus C prime square plus B square plus 2B C prime plus C prime square C prime square goes out and as a result we see that A square is equal to B square plus C square plus in this case it's plus in acute angle case it was minus 2B C prime which is equal to B prime plus the square plus C square plus 2CB prime as well because as I have just proven the multiplication of a side by projection of another side and the triangle is exactly the same as the other side by projection of the first side. So B C prime is equal to C B prime. So what's the difference again relative to the previous formula the plus sign rather than the minus and obviously you understand intuitively that if you have a right triangle then there is no this member and again for obvious reason because the projection would be equal to 0 because this would be the right angle but if you increase the angle over 90 degrees then it seems to be it's intuitively obvious that you are increasing this particular side so sum of square of this plus this should be greater sorry it should be smaller than just square of this one and we should add another member to be equal to A square. Alright, so these are basically four different theorems which are related to these angles and basically this particular theorem is it can be formulated in terms of trigonometry. I don't want to introduce trigonometry yet it's basically about the angles and certain functions like sine, cosine, etc. So instead of in those formulas instead of using projection instead of using projection A M of A B onto A C I can always use cosine of this angle. Then the formula would be much more symmetrical but again I would like actually to delay it until I will go through trigonometry in a more orderly fashion. So basically let me check if there is anything else for this particular lecture. I don't think so. Yeah, that's it. So now I recommend you to go to dunesor.com to the notes for this particular lecture and again try to prove all these well four if I'm not mistaken different theorems two for acute angles related to symmetry and related to square of one of the sides and two for obtuse angles also one related to symmetry and another related to square of one of the sides and try to get exactly the same formulas by yourself and that would be a very very useful exercise. Thanks very much until the next one.