 Hi, I'm Zor. Welcome to Unisor Education. This is another lecture on similarity. In this case I will talk about polygons. The last one was about triangles. It's a special case of polygons. Now I will try to expand it to some other cases. So let me just talk about general properties of similarity of polygons. First of all, if polygons has n sides or n angles, after the transformation of scaling, homotopy, the scientific term, obviously it will be transformed into another polygon and that other polygon will have also n vertices and angles and n sides. Why? Well, for obvious reasons that the straight line is transformed into straight line, which means segment is transformed into segment, point into point, angle, as you remember, into another angle, which is, by the way, equal to original one, as we have proved it before. So the number of vertices cannot be changed, straight lines will remain straight, so sides will be sides and angles will be angles after the transformation. So this is the first property. The n-sided polygon is transformed into n-sided polygon. You cannot scale a triangle into a square. Now the second property is what if you have an n-sided regular polygon. Now regular polygon, as you remember, is the one which has all interior angles equal to each other and all sides equal to each other. I'm interchanging using equal and congruent basically. It means exactly the same thing in this particular case. So how can we prove that regular n-sided polygon is transformed into regular one? Well, again, we were already mentioned that the angles are transformed into angles of equal size. All sides which used to be equal to each other in lengths of the original polygon, let's say it's an equilateral triangle, will be transformed into sides of another triangle after scaling. And what's important, there is a factor of scaling. Each segment, each side in this particular case, will have its lengths equal to the old lengths multiplied by the same factor. So these three lengths were the same multiplied by the same factor. They will be the same. So angles equal angles remain equal angles and sides are proportionally factored in the same factor using the same coefficient of proportionality, which means that the final sizes will also be equal to each other. So that's why if it used to be regular, all angles are equal, all sides are equal, it will be regular polygon after the transformation. Equology will be retained. Now, let's talk about different geometrical figures which we have already learned something about. For instance, we were talking about right triangles. Now, is the quality of a triangle to be a right triangle retained after the transformation of scaling? Yes, of course, because the quality is that this angle is 90 degrees. Now, after the transformation of scaling, all angles are preserved, which means the resulting triangle will also have 90 degree angle. So the quality of a triangle to be the right triangle is preserved. Now, the quality of a triangle to be isosceles when two sides are congruent to each other, equal in size, obviously will also be preserved because the final triangle will also have these sides equal in lengths to each other because they are equal to all sides multiplied by the same factor. So if these were equal, multiplied by the same factor, these will be equal as well. So the isosceles triangles are transformed into isosceles. Now, obviously, equilateral into equilateral, that's basically part of what we have already proven about inside the right polygons. What else? Parallelograms will be transformed into parallelograms because the quality of two lines to be parallel to each other is retained by the transformation of scaling because every line after the transformation of scaling is transformed into lines parallel to each other. So if I have two lines in a parallelogram, which are parallel to each other, the image will also be lines parallel to these. So parallelism is retained, which means parallelogram will be parallelogram after transformation. Rombos will be rombos because it's basically a combination of being a parallelogram and having all sides equal to each other in lengths. And again, the lengths is transformed proportionately for all sides of a polygon, which means that the final, the resulting figure will also have equal sides and will be a parallelogram, which means that it will be rombos. Square, obviously, will be converted into square. What else? Rectangles, again, all angles are 90 degrees and angles are retained by the transformation of scaling. Trapezoids. Trapezoid has two sides parallel to each other and the parallelism is retained as well. So all these different qualities of different geometric figures, which we were talking about before, will be preserved by the transformation of scaling by homotogy. Now, these are all kind of trivial facts about polygons. Now, just very, very slightly more difficult about regular polygons. I'm going to prove that two regular polygons, which have exactly the same number of vertices or angles or sides, there is always the transformation of scaling which transforms one into another. They are always similar. Okay, how can we prove that, let's say, if I have two triangles, for instance, how to prove that they always are similar? This is a particular case of this theorem and I would like actually to address it as an illustration that this is true. All equilateral triangles, which are n-sided regular polygons with n equals to 3, are congruent to each other, not congruence, are similar to each other. Now, how can I prove it well? For triangles, it's very simple, because these angles are always 60 degrees for equilateral triangles, and we know a theorem about triangles. If two angles of one triangle are congruent to two angles of another triangle, then they are similar. So in this case, obviously this is the case, so these two triangles, two equilateral triangles are similar. Now, how about squares? Squares are regular polygons of n-sided where n is equal to 4. So if you have this square and you have this square, how to prove that they are congruent to each other? Well, in this also a particular case, I will use the following technique. Let's have two diagonals here and two diagonals here. Now, I can use parallel shift and maybe rotation to basically position my square, smaller square inside the bigger square in this way. Now, why is it possible? Well, if I will move this point to this point and this point to this point, everything else actually will be in place, because if I will draw a line parallel to this and parallel to this and then parallel to this and parallel to this, I will have a square which can obviously very easily be proved, and the square will have exactly the same diagonal as this one. Now, I don't want actually to prove it in details, because I will I will be proving this in just general case for any number of vertices of the regular polygon. But you basically feel how I would probably be able to prove it. Now, I will actually prove it for just general case for n. Okay. Instead of a square, let's use something else. Now, first of all, let's say you have some kind of regular, not sure that regular, but let's consider it's regular hexagon. You kind of feel that there is a center of this figure, right, which probably is center of symmetry, etc. Well, let me prove it a little bit more rigorously. Forget about these three. I will draw three angle bisectors. So these two angles are equal to each other, and each one of them is half of the interior angle of hexagon. Now, these angles also are equal to each other, because this is bisector. And considering that this interior angle is equal to this interior angle, halves of these angles also are equal. So is this one. Now, how can I prove that three bisectors are intersecting in one line? Well, it's actually very easy. Let's just consider these two. Now, this is a triangle, and consider this side and these two angles. Now, let's draw a bisector here, and just, for example, it intersects in some other point, not in the same point. But if you will take a look at this, this triangle and this triangle are congruent to each other because sides inside the regular polygon are equal, congruent to each other, and all angles are exactly the same, and each one of them is half of the interior angle of the polygon. And so they are also equal to each other. So these two triangles must be congruent to each other, and that's why this particular segment must be equal to this. So because this is the side of this, it's also a triangle, and this is the side of this is also a triangle, and sides are equal because triangles are congruent. So that's why this bisector will intersect this in exactly the same point as this bisector will intersect this, because these triangles are congruent to each other. Obviously, as any other triangle, if we will put the angle bisector, so I mean, obviously, you feel that this is true, that if you will draw a bisector of every interior angle of regular and sided polygon, all these bisectors will intersect in one at one point. Yes, but this is a little bit more rigorous proof rather than just, you know, you're feeling about this. All right, fine. So we have something which we can call a center of the regular and sided polygon, and we will use this point to prove that any other, in this case, six sided polygon, let's say this one, is similar to this one. Now, how am I going to prove it? Well, to prove that these two are similar to each other, I have to build a transformation which contains some congruent transformations like parallel shift, rotation, reflection, maybe, and scaling. And as a result of these transformations, I will transform this into this. All right. So first, let's talk about congruent transformation, which will bring this particular sided polygon inside this one, and in such a way that the center will be moved into the center, and any one of those angle bisectors will coincide with corresponding bisector. Now, why would any other bisectors coincide with these? Well, for obvious reason that all these triangles are congruent to each other and have exactly the same angles, and if I will just draw a line from this, parallel to this line, and then from this parallel to this, etc., etc., etc. I will build a set of triangles. Each one of them is in a socialist triangle with a fixed angle at the base, this is the base, and these are two equal sides. So with these two angles exactly equal to any of these angles, for obvious reasons. So the whole n-sided regular polygon inside the big polygon will be congruent to this one, because of congruence of each triangle. Now, once I have built this particular regular n-sided polygon, and I know that vertices will lie on angle bisectors, now all I have to do is to use a scaling with this as a center, and the ratio between two bisectors, the big bisector and the small bisector, as a factor of scaling. Since all the big bisectors are congruent to each other and all the small bisectors are also congruent to each other, the ratio will be exactly the same, which means every point, this as well as this, as well as this and this and this and this, by stretching by the same factor as ratio between these two guys, these also will be converted into the corresponding point on the big n-sided polygon. So this is basically a construction, if you wish, of exact transformation which transforms one into another. As long as the number of sides, or the number of vertices is exactly the same, that's sufficient for them to be similar. And the proof is an actual construction of the scaling which will transform this into this. Okay, now, well incidentally, this angle is equal to 360 divided by n, right? Because there are n triangles here. Now, these two angles in sum would be equal to 180 minus 360 divided by n. Because the total measure of all three angles is 180 in this triangle, this one is 360 divided by n. So sum of these two is equal to 180 minus 360 divided by n. But don't forget that these two angles in sum would be exactly the same as interior angle because each one of them is half. These are bisectors. So this is actually the measure of any interior angle of n-sided regular polygon. But this is besides the point, has nothing to do with similarity in this particular case. Okay, next. Next is a special theorem which I would like actually to present as also a little bit less trivial. Well, the previous one was a little bit less trivial versus the theorems which I mentioned before that let's say the right triangle will remain right after the scaling. I mean, that's obvious because the angles are preserved. But the fact that two n-sided polygons, regular polygons are similar to each other, that's a little bit less trivial and requires some proof. Now, in this case, this is also a little bit less trivial theorem. If you have two triangles, both right triangles, and you have a proportional cotity and a hypophonosis, which means that a, b ratio to a prime, b prime, one cotitus versus another is equal to bc versus b prime, c prime. So a ratio between these two cotity is the same as the ratio between hypophonosis. Then these two right triangles are similar to each other. Now, how can we prove it? Actually, it's basically the same as we have proved something about similarity of triangles in more general case when we have, let's say, two angles or an angle and two sides proportional each other. So what I will do is the same. I will find a point m such that a m is congruent to a prime, b prime. These two. And draw a parallel to a hypophonosis, m m. Obviously, these two triangles are similar to each other. We have proved it in one of the previous lectures when I was talking about any triangle and a line parallel to one side. It cuts another triangle, which is similar to the big one, to the original one. Now, since these two triangles are similar to each other, their sides are proportional, which means a m divided by a e, or I can do it twice first. a b divided by a m equals to b c divided by m m, right? a b divided by a m like b c to m m. This is what proportionality means. This is basically the direct result of them being similar. Now, but here, considering a m and a prime b prime are exactly the same, so I can put here a b a prime b prime. But at the same time, a b divided by a prime b prime is the same as b c over b c prime. b c over b prime c prime. Now, if you consider these two, you have numerators the same, and that's why we have the denominators the same. So that's how we have proved that this is equal to this. Now, this was constructed as two congruent segments, right? So we have two right triangles with a calculus and hypotenuse correspondingly equal to calculus and hypotenuse, which means they are congruent. So these two triangles, a m n and a prime b prime c prime, are congruent to each other. And since a b c is similar to a m n, therefore, a b c is similar to a prime b prime c prime. That's it. That was easy. What's next? Now we have, okay, that's about a triangle. Triangle has a few major elements, like it has medium, it has altitude, and it has angle bisector. Now, what I'm saying is the following. If two triangles are similar to each other, then all these elements are also similar to each other. Well, let's prove it. I mean, you know, we don't really just give the information or facts about things. We prove things. At least we're trying to do it as rigorously as possible, just as a training exercise. Well, let's do it one by one. So let's start with medium, let's say. Okay, so we have medium and we have another triangle, a prime, b prime, c prime, medium, m prime. How can we prove that if these two triangles, a b c and a prime, b prime, c prime, are similar, then the proportionality actually is retained, which means that the factor this one is in this particular case longer than this one would be exactly the same as the factor between, let's say, a b and a prime, b prime, or b c and b prime, c prime, or a c and a prime, c prime. So let's say the factor between the sides is whatever, f. Let's say it's a b divided by a prime, b prime. Okay, that's the factor of scaling. Now, since a m is half of a c, similarly, a prime, m prime is half of a prime, c prime, we can definitely say that since this factor is the same, then this factor is the same. This is a c over a prime, c prime. This is half of a c over half of a prime, c prime. Haves can be reduced, right? But these halves means that this is equal to a prime, m prime, sorry, a m over a prime, m prime. So what we have just proven that halves of the sides are proportional to each other with the same coefficient of proportionality. But now let's consider a b m and a prime, b prime, m prime. Since they are similar, these angles are the same, and we have proven that the proportionality between a, we have just proven the proportionality between the sides, a b versus ratio to a prime, b prime is the same as a m to a prime, m prime, which proves according to one of the theorems about the similarity of triangles. We have a side angle side. Sides are proportional with the same coefficient of proportionality, which is the factor, and angles are equal. So a b m is similar to a prime, b prime, m prime. And what follows is proportionality with the same factor between b m and b prime, so medians are proportional to sides. Now let's talk about bisectors. Let's consider b m is a bisector, which means this and these are halves of a, b, c. But these and these, these are also two halves of a prime, b prime, c prime. And since a prime, b prime, c prime, the big angle of the triangle itself is the same as in this case, a, b, c, then halves are also the same. So that's why I use the same, the same sign to basically to signify that these two angles, h l or four angles are exactly the same of the same size, the same measure. Now what do we have? We have a b m and a prime, b prime, m prime having two angles exactly congruent to each other. And again one of the characteristics, one of the theorems about similarity of triangles was, if you remember, that if two angles of one triangle are correspondingly congruent to two angles of another triangle, then these triangles are similar. That's why a b m and a prime, b prime, m prime are similar to each other. And the same b m is proportional to b prime, m prime, with the same factor, with the same ratio of proportionality as a b to a b prime, a prime, b prime, same ratio. What's left is altitude. Okay, what about altitude? Well basically we really have to do exactly the same thing. We have to prove similarity of a b m and a prime, b prime, m prime. How can we prove it? Well very easily, since this is the right triangle and this is the right triangle, these are altitudes, right? b m is the altitude and b prime, m prime is an altitude. So we also have two angles congruent to each other, one because the triangles are similar and another because they're both right angles. So these triangles, a b m and a prime, b prime, m prime are similar in this case as well, which is why altitudes are proportional to the sides. So again, in two similar triangles, these elements like bisector, median and height, they are all proportional with the same ratio of proportionality as the sides. Actually it's true for other elements as well, for instance radius of inscribed circle or radius of circumscribed circle, but I would rather put it as a problem for some next lecture or something like this. I just want you to have a feeling that if two figures are similar, then every reasonably constructed element inside one would be proportional to element, corresponding element of an element. Now reasonably I mean bisector of the same angle or radius of circumscribed circle and stuff like this. Okay, and I think I wanted to have one more theorem as part of this. Yes, perimeters. Perimeter of similar polygons, not necessarily regular polygons, just similar. So if you have two similar polygons, let's say you have one trapezoid and another trapezoid, and let's say they are similar. Then their perimeters are similar as well. Why? Well, this is the simple thing. Let's call the lengths of these a, b, c and d and this will be a prime, b prime, c prime and d prime. These are lengths of the corresponding sides. Now we know that the sides are proportional to each other, right? So the coefficient of proportionality between all of them is exactly the same. So a divided by a prime is equal to b divided by b prime divided by c, c prime, d, d prime and equals to some kind of a factor of scaling f. Now the perimeter of the left one is a plus b plus c plus d. Perimeter of the second one is a prime plus b prime plus c prime plus d prime. Now a is equal to a prime times f, right? b is equal to b is equal to b prime times f. So is c prime times f and d prime times f, which is equal to factor out f, factor out the factor and what will be inside the parentheses a prime plus b prime plus c prime plus d prime, which is a perimeter of this triangle. So the perimeters are related to each other in exactly the same fashion as the sides, also proportional with the same coefficient of proportionality. And that would be it for today. Next lecture will be about circles. Don't forget Unisor.com website contains this material, this lecture and many other lectures. I do encourage parents to take a look at the site because parents can control the educational process of their children as well as group teachers, for instance. They can control on an individual basis the pace of learning of each student because each one could just do it at his or her own pace going through exams, lectures, exams, etc. Thank you very much and until the next time.