 Hi, I'm Zor. Welcome to Unizor Education. I continue talking about construction problems related to similarity. All right, let's start. Okay. If you have a segment AB, you have to find a point C outside of this segment on the same line in such a way that C A relates to C B as two given numbers. Well, if I'm looking for number M, which is smaller than N, then my C should be on the left side of the AB, because C A would be in this case smaller than C B. If M is greater than N, it should be on the right side from the AB, because then C A would be greater than C B. So let's just consider the case M less than N. Now, here is what I suggest. Now, if this is M times greater C A, if this is proportional to M and N, then obviously the segment AB should be proportional to N minus M. So from this, I can derive my construction method very easily. So this is my AB. Knowing M and N and N is greater than M, I divide AB into N minus M equal parts, and then go to the left, I add M such parts. That's my point C. Now, obviously the lengths of C B would be M plus N minus M, which is M parts, and the lengths of C A would be M parts. So the ratio would be whatever I need. Now, interestingly to know that N should not be equal to M, because then we cannot really divide AB into this number of parts that will be zero basically. We cannot divide the segment into zero equal parts. So for other cases, if M is not equal to N, everything is fine, including by the way the case was when M is equal to zero, because that means that I divide the whole segment AB into N parts. Now, M is equal to zero, which means C would be coinciding with A. And in case M is equal to N, there is no solution basically, because for obvious reasons, we cannot point to anything outside in such a way that these two segments have the same lengths. Okay. That's easy. Next. Inscribe a triangle into given circle by its one side and ratio of two other sides. Okay. Triangle is defined by a circle. It's inscribed into our second-scribing circle, side AC, and ratio AB over BC. Now, obviously if you have a circle and a chord, then this angle is defined, because any other inscribed triangle, which has the same chord in the same circle, will have the same angle on the top, because they're all inscribed angles which are supported by the same arc. So basically you have an angle. So from this, we derive the value of angle ABC. But now let's consider if you have a ratio of two sides and angle between these two sides, you can very easily build a triangle which is similar to the one which we need. Use this angle. On one side, put any segment. On another side, put the segment which is proportional to this one, and we know how to do that. Now, since you do have this type of a triangle which is similar to this one, similarity means all the angles are congruent. So this angle is the same as this one, right? So all you have to do to construct the point B if you have a circle and a chord A, construct this similar triangle and just have this angle, put it here, draw a line, and that's where you have your B. Simple. Let's go to the next one. Inscribe a triangle into a given circle where it's one side and medium to the other side. That's just slightly more difficult. Okay, so you have to construct a triangle if you know a circle, circumscribing this triangle. You know a C and you know the medium C-M. Medium means that A-M is equal to M-B. So that's what's given. Now, here is an important point. A-B is a chord originated at point A. Now, you have to remember from one of the prior lectures that locus of all midpoints of the chords originated at A. Let me use a different color. Let's say this one, midpoint, or this one, midpoint. So all of these points are lying on a circle which has O-A as a radius. O is the center of this. So if you have O-A, the radius O-A, as a diameter and put a circle around it, that's the locus of all midpoints of all the chords originated from A. So to find a point M, which is a midpoint of a chord and also on the distance C-M, given the distance from point C, what we have to do is firstly build this locus, this circle, around O-A as a diameter and then from C, using radius C-M, you draw another circle and wherever these circles intersect, these are two potential points M which are midpoints of the chords which we're looking for. One solution is from A we connect through M to B and another from A through M prime to D prime. So our problem has two solutions, two different triangles, ABC and AB prime C. In both cases, the medians C-M or C-M prime are the same, the given ones. So what's interesting here to remember that to find certain points you can actually find by intersecting two locuses. So point M is on the distance given distance from C which means it should lie on the circle around C of this radius and also it's a midpoint of a chord which means it's supposed to be on the circle which is built on O-A as a diameter. Okay, that's fine. Slightly more difficult than the previous one. Next, inscribe a square into a given segment. So just one side lies in the chord, okay. So here is our initial circle and a segment and we have to find, we have to construct this square. So it has, okay, it has two points on a given segment. A given chord and two points on the circle. Now, how to construct it? Okay, here's what I proposed to do. Let's use AB as a side of a square and build it this way. All right? And also have a midpoint here called X. If I connect AB, CD, if I will connect X to C and D, I'm actually stating that obtained intersections between these and the circle give me the P and N which I'm looking for. So the construction would be like this. If I have a chord AB, I build a square midpoint of AB connect to these points and wherever these connections are intersecting my circle, these are N and P. Now, why N, P, Q, N is a circle? Well, for very simple reason. Consider triangle X, C, A and X, N, M. Well, these two triangles are obviously similar because N, M is perpendicular, C, A is also perpendicular. And since these are two right triangles and the common angle are these, they are similar which means that M, N relative to a C is equal to N, X to C, X, right? N, X to C, X. But in the same ratio are based on the similarity of these two triangles, C, D, X and N, P, X, also similar because all again, all angles are equal so they're parallel. So similarity of this follows that N, P over CD is equal to N, P over CD is equal to N, X over C, X. N, X over C, X. Now, these are equal to each other which means these two are supposed to be equal to each other. And since M, N, since AC and CD are equal to each other, these are sides of a square and the fractions, the ratios should be equal which means M, N, N, P also are equal, M, N, N, P. So similarly, P, Q also included into this equation is the next thing. So that's how we prove that M, N, P, Q is a square. And the construction again, built a square on AB and having a point connect to the ends. The proof is very, proof is completely based on similarity of triangles. Okay, next. Inscribe a square into a given triangle such that one of its sides lie on one side of a triangle and two opposite vertices lie on two other sides. So it's, in a way, it's a similar problem. So if you have a triangle, any triangle, you have to inscribe a circle, excuse me, a square such that one of the sides of the square lies on the side of triangle AC. Well, how can we do this? Here is how it can be done using again similarity. Let's have the line which is parallel to AC and on this line have any square with any side. Now, this line is parallel to this one so I can draw from D prime line parallel to AB. So I will do this. And from E prime and draw line parallel to this and call these points A prime, D prime, and C prime. Now, what do we have? We have basically two similar triangles which we have, and we have already built this one and all we have to do is transform the square inside of this triangle to this. Well, but think about it. What do we know about the whole thing? Well, we know every side and we know everything about this triangle, about this square, which means if I want to find out what's the length of, I don't know, anything. Let's say AD, but I do know A prime, D prime, right? And I know AC and I know A prime, C prime. And since everything is proportional, I can say that AD over A prime, D prime is equal to AC over A prime, C prime. This, this, and this are all known segments, right? Because this we have constructed and this is given, this triangle is given. That's where I see. So that's how I differ in AD. Knowing the AD, I just put this point D and that's the beginning of my square, okay? So we are using a construction of a similar triangle and then just stretch it or shrink it to whatever the size we need. Okay, next. Okay. Construct a triangle similar to a given one around a given square. Okay. It's basically opposite. Right now I have constructed a square inside a triangle. Now I would like to construct a triangle around a square, but not just a triangle but a triangle which is similar to a given one. So if you have a square and you have a triangle which you would like your triangle to be similar to, how to construct the triangle which would be circumscribing this circle and similar to this one. Well, basically that's easy. Align these two lines. This is a given triangle and this is a given square. Now if you align these two, then draw a parallel line to this. Draw a parallel line to this and this is your triangle because all angles are equal, so it's a similar triangle and it's circumscribing a given square. Okay, next. Inscribe a rectangle into a given triangle such that one side lies on one side of a triangle and two opposite vertices on another and the ratio between the sides are m over m. Okay, it's kind of very similar. So instead of inscribing a square into a given triangle we have to inscribe a rectangle in the same fashion but in such a way that ABC triangle ABC is given and DGE over EF is m over m. So that's what's given. So it's not just any rectangle but a rectangle which has these two adjacent sides in the ratio of m over m. Well, I will do exactly the same thing as I did with the square. First, I built on a parallel line to AC. I built any rectangle with this ratio which means I use the base, I calculate the kite and this is my rectangle. Now I can use the lines parallel to AB and BC and I will get A prime, B prime, C prime and this is D prime, E prime, F prime and G prime. So I have the picture which is similar to the one which I have to build. So here I have a triangle and the rectangle is proper sides so I have to basically transform into this. Again, the same as with the squares. I know that AD over A prime, D prime is equal to, let's say, AC over A prime, C prime. This one are elements of the constructed triangle. This is also given and that's how we can construct AD and knowing AD, we can construct the whole rectangle. And the last but not least problem is exactly opposite. Construct a triangle, right, exactly. Construct a triangle similar to the one which we have here but the one which circumscribing a given rectangle. Well, exactly the same thing. Just make these two lines parallel to each other and then just draw the lines parallel to the one which are given. So this triangle for obvious reason is similar and it's circumscribing our rectangle, same as with the square. Well, that concludes this particular lecture. I will have more construction problems which I like and I do encourage you to do again all these problems just by yourself. And I also recommend to register as a student, have somebody or yourself if you want, as a supervisor or a parent registered and take exams, enroll into certain classes, certain topics or everything basically. Exams are important. I do recommend to register primarily for the purpose of taking exams. All right, that's it for today. Thank you very much.