 Hi, I'm Zor. Welcome to Unisortification. This lecture is a continuation of construction problems, everything about triangles, basically. And as before, I will be using certain things which I have already done before in the prior lectures. I'll use it just as known things, basically. Just for example, if, for instance, I'm saying, okay, let's construct the triangle equal to this one by three sides. I'm not going to explain how I do it. That's in the prior lectures. So let me go straight to additional construction works, construction problems, which I have. I hope you did spend some time with notes for this lecture where all these construction problems are written. And, well, I'm sure you have been able to construct it yourself. So just in case you have some doubts, I'll do it myself. All right, given a straight line and two points anywhere on the plane, constructed point on the given line, equidistant to given points, okay. So you have two points and some kind of a line on the plane. What I have to find out is the point M on this line, equidistant from A M should be equal to B M. I mean the length of these segments. Well, how to do it? I think it's appropriate to mention something which is called locus. Locus is a certain set of points usually which have certain property. Now, for example, what's the locus of points equidistant from two different points? Two points and I need all the points which are equidistant from these two. Well, I claim this is the perpendicular bisector to segment which connects these two points. Why? Well, it's very easy to prove. Let's say point M is equidistant from A and B, which means this is equal to this. So the triangle is isosceles and in the isosceles triangle the altitude and the bisector of this angle and the median are all the same line and that's why from M perpendicular to A B will always fall in the middle of A B. If I will take any other point M' and drop a perpendicular to A B again because these two sides are equal, the perpendicular will drop in the middle of A B. So no matter where my equidistant point is, it's on the same perpendicular which falls in the middle of A B. So the locus of all the points equidistant from two given points A and B is a perpendicular bisector to a segment which connects these two points. So what's the property this point must actually have? Well, it must be equidistant because that's what basically the construction problem is all about. Which means it should belong to the set of all the points equidistant from two given points A and B. Which means it should belong to a perpendicular bisector. But at the same time the construction problem is asking us to find a point on this given line which has this property. So basically we have an additional property. We have the property for the point M not only to belong to the perpendicular bisector of segment A B but also on this line. Well, if the point belongs to two different lines it's actually the intersection of these two lines. So if there is an intersection, that's a very important actually detail, then that's the point which we actually are looking for. So what we do is we connect A and B segment A B, then we construct a perpendicular bisector to this segment. And if it crosses our given line then the crossing point which is only one obviously because two lines can cross in one point only. So this point is the one which we have to construct. However, let's just consider what if this perpendicular bisector and this line do not have any crossing points. Which means they are parallel. So this actually also a case which we have to really think about. So if our two points are on the same perpendicular to this line, the perpendicular bisector will be parallel to this line and we'll never cross it. Which means there is no point on this line which can be equidistant from these two points A and B. No matter what point we have there is always a difference between the lengths of these two seconds. And by the way, if you remember one of the lecture was about perpendicular being the shortest distance from the point to a line. And the further from the perpendicular segment is from this given point M to this particular line. So this is perpendicular, this is the shortest and the further from the perpendicular our segment is the longer it is. So that's why if these lines are perpendicular, the line which connects A, B and the given line, then there is no equidistant line on the M. There is no solution to this problem. Okay, construct a point equidistant to three vertices of a given triangle. Alright, so if you have a triangle, I need a point equidistant to all these three points. Well, let's think about it. Again, let's approach this from the locus standard point. The locus of the points equidistant from let's say A and B is this line. The locus of the point equidistant from B and C is this line. And the locus of the points equidistant from A and C is this line. Now, I cross them in one point and that's obviously the case. But it needs certain comments. Why these three lines cross in one point? Because in theory, three different lines can cross this way, right? So this is not the case. This is the case. And let me explain why. Let me just assume in opposite that the line which is perpendicular bisector to A and C crosses this way. So I have three points M and P. Now, let me just try to prove that this is actually impossible. Look at this. The distance from the B M, distance from M to B is equal to distance from A to M. Because point M belongs to perpendicular bisector. At the same time, since point M belongs to this perpendicular bisector, distance from A and C is the same. So A M is equal to A C. Now, from this, sorry, not M, A C M or even better C M since I always put the verges in the front. Now, what's interesting is that B M, this distance is equal to A M because point M is on this bisector. And A M is equal to C M because M is on this bisector. So B M and C M are equal to each other because the equality is transitive, as we know. So B M is equal to C M. Which means what? Which means that the point M should also belong to the perpendicular bisector of the segment B C. Since it's equidistant. Which means point M should belong to this perpendicular bisector of the segment B C. So that's why my initial claim that intersection is of this type is actually invalid. Because the way how I draw it here, M doesn't belong to the perpendicular bisector. So basically this is not the case. What is the case is that all three lines cross in one point and this is the point at the distance from all three vertices of the triangle. From A to point M, the distance is the same as from B to M because M belongs to this perpendicular bisector. And this equal to this because it's this perpendicular bisector, etc. Which by the way means that if you would like to circumscribe a triangle into a circle, then the center of the circle is exactly here. Because these are supposed to be three ragi which are supposed to be of the same length since the circle is the locus of the points equidistant from one called the center. So basically if you would like to circumscribe a triangle, then you have to find out the center of this circle by drawing at least two, you don't need to really treat, but at least two perpendicular bisectors of triangle's sides. Okay, let's move on. Given an angle in the straight line that cross both its sides, construct the point equidistant to two sides of a given angle. Alright, now this is simple. Angle and line which crosses both sides, we have to find a point equidistant from both sides of the angle. Now back to the memory line, what is the locus of the points equidistant from the sides of the angle? Obviously it's bisector. From any point perpendicular to here or perpendicular to here are equal to each other. Because the triangles are right triangles with common hypotenuse and congruent acute angles. So we have to find the point which is both equidistant, which means it's supposed to be on the bisector of this angle, and line on this line. Well obviously that's the crossing of these two points, so the construction is draw a bisector of the angle and wherever it crosses our line, that's the point we have to, that's the point we're looking for. Okay, construct the point equidistant to three sides of a given triangle. Okay, we first made a construction of the point equidistant from three vertices. Now let's do distance from three sides. So we have to find the point where these perpendiculars have equal lengths. How to do it? Well, obviously if you do this and these are equal. Now these are perpendiculars. This is hypotenuse and these are tragedy. These are our legs. So triangles, this one and this one are congruent to each other. So these angles must be congruent too. So this is analysis of the problem. Now, therefore basically we have the solution. How to construct this point? Well, very simply. We just bisect this angle, then we bisect this angle, and then on the crossing we will have a point, and by the way if we will draw the third bisector, angle bisector. It will also fall in the same point, and the proof is exactly similar to the proof of the point equidistant from the vertices. So to construct this point which is equidistant from three sides, which means perpendiculars to the sides should be equal to each other, you just have to draw a couple of angle bisectors. And by the way, again, there is a circle involved here. If you will use this particular segment as a radius, then you can inscribe the circle into triangle. So the circle actually is tangent to all these three sides. We will talk about circle and its properties, and what's the tangent is, etc., a little bit further, but just know that for now that this particular construction of the point which is equidistant from three sides is actually a construction of a center inscribed circle. So there is an inscribed circle and circumscribed circle, circumscribed around the vertices, inscribed between the sides. Okay, given a straight line pq two points outside, but on the same side. Construct point on the line such that n goes pxm, okay, pq and points m and n on the same side. Construct the point x on the line such that n goes pxm and qxm, qxm are congruent. Well, that's simple. You just take the point m, reflect it relative to this line, and if you will draw a straight line from this point m' to n, now these are vertical angles, right? And these angles are congruent because m and m' are symmetrical relative to this line, which means they are on the same perpendicular, this is equal and this is common, so the triangles are right triangles and they are congruent by two legs, and that's why the angles are also congruent, and that's why this angle is congruent to this, because these are vertical and these are congruent to themselves. All right, that's simple. Construct the right triangle by one of its categories and sum of hypotenuse and the other categories. So we have the right triangle, now we have this category and we have a sum of these categories plus hypotenuse. Well, which means if I will continue this line and make this equal to this, then the whole segment obtained, so this is a, b, c and this is also c. So what's given is a and b plus c, these two segments. Now, how can we construct the original triangle? Well, considering we have a and b plus c, we can construct the big triangle, right? But now let's think about since, well, let me put some letters a, b and c and b. So if my triangle b, a, d, if you consider it, there are two sides which are equal to each other, which means this is a socialist triangle, which means that perpendicular bisector to a, b would be an altitude and bisector of this angle and the median of this triangle. So to obtain the point d, all I have to do is just take this segment and draw a perpendicular bisector. Now this is a triangle which I'm looking for because this part is equal to this part and this is given. So basically construct the big triangle and from its hypotenuse draw a perpendicular bisector to get this point and this is the triangle we are looking for. Okay, move on. Construct a triangle by side and then go it makes with another side and some of two other sides. Okay, it's kind of similar. So if you have a triangle and what's given is a, b, c, a, b, c and this angle. So construct a triangle by a side, let's say it's b, this side. An angle it makes with another side which is angle b, a, c and a sum of two other sides and a plus c. Okay, so these three elements are given for a triangle. What should we do? Well basically it's very similar to the previous task but the previous task was about right triangles and this one is about regular but we still have an angle which helps. So how to build this? Well let's just think about it. If you will, if you don't mind I'll make it slightly different, smaller. This is b and this is c and this is a. So if you will continue this line by the distance equal to a. Now what can you say about triangle a, b, c? You have a side and you have another side which is a plus c and you have an angle between them. So you can build this triangle very easily. Now this is a and this is a which means b, g, c is an esosceles triangle which means to find the point b we have to do the perpendicular bisector to this side. And here is our point b and here is our triangle. This one seems to be simple, right? It's exactly the same as with right triangles but instead of the right angle we have a given angle. Okay and this is my last one. Given an angle and two points, one on each side of an angle, okay? Construct a point equidistant from both sides of an angle and at the same time equidistant from two given points on an angle. Okay so we need the point which is equidistant from these two and equidistant from these two lines. Well again that's a very simple locus type of construction. What is the locus of points equidistant from two given points? So that's the perpendicular bisector, right? It's this line. And what's the locus of points equidistant from two sides of an angle? That's the angle bisector. So the crossing of these two lines satisfies both requirements. It's equidistant from these two points because it's on the perpendicular bisector and it's equidistant from two different sides of an angle because it's on the bisector. Well that's it for this lecture. These are very simple construction problems and they should be in your skill basically, in your repertoire if you wish, in your menu of available tools which you have to solve much more difficult problems. Now what is important is to find out what kind of problem this is. Is it a locus problem for instance which means you have to just draw one locus, another locus and maybe their intersection will be what you're looking for. Or maybe it's a construction problem which requires some kind of additional drawing which would help you to realize what exactly needs to be constructed. So maybe instead of constructing directly what you're looking for you have to construct something else. Like if you remember in the previous task where we had one side and some of two other sides and an angle. We basically constructed a bigger triangle and then we decided what to do, how to find the point of the triangle which we really need to construct. So analysis is a very important thing. So first you have to start with analysis of this construction problem and then if you do it right then you will realize how to do the construction itself. That's it for today and don't forget Unisor.com is your site for all kinds of interesting educational material and I would also like to encourage parents and supervisors to use this site to basically control the educational process because the students can be enrolled. They should pass certain exams. You as a supervising person can control actually whether they pass or fail a particular lesson or lecture or whatever. So it's a very good tool to better educate your students. Thanks very much. That's it for today. Bye-bye.