 Hi, I'm Zor. Welcome to Unizor Education. I will continue presenting mini theorems, which I started in the previous lecture. I think it's extremely important, and this is actually the whole purpose of this website, to teach you to solve problems. So all these mini theorems, construction problems in geometry, solving equations in algebra. So all these exercises, all these problems, which are presented on the website, and which I discussed in the lectures, accompanying the notes, are extremely important. And it's basically, again, the goal of this website, to teach you to solve the problems for whatever future profession you might choose. So consider this as just brain exercise, as you would exercise your muscles in a chain. So I have this list of problems, which I want to present as a subject of this lecture. Now, I deliberately did not really go through solutions of these problems before the lecture. So I might, you know, hesitate in some cases, because I'm solving these problems, basically, in real time, if you wish. So bear with me, if I make the wrong step, I will correct myself, so you will see the process of how I solve these problems. And whatever my solution is, is not necessarily the only one, obviously, you can come up with your own. But whatever I come up with, it's a good schooling. Anyway, so let me start. Prove that some of the lengths of two diagonals of a quadrangle is less than its perimeter, but greater than its half perimeter. Okay. So you have a quadrangle, and two diagonals. So what you have to prove is that some of these two diagonals is less than perimeter, but greater than half perimeter. So what I will do is I will use letters to signify the lengths of the sides, and either of diagonals will also be with letters. So I have this quadrangle, A, B, C, G, with sides having lengths A, B, C, and G. And you know what for, I think for my purpose, I probably have to break every diagonal into two pieces, like P1 and P2, Q1 and Q2. I anticipate that this particular problem would require it. All right, so we have two different problems inside this one. Number one, that some of these two diagonals is less than the perimeter. Okay. How can we prove that? Well, obviously, we have to use the triangle inequality for this particular case. Now, if you will take a look at, let's say, triangle ABC, AC would be smaller than AB plus BC, obviously, because this is a triangle inequality. So I can say that Q1 plus Q2 less than A plus B. All right. Now, similarly, with the same base, we can consider this triangle, ACD. And again, AC would be smaller than CG plus B8, which means Q1 plus Q2 would be less than C plus B. Now, let's talk about this diagonal. Considering the triangle on the left of it, ABG, we see that P1 plus P2, which is this side, is smaller than some of these, A plus D. And finally, this triangle BCD, again, the BD, which is P1 plus P2, would be smaller than B plus C. Well, actually, I think it's quite easy to see from here. Now, I have four different inequalities if I will add them up together. I will have Q1 plus Q2 twice, which is this diagonal, and P1 plus P2 twice, which is this diagonal. So I will have twice the sum of two diagonals. So it will be twice times Q1 plus P2 plus Q1 plus Q2. Now, it will be less than, if I add these inequalities together, the smaller part will give the smaller sum, and the bigger parts will give the bigger sum. I have A twice, I have B twice, I have C twice, and I have D twice. So I have A plus B plus C plus D twice, which is double perimeter, right? And obviously, we can reduce any inequality by two, which results in the first part of the theorem, mini-theorem, which we have to prove, that sum of two diagonals, Q1 plus P2, Q1 plus Q2, less than perimeter. Okay, so this part is easy. Let's think about the second part, that is greater than half perimeter. For half perimeter, I think we have to, well, we need an opposite direction, obviously. So diagonals should be greater than something. But now let's consider separately these four triangles. Now, these four triangles also can be used to apply the video triangle inequality. Now, A is smaller than P1 plus Q1, right? So A is smaller than P1 plus Q1. B is smaller than P1 plus Q2. B is smaller than P1 plus Q2. C is smaller than P2 plus Q2. And finally, D is smaller than P2 plus Q1. And again, add them together. On the left, I will have A plus B plus C plus D, which is perimeter. And on the right, I will have P1 twice, P2 twice, Q1 twice, and Q2 twice. So I have P1 plus P2 plus Q1 plus Q2 twice. So it's twice the sum of diagonals. And by reducing it by two, we have that half perimeter is less than sum of diagonals, which is the second part of this mini-theorem. Okay, that's it. Anyway, if you just think about how the whole problem was stated, I have any kind of a quadrangle. So basically, all these theorems about a socialist triangle, right triangles, et cetera, they're not really applicable. It's kind of, I think it's implied that the only thing which you can use to prove this inequality is the only inequality which you know, which is the triangle inequality, that in any triangle, sum of two sides is greater than the third side. All right, let's go on. Given an angle, p, m, q, p, m, q, and two points on each side, a and b on side, a, p, m, p, and a prime, b prime on a, q. Such that m, a is congruent to m, a prime and m, v is congruent to m, v prime. Okay, prove that crossing segments, a, b prime and b, a prime, they cross on a bisector. So this point is supposed to be on the bisector of this angle. And notice that this theorem implies easy way to construct an angle bisector. Well, yes, let's consider we prove this particular theorem. Then how can we construct the angle bisector? How can we basically find any point which lies on the bisector? Very easily, have a compass set to some particular distance and measure m, a and m, a prime. Then have another staging of the compass, measure m, b and m, b prime. Both statings are completely independent and can be anything, basically. And then just draw two lines from b to a prime to from a to b prime. Their crossing will be on the bisector. And then we can draw the bisector from n through this point. So it is an easy way to construct a bisector, angle bisector. But now let's go back to proving this theorem. Well, it looks obvious from here that triangles m, b, a prime, m, b, a prime is congruent to m, a, b prime. Now, why? Well, obviously, because this side is equal to this side, this side is equal to this side. That's just the way how we construct all these points, a, b, a prime, b prime. And angle, they share. So it's basically side, angle, side of one equal to side, angle, side of another. All right, so what actually follows from this equality? Let's think about it. Well, every element obviously is equal to the corresponding element, which means in particular that this angle of this triangle is equal to this angle. All right, that's quite obvious. What else is important? Well, we can also derive that this angle is equal to this angle. So angle m, b, a prime equals to m, b prime, a. That's from the equality of the triangles themselves. And since this angle is equal to this angle, and these are supplementary to these, this is supplementary to this, and this is supplementary to this angle. So that's why we have m, a prime, b congruent to m, a, b prime, a, b prime. And therefore, supplementary angles b, a prime, b prime, this one, b, a prime, b prime. This is supplementary to this one equal to this supplementary congruent to b, a, b prime. So now let's consider, let's put this point p. Let's consider two small triangles, b, a, p, and b prime, p, a prime. These two small triangles, this one and this one. Now, why are they equal? Why are they congruent? Well, because two angles, now, and this side in between is also congruent, because a, b is congruent to a, b prime, a prime, b prime, because it's difference between m, b and m, a. And this is a difference between m, b prime and m, a prime. So since a, b is equal to m, b minus m, a, which is equal to m, b prime minus m, a prime equals to b, to a prime, b prime. So these two small triangles are equal because of angle side angle, right? So triangle a, b, p is congruent to triangle a prime, b prime, p. Okay. So these triangles are congruent and that's why a, p is congruent to a prime, b. This is congruent to this. From this, we conclude that these triangles, this one, a, p, m, and this one, a, p, a prime, p, are also congruent. So triangle a, p, m is congruent to triangle a prime, p, m. Why? Because of three sides. This is common. These were congruent by definition, by the construction of these points, a and a prime. And these guys, we just proved that they're equal from these triangles. So we have three sides, which means these angles are equal. So this is a bisector. All right. Relatively easy theorem, finished. What's next? Given a straight line p, q and two pairs of points, symmetrical relative to this line. Okay. So we have line p, q and two points symmetrical, a and a prime. All right. And we have two other points also symmetrical, b and b prime. Okay. Prove that there exists a point m equidistant from all four points. Okay. We need to prove that there is a point m, which is equidistant from all four points on the line p, q. Well, actually it's quite easy. Because if you will take this segment and build a perpendicular bisector and call it m, the crossing of this perpendicular bisector to a, b. Now, let's think about this. These two distances will be the same. Why? Well, obviously, because these are two right triangles, let's call this point m. B and m and a and m are right triangles. Now, this is equal to this because it's bisector. So it's perpendicular bisector and this is the common leg. So we have two right triangles congruent to each other by two legs. Therefore, hypotenosis will be congruent as well. Now, considering these points are symmetrical to these, these two segments will also be the same. Why? This will be congruent to this because the symmetric, the symmetric means that they are on the same perpendicular and on the same distance. So this is straight, this is right angle and this is equal to this. And if you consider this triangle and this triangle, they have common leg. This leg is equal congruent to each other. That's why triangles are congruent and that's why this is also the same perpendicular. Now, in this case, it's also the same because B and B prime are symmetrical. That's why this triangle would be congruent to this. Exactly the same as the same logic as with this guy. So we have another hypotenuse. So this point is equidistant from all four. And how we have built it, we draw from A to B a sector and draw a perpendicular bisector of this segment. That's it. Given a straight line 2 points A, B on the same side, find a point which some length is minimal. Okay. This is also easy. So if you have line and two points A and B, how to find a point which is such a point M that A M plus M B is minimal among all the different M's which you can find on this line? Well, the answer is quite simple. You draw a symmetrical point A prime and connect with a straight line. So let's consider that A prime B is a straight line. Now, if you take any other point, then this particular distance would be equal to this particular distance. Let's call it M. Since A M is equal to A prime M, and A M is equal to A prime N, this is equal to this, and this is equal to this because A and A prime are symmetrical. So obviously, no matter where our point is, we have this congruence between A M and A prime M or A M and A prime M. So if this is a straight line from A prime to B, it will always be shorter than some of these two lines. So no matter where we put this M, if it's different from this M, we will have a triangle, and because of the triangle inequality, sum of A prime N plus NB A prime M plus NB would be greater than A prime straight line. So again, how to do the shortest distance, how to find the point with the shortest sum of distances to two different points, make a symmetrical point A prime and go with a straight line from A prime to B, wherever it crosses our line, that's the point with a minimum sum of two distances. Basically, there is a practical usage of this particular problem. Let's say you have two different towns, and you have a railroad. The question is, where should you put a station so that it will serve both towns and it will minimize the sum of two lengths. So people from one town and people from another town would have the smallest distance to travel altogether. Well, considering the population is the same, the point is exactly here. If population is different, then you might reconsider it and wait more towards the point which has a bigger population, but that is a completely different and much more difficult problem. Okay, let's move on. Given the q-tangle Pmq, Pmq, point A inside, find x on side mp, y on the mq, such that the perimeter triangle Axy is minimal. Well, it's actually the same problem, if you wish. Let's do it this way. What we do is we reflect the point A relative to this guy and reflect point A relative to another ray which makes the angle. So let's say this is A prime and this is A double prime. And then you connect with a straight line. Now, I claim that this is the real x and this is the real y. Now, if this is something like x prime and this is the y prime, these are my own points, I claim that the perimeter of a triangle Axy is smaller than Ax prime y prime. Why? Well, obviously because if you reverse this line, reflect this line Ax prime to here and Ay prime to here. So what will be? The perimeter of Axy is equal to a straight line from A prime to A double prime. But the perimeter of the triangle Ax prime y prime would be a combination of three segments, one, two, and three, which is not a straight line. And we know that the straight line is always the shortest distance between these two points. So that's the proof that you have to really reflect the original point relative to both sides of the angle and connect them with a straight line and this crossing will give you the triangle with the smallest perimeter. Well, that's it for this set of problems. Don't forget to look at Unisor.com, excellent site for all the exercises, problems, brain training, analysis, logic, and to the parents and supervisors can obviously control the process of education of their students by enrolling and controlling their progress by checking the exams, the scores on the exams, etc. I strongly recommend to use it. Thank you very much.