 Hi, I'm Zor, welcome to Unisor Education. I continue talking about circles, theorems about circles and I would like to present to you the lecture where a little bit more difficult one or two theorems will be proven. So out of whatever number of theorems I have, five, I think one or two will be a little bit more difficult. That's why I decided to call this lecture theorems, this is number two of this type, rather than mini theorems. Now it's very important for you to spend some time before you actually listen to this lecture. It's very important for you to be prepared for whatever is there. Simple problems are simple, that's okay, but more difficult problems, they require certain concentration. Try to prove all the theorems presented in any lecture actually yourself first before you listen to the lecture. It will be much more interesting for you to listen to whatever I'm saying, you will understand it better. And again, the whole purpose of this process of education which I'm trying to impose on you is to develop your creativity, your own analytical thinking. That's why you don't really have to maybe listen and remember whatever I'm saying. That's not very interesting and you won't actually use it anyway in your future. What's important is to prove it yourself first and then listen to whatever I'm saying. Alright, anyway, without further ado let me just start. In a circle with a center O given the diameter AB and quart AM, sharing an end point with diameter. Okay, got it. So there is a diameter AB and quart AM. Okay, BAM is 30 degrees, that's given. At end point of the quart AM in a circle draw a tangent, which intersects the diameter at point prove that triangle AMM is a socialist. Alright, well let's just think about it. Let's connect the tangency point AM with a center. As we know this is the right angle. Now, angle MAB is inscribed, supported by this particular arc, MB. Angle MOB is central, which is supported by the same arc, which means it's twice as big, which means this is 60 degrees. Now, if this is 60 and this is 90, this remains to be 30 because OMN is right triangle. 60 degree, 90 degree, 100, supposed to be 180 in sum, so this is 30. Now, since this is 30, AMM triangle has two angles congruent to each other, which means the corresponding sides are also congruent and triangle is a socialist. Simple. Given a circle with a center at point O, point P outside, and two tangents with points A and B. Alright, so we have a circle, we have two tangents, one and two, from point P. Now, these are obviously perpendicular, A and B. Alright, BC is a diameter, so this is diameter. BC is a diameter. Alright, from the segments AC and OP are parallel. AC and OP are parallel. Okay, actually it's very similar to the previous theorem. Look at triangles AOP and BOP. They are congruent, obviously. They are both right triangles. They have shared hypotenuse and these two radiuses are calculating. So, triangles are congruent, which means these two angles are congruent too. So, each of them is half of the central AOP angle. So, central angle OB, which is supported by this arc, is twice as big as each one of these two graphs. This is by sector, if you wish. Now, at the same time, angle ACB is also supported by the same arc as AOP, but this is inscribed angle, so it's equal to half of it, half of the central. So, POB is equal to half of the central angle and ACB is also equal to half of the central angle, so they are congruent to each other. Now, consider AC and OP two lines. CO is CB, whatever, is transversal, so these two angles, this and this, are corresponding angles and since they are congruent, lines are parallel. Easy. Now, this is an example and a previous of theorems which you definitely have to be able to do yourself without much thinking, I mean that should go really very quickly. Next. Two circles are intersecting at points A and B. From point A draw two diameters. AM and AM. Okay, prove that segment MN is passing point B. What I'm going to prove is that if I connect M to B and then B to N, then angle MBN is equal to 180 degrees. That's what I'm going to prove. That's why these two segments make up one straight line. And since there is only one line which connects to points M and N, this is the line which is actually going through B. Now, how can I prove that the angle MBN is equal to 180 degrees? Actually, it's very simple. If I connect A to B, angle ABM is right angle because it's supported by half a circle. Same thing with angle ABN is also right angle because it's supported by this half a circle. So 90 plus 90 is 180 degrees, so ABN is an angle which is 180 degrees which means ABN is straight line. Incidentally, we just proved that AB, the line which connects two points of intersection, is perpendicular to this segment MN. Next, consider triangle ABC inscribed into a circle. Okay, so let's start with a circle and triangle. ABC, okay, point P between B and C. Drop perpendicular. Alright, let me use a different color. So this is a perpendicular to one and this is perpendicular to another. Point K is AB and point M is on AC. Okay, now prove that angle B, P, C, KPM are concurrent. Okay, I think my drawing is not really too good. So let me just change it slightly. Because on my drawing they are definitely... Okay, let me try to do more precise but for this reason I probably choose another point. It would be better. Let's say here. So if I will draw a perpendicular to this point K and to this. So this is K and this is M. Once more and I need one more color. Now the theorem says that brown angle is supposed to be equal to a purple angle. Now it's quite visible. So KPM should be equal to B, P, C in size and measure. Alright, how can we prove it? Actually the proof is much faster than drawing. Look at the quadrangle A, B, P, C. It's inscribed into a circle and as you know inscribed quadrangle has a property that opposite angles in sum equal to 180 degree. Why? Because these are two inscribed angles A, A, B, AC and B, P, C to inscribed angles and some of them is supported by this R plus this R which is the whole circle and they are inscribed so it's half of a 360 which means 180. Alright, so angle B, P, C which I use letter alpha and angle KPM I will use the letter B. So as you see angle alpha plus gamma is equal to 180 degrees right? Because A, B, P, C is inscribed. Now if you consider AKPM it's also inscribable quadrangle because AKP is 90 degrees, AMP is also 90 degrees because these are two perpendiculars so sum of opposite angles is 180 therefore sum of other opposite angles which is gamma and beta or beta and gamma also 180 degrees because sum of all angles in a convex quadrangle as you know is equal to 360 degrees. So well that's basically it. I mean this is the end of the proof obviously alpha is equal to beta and that's what's necessary to prove this particular problem. That's it. Now the next problem is slightly more difficult and actually this is the beginning of the proof to the next problem. So what's the next problem? That's quite interesting. I will use the same drawing but I will also add another perpendicular. The perpendicular to the third side so we have perpendicular to AB, perpendicular to AC and I will continue with a perpendicular to side BC. So the next theorem states that these three points K, L and M are aligned from the same line, one straight line, which is called a Simpson line. Actually there was a person called Simpson, actually there are some other people who proved this particular theorem and this is one of the theorems which doesn't seem to be obvious, one of those. Three perpendiculars on three sides of any triangle from any point on the circle which is not a vertex. All these three bases of three perpendiculars are supposed to lie on the same line. Not very obvious but it's not very difficult to prove using whatever we have just proven before. Because let's think about it this way. How can you prove that the K, L and M are lying on the same line? Well, let's just connect K and L and L and M. If I will prove that angles B, L, K and M, L, C are congruent, well, B, C is a line, so if these two angles are congruent then K, L, M would be also 180 degrees. That's easy to prove. So we have to prove that these two angles are congruent. Well, let's think about it this way. Angle B, L, K. So we have to prove that angle B, L, K is congruent to angle M, L, C. That we have to prove. How can we do this? Okay, think about it this way. Consider B, P as a hypotenuse of two triangles, B, L, P and B, K, P. These are two right triangles of this because these are perpendiculars. So B, P is a common hypotenuse. So if we will use hypotenuse as a diameter, points B, L, P and K will be in the same circle. And look at this angle B, L, K, which we have to really prove that it's equal to this one. It's supported by the same arc B, K as angle B, P, K, right? Again, B, L, K and B, P, K in this circle are supported by the same arc. That's why they are congruent. So we can say that this angle is congruent to this angle. So far we have proven that this is B, P, K. Now, very similarly, if you use B, P, C as a diameter, P, M, C is the right triangle with this hypotenuse and P, L, C is the right triangle with this hypotenuse, which means P, L, M and C are lying in the same circle. And angles M, L, C is supported by the same arc as M, P, C. So this angle is the same as this angle. We have already proven that alpha and beta are the same. So this is also alpha. The purple and the brown, they're both alpha, they're the same. But now let's think about this way. What is angle M, P, C? This one, angle M, P, C. You can say that this is C minus B, P, M. So again, M, P, C is equal to M, P, C is equal to B, P, C minus B, P, M, right? So this angle is this angle minus this angle. M, P, C is equal to B, P, C minus B, P, M. Now, very similarly, this angle B, P, K is equal to M, P, K minus same B, P, M. Again, B, P, K is equal to M, P, K minus B, P, M. M, P, D, whatever. This M, P, K, M, P, K is this brown alpha. Now, angle B, P, C is purple alpha. So as you see from alpha, we subtract the same angle in both cases, which means this angle is equal to this angle, B, P, K minus M, P, C. M, P, K is this and M, P, C is this. That's why we have proven the equality of these two measures. Well, that's it. That's the end of the proof. It's a little bit more complex than the others. And again, that's why I called this particular lecture Theorem. I wanted to come up with this particular theorem. To tell you the truth, I was actually thinking for a day or so before I have proven this myself. I didn't read this proof. I mean, I'm sure there are some other better proofs or whatever. But anyway, this is one of the proofs, and I wanted to prove it myself. And that's exactly what I would like you to do with all these theorems, whatever I'm presenting. It's very, very important for you to spend time and to start thinking about all these theorems and how to prove them yourself first. If you listen to my lecture, you will just confirm that you are right or you might actually come up with a better solution than I did. All right, good luck. And don't forget Unisor.com is the website where all these is presented. And especially for teachers and supervisors and parents who would like to control the educational process of their children, this site enables you to basically enroll your students into particular courses or programs. You can ask them to go through exams. You can check the results of the exams. And you are in control basically by saying, okay, this has got good results. And consider this particular course completed or do it again, go through lectures again through exercises and then do exam again. So it's very much for this type of, well, home schooling or group schooling, if you wish. And whatever the material is presented is significantly different than traditionally presented in many schools because, again, I'm putting a very strong emphasis on problems and solving the problems because I think that's the only thing which is actually you will need in your future ability to solve problems. All these facts about mathematics, about theorems, whatever else, they will not be actually used by the real life, very, very rarely. But ability to analytically thinking, that's exactly what I would like to develop. This is the brain exercise. That's the most important part of it. All right, good luck, thank you very much.