 Hi, I'm Zor. Welcome to Unisor Education. Today I would like to continue talking about circles. But apparently the further we move down the geometry road, the more objects we are familiar with. And circles actually are very interesting objects and there are many very nice theorems about circles. And as you remember, the main point is basically developing your creativity, your logic, your abilities to rigorously think about certain statements which you are making. And circles is just adding one more object into this language, into this universe of different objects which we are talking about, which we are using, just as a brain exercise if you wish. So a bunch of other small brain exercises or mini theorems actually which I am more officially calling. Most of them are about circles but obviously some other elements which we have already learned will be involved in this. So this is about circles, mini theorems, it's number two. I have more actually. Alright, so let's just go one by one. Measure of an angle formed by two quarts, okay, so we have two quarts which are intersecting inside a circle. So let's call it a b a k is equal to half the sum of measures of central angles. Okay, so we have these angles and we are interested in measuring this particular angle. So this is the center of a circle and the theorem states that any one of these angles, and they are, by the way, congruent to each other because they are vertical, so any one of those is equal to half the sum of these two central angles. So this is the center of a circle. So this particular arc and this particular arc, I think I better take it to this way. So this particular arc a b and this particular arc a prime b prime are supported by, are supporting these two vertical angles which are equal to each other. And these two arcs support these two vertical angles. So the statement of this theorem is that the measure of any angle which is formed by two intersecting quarts let's say this angle is actually half of the measure of corresponding central angles which are supported by the same arcs. So if you take this arc and its central angle, this arc and its central angle, then the half sum of these two central angles is equal to this. Alright, now how can we prove it? You probably should remember that if you have an inscribed angle, then we have proved that it's half the measure of a corresponding central angle which is supported by the same arc. So we will use this theorem in this particular case. So we know about inscribed angle. Now this is a more complicated picture of two quarts actually crossing each other, but the theorem is, you know, in a way it's very, very much in tune with this one. So in this case we have two different arcs which support the angle. So half the sum would be the measure. Now how can we do it? Alright, let's do it this way. Let me wipe out the central angles and switch to inscribed angles. It would be easier. And then we use the fact that the central is twice as big as inscribed. Alright, so let's just connect these two things, these two points. Let's call this point M where two quarts intersect each other. Now AB prime B is an inscribed angle. And what we can say about this particular angle which we are talking about, AMB. We know that angle AMB is exterior to a triangle AB prime M. See this is an exterior angle. And that's why it's equal to a sum of two angles not supplemental with it. So it's equal to sum of AB prime B plus B prime AMB prime angle B prime AM or A prime. Doesn't really matter. So sum of this angle and this angle is equal to this one which we are looking to be measured. But now let's just think about AB prime B is an inscribed angle which is supported by this arc AB. And B prime A A prime B prime A A prime is actually an inscribed angle which is supported by this arc. So that's why this angle is equal to one half of, let's call this point O, this is a center. So angle A OB, the central corresponding angle which is supported by the same arc. And this angle is equal to one half of a corresponding central angle which is supported by this arc which is B prime O, A prime. Or B prime O, B prime. Same thing. And that's exactly what needs to be proven. This particular angle which is formed by two chords intersecting each other and supported by these two arcs is equal to half the sum of the corresponding central angle supported by the same arcs. That's it. So this is just a tiny bit more complicated case than just inscribed angles. All I was using is the theorem that the exterior angle of a triangle is equal to sum of interior angles not supplemented with it. Now these are two chords which are intersecting inside a circle. And that's why you have half sum of the corresponding angles, central angles supported by the same arcs. Now what if these chords do not cross each other inside a circle and they are intersecting outside? So this is A, A prime. This is B prime. And this is the angle which we would like to measure. So right now, actually the theorem is exactly the same but instead of edging two central angles which are supported by the same arcs, you really have to subtract. And here is Y. So let's forget for a second about the center. Let's just connect these two points. And again, this is an exterior angle to A prime and B triangle which means it's equal to sum of this angle plus this angle. So the angle A A prime B, this one, is equal to a sum of angles A and B plus this one A prime B D prime. Now the angle we are looking for is this one. Just solve this particular equation for A and B. So you will have A and B equals to the difference between angle A A prime B minus angle A prime B B prime. Well, this actually concludes the whole proof because A A prime B, A A prime B is an inscribed angle which is supported by this arc which means, I will not even write it down, which means it's equal to half of the central angle which is supported by the same arc. Now the angle A prime B B prime, A prime B D prime is inscribed which is supported by this arc and again it's equal to half of the central angle which is supported by the same arc. So our angle which we are looking for A and B is equal to half the difference between the corresponding central angles which are supported by the same arcs which are cut out from the circle. In a way it's really very similar to the previous theorems. If chords are intersecting insides is half the sum of the central angles which are supported by the same arcs. If they are intersecting outside then it's half the difference between the corresponding central angle. So you might actually remember it this way. Alright, so these are two very much related theorems to each other. They're all based on the fact that the measure of an inscribed angle is always half of the corresponding central which is supported by the same arc. These are just small complications to that particular theorems. They're used much much rarer but again our purpose is not to provide you with any practical knowledge just to exercise the brain. Okay let's go further. Angle between a tangent and a chord originated in a point where okay here it is. You have a circle, you have a tangent and you have a chord which has an endpoint exactly where the tangent touches the circle. Now as you remember the radius which is drawn in the same point where the tangent touches the circle is perpendicular to this tangent. That's the main property of the tangent. Now what's necessary to prove is if you put letters a and b and o here okay. Alright, so this is the arc which corresponds to this particular chord. So the theorem states that this particular angle is measured as half of the central angle which is supported by the same arc related to the chord. So you have a chord and in the endpoint you have a tangent and you are measuring the angle between the chord and the tangent. Now this particular arc is related to the chord and the theorem states that this angle is half of this angle. Now how can that be proven? Well again let's just use the fact that the radius is perpendicular to the tangent which means that this particular angle which we are looking for well let's call it alpha. This alpha is equal to 90 degree minus beta. That's obvious because this is the perpendicular. Now a o b is isosceles triangle because these are two radiuses so this is also beta. Okay now what can we say after that? If I will draw a perpendicular to a chord let's call it c then angle a o c a o c it's the right triangle if this is beta so this one is 90 minus beta which is alpha right because alpha is 90 minus beta. So angle a o c is equal to alpha well obviously a o c is half of a o b because these are equal congruent right triangles which means this is also an angle alpha and that's basically concludes the proof that this particular angle is equal to half the central angle which is supported by the same. This is alpha and this is alpha so this is two alpha and that's why it's twice as big as this one and the proof. As you see these proofs are really very very easy as long as you understand just a small thing for instance just draw an extra line which connects the couple of points or draw a perpendicular to a chord in this particular case and everything actually immediately becomes you know quite obvious because the whole proof was in one line or two lines so that's really very easy but what's really interesting is you have to know what additional construction what line to draw what perpendicular to drop which two points to connect to really make the whole theorem quite obvious so that's the most important part of it. Next among all chords that contain certain point inside a circle the shortest is the one perpendicular to a radius that contains this point okay so you have a circle this is the center and this is a chosen point inside. Now what this particular theorem says is that this chord it's not a good drawing I'm sorry I think this would be better okay so this is the radius this is AE this is all let's say M so and this is the point P through which all the chords are actually drawn so if you take any other chord let's say A prime B prime which is not perpendicular to all P it would be longer than AP so the shortest chord which is which contains a particular point inside a circle is the one which is perpendicular to a radius which goes through this point in this case it's OM so P is a given point and to draw a shortest chord through this point you have to draw a radius and then the perpendicular chord to this radius okay now let's think about how to prove it okay I'll probably change slightly the letter so it will correspond to whatever I have on the on the web so this is a point M okay AB the chord perpendicular that's right it's perpendicular to OM and A prime B prime is another chord which is not perpendicular to to OM so A prime B prime is not perpendicular to OM okay now what we do is we drop a perpendicular from the center to this new chord okay let's call this an M prime so OM is perpendicular to AB OM prime is perpendicular to A prime B prime now we know that um the radius which is perpendicular to a chord divides it in half it's a perpendicular bisector so basically A prime M prime is congruent to M prime B prime as well as AM is congruent to MB OM prime is perpendicular to this chord or M is perpendicular to this chord okay fine that's done what's next now let's consider the triangle OM prime M it's a right triangle and OM is um a hypotenuse which means OM is greater than OM prime because OM prime is a characters and OM is a hypotenuse now so what kind of a picture we have here we have two chords one of them is closer to the center than another and as we have learned uh from one of the previous lectures one of the previous theorems actually I think it was mini theorem one the closer the chord to the center the longer it is so A prime B prime is closer to the center than OM because OM prime is is smaller than OM that's why A prime B prime therefore A prime B prime is greater than AB so you just have to remember that previous theorem that the the shorter the distance from a chord to a center the distance being perpendicular obviously the longer the chord is and what element of a circle has the shortest distance to the center well obviously a diameter so if you draw a diameter doesn't really matter whether it's from this point or not diameter has zero lengths to the center that's the shortest as it can be and therefore diameter is the longest that's the longest chord okay that that's end for this particular problem now what was interesting in this particular problem what was important to to to lead us to a conclusion well number one a small construction which we made just a perpendicular to A prime B prime and what's also important is always remember what was before because the whole building of mathematics is built from the ground up so you know certain things and you can use it on the next floor in this case we were using this theorem that the closer chord to the center the longer it is okay theorem number five given the circle was a center O and chord AB center O and chord AB okay on equal distance on both sides from the midpoint M of this chord so this is the midpoint and probably this is as we know a perpendicular because the perpendicular bisector is actually crossing midpoint of the chord and goes through the center okay so on equal lengths from both sides of M inside a circle we choose two points C and D on equal side on equal lengths let's say this and this C and D we choose two points okay at each of these points we draw perpendicular to chord AB from C we call it CE and from D we call it DF okay prove that segments CE and DF are congruent so these two are supposed to be congruent all right so how can we prove that that let me first do a simpler task let's say our chord is not just any chord but a diameter it would be a little bit easier in this case now from the diameter on the diameter we have two points on equal distance from the center by the way M and the center O are coinciding right now right we draw perpendicular now it's quite obvious that CE and DF are congruent because CM and MG are congruent by the premise of the theorem by the condition because we actually use the same distance on both sides of M and these are radiuses so we have two right triangles which have congruent calculus and hypotenuse and that's why they are congruent and that's why GE and DF and CER congruent to each other so that's easy now how can we reduce this to this well simply let's draw a diameter parallel to a chord so diameter is parallel to the chord now it's obvious that um CG let's call these K and L CG L K is rectangle Y because these are parallel to each other these are parallel to each other because they're perpendicular to the same line AB and obviously since they are perpendicular these are right angles so CGL K is rectangle which means that this piece is equal to this piece now this part is exactly like this because these two segments obviously are congruent to each other because again these are all parallel lines these are equal so these are equal so basically we have reduced our problem to two simpler ones first that these particular pieces of the segments are congruent to each other because these are opposite sides of a rectangle and these two pieces are congruent to each other because we could just prove it for that case uh second to go so some of these and some of these would be congruent to each other so what's the lesson to to learn from this particular task well the complicated task can be reduced to two simpler ones right so by drawing the the diameter parallel to a chord i have basically broken two big segments into different into two parts each and prove the theorem for these two parts separately and then just sum them up with some sum them together okay next two chords are perpendicular to diameter okay so you have diameter C C prime so we have chord AB no AA prime sorry AA prime and let me check that's exactly what this is is AC A prime now that's the other way around i'm sorry it's not this diameter they're supposed to be perpendicular not parallel my fault diameter is this C C prime right AC A prime B prime C prime and B now it's correct okay so two chords are perpendicular to the end this is the center consider a segment that can take connects midpoints of AB prime and A prime B this is a midpoint and this is the midpoint prove that it's also uh perpendicular to the same diameter okay that's quite interesting actually there are many ways to prove this particular theorem but here is what i suggest let's just consider this diameter as an axis of symmetry is the picture symmetrical well let's just think about it since a diameter perpendicular to a chord is a perpendicular bisector A and A prime are symmetrical relative to C C prime same thing DB prime are symmetrical now if these points are symmetrical and these points are symmetrical then the segments which connect these points are symmetrical as well therefore their midpoints are symmetrical as well and since they are symmetrical the line which connects them is perpendicular to the diameter diameter is actually a perpendicular bisector of this particular segment which connects the midpoints so i have a approach from a symmetry standpoint that's an interesting approach i guarantee that it can be actually solved differently using some congruent triangles and basically using the same principle of symmetry but behind the scene and not really mentioning this uh because what is the symmetry symmetry is perpendicular and equal in length right that's what A and A prime symmetrical means they are on the same perpendicular to C C prime and on equal distance now from this you can obviously derive for instance congruence of this triangle and this triangle to right triangles and stuff like this so basically it's the same thing i have just shortened the way to explain all these problems by using a little bit more complex concept of symmetry but anyway it's quite quite a legitimate way to to to prove the theorem as long as you know that the corresponding points all are symmetrical to each other so what have i used if one point is symmetrical to another point and the third point symmetrical to the fourth point then the segment which is connecting which is connecting these points are symmetrical as well which is kind of a trivial symmetrical problem all right we can argue about rigorousness of this particular proof but it's kind of obvious that it can be made rigorous if we will just open up the concept of symmetry and prove each particular component of that concept the perpendicularity and the equal distance etc all right next given a circle with center o and chord ab continue ab by bc equal to radius so this is the radius this is the radius and we continue to bc so the bc is equal to radius well whenever you see something like this the isosceles triangle you know it just comes to my mind immediately as soon as you have these two segments equal with the common vertex most likely something will be necessary to consider like o bc as an isosceles triangle but anyway let's see whether i'm right or wrong i know it's just my intuition draw a second from c through a center of a circle c through a center a second okay so we are crossing the whole thing all right i was almost right this second intersects a circle at point g which is further from c and e which is closer to c prove that angle a od a od which is a central angle is three times greater than acg than acb all right how can we prove that so this is supposed to be three times greater than this well as i was saying o bc is isosceles triangle and that's why this is these two angles are equal to each other so that's obvious right okay what's next well recall that if you have two quads which are intersecting outside of a circle then the measure of this angle is half the difference between central angles which are supporting these two arcs one arc is ag another arc is ve so basically let's call this angle alpha so alpha can be measured as one half of a central angle which is angle a od minus central angle b oe but it's also alpha right i mean half of alpha i'm sorry i see something is not exactly right half of alpha so again this angle is measured as half the difference between two central angles one central angle is our a od and another central angle is same alpha because they are in in the isosceles triangle and two angles at base well let's just look at this obviously if you multiply it by two you will have two alpha equals to a od minus alpha from which three alpha is equal to a od and that's exactly what's necessary to prove simple isn't it so all you have to know is the theorem which was learned before about this angle between two quads intersecting outside of a circle measured as half of the central angle supported by the same arcs okay let's end of it and the last problem i have in this lecture given a circle with the center o and point m outside of this circle okay circle with the center o and point m outside of this circle all right consider all seconds that contain point m and intersect circle in two points okay this is one in the second one closer to m and one further so let's say this is p and q so the p is closer intersection point with the circle to m and q is further all right prove that a second that contains a center which is this second let's call this p prime q prime and these will be p and q so it's necessary to prove that among all the different seconds which are originated at m the second which goes through a circle has the closest point has the closer point of intersection closest among all other and the further point of intersection furthest among all other so m q is greater than m q prime but m p is less than m p prime so that's what's necessary to prove m q is greater than m q prime and m p is less than m p prime so that's what's necessary to prove now how can we prove it well um remember the triangle inequality if you have three points then the sum of distances along the angular line is greater than the straight line which connects these two points this is called triangle um inequality we will use in this particular case how very simple let's just connect these two points think of the line at the segment m q since all q and all q prime are two radiuses m q is equal to m o plus and instead of this radius i will use this all q prime this plus this now obviously since this is angular connection between m and q prime and m q prime is a straight line this would be greater than m q prime because of a triangle inequality so that's how we prove the first one now the second one is very similar o p prime m o p prime plus p prime m is greater than o m which can be represented as o p plus p m but all p prime and o p are congruent to each other so we can take out from this inequality and we will have that p prime m is greater than p m p prime m is greater than p m and that's the second one which is necessary to prove that's it okay that concludes my series of mini theorems number two number two there will be others dedicated to circles so as i'm saying there will be other mini theorems there will be other problems which are also related to circles and circles are really quite an interesting culture there are many very interesting problems to train your curiosity which i can provide everything will be obviously on the website on the unisor.com you are encouraged to spend as much time as you want on this website and obviously very important for parents who want their children to well succeed in developing of their creativity and other great qualities of the character please don't hesitate enroll your children into different programs let them take exam control them control their progress control their educational experience so everybody's welcome to unisor.com that concludes my lecture for today thank you very much