 Hi, I'm Zor. Welcome to Unisor Education. This lecture will be devoted to many theorems about circles. Most of them are very simple and trivial. One may be a little bit more difficult than others, but they're all interesting and well, let's just go through them. Okay. The theorem number one is about inscribed angles. So if you have a circle and you have an inscribed angle and this is a chord which supports this angle, now this is another angle which is supported by the same chord. Now and this is a central angle which is also supported by the same chord. So the theorem is that any inscribed angle is measured half the size of the central angle which is supported by the same chord. So this is twice as big as this. And consequently you can just follow from this logic another inscribed angle which is supported by the same chord since it's also is measured half of the central angle. It's supposed to measure exactly the same as the original inscribed angle. So all inscribed angles, all inscribed angles supported by the same chord are all congruent to each other and measured as half of the corresponding central angle. Okay, so this is the theorem. How can we prove it? First we will prove it in a very trivial case. So let's consider a chord, a diameter, and inscribed angle which is formed by a diameter and a chord. Not just two any chords but let's just consider that one of these chords is a diameter. Now why is it easier and why is it like trivial? Well because in this case the central angle which is supported by the same chord and inscribed angle they have come inside. Right? Since diameter is going through the center and the center is basically a vertex of a central angle we have this particular picture. So this is a central angle. ABC is inscribed angle or AOC is a central angle which is supported by the same chord AC. Now why is AOC twice as big as ABC? Well for a very simple reason AOC is exterior angle for triangle OBC which means this exterior angle is measured as a sum of two other angles not supplemental with it. But OB and OC are two radiuses which means that these angles are exactly the same and since AOC is equal to this plus this and they are equal to each other it means it's twice as big as one angle ABC. So in case one of the chords is a diameter then inscribed angle is half of the corresponding central angle which is supported by the same arc. Now what if our angle is not inscribed angle is not such as this one so it's not the diameter as one of the two chords. Let's say it's something like this A' so our angle is A'BC not ABC but we still can draw a diameter from B through O to get to the point A and consider two separate cases. This angle is half of this. Now this angle is half of this for the same reason absolutely because again this is exterior angle and this is equal to this because it's still a socialist triangle all our radiuses which means sum of these is equal to half of sum of these. So it's exactly the same result that the inscribed angle is equal to half of the central. All right there is one more case just for completeness but it's actually simple. What if our inscribed angle is such that it's not on both sides of the diameter let's say both chords are on one side of the diameter let's consider this way. If this is a diameter for instance our angle is this. How to prove that this angle inscribed angle supported by this arc is half of this central angle. Actually in the very very similar case in this case we have divided our angle A'BC as we represented as sum of these two angles. In this case you will represent it as a difference instead of a sum you will represent it as a difference between so this angle is equal to this minus this and both angles this angle and this angle are measured as half of correspondingly this and this. So the difference between them is measured as half of the difference between these two which is exactly the central angle supported by the same arc. So in any case wherever one of the two chords is located on the same side of a diameter as another chord or on different sides of the diameter no matter where it is the angle is represented either as a sum or as a difference of correspondingly inscribed angles with one of the sides coinciding with the diameter. And the theorem is true for all cases. Now as a consequence from this you obviously understand that any other angle which is supported by the same arc or the same chord is still equal to exactly the same half of the central angle which means that let me just take this away and this away so we have an inscribed angle and the central angle which is supported by the same arc and then if you will take any other angle which is supported by the same arc inscribed angle it's also exactly equal to half of the central which means these are equal to each other so all inscribed angles supported by the same arc the same chord are equal to each other and equal to half of the same central angle which is supported by the same arc and as another consequence let's consider all angles which are supported by half a circle so our central angle which is supported by the same half a central half a circle is 180 degrees right which makes this angle equal to 90 degrees half of this right since inscribed angle is always half of the central angle supported by the same arc if this is 180 this is 90 so any angle which is supported by diameter by half a circle is equal to 90 degrees so all these angles are 90 degrees because they're all half of 180 degrees so they're all congruent among themselves and all equal to 90 degrees all angles supported by them okay so this is my first mini theory it's actually mini it's a little longer but it's it's really proof itself it's really trivial okay another even simpler if you have a chord and you have a diameter which is perpendicular to this chord then this diameter bisects the chord now how can that be proven well obviously when you want to prove congruency of certain things you have to include them into some kind of triangles right so obviously we will use this chord AB this is a diameter and we know it's perpendicular to the chord now we know that AO and bo are radiuses which means they are they are congruent to each other which makes a triangle AOB isosceles and we know that in an isosceles triangle altitude to the third side is also immediate as well so if this is the perpendicular it cuts it in half inverse theory is also true if it cuts if diameter cuts in half the chord then it's a perpendicular to this chord exactly from the same property of the socialist triangles when the medium is an altitude and the bisector of the angle as well so these are two congruent animals as well okay next next is a symmetry well you feel that diameter is kind of an axis of symmetry right for for a circle let's prove even a stronger theory let's draw two parallel chords and diameter perpendicular to these two chords then the statement of this theory is that these arcs are symmetrical to each other what does it mean actually well it means that if you will take any point here draw a perpendicular to the axis of symmetry which is our diameter and continue with the same lengths you will hit the point on another arc and vice versa now why is it true well because precisely because of the previous theory because whenever you draw a line which is perpendicular to a diameter you start it on a point of a circle you end it on a point of a circle you basically make up a chord which is perpendicular to diameter by construction and because of the previous theory divide it in half which means if you prolong it on the same distance you get exactly that point so these two points are symmetrical and same for any pair of points in between these two initial parallel chords so every point here has symmetrical point here so this is a theory if you have two chords then the arcs corresponding arcs which are cut from from the circle between these two chords are symmetrical to each other which actually leads to another a simple basically statement that if you don't have any chords then the left part of a circle is symmetrical to the right part of the circle i mean i conditionally would call it left or right it can be up or down or whatever so in any case if you have a diameter any diameter it would be an axis of symmetry for for a circle for exactly the same reason i just mentioned to you so if you don't have any initial chords you just have a diameter then again you start from any point you draw a perpendicular you continue by the same size and you obviously hit a circle somewhere and you know that since this is a perpendicular to a chord these two supposed to be equal so that's why these points are symmetrical so any diameter is an axis of symmetry for a circle this diameter this diameter this diameter any diameter is an axis of symmetry circle is very symmetrical another simple theory if you have two congruent chords then they are positioned on the same distance from a center so distance from a center is obviously along the perpendicular now how can we prove that these two perpendicular are equal in lengths if the chords are congruent very simple again triangles all these are reduces and if this is congruent to this you have two triangles congruent to each other by three sides and that's why their altitudes are congruent as well now a little bit more difficult theory is it's actually growing from this one okay if chords are equal in lengths then they are equal in the distance from a center what if one chord is greater than another longer than another well you obviously feel that the longer the chord the closer is supposed to be to the center the longest chord which is a diameter is on a zero distance from a center it passes through center right and the shorter chords are on a further distance from a center okay we would like to prove now this is a little bit more involved and before addressing this particular theorem i would like actually to recall a couple of other properties of triangles okay one property is this let's say you have two triangles one and two with these congruent sides now it actually has been addressed in one of the lectures before but i will repeat it right now that if this angle is greater than this between the corresponding with congruent sides then this side opposite to this bigger angle is longer than the side which corresponds to a smaller angle how can we prove it well let's just take this triangle and super super position it onto this triangle because these are congruent to each other this point will go to this and this point will go to this and this point will go to somewhere here so if this is abc and this is def then dgf is congruent to abc by construction basically so this angle is the same as this one even if it doesn't look like that in my drawing all right so how can i prove that ef is smaller is longer than gf okay here is how look at the angle e egg by the way this is exactly the same size ab is equal to dg okay now angle e dg i would like to draw a bisector of this angle okay and this is point which i call let's say k now ek and kg are congruent to each other they are equal in lengths why well because the triangle d ek dgk what these two triangles have in common aside then these two are congruent eg and gd and the angle is it's a bisector which means these two angles are congruent to each other as well so d ek triangle and dkg are congruent which means this is congruent to this now we will use a triangle inequality f k g f k plus k g is greater than f g correct triangle inequality f k plus k g greater than f g but since k g and k e are the same and what is f k plus k e f k plus k e that's ef so that's our side and this is exactly what needs to be proven that ef is longer than gf so as long as this angle is bigger this side is bigger provided that two other sides are correspondingly congruent to the other two sides of the other triangle so again if you have a situation of two triangles with two pairs of congruent sides then depending on the angle between them the size of the third side will be bigger against the bigger angle okay that's one of the properties which we have already discussed before in one of the lectures about triangles i just wanted to remind it to you and it's an interesting little proof anyway so it doesn't i don't mind actually to repeat it it's an interesting proof now i would like to prove another theorem mini theorem which i will use later on and this is about right triangles now if you have a bunch of right triangles with common hypotenuse so these are all right angles so they have common hypotenuse now my statement is that the bigger this angle is the bigger the leg against this across this this angle this angle is bigger than this so this side is bigger than this now how can i prove that well actually i will use something which i have already explained before namely that all these right angles the vertices of all these right angles are lying on a circle remember all angles inscribed into a circle supported by diameter this is a diameter so i'm using my hypotenuse as a diameter so all these triangles are right triangles because they are supported by diameter so the locus of all the vertices of the right angle of all the triangles with the same hypotenuse are making a circle now i will use this to prove that against bigger angle this bigger than this one lies bigger leg how can i prove it very easy since any inscribed angle is half of the corresponding central angle i know that this if this inscribed angle let's use letters now inscribed angle CAB is greater than GAB now what's the corresponding central angle for CAB which is supported by this arc by this chord the corresponding central angle is COB right now what's the corresponding angle for GAB well it's supported by this arc so the angle would be this now in the same way as against in the same way as this angle CAB is greater than this angle GAB their central corresponding central angles which are each one of them is equal to double inscribed angle also will be one will be greater than another so COB will be greater than DOB so what i actually have accomplished i have reduced my more i would say difficult task to prove that the leg against bigger acute angle is bigger than the leg across the smaller angle i have actually reduced it to another problem related to circles if i have a circle and i have a central angle then the bigger central angle the bigger the chord like in this particular case or i should actually probably draw it a little bit more like i have in this drawing so i have a circle i have one central angle which is BD and i have another central angle C so this angle is bigger than this one and i would like to prove that this is bigger than this which is kind of obvious right okay so let me take this away and i can draw this again too and let's discuss this particular topic so you have two central angles so angle AOB and angle AOC so um now why is the angle the bigger angle corresponds to a bigger chord that's an interesting actually topic i kind of assumed that this is an obvious thing but right now i'm just thinking that maybe it's not that obvious let me think about this a little and how can we actually prove that this is the case draw it slightly differently okay just bear with me and let's think about how to prove that this particular chord chord corresponding which corresponds to a smaller angle central angle is smaller exactly um the one which the theory in which i have just proven before remember if you have two triangles which have two pairs of congruent sides then depending on the size of this angle the bigger angle corresponds to a bigger opposite side well here we have exactly the same thing we have triangle AOB and AOC AO in both cases is the same and OB is also the same so in this case we have even more complicated case when the sides are different in this case all four sides are the same this and this and this and this and that's why by this theory AB is shorter than AC because the angle AOB is less than angle AOC okay so we have proven two different things so this is a one thing before and the second thing i have just proven that if you have a right triangle a bunch of right triangles basically then the bigger angle corresponds to a bigger opposite leg now we can actually approach this problem which i wanted to do in the very beginning which is the bigger chord is supposed to be closer to a center now how can we prove this so we have two chords one is bigger than another now i would like to prove that this distance is less than this distance all right so we will do these triangles as usually one chord is AB another is BC so AB is greater okay oh by the way i did not really mention before with these right triangles the theory is true from both sides bigger angle corresponds to bigger side the bigger side corresponds to bigger angle and it's very easy to prove because if you consider the opposite what if the angle is not really bigger against the bigger side then it's smaller or equal if it's equal then triangles are supposed to be equal but if it's smaller then the corresponding side supposed to be smaller so the inverse theory is very easily easily proven so in this particular case we have a very similar situation we have let's call it m and n we have triangle O and B and let's consider and compare it with a triangle m O B you see they have they are both right triangles right because these are perpendicular and they're both sharing the same hypogenesis O B which means that if this chord is greater than this half of this chord is greater than half which means that this angle is greater than this angle and if that's true then the corresponding the other angle of this right triangle O and B if this angle is greater than m O B then this angle is smaller than this angle because m B O plus m O B is 90 degrees right because there are two acute angles of a triangle so if this angle is greater than this then this angle is smaller than this because some of these should be 19 some of these should be 19 and now we are using this theorem again that if you have two right triangles with the same k-partinals against smaller angle you have a smaller casualties that's why this is smaller than this so the bigger chord corresponds to a smaller distance to a center and that concludes my last mini theorem which I wanted to present today as usually I address you to unizord.com as the source of many interesting facts about mathematics especially parents they can control the educational process of their students and well welcome to unizord.com and the mathematics are basically at your fingertips thank you