 So, we say that 1A and U is 1 by 12 of carbons that are on mass. Why 1 by 12? Because you do not know that the mass of electron is 1 by 12 of carbons. So, we say that 1 by 12 is 1 by 12 of carbons. So, we say that 1 by 12 is 1 by 12 of carbons. Mass of proton, mass of proton is roughly equal to mass of electron. Which is 1.67 into 10 is equal to minus 27. Now, tell me how many times 1 electron of proton is heavier than 1 by 12? So, when you talk about the mass of electron, you know that it is on mass. Yes, marginal mass of proton and mass of electron. So, that is why, you know what is mass number? Number of proton. What is mass number? It is mass of proton plus mass of electron. Do you consider electron and mass? Why? Because when you talk about mass, electron is heavier. So, C12 has proton 6 and electron 6. Should I start guys? There are only 4 people. I thought there would be all 12 people. That is why I started class in YouTube. Let me wait for few minutes. Who has already joined is Sanjana Ritu Tanay Shraddha Vaksha. Anchita has also joined. Anchita has only one proton inside. So, hydrogen, mass will be approximately 1 in 1. Similarly, chlorine, chlorine, sodium, two like that. So, what does it look like? It looks like hydrogen is 0.99997. Because even though the proton and the mass, they are like electrons. So, if you have an electron, you never put a proton. Ruti has also joined. Okay, so I am starting with circles. Anyone please let me know as far as I remember I have done circles a little bit in the class. So, do you want me to start in the beginning or do you want me to do it? From where I have left and second thing how much more is left out. So, just go through your syllabus and just check what is left out. Because your first exam is on 18th February. So, I would like to finish it by from beginning. Just let me know which topics are left out. Okay, I am starting from beginning. Don't worry, I am starting from beginning. So, you will see the charted off view of the element. And you can point out there. So, here is an X class. Ruti, only circles is left out. This is silver 18 plus C. C, when is U plus N U plus. Can be U plus C, U plus N U plus N U plus N U plus. Okay, so here is 3 plus U plus, and then C can be U plus and then U minus. Okay, if you remember, there is. any topic which topic Sanjana and Max want I'll finish it off probability okay I'll finish it off probability what I want you is that Sanjana what is IZ okay fine so let's do one thing statistics and probability okay let's do one thing where I meet you next week I love finish of circles and then I'll take an extra class perhaps next week on Sunday or something so that day I did probability and statistics what I want now is all of you get your RS Agrawal with you because I'll be directly referring to the theorems which I'm one suppose this is my circle so this is known as center and center is denoted by what any point on the circumference what is circumference so the perimeter of the circle is known as circumference so complete perimeter this circular arrangement is known as your circumference any point distance between any point on the circumference and the center is known as radius R and it is noted by a small letter R what is a chord chord is distance between any two point on the circumference if you draw a line between any two points on the circumference that would be chord of the circle so what is the biggest chord of the circle biggest chord of the circle would be diameter because it will be passing through the center and joining the line on the other side so the biggest chord is diameter not diameter is two times of radius so what is a second so second is extension of a car so any line which crosses or which intersects two points on a circle is known as second of the circle so it's an extension of the car with just that I have extended this car suppose the car is maybe and I extend it to CD and this CD intersect the intercept the circumference at points and so CD becomes second now what is a tangent so tangent is something like any line which touches the circumference of the circle at only one point so a line that intersects the circle exactly at one point is known as something like this is known as tangent so this line P and and this is Q is known as PQ is known as tangent of the circle I want all of you to be with your RS at a while because I'll be taking theorems of RS at a while directly now next next thing is interior of the circle so any reason which is inside of the circles so this this reason will be known as interior of the circle and whatever is outside this circle is known as exterior of the circle so the way to define interior of the circle is any points suppose you take the center O and you take point P if distance between any point P and center O is less than R that would be interior of the circle if you take any point Q such that distance OQ is greater than R that would be exterior of the circle so that's how it is now what is circum centric circles circum centric circles are those circles which have same center and different radius so try to understand it here this is the point this is my circum center this and I have different radiuses so this is R1 and this is R2 so these two circles suppose this is circle C1 and C2 so I'm writing circles C1 which has radius R1 and circles C2 which has radius R2 would be known as circum centers because circum center is by property those circles which have different radiuses but same center so wherever I am talking about circum center you should be having these concepts in mind next topic is arc of a circle so arc of a circle is a portion of circumference so a portion of circumference is known as arc of the circle so suppose this is a this is B and let's take this point as P so this A P B which is a part of complete circumference is known as arc of the circle and arc of the circle is denoted by this arc sign over the alphabet so this kind of sign above any particular alphabet or something like that would be known as arc of the circle now any arc of the circle would be making some angle with the center so see this arc A B if I if I ignore P for the sometime and I write arc A B arc A B is making angle A O B at the center and suppose this this I assume to be theta degrees so this theta is known as or angle A O B is known as degree measure of an arc so wherever I write like this A B and I write angle A O B is equal to theta you should understand that A B is an arc where this arc A B is making angle theta on the center and if I use the terminology degree measure of an arc you should understand that arc is making that much angle on the center now what do you mean by congruent arc so congruent arc makes those arcs will extend the same angle of the center so suppose I have an arc here in the same circle I am drawing and this is P Q and this P Q is also sustaining angle theta on the center so arc A B would be congruent to arc P Q there sorry arc P Q so as soon as I use word congruent with the arc you should understand there is only one condition for arcs to be congruent there has to be same angle substituted by those arcs on the center so that becomes congruent arc now semi circle we all know half of the circle is known as semi circle or when the arc substance 180 degree on the center that arc is semi circle so you see here if I draw this circle for you and if this is the arc I am considering so this arc A B if I take arc A B I mean this complete thing and if I look at this angle substituted on the center this would be 180 degrees arc A would be angle is 180 degrees arc A B would be known as semi-circle okay now what is minor arc and major arc so try to understand minor arc means angle of the arc substantive or degree of arc is less than 180 degree and if degree of arc is 180 degrees so let me write major arc here so if degree of arc is greater than 180 degree that would be known as major arc so let me draw a circle over here this kind of suppose this is the arc from here to here and this is A B so and this is this is the point so this is this this theta is less than 180 degree this would be minor arc and you look at the other side of it this side of A B is this is sustaining this angle which is like more than 180 degree so this is greater than 180 degree so that would be major arc wherever you are not understanding please raise doubt keep on listening is so now let me move to society coordinator later now let me move to theorem so first theorem is theorem one is congruent circles and what do I mean by congruent circles so if so you go to RS Nrwal I'm just reading it for you the congruent circle means if two circles have same radius then those circles would be known as congruent circles so here it is there is a circle here this is I'm assuming it to be more circle C1 and radius is R and suppose I'm assuming this to be S center to be S and this is circle C2 and such that this radius is also R so circle C1 is congruent with circle C2 now try to understand I told that arcs are also congruent and I told that two arcs are congruent when the angle theta is same for those arcs you need to understand difference between two different arcs and two different circles for two different arcs I'm assuming arcs to be in the same circle for two different circles as the language itself suggests I'm drawing two different circles so don't get confused that how radius is becoming equal and how theta is becoming equal if theta is equal for arcs it is between the same circle congruent circle means I have two different circles all together so this is how it is now let me move second theorem so all of you please write till now whatever I have explained you have understood everything I don't know what you are doing at home all of you please respond have you understood everything now let me go to theorem and theorem 2 tells me that so now the theorems that I will be talking would be about the chords and its properties first theorem is which is theorem 2 so I am writing p2 equal chord 10 equal angle at the center so suppose this is chord a b and suppose this is chord cd so suppose this is angle theta 1 and this is angle theta 2 I have to prove that I I tell you that a b is equal to cd this has been given and I have to prove that theta 1 is equal to theta 2 now what I do is that try to understand I take two triangles triangle a o b and triangle c o b I know o a is equal to od why because they are radius o b is equal to oc why because they are radius and a b is equal to cd which has been given to me it means that I have side side side congruency now by cpcd corresponding parts of congruent triangles are equal angle opposite to equal side so opposite to a b theta 1 is there so I am writing here opposite to a b and opposite to cd theta 2 is there so I am writing here so theta 1 would be theta 2 by cpcd so that is how you have to prove it now what you have to do I am writing Anirudh why are you late I am just going to ask that Anirudh is not there I am always looking for you you can't bunk my classes and who is not there Disha is not there Shreya is not there Aditi is not there if they are there just give your names because I am not theorem 3 you have to prove it that if angle angles subtended by two cards at the center of the circle is same then the cards would be equal just prove it I am giving you one minute of time it should not take more than one time so it is important for us to understand beyond the gas equation as well because at times you are dealing with relation sorry reaction in a liquid so I am very happy so I will just for the sake of confirmation I will do it suppose this is a b and this is cd and this is center 4 theta 1 is equal to theta 2 has been given and I have to prove that a b is equal to cd now what I need is I will say that I take triangle a o b suppose this is o and triangle c o d so I will say that o a is equal to o d because it is radius and o b is equal to o c because it is radius and angle a o b is equal to angle c o d it has been given so it is side angle side side angle side congruency and again by cpct sites opposite to equal angles would be equal so a b which is opposite to theta 2 would be equal to cd opposite to theta 1 so this is what you have to do all of you did it so wonderful now let me go to theorem 4 so theorem 4 is you congruent if two parts of a circle of congruent then corresponding parts are if you talk too much about how the sorry I have written parts that is chords are equal now guys try to understand its extension of the its extension extension of the third theorem that we just proved so what happens is if two arcs are equal first of all try to understand the relationship between chord and arc so suppose I take this chord arc a b so arc a b is this portion of the circle is it okay a part of circumference now if I draw a line from here so arc a b is related to this chord a b because chord a both arc a b and chord a b would be sustaining same angle theta on the center so if somebody is telling you that arc a b is sustaining an angle theta on the center or you should know that he's also talking about the corresponding chord so arc of chord of an arc would be sustaining same angle on the center so he is saying that arc of the two circles are congruent it means that he's saying that suppose this is arc a b and arc a b q and it is sustaining the same angle theta over here he's also telling me that chord a b and chord p q is sustaining same angle on the theta here which and if I assume that a b and p q are sustaining same angle theta on the center this theorem four turns out to be theorem three now how do you prove it you have to prove the same thing that a b is you have to prove the same thing that ss congruency would be here so I write o a is equal to o q o b is equal to o b and theta is equal to theta a o b so by ss congruency a b comes out to be equal to p q I just wanted to explain one thing that every arc has a corresponding chord which arc is the length on the circle chord is a straight line between the two points so that that becomes corresponding chord and if arc is making angle theta the chord will also make the same angle this thing you need to understand and if this theorem was an extension of theorem three so I hope you all understood it okay now let me go to theorem five theorem five says that it's the same thing I'm not doing it you have to make two chords okay just do it you just go and see this theorem in in in in rs lorval just go through it don't solve it just go through it theorem five says that if and if it not theorem five in your book just just search for it if two chords of a circle are equal then their corresponding arcs and when I talk about arcs here I'm talking about three I have taught you three kinds of arcs minor major and semicircular so I'm talking about all three of them or congruent so you have to do this yeah okay just try to solve it so you'll have a relationship how many atoms are there so that is the significance of more you'll know the number okay but when it comes to the express state if I need this because it's a gas law are you done man you'll take it you'll take some time so we never use this on top but we use concept of more what is the concept of more that one more that substance if I say one more of one more of anything means that many number of that substance so now more is so by the use that see it might be molecular mass also just try to understand properly how it is doing for major angle and semicircular arc there is nothing different in that just done okay one minute take one minute no issues so one amu and one and similarly the molecular mass will be one okay for oxygen is 16 the atomic mass 16 grand 16 gram will be the mass of one more of oxygen theta 32 gram will be the mass of one more of oxygen molecule so this is the more concept in your servers but more okay are you done the school after school will take for two hours if school is not there it will take some more time then I mean can you cover that to yourself huh so now tell me look at that reaction okay fine understanding anyone if you're not understanding please let me know I will solve it I have no issues if one it's a huge number Can you find out the number of holes and the number of holes in 52 grams of premium? Okay, I am solving it then. Let's wait for a moment. So, try to understand. I am first of all taking it for... I have been given that if two quads of a circle are equal, then the corresponding arcs are congruent. So quads, first of all, are making diameter and suppose this is at 12.04. So this is A, this is B, this is C, this is D. And I have been given that A B is equal to C D. I have to prove that arc A B is equal to arc C D. So what I will do is that I will simply write that A B and C D are diameters. So if A B and C D are diameters, arc A B and arc C D are semi-circles. Now, if they are semi-circles, it means that quads would always be congruent. If they are semi-circles, they are making 180 degree arcs. It means that arcs are congruent. If arcs are congruent and quads are equal, then arcs will also be... So I can write over here that arc is congruent and if A B is equal to C D, quads are equal, then arc A B would also be equal to arc C D. This is it. Now, I will go and then these become minor arcs. So I am moving it, let me do minor arcs. Now, if I make it for minor arc, suppose this is my circle and this is A B and C D and it has been given that A B is equal to C D. I am taking minor arc, suppose this is P and this is Q. So I am taking minor arc A B B and minor arc C D. This is sector O. So what has been given to me is the same thing actually. I just draw a line from here and say that O A is equal to O B. O B is equal to O C and angle A B is equal to C D. So this is side-side-side congruency. If this is side-side-side congruency, it means that this angle, angle by C B C D, angle A O B is equal to angle C O D. It means that corresponding quads of the arc are making same angle and if A B is equal to C D, then corresponding quads A B B would be equal to corresponding quads C Q D. Similarly, you can do it for major arc, so I am not wasting time over there. Now let me move to another theorem. The class may extend, I don't know, maybe it can be of 3 hours also, so don't worry about it. Leave all your plans and sit there and enjoy the class. Theorem 6, the perpendicular from the center of a circle to quads. I have a number of particles to the chord, bisects the chord. So nothing you have to do. You draw a chord here and this is O, this is A and this is B. You draw a perpendicular from here. What I need to do is that I need to draw these two lines and then I will have two triangles. So I take triangles about this point is B. So I take triangle O, I take triangle O P A and triangle O P B. So I, in this two triangle O P is equal to O B. And O A is equal to O B, there are degrees and angle A P O is equal to angle B P O because it is perpendicular. So this is 90 degrees. So what kind of congruency I have? I have RHS congruency. So if I have RHS congruency. So it means that it means that this angle and this angle are equal. And how will you say that these A P is equal to P B you have to prove. So how will you say that A P is equal to P B? So this is 90 degree and suppose this is theta. So this will be 90 minus theta in both of them. So angle of, sorry, side opposite to same angle would be same. So you see here A P comes out to be P B because angle at both the places is same. 90 minus theta and 90 minus theta. You should also know that it can be the same theorem which says that the, so it becomes perpendicular bisector actually. And this perpendicular bisector also acts at angular bisector of the center substituted by this arc on the center. So there are two, three correlations. So now things are saying that the theorem number seven is telling me that if there is slight change in theorem number seven is saying that if it is equal. So theorem seven is saying that if this and this is equal, if a chord bisects the, if a line drawn from center of the circle on the chord bisects the chord then it will be perpendicular. Now how do you prove it? So you know that you can write it, this is A, this is B, this is P. You can write it as O A is equal to O B. You can also write that O P is equal to O P and you can write that A P is equal to P B which has been given to you. So this is side, side, side, congruency. Now you also know that this angle because angles opposite to equal sides would be equal. So O A is equal to O B. So angles opposite to this side and this side. I mean this angle and this angle would be equal. So angle O P A is equal to angle O P B and I know that O P A plus angle O P B is equal to 180 degree and these are equal. So you can say that O P A is equal to 180 degree or angle O P A is equal to 90 degree as simple as that. So that is how you have to do it. Now let me move to another theorem. So these two theorems have proved that if perpendicular drawn from the center of the circle to the chord it will be a bisector and if line drawn from center of the circle to the chord bisects it then it will be perpendicular. So these two theorems I have solved it simultaneously. Now there is a corollary to it. Can you please look at it? Go to your book. There is a corollary to it. Just look at it. The corollary says that the perpendicular bisector of the two chords of the circle intersect at the center. Just have a look at it. All of you, if you are, all of you have a look at it. Are you doing that corollary? Any doubt in that? Yes sir. Have you done it? Should I move ahead? Okay if you are not saying anything anyway I have to move ahead. Theorem 8 says to me that equal chords of a circle are equidistant from each other. Equidistant. It's like a fix. It's just on the merit then it is a point of interest. If you are paying like this then it will be just a fix. So A, B, C, D, again I have drawn this. Fine. Now what I need to do is that if I want to find out the distance of this chord from the center, distance is decided by perpendicular drawn from a point to the chord. I am drawing a perpendicular from here and suppose this is M, this is N. I have to prove that M is, sorry, O M is equal to O M. How I do it is that I will prove that. I take triangle A, M, O and triangle C, M, O. So I know that O A is equal to C because they are radius. I also know that if a perpendicular is drawn from the center to the chord it bisects it. So I know that A M is equal to C M because A B is equal to C D. So half of A B will be equal to half of C D and half of A B is A M, half of C D is C M. So I am writing here A M is equal to C M and I know that angle M is equal to angle M. So I have RHS congruency. By RHS congruency I know that this angle is equal to this angle. So angle A is equal to angle C and side opposite to it which has O M is also equal to O M. So that's how you have to prove it. I hope you understood it. Where are you going? This is down. Now let me go to Theorem 9. Theorem 9 says me that the chords of a circle, you saw this. I am writing it as the chords circle. The chords of a circle. The chords of a circle. And it's a dynamic loop where we are going to loop the chords up to the mark. If they are just in the loop on the passing stage, that's what I am trying to do. Let me just tell you. And then every month if suppose a student is not able to loop the chords up and suddenly then we will replace one student with a new one. So it's a dynamic loop. Just solve this. If I just R2 set it because many of the current R2 set it because right now we have classed then we will be running as we did. We have not approached them. I want all of you to write whether you have done it or not. Just don't sit and enjoy. Have you done? Four people have written they have done. Which are Shraddha, Dhruti, Tanna and Ritu. I need answers from Anjita, Varsha, Sanjana, Aniruddha. There are only seven people here who left. Okay, so five people have answered. Now I am moving to the other one. Theorem 9. So theorem 9 we have already done. Theorem 10 we will do now. There are too many theorems. Some 18 theorems are there. I want to touch upon each one of them because many a times direct theorems are told to be proved in the examination. Theorem 10 is a bit theoretical. Theorem 10 tells me that there is one and only one circle passing through three given non-colinear. Just try to prove it once. Take two to three minutes and try to prove it. If you are not able to do so, I will prove it. So let me give you three minutes for this. It is 11.30. At 11.33, I will start solving. You can take help of your book also. No, it is not too easy to understand it. Okay, Dhruti, I solved. Don't worry. So to solve this particular theorem, you need to realize one thing and that I will discuss now. It is this. Please understand it. So what I am trying to say is just look at here properly. Suppose these are the two points A, B, and C, D and I am drawing a perpendicular bisector like this. Now any point, I am writing here, you should also write any point on the perpendicular bisector of a line is equidistant from the two ends. So you take any point O, this OA will be equal to OB. Why? Because let's take this P, this is perpendicular bisector. So it means that AP is equal to PB and OP is equal. So you can say that OP square plus AP square which is also equal to OP square plus BP square because AP and BPR equal. And that would be equal to here it will be equal to OA square. Here it will be equal to OB square. So it means that OA is equal to OB. Now this point O can be anywhere. This length would be same and this is already same. Hence this hypotenuse will always be same. So you should always know that any point on the perpendicular bisector of a line is equidistant from the two ends of that line. Now what we do is in this particular case, we will take three points. Suppose this is point A, this is point B, and this is point C. So this is AB and C, draw this, this is construction which I am doing. You should do it with your compass and all those things. So I am drawing AB, I am joining AB, join AB and BC. Now you draw a perpendicular bisector of AB. So perpendicular bisector of AB would be something like this. And you draw perpendicular bisector of BC. So perpendicular bisectors of AB and BC will intersect at point O. So what I am doing is I am drawing perpendicular bisector. Suppose this is PQ and this is RS and perpendicular PQ is perpendicular bisector of AB. And RS is perpendicular bisector of BC. So O is intersection of PQ and RS. Now what happens if I draw these lines. So I can say that OA is equal to OB. Because any point on perpendicular bisector would be equidistance from the ends of that line. And similarly OB is equal to OC. So OA is equal to OB is equal to OC. Now what I say is that let's make this as radius of the circle and mark take O as center and draw a circle like this which passes through these three points. So ABC, you will be able to show that with ABC there is only one circle. Now if you try to take any center O dash, this O dash would be coinciding with this O because to make radius equal from these three points, O dash has to coincide with O. So you will have to write that COR would be congruent with CO dash S. So that's how you prove it. There is nothing else in it. Did you all understand it? Okay, so next theorem we will prove now. Did you understand this? Okay, so now look at here. Okay, so next one is theorem 11 is let me finish off the theorems quickly so that I can move to questions. Any two chords of a circle show that the one which is larger is nearer to the center. This is AB and suppose this is CD and this is O. So suppose this is OE and this is OF, you have to show that OF is greater than OE. Solve it for two minutes then I will solve it. What are you doing? I am just testing. You will have to do it yourself. You will have to put the indigil powder in the group. If you put it in the group, then it will be easy to solve the problem. You will have to put indigil powder in the group. You will have to call the parent. But you will have to do this puzzle. You will have to do a school trip. Today we are going to Pongalonga. Pongalonga is 15th. It will be a celebration. You will have to do it today. Today is Saturday. Who is the dad? Ankur, Ankit and Rekha. I was the Maid. I didn't even have a chemist class. There is no Rekha. There is no Rekha. There is no Rekha. He didn't even talk to me. When? From 10 am. When did you go? Done. Okay, Tame is done. Okay, I will solve it. Few people are saying me to solve. So what I know is A B is given. A B is greater than C D. So A E would be half of A B and C F would be half of C D. Now O A is equal to O C. So let me take these two triangles. I am taking triangle O E A and I am taking triangle O F C. So let's apply Pythagoras theorem in it. So by the Pythagoras theorem, O A square is equal to O E square plus, sorry, one second, not plus. Let me write the Pythagoras theorem correctly. And then here O E square plus A E square is equal to O A square. On the other side in O F C, O F square plus F C square is equal to O C square. Now I know that from here, from here I know that these two, I know that if A B is greater than, if A B is greater than C D, then A E would be greater than C F. So what I can do is that, what I can do is that from here I can write because O A square is equal to O C square, O E square plus A E square is equal to O F square plus C F square. Now if I replace C F with A E, so what am I doing? I am, if C F is replaced by A E, so C F is lesser than A E. So if I put a bigger number here, then I'll have to put a greater sign here. So O E square plus A E square here is lesser than O F square plus A E square. Why I am putting this greater sign over here is, I am not putting this equal to sign. I am putting this greater sign because I am replacing C F with A E and A E is a bigger number. So this A E and A, sorry, this A E and A E gone. So O F square is greater than O E square or O F is greater than O E. This is how you have to prove it. You just write it down and I will move to theorem 12. Let me move to theorem 12. Theorem 12 tells me that if any of two chords of a circle show that the one which is nearer to the center is longer. It's exactly the opposite of the theorem. You just have to prove it the same way. I hope all of you will prove it in next two minutes. It's 11.49 in my watch. Just solve it in two minutes and write your comments. Done. Okay, three people have already done it. From next class onwards I am taking it on Google handout because it's good for me to hear all of you. Okay, five people have already done it so I am not solving it. If anyone wants me to solve it please write it down and I will solve it for you. Please write it down and I will solve it for you otherwise I am moving to another theorem. I have to solve one, two, to solve four, five more theorems. Okay, as nobody is responding I am moving to another theorem. Now I am moving to congruent circles. So theorem 13 is if two arcs of a congruent circle are congruent then the corresponding chords are equal. If two arcs of congruent circles are congruent then the corresponding chords are equal. How do you prove it? I will prove it for you, don't worry about it. so I am taking two circles which are congruent so this is O and this is O- this is AB this is CD so I know that this circle is two congruent circles so I am writing circles, circles are congruent when their radiuses are equal so I am writing OR and this is this is another circle so give me a moment and I have another circle which is O-R and they are congruent to each other now I know that AB it has been given to me that called AB is congruent to called CD so what I can do is that first case AB and CD are minor circles it's minor arcs sorry if they are minor arcs so let me take triangle OAB and triangle O-CD so OA is equal to O-C OB is equal to O-D they both are radius and I know that angle A OB is equal to angle CO- because if the chords are congruent chords becomes congruent only when they form same angle on the center so by this I can prove that this is SAS congruency between the two of them and hence AB would be equal to CD similarly for the major arc you can you can you can take it on the other side and you can do so so in that case also what happens here is here arc AB is equal to arc CD for major arc you can write as arc BA opposite from here to here arc BA is congruent to arc sorry arc DC so from here to here and everything remains same and you can prove it the same way so that's how you have to prove this next theorem that you have to prove is theorem 14 is if two chords are congruent of congruent circles are equal then corresponding arcs and when I say arc I talked about all three arcs major minor and semi-circular just solve it okay Dhruti has done it okay so then I will solve it let's give others to us okay so let me solve as few few people have already solved it people are asking for it so I'm solving it if I'm talking about two congruent circles I'll make two congruent circle this is my C O R and this is my C O dash R and what I need to do is that first I'll assume for minor arc so what I'm doing is I'm taking minor arc AB here and I'm taking minor arc CD here so I draw a line like this and then I make a corresponding chord and I draw this and like I and I make a corresponding chord here now I know that way is equal to O dash C and AB is equal to O dash D and AB is equal to CD which has been given to me it means that it is SAS congruency if it is SAS congruency then angle AOB is equal to angle C O dash D it means that both the chords and if chords are making corresponding angles this is by CPCT you can write here so it means that measure of chord AB is equal to measure of chord CD or AB is congruent to CD that simple as that now if I have to do it for minor arc sorry major arc it would be exactly the same thing just that you have to select the arc differently case 2 I'm taking that it is for major arc so for major arc what I do is I'll take AB and CD or CD so AB also can be major arc and in that case BA becomes minor arc so it's actually the same thing so you can write that AB is equal to CD and you can you can say that when this is theta because I have already proved in case of minor arc that this and this are equal so if I take it for BA and DC so you can say that this will be 360 minus theta and this will be 360 minus theta hence angles made by measure of angle made by BA is equal to measure of angle made by DC and if they are equal hence arc BA would be equal to arc DC as simple as that so this is for major arc and for semicircle you can directly say that it makes 180 degree in both the cases hence they would be equal because the diameters are equal so this is how you had you had to prove it now let me move to theorem 15 and what I want you because I have proved the most of the theorems in the other part it's just that now I'm using two things so what you need to do is that theorem 15 to theorem 18 in your book you go through it and just let me know if you are not getting any any theorem and I'll start solving questions one hour I take questions from now so I give you three four minutes to solve it please do it okay I've got done message from three four of you so should I go to the questions now I'm done with theorems of this section okay so now let me start with the questions so fast question is and I'm not doing easier questions so this is the first question this is CD record and this is AB this is a part and this is EQ and this is center O so I'm saying that R is equal to 5 centimeter AB is equal to 8 centimeter and CD is equal to 6 centimeters O P parallel to AB sorry perpendicular to AB and OQ perpendicular to CD determine the length I've written it Shraddha on the board okay for this question try to understand this is 90 degree this is 90 degree so O P square is equal to OA square minus AP square OA is 5 so this is 5 square and AP will be half of AB so minus 4 square so 5 square minus 4 square is nothing but 3 square so OT comes out to be 3 similarly OQ square is equal to OC square minus CQ square so this is 5 square and this is CD is 3 so CQ will be 3 so 5 square minus 3 square so this will be 4 square so OQ will be equal to 4 so PQ is equal to OP plus OQ and that is 3 plus 4 that comes out to be 7 so this is how you can solve this question now next question next question is next question is I'm writing here two chords AB and AC of a circle are equal prove that and I've told you this I've told you that how to prove this right I'll prove it prove that the center of the circle lies on the angular bisector of angle BSE the solvent okay so I will solve this question just look at here so suppose I have two chords AB and AC so A is the common point of the two chords so this is B and this is C AB and AC and the circle is C OR so this is my point O and I am drawing a line from here to O so suppose this is this point is D and this is 4 and this is my chord BC now O lines on line AD so I have to prove that AD is bisector of angle BSE so I know that AB is equal to AC and suppose this point is M so what happens is let me take triangle BAM and triangle CAM so in this particular triangle AB is equal to AC and this angle is equal to this angle it has been already given to us so BAM is equal to angle CAM and AM is common among both this is SAS congruency so you can say that from here BAM is equal to CAM and angle BMA is equal to angle CMA and addition of both these angles BMA plus angle CMA is equal to 180 degree so this 2 would be equal to 90 degree so it means that AM is perpendicular bisector of BC chord so if AM is perpendicular bisector of BC chord so AD would also be perpendicular bisector because AM is a part of it now what can be done is that in this what I assumed initially that O lies on AD so you can you can you can solve it from the other side also you can say that these two angles are equal and if these two angles are equal you will have to prove that this line O lines on AD so as soon as you prove that AD is a perpendicular bisector you know that AD passes through O and and and the theorem gets proved so that's how you can do it. Did you guys understand it or do I need to solve once more? Did you guys understand it? Try to understand I'll again say that the center of the circle lies on angular bisector of BC as soon as it has been told that this is angular bisector of BC this has been given to me AB is equal to AC has been given to me because here only it has been written and AM is common so this SAS congruency is proved if SAS congruency is proved I can say that this and this is equal and this angle is 90 degree it means that this line AM which is a part of AD is perpendicular bisector of the card perpendicular bisector of the card will always pass through the center that's what you have to do and this is this is what you have to this is what you have to do so let me go to the next question the next question which I'm giving is I'm writing the next question if two circles intersects in two points prove that the line through the center is perpendicular bisector of the common card and the previous question you are not supposed to prove that AD bisects angle BC you are supposed to prove that AC passes through the center the perpendicular bisector of angle BAC passes through center it was already given that sorry angular bisector of angle angle BAC passes through center so it has already been given that it is angle bisector and that is why I assume that BAM is equal angle BAM is equal angle CAM and as soon as I took that I was able to prove that the triangles are SAS congruent hence I was able to prove that AD is perpendicular bisector of the card if AD is perpendicular bisector of the card any perpendicular bisector of any card will pass through center that is what the observation was when when when I finish the session you can go to that question and you can you can find it out what about this question anyone wants me to solve this question okay let me solve this question for you so I will take a bigger circle like this and I will first let me draw the slide and then let me take a smaller circle so this is O this is O dash and this is the common card common card I am assuming to be AB and this point I am assuming to be M let me draw a line from here to here this is the radius R this is the radius R and let me draw these two points also this is the radius S and this is the radius S I have to prove that O O dash is perpendicular to line AB so construction I have already done I am taking triangle O A O dash on the upper side and I am taking O B O dash on the lower side so O A is equal to O B because they are radius O dash A is equal to O dash B because they are radius and O O dash is equal to O O dash hence O A O dash is congruent to triangle O B O dash if this is congruent then what is possible then I will prove that angle A O O dash is equal to angle B O O dash this and this because they are opposite to equal sides and angle AOM because this is part of it because it's the same thing rather than O dash you can write it and angle B O M are equal if angle A O M these two angles are equal rather that O dash I am taking this I am taking another point on the same line so M is see O M is part of O O dash so this O dash can be replaced with M so now let's take triangle AOM and triangle which triangle I should take B O M so if I am taking this I know that O A is equal to O B I also know that I have just proved here that A O M is equal to angle B O M and I know that O M is equal to O M so O M is equal to O M so these two triangles A O M is congruent to triangle B O M and if this is two then what happens is A M becomes equal to B M and angle O M A becomes equal to angle O M B and I am writing here angle O M A plus angle O M B is equal to 180 degree and the two are equal hence angle O M A would be equal to angle O M B that is equal to 90 degree so as soon as you prove this the line O dash become perpendicular to A B because M is part of line O dash this is how you have to solve it let me take another question question question is prove that the perpendicular bisector of the cot of the circle the perpendicular bisector of the chord of the circle always passes through center just do it Dhruti I will take one or two more questions and then I will finish off maybe maximum by 115 okay have you done this question Tana has done it great okay Varsha has done it Dhruti has done it and Sanjana has done it I told Varsha it's still 115 maximum and doing it because I need to finish it fast I'll take one or two more extra classes so maybe next week I'll have one class at center and one class on Sunday next Sunday I'll have an extra class for you guys and that class also would be for three hours okay so as most of you have done it anyone who wants me to do it otherwise I move to next question so no one is saying me anything I'm moving to next question the next question is that I want you to solve is this so this is a circle which has been made and there is a chord AB there is another chord CD and they cut at E so angle so I'm writing here equal chords AB and CD cut at right angles at E now this is the center O and I draw a perpendicular here OM and I draw perpendicular here OM if OM and OM are midpoints of so as soon as I say midpoints you should understand that midpoints of AB and CD which I'm indirectly though it has not this is OM which indirectly which has not been written in the question OM would be perpendicular to AB and OM would be perpendicular to CD that you have to assume because so prove that prove that OMEN is a square so this is the last question so this question is okay anyway this is the last question you can leave or you can wait for two minutes anyway I will finish it off in two minutes so try to understand you join this OE do you just wait for this question I'm just finishing it so OM is perpendicular to AB because they are midpoint M is the midpoint and OM is perpendicular to CD so in this particular triangle let me take triangle OM E and triangle OM E OM is equal to OM because perpendicular drawn to the equal cuts would be equal and angle M is equal to angle N which is equal to 90 degree and OE is equal to OE it means that I have RHS congruency if I have RHS congruency then M E would be equal to NB I have already proved it and OM is equal to OM so sides has already been proved as as equal now if I prove that all the angles are 90 degrees so I know that angle M and angle N are 90 degree angle E has already been given to me as 90 degree so angle O has to be 90 degree because in a quadrilateral the total angle is 360 degree so all angles are 90 degree and all sides are equal hence OMEN is a square that is how you had to prove it so okay what I'll do is I what I'll do is which chapter is this can you write me the chapter name in your chapter number in your book please just write the chapter number in the book okay so circle exercise 12a is your homework exercise 12a is your homework just finish it off okay so thank you so much for joining the class let's meet next week and finish off the circles