 So I've got four people here. So, I'll wait for two more minutes and then I'll start. Okay, Tanay is also here. Is my voice clear guys? Okay, so I'll finish off circles today. The first part of the circle we have already done it. I am starting with one question. Please solve this question. The question is, there is a circle like this and there is an external point E and then we have points A, B, C, D. This is the center of the circle O and I draw perpendicular from here, which is point L and one perpendicular from here, which is point M. And you have to prove that, you have to prove that B E is equal to D E. Second one is A E is equal to C E. O L is perpendicular to A E and O M is perpendicular to C D. Just solve this question. After doing this, let me know who all have done it. Done. Guys, let me know in the chat box, please. Done. Yes, A B is equal to C D. They are equal chords. One hint to you guys is, if you have to prove that B E is equal to D E and I have given that A B is equal to C D. So half of A B would also be equal to half of C D. So it means that B L will be equal to M D. So if you prove that E L is equal to E M somehow, what will happen is that from there, here you can subtract B L and here you can subtract M D and you can prove that B is equal to D E. And once you prove that B is equal to D E, you add A L and C M which are of equal length and you can prove that A is equal to C D. So you just have to prove here that E L is equal to E M. If you can prove that, that will suffice. For that, you need to do a construction here. So this O E you will have to match. Of course, you know that you will take two triangles. Triangle O L E and triangle O M E. Tanah has done, Sraddha has done. So I am doing it O L E and O M E. I know that if chords are equal, perpendicular is drawn to the chord from the center of the circle would be of equal length. So O L is equal to O M. I already know it. O E is common and angle L is equal to angle M. So this is by theorem. This is common. This has been given. Hence, I have RHS congruency here. If I have RHS congruency here, then I can prove that by C P C T EL would be equal to E M. Now if EL is equal to E M, so EL minus BL would be equal to E M minus MD as we already know that BL is equal to MD. Now EL minus BL is EB and E M minus MD is ED. Now what happens is this has been proved. Now once I have proved that EL is equal to E M, what I will do here is in EL I will add EL and in E M I will add C M. So EL is equal to C M. So this becomes A and this becomes C. So now let me go to... I'll give you one more question and then I'll move to another topic. Apart from circles, is there any topic left out? I don't think so. We have covered up most of the things. Statistics is left out. Okay, I'll take one class on Thursday and I'll finish statistics that day. Very easy topic. Any other topic other than statistics? This is a circle and this is my center O. And what I do here is I draw a line OQ. I draw a line OP. I draw a chord AC. I match this, join this QC. I draw another chord AB and I join this PB. I given is AB is equal to and this point is N and this point is M and this is 90 degree obviously. AB is equal to AC. Then OM is perpendicular to AB and OM is perpendicular to CA. O is center of the circle. AB is equal to QC. Okay, probability and statistics are the two topics. Done. Let me know when you are done. Okay, three people have done till now. I'll wait for one more minute. What I'm thinking of taking a class on Monday to cover up statistics online. Is there any problem with anyone? Varsha because I'm not free tomorrow. I have to go to TISB. I'll be unable to come to Raja Ji Nagar because I get free only by 2230. It'll be very late for me till the time I reach Raja Ji Nagar. Anyway, on Thursday I'm coming to Raja Ji Nagar next week. I'll take an offline class on Thursday and this week I'm planning two classes. One day after tomorrow on Monday and one on Thursday. Thursday's class would be offline in the school. I can come but I'll be very late because I leave TISB by 2 o'clock. That's why Monday's class I'll keep online. Tuesday, Wednesday and Saturday would be as you have classes in the school. Thursday I'll surely take a class there. On Thursday I'll take a class on Thursday in the school. That's what I'm saying. On Monday let me take an extra class. I'm not saying I'll not take class in the school. I'm only taking an extra online class because it will not be possible for me to come on Monday. I'm taking two classes this week. One on Monday online. Let me solve this question enough time. What this question is all about is if I have to prove that PB is equal to QC I know that I have to take triangle OPB and sorry triangle MPB and triangle NQC. What I need to do is that I already know that if this angle is 90 degree this angle is 90 degree. You look at here I will first take angle only. Angle PMB is equal to angle QNC that is equal to 90 degrees. Now what about sides? I will take these two sides, MB. I am saying that AB is equal to AC so half of AB will also be equal to half of AC and you see here half of AB would be MB and here half of AC would be NC so the same I am writing here MB is equal to NC. Now what I need to do I know that try to understand one thing if AB is equal to AC equal codes are there then perpendicular drawns to equal codes from center OM would be equal to OM. Now I know that OP is equal to OQ which is radius so if I take OP minus OM that should be equal to OQ minus OM. So I am writing here OP minus OM which is equal to PM and that is equal to OQ minus OM and that is equal to NQ so what I have proved is I have proved that this side and this side is equal and this angle is equal so I have side angle side congruency and with side angle side congruency I can prove that this PB by CPCT Anirudh if you are not feeling well you can go no issues PB is equal to QC so now I will do a bit of theorems from the other part of the circles so it is something like this so now the topic is results on angle subtended by so I am taking first theorem in it and first theorem is the angle subtended by an arc of a circle at the center is double the angle subtended by it any point on the remaining part of the circle this is a very important theorem with respect to concepts of circle so what I do is I take a circle like this I have to prove this so I am taking a circle C with the center O and radius R now what I am doing is I am taking an arc AB now arc AB is congruent to or arc AB is synonymous with chord AB so I have to prove that and I am making an angle here so this is C so I have to prove that angle made by this arc AB at center O so angle A O B is two times angle made by this arc AB on any point on the circle so two times ACB so first I am assuming that AB is a minor arc so what I do is now you look at here I am drawing a line passing through from here I am drawing a line passing through CO which crosses here at somewhere this is this I am saying B so you look at here if I take this particular triangle so in triangle A O C this angle becomes exterior angle so exterior angle A O D is equal to angle O AC two interior angles this and this O AC plus angle O C A and on this side in triangle B O C the exterior angle B O D would be equal to these two interior angles so B O D would be equal to angle O B C plus angle O C B now I know that see this is radius and this is radius so what happens is this angle would be equal to this angle and this angle would be equal to this angle so what I am saying over here is angle O AC is equal to angle O C A because this triangle A O C is an isocellous triangle and why it is an isocellous triangle because O A is equal to O C which is equal to radius R and this triangle is also an isocellous triangle and why this is an isocellous triangle because O B is equal to O C and which is equal to radius R so these two angles are same so what I can write is A O D is equal to two times angle and I have to prove this O C A so I am taking the upper angle O C A and angle B O D is equal to two times angle O C B now I will add these two angles angle A O D plus angle B O D is equal to two times angle O C A plus angle O C B and O C A and O C B is nothing but this complete addition of these two angles is nothing but angle A C B and A O D plus B O D is nothing but two times angle A sorry not two times angle A O B so angle A O B is equal to two times angle A C B so this is how this has to be proved now let me go to another theorem theorem two is the angle in a semicircle is a right angle so I draw a circle and then I draw a diameter and this is point O this is point A this is point B and then I draw an angle here I have to prove that angle A C B is equal to 90 degrees so try to understand how I prove this I know that by theorem one which I just discussed angle made by God on circle on center of circle is equal to two times angle made on any point on circle so this particular arc AB is making 180 degree angle here and if it is making 180 degree angle here then the angle two times angle A C B would be equal to angle A O B so angle A C B would be equal to half of angle A O B and angle A O B is equal to 180 degree so two one by two into 180 degree is equal to 90 degree so this is how you we prove that angle made in a semicircle is 90 degree now let's move to theorem three and theorem three tells me that the arc of a circle subtending a right angle at any point of a circle in alternate segment is a semicircle this is center O and I have I will draw this line AB this is my point C so I know I have to prove that arc AB is a semicircle and I know this angle is 90 degree given is angle A C B is equal to 90 degrees so what I will do I will find out the angle opposite of this A O B would be equal to two times angle A C B so two into 90 degree which is equal to 180 degree after this I have I just have to write that A O and O B falls on same line AB that's when the angle would be 180 degree hence AB is a diameter quad AB would be sorry arc AB would be a semicircle so this is what you have to prove here now let's move to another theorem this is theorem number 4 so this is the second last theorem that I need to prove so I am proving for minor theorems you just go through it once from your book and prove it for I am proving for minor sectors you prove it for major arc and semicircle so look at here angles in the same circle angles in the same sector segment of a circle are equal so let's draw it and let's draw an arc AB and suppose these are the two points sorry this is the center and these are the two points C and D so let me draw a line like this a line like this so angle to prove that angle ACB is equal to angle AB this I have to prove now how I will prove so I will prove like this angle AOB is two times angle ACB I also know that angle AOB is two times angle ADB why I am writing like this because both these angles have same arc arc AB so that is why I can write that two times angle ACB is equal to two times angle ADB and this two and two gone you get angle ACB is equal to angle ADB so this is what it is now this particular line can be anywhere so if I make this line diameter and I have two points suppose this is point O I have two points ACB I have a point D this and this this is equal to 180 degrees so I can write that angle ACB is equal to 90 degrees because we just discussed the theorem which tells us that angle in a semi-circle is a right angle so ACB is 90 degrees that is equal to angle ADB so we can prove it for semi-circle for major arc also you can prove it so suppose I do it like this so I take one point here C and one point here D and this is A and this is B so what I do is that I make this line like this this is A and I make one line like this this is center O so what I can write over here is again AOB is equal to two ACB same thing you will write here just you have to make a different figure all together same thing which you have written for minor arc the same thing you will write here and write that these two cases are same now let me go to the last theorem of this topic and the last theorem of this topic tells me which is theorem 5 that if a line segment joining two points subtends equal angles at two other points lying on the same side of the line segment then four points are cons cyclic it means that lie on the same circle cons cyclic means lie on the same circle so what do I need mean I what I mean by this statement is given to me is AB is a line segment let me draw first and then I will explain both this AB is a line segment and C and D suppose a point C is here and the fourth point I am not assuming on the circle I will check a different point D dash here and try to prove that these two points D and D D dash are on the same circle so D and DS are coinciding with each other so AB is a line segment and C and D are points AB a line segment C and D are two points on same side of AB now I know that I have to prove that AB CD points lie on same circle so for that what I have done is let me take this point here so that I can make this same line so this point D is there so and let me make this so what I do is point D false on circle then angle ACB is equal to angle ADB but by figure or but by diagram angle ACB is equal to angle A dash B so these two things are only possible so which is saying that ADB is equal to angle AD dash B which is only possible D is equal to D dash or D and D are coinciding and hence we can write that AB C and D are on same circle so now let me move to two three questions and let me give you questions so one question is this this is my circle this is called AB this is center O so I have this this angle I am assuming to be Z then I have something like this then I have something like this so this angle I am assuming to be Y and this angle I am assuming to be X this is ACE and this is point F this is point D so you have to prove that angle X plus angle Y is equal to angle Z this is angle 1 this is angle 2 this is angle 3 and this is angle 4 and this point is point O O is center of circle so if you are done just let me know just let me know when you are finished okay three people have done Shraddha, Tanay and Sanjana I am waiting for one more minute so that others can give me answer okay okay so I will start solving this question now look at this angle this angle is exterior angle if I take triangle ACF so exterior angle Y is equal to angle 1 plus 3 and I know that angle now let me take triangle AED so in that exterior angle 4 is equal to angle 1 plus angle X now let me add this angle 1 plus angle X plus angle Y angle 1 plus angle 3 plus angle 4 now angle 1 and angle 1 is gone, angle X plus angle Y is equal to angle 3 plus angle 4 and angle 3 and angle 4 are equal so I write it 2 times angle 3 and 2 times angle 3 would be equal to angle Z so I write that angle X plus angle Y is equal to angle Z now let me give you another question and the question is prove that midpoint of the hypotenuse of a right angle is equidistant from its vertices let me know once you are done ok one more minute nothing more than that ok versile solve it ok 2 people have given me the answer what are others doing ok let me solve it so what I do is that I draw a circle and I know that and in the other circle I take AED as inside and I draw a line from here and this is 90 degree what I have to prove that OA is equal to OB is equal to OC so for this proof these 2 points if I have to prove I just have to write that BAC is equal to BA dash C in both the cases is equal to 90 degree it can only happen that when A and A dash coincide on each other so if I have written that then what happens is if you look at here exterior angle if I take triangle BA dash C we can also write that this angle A dash is equal to angle B plus C because B plus C if A dash is 90 B plus C would be 90 and what happens is any exterior angle is equal to its interior opposite angles so it means that A dash and A will have to coincide when they coincide then the circle passing through these 2 points BC will also pass through this point A it means that BAC fall on same circle and if BAC fall on same circle then OB would be equal to OC because this would be radius of the circle so that is how this needs to be proved what I have tried to do is I have taken I have not taken all 3 points on the circle I have taken any other point A dash instead of A I have tried to prove that this A dash and A coincide with each other and then I have told that as BAC is on the circumference of the circle then OB would be equal to OC and that would be equal to OA because any point on the circumference of the circle would be equidistance from the center because it will be radius of the circle so that is how you have to solve this now let me go to another question and another question is prove that the circle in a segment smaller than a semicircle is greater than a right angle do it quickly let me know when you are done done so Tarnay has done it Sanjana has done it okay Varsha has done it good okay so enough time for this what I do is I draw a circle like this and this is my center now segment should be smaller than a semicircle so this is semicircle I am drawing a line here and then I am drawing an angle here so suppose this is A, this is B, this is C I have to prove that C is greater than 90 degrees so let me go here so I know that angle subtended by an arc of a circle at the center is double the angle subtended by it at any point of the circle so you look at here if I take a point B over here BDC BDC C subtended angle BOC at the center and BAC angle BAC at point A now I know that angle BOC is a reflex angle why because why I write this because angle BOC would be greater than 180 degrees why it will be greater than 180 degrees because if I draw a line like this and if I take this angle EOD or EOF this angle EOF is 180 degrees so anything which is going beyond in the other segment would be greater than 180 degrees so to BAC BOC is equal to 2 BAC and as BOC is greater than 180 degrees 2 BAC would be greater than 180 degrees hence BAC would be greater than 90 degrees so this is what you had to prove now let me give you another question of this topic we have solved 3 questions we will solve 2 more questions of this type and then I will move to cyclic quadrilaterals and then I will finish the class so now what I do is let me draw the figure for you so this is AB this is CD this is point E and I join CB this is AO this is ODE AB and CD are given data AB and CD are chords and they intersect at point E you have to prove that angle AOC plus angle BOD would be equal to 2 times angle AAC just prove it and let me know once you are done let me know once you are done let's Radha is done Ritu is done one more minute Radha is done most of you are done so let me solve it so you look at here what I do in this question is what is AOC AOC means this angle so let me draw this line first this angle is AOC if this angle is AOC it means that I am talking about this particular so arc AC is making angle AOC so angle AOC would be 2 times the same arc AC is making angle ABC so it will be 2 times arc ABC and angle BOD would be equal to you look at here when I talk about BOD so let me draw a line here BOD and I am talking about arc BD and arc BD will make 2 times angle the same arc BD is making this angle 2 times BCD so I added AOC plus BOD is equal to 2 times angle ABC plus angle BCD now try to understand angle ABC because if false online AB I can write angle ABC as angle ABC I am not doing anything instead of complete AB I want to take help of triangle ABC so what I am doing is because why I want to take help of triangle ABC because I have to find this angle so if I find this angle CEB angle AC look at here angle AEC is equal to 180 degree minus angle CEB so this angle plus this angle is 180 degree now I will find out these 2 angles so EBC plus you see here this angle BCD can be written as angle BCE now in triangle EBC triangle EBC C plus triangle BCE plus triangle plus angle EBC plus angle BCE plus angle CEB is equal to 180 degrees so these 2 angles would be equal to 180 degree minus angle CEB so what I am writing here is this is equal to 180 degree minus angle CEB now this angle CEB from here can be written as 180 degree minus angle ACE so I am writing CEB as 1 second so this will be 2 times 180 degree minus 180 degree minus the CEB can be written as 180 degree minus EB sorry 180 degree minus CEB can be written as 180 degree minus AAC once again I made mistake here once again guys so this can be written as AEC so I have written here AAC so now this 180 180 is gone this can be written as 2 times angle AAC so this is how this question has to be done now let me give you another question from this topic and this will be last question from this topic then I will move to cyclic quadrilaterals so now this is a circle and I have AD here I have BC here I join these 2 points 2 points make this I say that this is point E I say that BD is perpendicular to AC this has been given this is center O this point is center O and this angle is X degrees this angle is P degrees this angle is R degrees this angle is Q degrees so you have to find out angle P angle Q and angle R in terms of angle X and let me know once you are done done guys 1 or 2 more minutes nothing more than that ok ok so few of you have already done it so let me do it what is this angle X angle X is angle subtended by AD sorry arc AD so this arc AD is equal to this angle AOD would be equal to 2 times angle ABD this this angle ABD so this is equal to half X degrees now try to understand if this is perpendicular angle AEB would be equal to 90 degrees so I know that this angle P plus this 90 degree plus this angle which is half X is equal to 180 degrees so angle P is equal to 180 minus 90 divided by half X so P is 90 minus half X now this angle Q would be angle ACD also would be half of angle AOD so it means that Q would be equal to X by 2 now I have to find out angle R what I will do is that I know that angle ABE plus angle you look at here this particular line AC is diameter why it is a diameter because center O falls on it and if it is a diameter this angle would be 90 degrees so I am writing here angle ABC is equal to 90 degrees and angle ABC is angle ABE plus angle CBE and that is equal to 90 now you know this angle ABE is 1 by 2 X so angle CBE which is angle R here is equal to 90 degrees so angle R is 90 degrees minus 1 by 2 X 90 degrees so that is how this has to be done now let me prove 3 theorems on cyclic quadrilaterals so theorem 1 is the sum of either pair of opposite angles of a cyclic quadrilateral is 180 degrees or you can write the same thing that opposite angles of a cyclic quadrilateral and I am writing short form is supplementary to each other so suppose I draw a circle over here and this is ABE CD the center O I have to write prove that angle A plus angle C is 180 degree and which is also equal to angle B plus B before I discuss this you all know that a cyclic quadrilateral is that quadrilateral which all the points the vertex all the vertexes of the quadrilateral would be falling on the same circle or circumference of the same circle so if I again repeat what I am saying now what I am saying is a cyclic quadrilateral is that quadrilateral of which all vertexes fall on the circumference of same line so what I do over here is I have joined line AC and BD and I know that angle ACB would be equal to angle ADB why because they are in same segment prove that in a circle angles in the same segment are equal similarly angle BAC would be equal to angle BDC similarly angle so if I add it ACB plus BAC is equal to angle ADB plus angle BDC now angle ACB this angle and angle BAC this angle is equal to angle ADC so angle ACB let me write here angle ACB plus angle BAC is equal to angle ADC what I do over here is and why I am writing this angle ADC because you look at here I have angle ADB and angle BDC so this is angle ADC ADB and BDC I write here this is equal to angle ADC so angle ACB plus angle BAC is equal to angle ADC what I do is that I add on the both side angle ABC so that I complete this triangle and I add here angle ABC on this side also so what I get here is this is equal to 180 degree because this is complete triangle ABC so this would also be equal to 180 degree so what is angle ADC angle ADC is angle D and angle ABC is angle B angle D plus B comes out to be 180 degree now I know that angle A plus B plus C plus D is 360 degrees so B plus D I have already told that it is equal to 180 degrees so angle A plus C would be 360 minus 180 which will give you 180 degrees so this is how you have to solve this now let me move to another theorem which is if a pair of opposite angles of a quadrilateral it is a converse of the first theorem its supplementary then it is a cyclic quadrilateral and for this you know that I have to prove that four points are on same line sorry same circle so one of them I will not take on circle what I will do is that I will draw two lines AB this is the line so I am taking this this is point C so I am matching this and this is point D I am taking here and I will join this on the same point on the same line I will take a point D dash and I will try to prove that D and D dash are coinciding on each other wherever you have to prove that two or three points or four points are on same circumference one of the point you take outside or inside of the circumference and on the same line on the circumference so any other point and so that those two points are coinciding so what I will write over here is angle ABC plus angle ADC is equal to 180 degree it has been given to me that opposite angles are supplementary but at the very same time ABC plus AD dash C is also equal to 180 degree because by diagram it I have made like that and this is only possible when angle ADC is equal to angle ADC which tells me that this is not possible since exterior angle of a triangle can never be equal to its interior opposite angle so this exterior angle if I draw a line this exterior angle can never be equal to of a triangle can never be equal to interior opposite angle so this cannot be equal to this so this this 90 degree cannot be equal to or this angle cannot be equal to this angle because so this is not this is only possible when D and D dash are coinciding so I have proved that all ABCD are falling on the circumference hence it is a cyclic quadrilateral now theorem 3 if one side of a cyclic quadrilateral is produced the exterior angle is equal to the interior opposite angle and this is nothing but what I am saying you is that I draw a cyclic quadrilateral like this this is ABC and D and if I produce this line BE I am saying that angle CBE this exterior angle would be equal to angle D so I can write that angle B plus angle D is equal to 180 degree and angle CBE is equal to 180 degree minus angle B so angle D would be equal to 180 degree minus angle B from here so this will be equal to angle CBE that is what you have to write over here so this is how it has to be proven these are the 3 theorems which are required for cyclic quadrilateral I have proved everything now we will solve 4-5 questions and then we will wrap up the session so the first question which I am giving you is prove that every cyclic parallelogram is a rectangle let me know once you are done okay 3 people have done there is nothing in this question just draw a parallelogram here and this is A B C D parallelogram I know opposite sides are equal if I prove that all angles are 90 degree I will be able to prove that it is a rectangle so opposite sides are equal so I write that AD is equal to BC and AB is equal to CD I have written this now angle A plus angle C is equal to 180 degree but in a parallelogram opposite angles are also equal so angle A is equal to angle C hence A is equal to C that is equal to 90 degree and same I can write here B plus is equal to 180 and B is equal to B is also equal to B so this is also equal to angle B and D so I prove that all angles are 90 degree and opposite sides are equal hence ABCD is a rectangle now let me go to another question this angle is 84 degrees and this angle is 80 degrees you have to find out angle QBC and angle BCP I take class till 7 sorry 545 so 20 more minutes okay Shraddha has done it Rana has done it 2-3 people have done it okay most of you have done it still I will do it so this line PQ has been drawn now try to understand in if I have to find out QBC this triangle so first I am trying and then I have to find BCP so this angle so what I am trying to find first I will finish it on this side so this angle would be 100 degrees and this angle would be 96 degrees because in a cyclic quadrilateral opposite angles are supplementary so this angle would be 80 degrees and this angle would be 84 degrees again this angle would be 100 degrees and this angle would be 96 degrees so all the angles I have written over here there was nothing in this question now I am drawing another circle this is A this is B this is Q this is C this is D this is P I will match this P like this so this angle is 60 degrees and sorry this angle is 80 degrees you have to find angle DPC and angle BQC let me know when you are done ok Ritu has done it anyone think Sanjana Sanjana is saying I need to solve Ruti I think has done it so look at here angle P DC this angle how will I find angle DPC angle DPC is nothing but angle A plus angle B plus angle P is equal to 180 degrees and angle P would be equal to 180 degrees minus 80 minus 60 there is nothing else in this so this will be equal to 40 degrees because in triangle ABP this is 60 this is 80 so this is 40 there is nothing in it now I have to find out angle BQC this one I have to find out so what is this angle this angle is 80 so this angle would be 100 this angle so in triangle QAD angle A plus angle D plus angle Q is equal to 180 degrees and this is 60 this is 100 is equal to 180 degrees so angle Q is equal to 180 degrees minus 160 degrees which is equal to 20 degrees write down question is in an isosceles triangle ABC AB is equal to AC a circle passing through point B and C intersects AB and AC at points D and E prove that this is parallel to BC are you guys done okay I got two people saying that they are done with it so what I do is I draw a circle and here is my called BC and this is point A and I draw a line which is ED so I know that AB is equal to AC and then ABC is equal to angle ACB now side BD of the cyclic quadrilateral is produced to meet at A so of cyclic quad is produced to meet at A this is the case so angle AD is equal to ACB why? because try to understand if this angle is X this angle is 180 minus X and so this angle will be X so in a cyclic quadrilateral we just proved that theorem that the exterior angle of opposite angle would be equal to this interior angle AD is equal to ACB and this angle and this angle are equal now try to understand this angle is also equal to this angle so I can write it as this is equal to ABC so AD is equal to ABC so this is X and this is X so these are corresponding angles and if these are corresponding angles it can only happen when BC is perpendicular to DE so this is how you have to solve it one more question and then I'll finish of the session the question is prove that the altitudes of a triangle are concurrent are you done? few people are saying that they have done it I will have to solve it so see what I am doing is let's take a triangle ABC BE is perpendicular to AC CF is perpendicular to AB and they intersect at point O and this is BD I have to prove that what I am doing I am drawing two perpendicular so even is CF perpendicular to AB and BE perpendicular to AC what I have to prove is AD is perpendicular to BC why because I have shown that all the points are passing from O so BEC is equal to angle BFC which is equal to 90 degrees now you try to understand if I draw a triangle like this this will not be possible until and unless I prove that these points E and F because they are making 90 degrees this will not be possible until and unless I prove that this E and F are on center of the circle because making this 90 degree will not be possible until and unless they are in look at this if this is a line and this is making 90 degree angle here in a circle this point has to be on the circumference of the circle so that is why I will have to make a different line where I will prove that this is BCF E and I will draw this I will draw this and this would be somewhere here I will draw something like this this is E not this is E and I will draw a line and then I will draw a line from here and I will write that because what happens this the CF is perpendicular here this is not possible so what I am saying is angle BC or sorry quadrilateral BCEF is a cyclic quadrilateral so if it is a cyclic quadrilateral angle ECB plus angle BEF is equal to 180 degree and similarly angle ACD plus angle BFE is equal to 180 degree so why this is possible because angle ECB is also equal to angle ACD this is A so to solve this I will write angle ACD and BFE can be written as angle BFO which is here plus angle OFE which is equal to 180 degree so I know that angle ACD and BFO is 90 degree so 90 degree plus angle OFE is equal to 180 degree so angle ACD plus angle OFE is equal to 90 degrees so what I am proving that EF so from here I can write that I will have to make some space here so from here I can write that angle OFE would be equal to angle OAE because they are in same segment so now I can write that angle ACD plus angle OFE is equal to 90 degree and this OAE is equal to DSC which is also equal to 90 degree because angle OFE is equal to angle DSC but in triangle ACD I know that angle ACD plus angle DSC is equal to 180 degree so angle ACD plus this angle DSC is equal to 180 degree minus angle ADC so I can write this as 180 degree minus angle ADC is equal to 90 degree so angle ADC is equal to 180 degrees minus 90 degrees which is 90 degree so ADC this angle is 90 degree hence AD is perpendicular to BC this is what we had to prove now I am wrapping up the session if you have any doubt you can ask me in next class next class I will be taking online on Monday so I will post the timings in in the group or let me post the time in the group what is the time when you all are comfortable are you comfortable from 6 to 8 kind of timing it's 6 to 8 pm fine for Sunday's class sorry Monday's class okay fine so next class will be on Monday and I shall be taking that class fine thank you so much for joining the class