 Hello and welcome to another session on gems of geometry and today we are going to discuss another very interesting lemma and The lemma is something like this that if a b c and b are four points Okay, such that a b is equal to bc and bc is equal to cd and Angle a b c is equal to angle b c b then The four points a b c and d are consignally that is They lie on the same circle now. It appears here that they are you're lying on the same circle. So let me first switch this off So now you can see there are four points a b cd and It's given you can see I have mentioned the values of the length of a b b c and cd So a b is 5.42 b c is 5.42 and cd is 5.42 in this configuration And angle a b c you can see this is 151.43 degrees and b cd is equal to 151.43 degrees So whenever that happens The four point a b cd will be consignally But there is one constraint the constraint is these angles a b c and b cd must not be less than 60 degrees They must be greater than 60 degrees only then this will hold Okay now so This is for first part of the lemma the other part of the lemma says that Not only these four points, but there is one more point e such that The angle a e d Right angle a e d is thrice the angle a o b now While we are dealing with the proof you will see that triangle a o b b o c and c o d are congruent triangles Okay, such that these vertex angles that is angle a o b b o c n c o d are all Equal okay, so I can show you here. So if I Take these Angle so let me show you the angle a o b right Let me a O and b so you can see this is 28.57 b O See that's also 28.57. Let me just separate them out. Yeah, so Delta and epsilon These two angles right and then finally if you see C o and b that's also Gta 28.57. So all three are equal now these three are equal So second part of the lemma I was discussing was this that if there is a point e on the other side of ABCD, right on This side as in you know on the side facing the arc This are ABCD, which will be there. So if you find a point e such that a e d is thrice Alpha thrice alpha where alpha is or where two alpha is this angle a o b Okay Then this point e also lies on the same circle once again What does it mean? So if you find a point e such that a e Be is thrice alpha now you'll ask. What is alpha? To alpha is the angle a o b that is if a o b is to alpha then a e b is three alpha correct if such Point exists that is if you can find a locate such point e Then that e point is also on the same circle on which a b c and d are lying So let me now switch on the circle. So this is a circle Okay, now what I'm going to do is I'm going to vary the positions of dcb a and c whether e Lies on that circle or not So I'm going to change the length as well as the position of the line segments a b b c c d Okay, so here is I'm changing the angle first. So if you can see and I've changed the angle ABC and BCD are still same But then you know that the positions and configuration has changed and You can see all of them are still lying on one circle. So they are consicclic, isn't it? So whatever be the position of a b c and d they all lie in one circle Okay, so let me also change the length of the site so that it becomes more valid So I'm changing the length as well as the position and you can see They do not leave the circle, right? So they are all on a circle, right? So clearly a b b c c d Are equal then and angle a b c and b c d are equal then the The point a b c d are consicclic As well as there exists a point e such that if angle a o b is to alpha Then a e d will be three alpha, right? That's what This particular lemma suggests now in the second part of this video we'll try and prove this lemma In the previous part of the video we had seen the demonstration of a very interesting lemma and This part we are going to prove that lemma. So what was the lemma? Let's first Discuss that so it's given that a four points a b c and d Satisfy the conditions what conditions first of all a b is equal to b c is equal to cd Okay, so you can see in the figure a b is equal to b c is equal to cd And this angle a b c is equal to Bcd angle, okay, if these conditions are met then A b c d are consicclic points first of all will prove that and then we'll also prove that if there exists another angle Three alpha, right or a point if there exists a point e on the other side of this arc a b c d Such that angle a e d is thrice alpha Okay, and what is alpha here alpha is nothing, but this angle which the Each of these a b b c and cd are making at the center of that particular circle on which they lie Okay, then then that point e also Is consicclic to the same a b cd. Okay, that means he also lies on the same circle So there are two parts once again first we have to prove that if a b is equal to bc is equal to cd And angle a b c is equal to angle b cd and both of these have to be more than 60 degrees. There is a reason why we are doing this. We will try and explain it later so If that conditions conditions are met then a b cd are consicclic. So let's first prove that Okay, so what I've done is I have drawn a circle No, sorry, I've drawn a b is equal to bc is equal to cd and I made sure that angle a b c is equal to angle b cd Now what I've also done is construction some construction has to be done. So the first construction is ob and OC ob and OC are Internal angle by sectors angle by sectors Internal angle by sectors of angle a b c and angle b cd Okay, and they meet it. They meet at oh, they meet at Oh Okay, so if you look at triangle boc Boc boc. Sorry is what do we see here? Boc is an isosceles triangle. I saw Celestrangle why? If you see angle obc obc is Equal to angle ocb. Why is that because both of them are equal to half of Angle a bc or half of angle bcd whichever because both of angle a bc and bcd are same given Okay, so the moment this becomes isosceles then therefore We can say ob Is equal to oc? Okay ob is equal to oc Right now similarly triangle Oh Yeah, oh a b o a b is congruent to triangle oh DC oh DC why Because first of all ob is equal to oc just proved our and What else? Angle a b o Or ob a whichever way a b o yeah a b o is equal to angle dc oh and both are equal to half of angle a bc or half of angle Bcd is it because this angle is 50% or half of a bc and a bc and bcd So these two angles are same a bc and bcd are Same so 50% of them that is half of them will also be equal So ob is equal to oc and a b o is equal to dc o and another one is a b is equal to bc correct, it's given that means by by Side angle side congruence criteria right triangle. Oh a b is congruent to triangle odc Right, so that means by cpct. We can say oh a is equal to Od oa is equal to od and what else? We can also say That angle o a b is Equal to angle Oh dc correct odc also angle a ob a ob is equal to angle cod These are the things which we can infer from it now so after this We can also prove that triangle Oh a b. Oh a b is congruent to triangle Oh cb. Oh Cb Oh a b is congruent to ocb and how's that? Ob is equal to common both sides have ob then we have angle oba Is equal to angle obc Is it obc and which is equal to half of angle a bc right because ob is the bisector There and also a b is equal to bc Okay, therefore by which criteria again s a s s a s triangle oh a b is congruent to triangle ocb Okay, that means what we can say oh a is equal to oc now previously we had proven that oh a is equal to od and Also, we had proven that ob is equal to oc. So what does it mean? It means Oh a is equal to ob is equal to oc is equal to od right So it seems or it appears that oh a a b bc and cd are the chord with a center Oh and radius is Oh a ob oc od, right? So all is this equal to r Okay, so hence we say that a b c b lie on a circle with center oh with center oh and radius Is equal to a b sorry oh a or equal to ob or equal to oc or equal to od right? Hence we prove that a b cd are consiclic a b cd are consiclic hence this circle c is these you know, let's say small c is the The given circle so hence now what how do we prove that? You know there is a point e Which is having an angle? Let's say a e d is 3 alpha Right, so I have considered a ob is equal to 2 alpha boc is equal to 2 alpha and Cod is equal to 2 alpha because now all of them are equal because there are three congruent triangles in the middle So if these are 2 alpha then There point there exists a point e on the other side of this arc a b cd where if that is 3 alpha Let's say angle a e d is 3 alpha Then e also lies on the same circle Okay, and that will be so and it can be very easily proven Why because let's say a a ob is 2 alpha Definitely, there is a point e on the circle where a e d is alpha is it? So if you look at this circle carefully, so what is this circle? Let me draw the circle once again and Okay, so let us say there is a circle this circle and we had So let me draw a circle once again Okay, this is a circle and we saw that this is the center o and let's say this point was a This point is b This point is C. So all of them are equal a b bc CD, let's say and and We have a point over here, right? Let's say e So if you see if this is 2 alpha and this is o clearly this point Or this angle rather is alpha Why because the angle subtended by an arc or a chord at the center is double or twice that its substance at the others part of the circle, right and then similarly you see This also would be alpha by the same logic because this is 2 alpha and finally This one also is alpha Correct, so if you see there is a there is a point e such that angle a e D is 3 alpha if Angle a o b is equal to angle b o c is equal to angle COD is 2 alpha Isn't it? So this will this point e exists definitely if and obviously this point has to lie on The circle then only this condition of the angle subtended at the center is twice that of angle subtended at the segment Is true, correct? So hence e is nothing but e are such points Where this condition is satisfied and obviously that will be satisfied only when the point e lies on the circle as well So hence we prove that a b c and on b as well as e All of them are consicclic points Consicclic points now we are going to use this lemma in proving Morley's trisector theorem whose Demonstration or you know validation. We just tried in the previous video in the next video. We'll try to prove Morley's trisector theorem and we'll be using this particular Lemma there to prove Morley's trisector theorem. So I hope this theorem is understood or lemma is understood by all