 Now session that we had, we discussed about our in-circle radius, right? Small r. And we had seen several formulas for small r. So I'll be continuing with that same topic. And today we'll be introducing few more circles to you and their radius. So let's get started. Maybe we'll not be able to finish the chapter in today's session. So I would be requiring one more session after this. May not be a very long one. Maybe one hour, one, one and a half hours will be required. Maybe after today's session. So after today's class, one more session of properties of triangles and then we can start with our binomial theorem as well. Okay. Anyways, let's get started. So when we talked about the in-circle, in-circle, we talked about the radii of the various formulas for the radius of the in-circle, right? And these were the formula that we had seen, delta by s, right? We had also seen s minus a tan a by 2. We had also seen s minus b tan b by 2 and s minus c tan c by 2. Okay. And apart from that, we had also seen a special formula relating the in-circle radius with the circum-circle radius. And that formula was 4r sin a by 2, sin b by 2, sin c by 2. So this was what we had done in our last session of finding the radius of the in-circle. Okay. Now very interesting inequality exists between small r and capital r. Okay. A very important relation between, relation between, between small r and capital r. Okay. And that relation is capital r is greater than equal to twice small r. Okay. That means to say that the circum-circle radius is greater than equal to twice of the in-circle radius. Now, a million dollar question here is how? Why is this relation true? Can we prove this relation? Okay. So are you proving this relation pretty quickly for you? For this relation, for proving this relation, I need to prove one very important inequality, which in general you should be also aware of. So for proving this, we'll first prove that. Okay. In any triangle, in any triangle, in any triangle ABC, cos a plus cos b plus cos c will always be lesser than equal to 3 by 2. Okay. And from here, we'll derive one more identity that sin a by 2, sin b by 2, sin c by 2, this expression will always be less than equal to 1 by 8. By the way, if you get the first one, the second one becomes very obvious from there. Okay. So let us derive the first one first. Why is cos a plus cos b plus cos c lesser than equal to 3 by 2 for any triangle ABC? Okay. So let's prove that. And via this proof, we will be able to get to our result that capital R is greater than equal to 2R. Okay. So first we'll see this proof. From here, we'll see this proof. And from all of them, we'll, you know, reduce that your capital R will be greater than equal to 2R. Okay. So let's prove this. So first of all, I'll call this expression cos a plus cos b plus cos c as let's say x. Okay. So let's say this expression is x. Okay. Now, we have only learned our transformation formula. So I mean, many of you would not have, you know, remembered it, right? How many of you actually remember that this could be written as 1 plus 4, sin a by 2, sin b by 2, sin c by 2. Anyways, don't worry about it. If you have not, if you have not remembered it will be anyways, you know, proving it in some time. But before that we'll do some small activity here. I will use the transformation formula cos a plus cos b on this particular term. Okay. So cos a plus cos b, we all know is becoming 2 cos a plus b by 2 into cos a minus b by 2. And let's also do one small thing. Instead of cos c, we will write 1 minus 2 sin square c by 2. Okay. So let's say this whole thing is an x. Now, dear students, please remember we are dealing with a triangle. So we are dealing with a triangle means a plus b plus c is a pi. That means a plus b is pi minus c. So a plus b by 2 is pi by 2 minus c by 2, which means cos a plus b by 2 is actually sin of c by 2. Cos of a plus b by 2 is sin of c by 2. So what I'm going to do here, I'm going to just write this as 2 sin c by 2 cos a minus b by 2. And of course, we have the same stuff over here. Now, all of you please pay attention. I'm going to now create a quadratic in sin c by 2 from here. Okay. So let me create a quadratic in sin c by 2. Let's take everything to the left hand side. So minus 2 sin c by 2 and you will have x minus 1. Okay. Now, all of you please pay attention. This is your small a. Okay. This is your small b. Okay. And this is your small c. Right. Now, we all know that in a quadratic equation, if the roots are real, the discriminant is always greater than equal to 0. So here also, if you see, sin c by 2 would be a real term. If you're talking about any angle from a triangle, sin of half of that angle, any angle you take it a triangle will always be a real value. Isn't it? So from here, can I say that? So let me just wipe this off because this is of no use to me now. Yeah. So from here, can I say that the discriminant of this particular term should be greater than equal to 0 because your sin c by 2 term would always be real. This term is a real term. Yes or no? So discriminant is what b square minus 4 a c. So b square, b square. Okay. Minus 4 a a c. This should be greater than equal to 0. Please note, b is your this whole term. So b square will be 4 cos square a minus b by 2 minus 4 a a is 2 and c is x minus 1. Okay. So from here, you drop your four term throughout. So it will become this. Okay. Now, all of you please pay attention. Something very interesting I'm going to do here. So from here, I can say 2 x minus 1 is less than equal to cos square a minus b by 2. Correct. Isn't it? So from here, I can say that this inequality will come out. Now, always remember cos of any angle and if you're squaring it, it definitely has to be lesser than equal to 1. Correct? So which implies twice of x minus 1 will always be less than equal to 1, which means x minus 1 will always be less than equal to half. That means x will always be less than equal to 3 by 2. Therefore, for the triangle, I can say cos a cos b cos c will always be lesser than 3 by 2. It's a very, very important inequality, by the way. Recently, I was, you know, taking the crash course for your seniors who are going to appear for the JEE main exam. And I'm sure exactly one year from now, you will be also attending the same crash course and preparing for some of the most awaited competitive exams in India. And there, I asked people whether they are aware of it. Trust me, 40% of the people did not know this inequality and it is supposed to be a very important inequality. So no doubt, in any triangle A, B, C, cos a plus cos b plus cos c will always be lesser than equal to 3 by 2. Any questions? Any questions in this entire derivation? Anything that you would like to copy? So basically, I use my discriminant situation in a quadratic equation to come to this inequality. To give you a picture, very important, you know, information which I'm giving you, many of the inequalities actually come from your discriminant of the quadratic equation. Okay, and you would be surprised to see that it finds a lot of its applications in solving many questions. Many of the inequalities also come from your AMGM fact. But here we cannot use AMGM because you may have, see, AMGM is applied to positive terms, isn't it? So cos a, cos b, cos c may be negative also because if it is an obtuse angle triangle, any one of them can be negative as well. Therefore, AMGM inequality is not helping out in this case. However, AMGM inequality, AMGM HM inequality to be more precise, that is also one of the very heavily used, you know, situations in order to prove or find many inequalities. Is it find any questions with respect to this proof? Now, coming to the second proof, the first proof we have already done. Okay, this is already done and tested. Coming to the second proof as to why sin a by 2, sin b by 2, sin c by 2 will always be lesser than 1 by 8. This proof you have already seen. So I'll just do the proof of 2 over here on the top, proof of 2. So we have already seen cos a plus cos b plus cos c is lesser than equal to 3 by 2. Now start using your transformation formula the same way as we did. However, I will be not writing the whole step once again. I'll be just quickly saving my time. Okay, so this was the inequality that we had written. So same thing I'm copying. I'm not doing anything differently. This left hand side what I wrote, I had already written over here. Yeah, look here. Here, I've already written over here. So that's why I don't want to redo the same stuff. Okay, so now here what I'm going to do is I'm going to keep this one aloof and from this term and this term, I'm going to take a minus 2 sin c by 2 term common. Okay, so that leaves me with, oh sorry, let's take plus. Yeah, so that leaves me with cos a minus b by 2 minus sin c by 2. Correct? So far so good. So far so good. Any questions so far? Okay. Now, this term, which is your sin c by 2, can I not write this again as cos a plus b by 2 because after all we are dealing with a triangle. Correct? Yes or no? Now again, use your transformation formula cos c minus cos d is 2 sin c plus d by 2 c plus d. Please note that c plus d will give you a. So c plus d by 2 will be a by 2 into sin d minus b c by 2. So d minus c will be b. So b by 2 will come over here. Okay, so from here what do we see is that 1 plus 4 sin a by 2 sin b by 2 sin c by 2 is lesser than equal to 3 by 2. Yes. Correct? Which clearly implies that sin a by 2 sin b by 2 sin c by 2 will be lesser than equal to by the way if you want to put a 4 over here for the time being you can do that. So subtract the 1 on the right and again divide by 4. So this I'm doing it in the same step rather than writing it once again. So this proves the second inequality. Okay, so how these two inequalities will help us to reach our original inequality that we started our class with where I'm claiming that my in-circle radius is more than equal to or greater than equal to twice of the in-circle radius. So now everybody please pay attention. First of all, if you want to copy anything from here, please do so. I can give you around 30 seconds to copy it. All right, now coming back here. Yeah, coming back here, we have already seen that our r is expressed as 4r sin a by 2 sin c by 2. Correct? r is expressed as 4r sin a by 2 sin b by 2 sin c by 2. So from here, can I say that sin a by 2 sin b by 2 sin c by 2 is nothing but small r by 4 capital R, which just now you proved is lesser than equal to 1 by it. Correct? So if you take this inequality and try to find what is our relationship between capital R and small r, you will automatically see that 2r is lesser than equal to r, which means r is greater than 2r. Hence proved. So while proving this, we proved three important inequalities. One is in for any triangle ABC, and remember this, maybe one year down the line when I'm taking the crash course for you and if you forget this, then I mean, when you have to recall it, that's it. That's all I can say. Yeah, so we learned three inequalities. One is your cos a plus cos b plus cos c is lesser than equal to 3 by 2 for any triangle ABC. Okay? And from here itself, you get the result that sin a by 2 sin b by 2 sin c by 2 is lesser than equal to 1 by 8 for any triangle ABC. And from here, from these two, we end up getting this result that capital R is greater than equal to 2r for any triangle ABC. Please note this down. Okay? Any questions? So we are now going to move towards our next type of circles. So I think we had already discussed circum circle, in circle, now the time for the third type of circle that we are going to associate with the triangle. So can I go to the next slide? If you have any questions, any concerns, time to ask is right now. Okay, no questions. Fine. Then let's go on to the next type of circles, which we call as the scribe circles. Okay, now we'll be talking about three types of scribe circles. Okay, so as scribe circle opposite to a vertex a scribe circle opposite to vertex b and as scribe circle opposite to vertex c. So altogether, we'll talk about three types of scribe circles. Now, first of all, what's an scribe circle? Let us try to figure that out. See, let's say this is our triangle. I'm just making a very small triangle, because if I make a bigger triangle, my circles also will come out to be very big. Okay, so let's say this is our triangle. Okay, abc. Okay, now what I'm going to do here is in order to make an ascribed circle opposite to vertex a. So what I'm going to make now here, what I'm, what construction I'm going to do now here, I'm going to make a circle ascribed opposite to vertex a. Okay, so what is a scribe circle opposite to vertex a? So it's a circle which basically will be touching your bc line ac and ab line extended. So what I'm going to do, I'm going to just make a quick extension of ab and ac. Okay, just a quick extension. Now let me, let me bisect internally the angle a. Okay, so this line, blue line that you see, this blue line is the internal angle bisector of a. That means this is a by two. This is also a by two. And let me externally bisect the angles b externally means this angle you are bisecting. This angle you are bisecting. Okay, by the way, this was b angle. So this whole thing was pi minus b. So each one of them will be pi by two minus b by two. This one also pi by two minus b by two. Okay. And similarly externally bisect angle c as well. That means this angle is equal to this angle and each one of them will be equal to pi by two minus c by two. Wherever they meet, this point is called I one, right? Now this I one will be such a point which would be equidistant from ab extended and bc. Do you all agree? I one would be a point which would be equidistant from ab extended, ac extended and bc. That means if I drop a perpendicular from I one onto these three sides, they will be of equal lengths. Okay. Maybe my diagram is not very accurate. But these green lines that you see my dear students, these green lines are of equal length. Okay, this is equal to this is equal to this. And this equal distance is called R one. Okay, R one by convention, we call it as R one. So in the light of this entire diagram, we can safely conclude that if somebody takes the center of any circle at I one and radius as R one, so center at R I one and radius as R one, he can draw a circle in this manner. So I'm just drawing a circle for you. Orange one I will draw. Oh, so sorry. Okay, so this is a circle. Maybe my circle, yeah, may not be very accurate. I'm just trying my best to fit it. Yes, okay, almost there. Almost there. Yeah. Yeah. So this circle that you see, this circle is called the circle ascribe opposite to vertex A. Okay, vertex A. So its center is I one. Okay, and radius is R one or sometimes people call it as RA also. So for this circle, your center will be at I one. And radius will be R one or RA. Okay, some books will call RA, some books will call R one, both are fine. Okay, now using the same approach, you can actually also draw a circle, which is opposite to your vertex B. Okay, that means let's say if you extend this line, I'll just quickly draw one so that it doesn't take much of your time. Yeah, so let's say I just draw this extension. I draw this extension. Right. I again internally bisect the this angle B. Okay, externally bisect these two angles. Okay, so with this as a center, which we call as I two and the perpendicular distance that we drop from I two onto the three sides, a BA extended and you can say AC and you can say BC extended they will all be of equal length. So these three will be of equal length. And this length is called R two. This length is called R two. So with I two as the center and R two as the radius, you can sketch a circle like this. Sorry, circle, circle. So this is a, this is a circle with I two as the center and radius as R two. So this is called the Scribe circle opposite to Scribe circle opposite to vertex B opposite to B. Okay. Okay. It's center is written as I two center is written as I two and radius is called a radius is mentioned as R two or sometimes some books will also write it as RB also just to show that it is the they write a subscript of B just to associate it to a Scribe circle opposite to B. Okay. In the same way we can do this exercise by drawing an Scribe circle opposite to opposite to C as well. Okay. So for that we need to extend this line. We need to extend this line. Okay. And we need to internally bisect this angle. Okay. Externally bisect these two. Okay. This center will be called as, this center will be called as I three. Okay. And from here, if you drop perpendicular from the three sides, they will be of equal length. This, this and this, they will be of equal length. And these equal length values are called R three or RC. So you can call it as R three or RC also is called. Okay. And if you make a circle having the set, I always draw a line when I have to draw a circle. Yeah. So if you draw a circle with center at, with center at, yeah, yeah, with center at I three and radius as R three, you get this as your Scribe circle opposite to C. So this is called a Scribe circle opposite to opposite to C. Okay. So it's center is written as I three. And it's radius is written as R three or RC, RC or RC. Okay. By the way, one of very important thing that I would like to tell you is that this line that you have over here, I mean, I'm just correcting this figure slightly. This would be a straight line actually. Okay. So this thing would be a straight line. This whole thing would be a straight line. Okay. Anyways, it may not look very okay. So if you see I one, I two, I three, that itself will form a triangle. We'll talk about that triangle a little later on. But as of now, this triangle that you see, this triangle is called, this triangle is called the x central triangle, x central triangle. We'll speak about it a little later on in our chapter. But right now, we're just focusing on our Scribe circles. So again, I will repeat how is an Scribe circle made? So let's say if you want to make an Scribe circle opposite to a, then internally bisect angle A, internally, internally bisect angle A, externally bisect angle B, externally bisect angle C, wherever they meet, that would be your center of the Scribe circle opposite to A, that is called I one. Now the perpendicular drop from I one on to BC, AB extended, AC extended will be equal in length and that length is called R two, sorry, R one, R one or RA, which is your radius of the Scribe circle. So with center as I one and radius as R one or RA, whatever you want to mention, you can basically sketch a circle and that is called the Scribe circle opposite to it. Same goes with B and C as well. Now, let me give you some data related to I one, I two, I three coordinates and some data related to the values of R one, R two, R three. Okay, so let's take that. So for the next half an hour, we'll be discussing that. I hope nobody has any kind of a doubt with respect to how are these Scribe circles made? Okay, now, please note, the coordinates of these, the coordinates of these Scribe circles, many books also call it as eccentric circles, eccentric circles, Scribe eccentric, both words are fine for the Scribe circles. So I one coordinate, please note down, I one coordinate is given by minus AX1 BX2 CX3 upon minus A plus B plus C comma minus AY1 BY2 CY3 by minus A plus B plus C. Okay, so this is the center of the circle Scribe opposite to A. Okay, so since it is described opposite to A, you can see only A is negative, rest B and C are positive, here A, B, C are respectively your sides of the triangle. Okay, sides of the given triangle ABC, which you already know. We have been using ABC since our sine rule, cosine rule and all the same ABC here. Okay, for I2 it will become, now again, see the pattern. Since you're writing the center of the circle Scribe opposite to vertex B, B will be kept negative. Okay, as you can see, B is kept negative. Similarly, for I3, C will be kept negative. Oh, I'm so sorry. Yeah, I thought I forgot to put a minus for the B, but I'd already done it. Yeah, so this will become your coordinates for I3. Is it fine? Any questions? Now, how does this actually come? It's very simple. If you see I1, in case of our in-circle, we realize that I1 used to divide the join of A and the opposite point. So let me just, you know, make it quickly here. We had learned that in case of our in-circle, okay, in case of our in-circle, whenever we had made an in-circle, in-circle, what did we do? We had basically bisected these, these internal, these angles internally, right? Okay, so let's say this is our I1 position. I'm just making it on the, on the screen, or sorry, I position. Okay, not I1, I position. So we had seen that, let's say this is your A, B, C triangle, and this is your point D, then this BD is to DC. This BD is to DC was nothing, but C is to B, right? And AI is to ID is actually B plus C is to A, right? So in case of an in-circle I, in case of an in-circle I, this was the ratio that we had seen, right? In case of our ascribed center opposite to vertex A, let me just read all the figure, okay? So I'll not, I'll not make the complete diagram, okay? So let us say this was your, this was your ascribed center I1, okay? And this was your D point. This was your D point, okay? So D divides the join of B and C in the ratio C is to B, just like we had it for our in-circle, but AI1 is to I1D. This ratio, this ratio, please note this down, very, very important. This is B plus C is to negative A. Why negative? Because you have to go in a reverse direction. See, AI1 is in this direction, down direction AI1. I1D will be in opposite direction. So because there is a reversal of the direction happening, this is the ratio of your directed lens. So the sine of one of them has to be considered to be negative. So if you're considering this to be B plus C, the other will be minus A. Or you can do other thing, you can consider this to be minus B plus C, the other to be plus A. But you have to keep the signs to be negative. That is what is important, okay? And because of this, you end up getting a minus sign over here. Is it fine? So this is something that you should all be knowing so that you can save a lot of your time if a direct question is asked on this. Now, what is important is our radii of the ascribed circle. So we'll now be talking about the radii of the ascribed circle. So first note this down. I think NPS Kodamangala has some exam or something, right? I can see that. Some UT or exam is going on for any of the schools. I can see for HSR also. I don't know. I may be wrong also. All right. So let's talk about the radii of the ascribed circle. A radii of the ascribed circles. Ascribed circles. So I'll begin my journey with R1. Okay, let's talk about R1 or RA, whatever you want to call it. So there are actually three formulas for radii of a radius of a ascribed circle. So I'll start with the first formula, delta by s minus a. So please note down the radius of the ascribed circle opposite to vertex a is delta by s minus a. Okay. How do we get this formula? What is the proof for this formula? Let's try to derive it. So for that, I'll quickly make a diagram of a triangle. Okay. And let's internally bisect this angle. Okay. I'm just making the whole construction once again, just to prove this is it. Let's extend this further. Let's extend this further. Okay. And again, internally bisect this angle. Internally bisect this angle. Okay. So this is your position of your I1. I'm not drawing the circle because it's just a waste of time to draw the circle. This is good enough to find our answer. So let's say this is our I1. Okay. Now, all of you please understand that if you drop perpendicular from I1 onto, onto our three sides, they will be equal to the radii of the ascribed circles. So this is your R1 is also R1. Correct. They will all be R1, R1, R1. Right. Now, all of you please pay attention. If you look at the figure, can I say area of the triangle ABC, area of the triangle ABC will be some of area of the triangle AI1B plus area of the triangle B, sorry, AI1B, AI1C. But when you do this, please realize when you add these two areas, I'll just show you by shading them. If you add these two areas, you will actually end up getting, you will end up getting this area as extra. So as you can see here, this blue area has come out to be extra. Correct. So that is actually the area of BI1C. Is it? So I have to subtract the area of BI1C and whatever will be left will be your area of this given triangle. Yes or no. Yes or no. Correct. So what is the area of the triangle AI1B? Let us try to talk about that. So for AI1B, so for AI1B, this is the base, which is called small C. And this is the height, which is your R1. So let me call area of the triangle ABC by Delta that we normally do. And area of AI1B will be half base into height. Base is small R and height is R1. Do you all agree with me on this? Any concerns with the area of the triangle AI1B? Okay. Everybody's happy. Fine. What about AI1C? Can I say this will be the base? Okay. And this is the height. Now don't be surprised. Height is outside the triangle. That's fine. I mean, for an obvious angle triangle, we can have the height of the triangle outside the triangle as well. Right. So it is half BR1. And what about area of BI1C? That's again half base. Base is your VC length, which is A. And again, height is going to be R1 again. In short, you get Delta as half R1 B plus C minus A. Okay. Now all of you please recall that B plus C plus A is actually twice the semi-perimeter. Isn't it? So B plus C minus A will be 2S minus 2A. Correct. So from here, can I write this Delta back as half R1? Instead of this, I can write 2S minus 2A. Yes or no? Just cancel your two factor and you'll have something like this. And by the way, this is another formula for area of a triangle that I was talking about in the last session that will be studying various formulas for area of a triangle. And one of them is right now on your screen. Okay. So I think this is the seventh or eighth formula that we have already seen. So from here, I can say R1 will be Delta by S minus A. And that's the proof that we wanted to do. So R1 is Delta by S minus A. Is it fine? Any questions? Please make a note of it. And I'll give you two more formulas for radius of the S-cribe circle opposite to A. And similarly, we'll also write down the expression for the radii of other S-cribe circles. Okay. So let's see another formula. Formula number two. R1, another formula that you will be studying is S stand A by 2. A very simple looking formula, but we'll be proving it as well. So what is the proof for this formula? So while proving, we will learn something very interesting also. So let me recall my previous figure into this. Oh, I have almost corrupted it. Anyway, I'll make it once again. Let's make it again. Please ignore the sounds which are coming. There are a lot of constructions going around. So you may hear some hammering sounds, hammering sounds here and there. Just give me a second. Yeah, sorry. So how do you prove this formula? Let's talk about it. See, this is again a simplistic case and we'll get to learn something very interesting over here. Again, let me make a triangle here. Okay. Yeah, let's say we extend this a little further. This extend further. Internally bisect this externally bisect these two angles and let's draw a circle itself. Let's draw a circle. Okay. So this is a circle which is ascribed opposite to your vertex A. Oh, sorry. Let me redraw it again after pushing this. Now, all of you, please pay attention. Very, very interesting fact I'll show you. This is half of your angle A. That's obvious. Yes or no? Okay. Now, you would also be realizing here one thing is that this is, let's say I call this as a point. I call this as a point P. Okay. And I call this as a point. Let's say here Q. Okay. So P is the point of tangency of this line AP onto this circle. Correct. So can I say a couple of things here that BP will be equal to BQ. Correct? Yes or no? Because you can treat them as if there is an external point and you're drawing two tangents of B onto that circle. So BP and BQ will be of equivalent. Same, let's say I call this as our point. Let's say this point is our point of tangency. So can I say CQ will be equal to CR? Right? Because they are tangents drawn from an external point C onto this Scribe circle. Right? And not only that, AP is also equal to AR, isn't it? AP will also be equal to AR. Yes or no? Do you all agree with me on this? Correct? Now, let's listen to my analysis over here. Can I say the perimeter of the triangle will be AB plus BC plus CA? Okay. So 2S, which is your, S is your semi perimeter. So 2S will be your entire perimeter will be nothing but AB plus BC plus CA. Correct? So can I say AB, let it be as it is, BC, I'm going to break. BC, I'm going to break it as BQ plus QC. BQ plus QC. And CA, let it be as it is. Okay. Now, instead of, instead of BQ, can I write BP? And instead of QC, can I write CR or RC, whatever you want to call it? Okay. RC or CR, whatever you want to call it. Now, all of you please understand, AB plus BP will actually become AP. Isn't it? And RC plus CA, RC plus CA will actually become AR. Correct? And since both are equal, let's say I call them as X for the time being. So you have 2S is equal to 2X. So X is actually equal to S, which means this length AP is S and this is also S. Please note down. Right? So A2P is actually your semi perimeter of the triangle. Now, from here things become very obvious when I drop a perpendicular from I1. Okay. Please note that this is R1 and this is S. So from here, can I say tan A by 2, tan A by 2 is R1 divided by or you can say I1P divided by AP, which is nothing but R divided by S. So from here, can I not comment that my small R is S tan A by 2. So this is one way to prove it, which is actually a geometrical way to prove it. I could have also proven it by the use of my half angle formula of tan, which I will also do in some time. But first note this, this is a very interesting observation, especially the fact that the distance of A from the point of tangency of this S-cribe circle, this length is S. I have seen at least 2 questions in the past year asked on this fact. Okay. So kindly make a note of this and if you have any questions, please do ask. Please do ask. All right. So I'll be also proving this formula by the use of half angles alternately. So let me write all it. So let's prove that S tan A by 2 will actually give you R1. Okay. R1 is what? R1 is actually delta by S-A. Okay. So we'll start with my left-hand side. We all know that tan A by 2 formula is actually sine A by 2 divided by cos A by 2, which you have already done before. So I'm not, you know, rewriting it. This is your formula of tan A by 2, right? Now here what I'm going to do, very, very interesting fact, I'm going to multiply with S S minus A both in the numerator and denominator, S S minus A both in the numerator and denominator. So this becomes under root of S S minus A S minus B S minus C and down in the denominator, you will have S S minus A, correct? Which is clearly a case where this will be delta on the top and S minus A is obviously S minus A, which means you've got your right-hand side, which means this is your R1, isn't it? But this is not a derivation, actually. This is actually proving that S tan A by 2 is equal to R1, right? So for that, you need to know the result. Okay. But this is your derivation, that how S tan A by 2 becomes your R1. Okay. So if let's say this formula was not given also, from this triangle, I could have made out this relationship. Okay. So one is a geometrical way, another is by the use of your half-angle formula, half-angle formula. Okay. So if you have understood these two, let me put forth the third expression for R1 and this is something which you will prove. Okay. So if you have copied this down, can I go to the next slide? Can I move on to the next slide? Okay. So this is actually a formula for R1, but I would like you to prove it. Prove that R1 is 4R sine A by 2 cos B by 2 cos C by 2. Please note the difference. Remember your R, which was your in-circle radius that had 4 capital R sine A by 2 sine B by 2 sine C by 2. But here we have 4 capital R sine A by 2 cos B by 2 cos C by 2. So I'm giving you one minute. Anybody, by use of your half-angle formula, you can easily prove from right to the left, right? So you need to prove that you need to prove that 4R sine A by 2 cos B by 2 cos C by 2 is your delta by S minus A. Can you prove this? Please try it out. I'm giving you around one minute. Okay. Simple. So let's start with sine A by 2 formula. Sine A by 2, we already know it's S minus B S minus C by BC. Okay. Cos B by 2 formula is SS minus B by AC and cos C by 2 formula is SS minus C by AB. Now I'm sure most of you would have forgotten these formulas as well. It is very obvious. People do forget. It's not a new thing for me, but you can only remember it through periodic problem practice, right? See, once you practice, you'll remember for at least one week, but after that, you'll forget. So there has to be a periodic testing. That's why I always keep telling you that give yourself assessment tests every Sunday, Saturday, whenever you're free. Anyways, let's discuss this out. So from here, if you see the denominator would be, if you see these denominators, they will be A square, B square and C square. So you can safely pull it outside the under root sine. So it'll be 4R by ABC. And within this under root sine, you'll have S square, S minus A square, S minus B square, S minus C square. This is as good as saying, this is as good as saying, am I missing out anything? Oh, I'm so sorry. There is only S square, S minus B square and S minus C square. Why did I introduce this? I don't know. Okay. There's only S square, S minus B square, S minus C square. Now, I will purposely now introduce S minus A square and divide it by S minus A. Okay. So this is nothing but 1 by delta, which I'll write it later on. This is 1 by S minus A. Okay. And this term is actually delta square. Isn't it? So 4R by ABC, we have already seen in our last class formula that is 1 by delta. This is 1 by S minus A. And this is delta square. Just cancel off one of the deltas, giving you delta by S minus A, which happens to be your right hand side, which happens to be your R1. Right? This happens to be right hand side, which happens to be your R1. Okay. So this is yet another formula that if you want or if you require, you can use for your S-cribe circle radius opposite to vertex A. So without much waste of time, I will also give you the formula for the S-cribe circle radius opposite to vertex B and S-cribe circle radius opposite to vertex C as well in the same way. So let me, let me summarize that. So let me summarize your S-cribe circle radii. I hope you have all noted this down, all copied this down. If you have any questions, you can definitely ask. Okay. So let's not talk about, let's just summarize R1. So what we did under R1, we had three formulas. One was delta by S minus A, other was S tan A by 2. And the third formula, which we just saw in the previous slide was this. Okay. Now anybody, I mean, can you see the pattern here and if you see the pattern, can you tell me what could be the corresponding expressions for the radius of the circle S-cribed opposite to vertex B? That means R2. So what will delta by S minus A become for R2? What would be the synonymous term for that? Anybody can tell me that. If it is opposite to vertex A, there was an A coming over here. If it is opposite to vertex B, what should be coming? Tell me. Right, shall I say? Delta by S minus B. Right. Similarly, this will become S tan B by 2. So if you remember one, you will happen to remember the other as well. Now see, there's a pattern in the third formula also, whichever vertex opposite to which you are drawing your ascribed circle that will have a sign. Let's all will have a cost cost. Okay. So now if you're drawing or if you're writing the expression for the radius opposite to vertex B, you will have cost of A by 2, but you will have sign of B by 2. Okay. So something like this. Yes or no? Similarly, R3 will become delta by S minus C, S tan C by 2 and 4R cos A by 2 cos B by 2 sin C by 2. Okay. So this is basically your radii of your S-type circles. Please make a note of it. You may require any one of these formulas to solve a given question. All right. So let's take a question based on the same. Let's take a question based on the same. If you have copied it, if you want me to drag the screen to the left, there you go. Should we take a question now? Okay. So let's take a question. Are it equal to R2 plus R3 plus R where R1, R2, R3 and R have their usual meanings, which we have already seen. R1 is the radius of the circle S-type opposite to A. R2 is the radius of the circle S-type opposite to B. R3 is the radius of the circle S-type opposite to C. And small r is your in-circle radius. Then prove that this triangle is going to be a right angle triangle. And also tell me it is right angle at which angle? Take a minute. See as per our given situation, we have R1 as delta by S minus A. This is given to you as R2 plus R3 plus R. Okay. So this is your expression in terms of delta. Do one thing, remove your deltas from everywhere. Delta is not required and send this to the other side. So let's say I directly write both the steps or I directly do both the steps at the same time. But just take an LCM over here. So this will be SS minus A upon SS minus A. Here you will have S minus C plus S minus B by S minus B S minus C. Let's simplify this a bit. So on numerator you will have A by SS minus A. Here on the numerator you will have 2S minus B plus C. Please note, 2S minus B plus C is actually small A, isn't it? So you can write this as A by SS minus A upon A by S minus B S minus C. So from here I can make a small conclusion that SS minus A will be equal to S minus B S minus C. On opening the brackets over here, you end up getting something like this. S square, S square will go off. And from here, by the way, many of you would have recognized here that's why are you doing so much? Okay, this also is not required. So from here you can say S minus B by S minus C by SS minus A is 1. If you under root it, this will also be 1 only. And this is an expression of tan A by 2. So from here I can say A by 2 is 45 degrees, which means A is going to be 90 degrees. So it's a right angle triangle. It's a right angle triangle, right angle at A. Okay, right angle at A. Right. Is it fine? Any questions? Any questions? Any concerns? Okay, all fine so far. Okay, great. Now here, one thing that you can actually extend from here. If you had R2 as R1 plus R3 plus R, then it would have been a right angle at B. Right angle, but right angle at B. Okay, so this is one takeaway. And if you had R3 equal to R1 plus R2 plus R, then it would have been a right angle, but right angle at C in that case. Okay, so this is how you can extend your, I know one question to get two results from there, two more results from there. Is it fine? Any questions? Any questions? Any concerns? All right, so time to move on to the next important concept, which is called the concept of an orthocenter. By the way, everybody knows orthocenter. Orthocenter is represented by H. Many books will call it as O also, but I will be using the notation H. What's an orthocenter? Can anybody tell me that? How do you make an orthocenter? What is the orthocenter of a triangle? Orthocenter is the meeting point of the altitudes exactly. So if I drop perpendicular from A onto the opposite side BC, from B onto the opposite side AC and from C onto the opposite side AB, they will be concurrent at a point. They will be concurrent at a point that is called the orthocenter. Okay, now few important concepts that we'll be discussing about it. One is what is the coordinate of the orthocenter? Okay, we'll talk about that first. Second thing we'll talk about the critical distances over there. That means how far is the orthocenter from the vertices? That result is also very important by the way. Okay, it is used in a lot of provings as well and questions have also been framed. And of course, we'll also talk about how far is the orthocenter from the respective sides. So before I go to the coordinates, I would like to actually do the distances of the orthocenter from the vertices. So here I would give you the result and of course we'll be proving the result as well. Okay, the distance of the orthocenter from the vertex A is actually given as 2R cos A. Distance of the orthocenter from vertex B that is given by BH is given by 2R cos B and the distance of the orthocenter from vertex C is given as 2R cos C. Now, how do we get this result? Let us prove it. Let's prove any one of them. Let's prove the first one. Okay, so how do you prove that AH is equal to 2R cos A? Okay, so to prove that AH is equal to 2R cos A. Now, very simple. Look at this diagram. Look at this diagram. In this diagram, this angle is angle A and this side is your small c. So let's say this is your AP. This is the foot of the perpendicular. P is the foot of the perpendicular. So from the diagram, ABP. So in triangle ABP, you can say that AP length will be C cos A because this is the right angle triangle. This is your hypotenuse. This is your angle A. So the base of the right angle triangle will be C cos A. Simple. Yes or no? Okay. Now, what is this angle? If this is 90 degree and this is C, this angle will be 90 degree minus C. Okay, this angle. Okay. So angle HAP is 90 degree minus C. Now, all of you focus on the triangle, HAP. In this triangle, can I say AP divided by AH. AP divided by AH will be cos of 90 degree minus C. Based by hypotenuse is cos of that angle. So from here, can I say AH is going to be AP divided by cos 90 minus C, which is sin C and AP is nothing but C cos A. So it becomes C cos A by sin C and we already know that C by sin C is actually 2R. C by sin C is actually 2R. Isn't it? So there you go. We have proved this formula that AH length is 2R cos A. Okay. Similarly, you can prove BH, CH as well and please note this down. This is important. I would not have asked you to note this down had this result been just any ordinary result but it has got a lot of uses also. Okay. If you remember it, you save you a lot of time. Okay. So let me just show you the proof if at all you want to copy it. Yeah. Great. The next important thing that we need to remember is the distances of the orthocenter from the respective sides. Okay. So let me just write distance of H, distance of H from BC, distance of H from AC and distance of H from AB. Okay. And please note down the distance of H from BC is 2R cos B cos C. Distance of H from AC is 2R cos A cos C and distance of H from AB is 2R cos A cos B. I'll be proving just one of the results. Maybe the first one. So first note this result down and then I'll be proving the result. Okay. So for this, you need to look at the diagram once again. Okay. So prove that the distance distance of orthocenter H from BC is 2R cos A, sorry, 2R cos B cos C. Okay. So look at the diagram once again. So distance of H from BC, let me call this point as R for the time being. Okay. So I have to prove that HR is 2R cos B cos C. Now that is very simple actually. All of you please answer this question of mine. If this is angle B, what is BR length? If this is angle B and this is length C, what is BR length? You'll say, sir, C cos B. Yes or no? So BR length is going to be, so focus on the triangle. In triangle A, BR, in triangle A, BR, BR length will be, BR length will be C cos B. Yes or no? Right. Now, what is this angle? What is this angle? Now again, this is 90 degree and this is C. So this angle is 90 degree minus C. Yes or no? So can I do one simple trigonometry here in triangle HBR? So focus on in triangle HBR. In triangle HBR, can I say HR divided by BR is tan of 90 minus C. HR divided by BR is tan of 90 degree minus C. Please have a look at this diagram. HR divided by BR is tan of this angle. Correct. So HR which is supposed to be found out, BR which is C cos B is equal to cot of C. In other words, HR is going to be C cos B. Now cot C, I'll write it as cos C by sin C. And we already know that C by sin C from our sin law that is 2R. So HR length will be 2R cos B cos C and that's what we wanted to prove and prove. Okay, this is what we wanted to prove. And again, these three values are to be noted down because they are used. Okay, please make a note of this. They are used at several places. Okay, so I've already written everything in detail fashion. If you have any questions, any concerns, still you can please ask me. Okay, great. Now comes the concept of coordinates of the ortho center. So I'll be again redrawing the diagram once again and we'll be deriving the coordinates of the ortho center from there. So let me make a triangle first. Okay, so this is let's say our triangle. Okay, let me draw perpendicular. I mean, two of them are sufficient. You don't have to draw enough perpendicular wherever two meet, the third will also meet at the same point. So don't draw all the perpendicular. Only two are good enough. Okay. Now I have a very small question for you all. Okay, just a small question. If this is your point D. Okay, what is BD is to DC? What is BD is to DC? Let me name the topic name here. Topic name is coordinates of the ortho center. Coordinates of ortho center. Okay, what is BD is to DC? Can somebody tell me what is the coordinates of, sorry, what is the ratio of BD is to DC? Now many of you would be wondering, sir, why do we need that? Why do we need the ratio in which D divides the joint of B and C? Simple. See, my agenda is this. My agenda is if I know BD is to DC, I will be able to know the coordinates of D. Correct. Right. If I know BD is to DC, I will know the ratio in which D is dividing the joint of B and C. So I will be able to know the coordinates of D. Now, I will then find AH is to HD and since A coordinate and D coordinate would be known by then, in fact, A is already given and D would have been known by then, I could get the coordinates of H also from there. So this is my thought process. Now, let us execute. Okay. So first, BD is to DC. I'll say very simple, sir. BD length we had already seen in our previous stage that this is C cos B because this is C and this is angle B. So this length will be C cos B. So BD is C cos B and DC will be, DC will be, let's say this is your C and this is your small B. DC will be B cos C. Okay. And you would have seen that in your projection law as well. So DC will be nothing but B cos C. Anybody has any doubt with respect to these two values? Can you doubt? Simple trigonometry I have applied, nothing. So what is BD is to DC? BD is to DC will be C cos B by B cos C. Now, all of you please pay attention. In fact, I have got my answer, but I will write it in a very fancy way. What fancy way I will write it? C here. Very simple. This C I will write it as 2R cos C. Sorry, 2R sin C. Okay. By sin law. So this C, I can actually write it as 2R sin C by sin law. Okay. And this B, I will write it as 2R sin B as per the sin law. Okay. So we also know that B is 2R sin B. So these two come from your sin law. Right now. So 2R2R goes off. Sin C by cos C will become tan C and sin B divided by cos B will become tan B in the denominator. So this will become tan C by tan B. In other words, the ratio in which the ratio in which D divides the join of B and C is tan C is to tan B. Is it something which is worth noting? Yes. It is something which is going to be useful if at all a question requires it. So if somebody asks you that if you draw a line connecting a vertex A to the ortho center and extend it further, that means this is an altitude. What ratio does the feet of the altitude or foot of the altitude divide the site? So it is in the ratio of tan C is to tan B. Please note the way I have written it is not tan B is to tan C. It is tan C is to tan B. So if I know this ratio from here, can I not figure out that D coordinate will be, all of you please focus, tan B into X2. Okay. Tan C into X3 upon tan B plus tan C. Okay. Same will go for the Y coordinate also. Y2 tan B, Y3 tan C upon tan B plus tan C. Correct. Any questions, any concerns so far in getting your coordinates of D? Any questions, any concerns so far in getting your coordinates of D? Fine. No problem with this. All set. Okay. So let me redraw this whole figure once again in the side because I'll be needing it. So let me just take a snapshot of this figure and let me just drag the screen to the right and post my diagram once again here. There are a lot of extra things which I'll be raising off, not to worry. Sir, you erase the diagram only, sir. Okay. No problem. I'll bring it back. Yeah. So let's erase these extra things that we don't want. This we don't want, this we don't want. Okay. And of course, this we don't want. Now, having got the D coordinates, can I find out the ratio in which H divides the join of A D? Can I get A H is to HD? Now, this is something which is related to whatever we did a little while ago. A H. What was A H? 2R cos A. Don't forget that. Yes or no? What was HD? What was HD? 2R cos B cos C. Yes or no? So can I say A H is to HD will be nothing but 2R cos A upon 2R cos B cos C. So 2R 2R goes off. Yes or no? Any problem, any concern so far? Okay. Now, I've actually got the ratio but I would still do some manipulation with this result. I will make it a little bit fancy to look at. So what I'm going to do this result cos A by cos B cos C. I'm going to multiply with sin A on the numerator and denominator. Okay. Let's multiply with sin A. Okay. Now, we all know that sin A can actually be written as sin B plus C as well, isn't it? So only the numerator sin A, I would write it as sin B plus C. Not the denominator one. So only the numerator sin A, I'm writing it as sin B plus C. Okay. So far so good. No issues. So this result I have rewritten it like this and this result I have written it like this. So far, good. Nothing, no problem. Right. Now, what is sin B plus C? We already know from our compound angle identity that sin B plus C is sin B cos C plus sin C cos B. Okay. Multiply it with cos A and this is already sin A cos B cos C. Okay. Now, let's expand it and write it. So when you expand it and write it, you end up getting sin B cos A cos C sin C cos A cos B divided by sin A cos B cos C. Okay. Now, divide both numerator and denominator by divide both numerator and denominator by cos A cos B cos C. So if you divide this by cos A cos B cos C, this will give you tan B. If you divide this by cos A cos B cos C, this will give you tan C. And if you divide the denominator by cos A cos B cos C, it will give you tan A. Right. Yes or no? So what does this tell? This tells us that the ratio in which the ortho center divides the join of A D is tan B plus C, sorry tan B plus tan C is to tan A. But before I go on to do that final analysis, final touch up, please note down anything that you would like to note from here and ask any questions that you have. Fine. Now, everybody's happy. No issues. Okay. Great. So let's move on to the next page where we'll be doing our final touch up. Okay. The final touch up of our answer. So now that, now that again, let me call the figure once again. So I'll save my time by using this figure again. Let me erase unwanted stuff. Yeah. Now, all of you please pay attention. This point that you have D, this point is X3, sorry, X2 tan B plus X3 tan C upon tan B plus tan C. And this point that you had, I mean, sorry, the Y coordinate that you had was Y2 tan B, Y3 tan C upon tan B plus tan C. And this point that you had was already X1, Y1. And this ratio, this ratio was figured out by us in the previous slide that this ratio is tan B plus tan C is to tan A, yes or no? Tan B plus tan C is to tan A. So can I now apply section formula on AHD? Right? So if I apply section formula on AHD, what I will be writing, I'll be writing H coordinate as no doubt tan A times X1 plus tan A times X1 plus tan B plus tan C times this coordinate, which is your X2 tan B X3 tan C by tan B plus tan C. Okay? Whole thing divided by tan A plus tan B plus tan C. By the way, similar will be written for your Y as well, which you want. I can write it down just quickly without taking much time. So this is for your Y coordinate. Okay? Okay? This whole thing divided by tan A tan A tan B tan C. On simplification, on simplification, this is going to yield, by the way, this is going to get cancelled off from here. This is also going to get cancelled off from here. Okay? So on simplification, this yields H coordinate as X1 tan A X2 tan B X3 tan C upon tan A plus tan B plus tan C. Okay? Y1 tan A, Y2 tan B, Y3 tan C, Y3 tan C upon tan A plus tan B plus tan C. Now, this formula, I mean, let me tell you very frankly, and you will curse me for saying this, that this formula is hardly used. Yes, sir, you derived so much, three pages you took to derive this, and now you are saying this formula is hardly used. Sorry to break that information, but yes, I have not seen any use of this formula so far in my life. But what is important, and let me emphasize that, what is important is the process involved. You finding A H, you finding H D, you finding the ratio in which D divides B and C and you finding the ratio in which H divides A and B, those facts and figures are important. The end result is not important. What is important is the means that we underwent to get that result. So means is important here, not the end. Are you getting my point? Okay, so here you should enjoy the means, not the end result. Is it fine? See, because see this formula becomes impractical to be very frank. So if somebody gives me the coordinates of a vertices of a triangle, I will rather find out the equations of any two of the altitudes and find out their point of intersection. Why will I sit and find A, B, C and then find tan of A, tan of B, tan of C and why would I then use this big formula? A simpler approach would have been to get your equation of A D, get your equation of the other altitudes, see where they intersect. Right, that is a practical way of solving it. So that is not important here. The important point here is in this entire process, what did we learn? We learned a lot of things. We learned the ratios, we learned the distances. That is more important. Okay, please note this down so that we can move on to our next concept. So in our next concept, I hope all of you have copied whatever you wanted to. If you still have something to copy, I can wait for a couple of more seconds here. Copy done. Just write it done on the chat box so that I know you're done and it'll also tell me how many of you are paying attention in the class. Okay, only 1% has been done. Very good. 2% has been done. Good. Let's see whether it is 1 or 2. I'm there for all of you to benefit out of me. Okay, now next is the concept of orthic triangle. It's actually a special case of something which we call as peddle triangle also. Peddle triangle. Okay, orthic triangle. I mean, there is actually a difference between these two. I will talk about that difference a little later on. So orthic triangles, but if you see your books, if you refer to your standard books of J preparation, they will use the word peddle triangle also sometimes. So peddle triangle, orthic triangle, we'll discuss what are they first of all. Okay, and actually there is a difference between these two that also we will talk about. Okay, so what's an orthic triangle? Orthic triangle is basically a triangle obtained by joining the feet of the perpendicular. So let's say this is a triangle. Okay, let me just make a triangle. Okay, and let's say these are the feet of the perpendicular, which I am dropping onto this triangle. And of course, we know that the meeting point is the orthocenter. So these are your altitudes. Okay, let me name them as KLM. So KLM are your feet of the perpendicular dropped from the opposite vertices. If you connect a triangle or if you make a triangle by connecting the feet of these altitudes, then you get a triangle which we call as the orthic triangle of ABC. This is called the orthic. The word orth has come from the word perpendicular, ortho-gunnel. Okay, so orthic, that is why we also call it as orthocenter. Ortho means perpendicular. Later on also you'll use this word ortho-gunnel. Okay, so orthic triangle of triangle ABC. So this green triangle that you see that is called the orthic triangle of triangle ABC. Now what is peddle triangle on the other hand? Peddle triangle is actually a more sophisticated case. You can say more generic case. So if you have a triangle, if you have, let's say a given triangle, let me make it in white. Okay, and you have been given any point P inside the triangle, any point. From this point P, you start making perpendicular to the sides of the triangle. Oh, sorry, sorry, sorry, sorry. I have to make a line. Yeah. If you stop dropping perpendicular, some P onto the sides of the triangle. Okay, so again, let me call them as some name, maybe, what name can I give? M N L, maybe, no, not L, M N K, no, not K, M N Q. Okay. So this triangle that you will be forming, this triangle will be called the peddle triangle of P with respect to this triangle. So this triangle will be called the peddle triangle, peddle triangle of P with respect to the triangle ABC. Okay, so there's a difference. That's what I wanted to highlight. But in your many competitive books, they will directly use peddle triangle for everything, which is actually not correct in my sense. You may say that sir, orthic triangle is actually a peddle triangle when the point P is actually your ortho center of the triangle, right? That's what you can say. So you can say orthic triangle is a special case of a peddle triangle. So that is why they normally use peddle triangle without any rope took, without any second thought. Okay, so our main focus will be this triangle, orthic triangle. We'll be not talking about peddle triangle. Okay, so orthic triangle has got certain properties. So we'll be talking about those properties, we'll be proving it also one by one. The first thing that we'll be talking about orthic triangle is the orthic triangle in center. So in center, in center of, are you? Yeah, in center of KLM, which is your orthic triangle, is your ortho center of ABC. Okay, so ABC and KLM. So ABC's ortho center is KLM's in center, right? I'll be proving them one by one. Don't worry. Second important thing that you should be knowing is that angles, these angles which I'm writing down, please note it down. The angles of the orthic triangle, let's start with the angle K. So I'll write it as MKL. Okay, that angle is 180 degree minus 2A. Angle KLM will be 180 degree minus 2B. And angle LMK will be 180 degree minus 2C. Don't worry. I'll be proving each one of them one by one. Don't worry. So I will also prove that it is the in center of KLM is ortho center of ABC. And I will find out these three angles also. Okay, another important point to be noted about the orthic triangle is the length of the sides of the orthic triangle. Okay, so note that down also. So LM length, LM length that is the side length LM that is given by A cos A, where A is the side length of the BC side BC of the ABC triangle, which we already know by convention. Okay, so LM, LM will be A cos A. Okay, now MK, MK would be B cos B. MK would be B cos B. Okay, and KL will be C cos C. KL will be C cos C. Okay, again, these are all things you'll be proving. Don't worry about it at all. Don't worry about it at all. We'll be proving them. Okay. Third thing that we'll be proving is that the radius of the circum circle, the radius of the circum circle opposite to or sorry, radius of the circum circle of the orthic triangle, the orthic triangle KLM will be half the radius of half of the radius of the circum circle of your triangle ABC, of your triangle ABC, of your triangle ABC. Okay, so all these results we'll be proving one by one. So please note this down. Number one, in center of KLM will be ortho center of ABC. Number two, no down, no down the three angles of the orthic triangle, no down the three lengths of the sides of the orthic triangle, and also a quick relationship between the circum circle of the orthic triangle radius and the radius of the circum circle of ABC. By the way, circum circle of orthic triangle is also called the nine point circle, nine point circle. Just a name. There's a reason why it is called nine point circle. We'll discuss about it in some time. Okay. All right. So let's go to the proof one by one. So first I have to prove that the in center of triangle KLM is ortho center of triangle ABC. Now what I have to show that this point H or this point KH or this line KH is bisecting this angle. Okay. And KM is bisecting this angle. And KL, sorry, HL is bisecting this angle. If I'm able to show that, will I be able to show that H is the in center of KLM? Yes or no? So how will I show this? Let's try to understand. So here my job is to show that KH is the internal angle bisector of K, LH is the internal angle bisector of L, and MH is the internal angle bisector of M. If I'm able to show that, my job is done. That means I will be able to convincingly say that the in center of KLM is the ortho center of ABC. Okay. So let's do that. The proof is very simple. Maybe I'll redraw the figure once again if you want. I mean, it is not very simple actually, but I don't want to scare you. I mean, it's simple, but you have to be a very good observer for that. Okay. I'm again drawing the figure once again. See, this is 90 degree, correct? Okay. This is also 90 degree. And let's name the figure once again ABC and this was your H and this we had called it as KLM if I remember properly. Right. Now, something I would like to know from you rather than me telling you, do you see a quadrilateral here? Quadrilateral, I can say MBKH. Okay. What kind of a quadrilateral is this MBKH? Sir, what do you mean by what kind of quadrilateral? I mean, is it cyclic or non-cyclic? What kind of a quadrilateral? I mean, would you categorize the quadrilateral MBKH as a cyclic or a non-cyclic? That's what I want to know from you. What do you think? Is it cyclic? Cyclic means can you draw a circle passing through B, M, HK or M, B, HK? Okay. So, Prathej is saying it is cyclic. Why does cyclic Prathej? What is the characteristic of a cyclic quadrilateral? How do you recognize that a particular quadrilateral is suitable enough to be fitted within a circle so that the vertices of the quadrilateral lies on the circle? Or in other words, how are you sure that you can draw a circle passing to the vertices, all the vertices of a quadrilateral? Of course, all quadrilaterals may not be cyclic. So, how do you identify that, hey, this quadrilateral can be cyclic? Come on, I'm asking you a class 9th question, 9th or 10th, maybe. See, agreed it's a cyclic quadrilateral, but why? Why it's a cyclic quadrilateral? It was simple. In cyclic quadrilateral, the opposite vertices are supplementary. That means, this angle and this angle should add up to give you 180. In whichever quadrilateral, the opposite vertices angles add up to give you 180. That is a cyclic quadrilateral. Yes or no? Do you recall that fact or should I prove it for you? Anybody who wants to prove for that? Nobody wants. Okay, so everybody's happy with this thing. Okay, yes, it is cyclic because opposite vertices, opposite angles of this quadrilateral add up to give you 180. Okay, that means I can always draw a circle. Whenever I'm trying to draw a circle, I end up drawing a line. Okay. That means you will end up drawing a circle passing through, passing through, passing through, B or M, B, K, H, agreed. Yes or no? Okay. And let's also draw the triangle which we were supposed to draw. So this was our, this was our orthic triangle. Correct. Yes or no? Okay. Now, all of you please pay attention. Do you see that this angle and this angle will be equal? Do you see these two angles will be equal? If yes, why do you see that angle? Both these angles should be equal. This angle which I'm showing with the tick mark and this angle which I'm showing with the tick mark, are they equal? Do you agree with me? Why? Why are they equal? Just because I told it is equal. Why are they equal? Because they both lie on the same chord image, isn't it? So in a circle, you have already learned in class 10, angles subtended by the same chord on the circumference. Okay. I mean angles subtended by an arc on the opposite arc, right, will be same. Okay. So if this angle is A and this angle is 90 degree, do you agree that this will be 90 degree minus A? So this angle will also be 90 degree minus A. Sorry for writing in that small space, but I'll rewrite it again. So can I say angle MKH will be 90 degree minus A? Do you all agree with me on this at least? Correct. Now, by the very same logic, can I say quadrilateral, let me write it in white. Can I say quadrilateral, quadrilateral LHKC, LHKC. This will also be cyclic quadrilateral. Why? For the very same reason that their opposite vertices are supplementary or the opposite angles of the vertices are supplementary. This is also 90 degree. This is also 90 degree. Right. Which means I can actually draw a circle passing through HLK. Where is my circle? Let me drag this figure to this side because yeah, I am unable to fit a circle, but I will try my best to make a circle here. Okay. Okay. Just assume that. Yeah. Just assume that. I mean, I've done my best to fit a circle over here. Yeah. Correct. So this is a circle which is passing through LHKC. Okay. Right. Now, what is this angle? Can I say this angle, which I'm showing with a dot and this angle, which again I'm showing with the dot, they will be equal because they lie on the same chord HL. And this angle is again, 90 degree minus A. My dear, please note that because this is angle A. This is 90 degrees. So this is 90 minus A. So this angle is also 90 minus A. This angle is also 90 minus A. Isn't it? In other words, in other words, angle HK, HKL, that is also 90 degree minus A. What does it mean from these two results? Let me call this as result one and result two. From result one and result two, what do we conclude that your total angle, sorry, your HK line or KH line bisects the angle MKL. Isn't it? So if there is a line internally bisecting an angle, in the same way I can find out the others as well, they have to meet at the in-center. Isn't it? Okay, so I'll just quickly write down the other values also. So this angle would be 90 minus B. Let me write it here, 90 degree minus B and this angle is also 90 degree minus B by the same logic. This angle will be 90 minus C and this angle will also be 90 degree minus A. Isn't it? Just like this was 90 minus A and this was also 90 minus A. Okay, so what is happening? KH line, LH line, MH line, they are internally bisecting the angles of this orthic triangle. That means they should meet at the in-center of the orthic triangle and that in-center actually happens to be the ortho-center of ABC and that proves that proves our first statement over here that in-center of KLM is ortho-center of ABC. Got it? And not only that, it also gives us the idea of angle MKL. MKL will be twice of 90 degree minus A, so it is going to be 180 degree minus 2A. Similarly angle, so this is proved. Similarly angle KLM, KLM, that will be twice of 90 degree minus B. That's nothing but 180 degree minus 2B. And finally the angle LMK, angle LMK will be twice of 90 degree minus C. That's nothing but 180 degree minus 2C. So this proves our second result as well. Is it fine? So this whole angle will be twice of 90 minus A. This will be whole of twice of 90 minus B and this whole will be twice of 90 minus C. Are you getting my point? Should I make the figure once again for you to realize or is it like clear from the figure? Okay, so I'll just make the KLM figure. See this was your H. Okay, so H was internally bisecting these angles. So this was your KLM. Okay, so this was 90 minus A and 90 minus A. Okay, 90 minus B, 90 minus B, 90 minus C, 90 minus C. Now coming to the sides, so we had to prove that the sides is A cos A, B cos B and C cos C. Let's do that also. Quickly it will not take much of our time. Let's go to this figure. Okay, let's go to this figure. I'll be doing the proof here itself. It will not take much of our time. See, we had already figured out that, we had already figured out that this length, this length, I mean, while we were finding the distance of the orthocenter from H, we had figured out that this length will be C cos A. Am I right? Do you all remember this? If no, you can find it very easily yet again. This whole thing is C. Okay, and this is angle A. So this is the base of that triangle AML. So that is C cos A. Correct. Now let's say LM is X. Okay, LM is X. Who will tell me what is this angle? Who will tell me what is this angle? Anybody? What is that angle which I have shown with that arrow? Guys, not a rocket science. You have already done half the work. Okay, so let me help you out. So this was 90 minus C. This angle was 90 minus C and this whole angle is 90 degrees. So can I say this is angle C only? So if this is C, you will be happy to know that this is B. Okay, so this triangle A, B, C. Okay, I mean, this triangle ALM is actually similar to ABC because this angle is B which matches with this angle and this angle is C which actually matches with this angle and this is the common angle. Okay, anyways, that is something which is not going to be utilized for us, but it is a good to know information. So this angle is C, this angle is B and this angle is already A. Now, apply sign rule on triangle AML. Apply sign rule on triangle AML. So if you apply sign rule on triangle AML, can I say X divided by sine A will be C cos A divided by sine of C? Yes or no? Yes or no? Any questions? Okay. Now, C divided by sine C is 2R. Send the sine A to the other side. So it will be something like this. By the way, this is another formula that you can remember, but 2R sine A is actually small A and that's what we wanted to prove. This is our A cos A. Okay, so please note down that the length of LM is A cos A. Similarly, you can find out the length of MK which will be B cos B and LK which will be C cos C. Any questions with respect to this proof? Please do let me know. So A cos A or you can also write it as 2R sine A cos A. You can also write it as R sine 2A. All of them are same. Similarly, here also you can write 2R sine B cos B which is nothing but R sine 2B. Same thing. This is 2R sine C cos C or you can say R sine 2C. Same thing. No difference. So three things out of the four we have already proven. Now, the last thing that we want to prove is the radius of the circum circle of the orthic triangle which we call as the 9 point circle is half the radius of the circum circle of ABC. By the way, this I would like you to take as a question. Okay, so now we have to prove that. By the way, you have all noted it now because I want to move to the next slide. Everything is copied? All right. So this thing I want you to prove that radius of 9 point circle 9 point circle is half the radius of circum circle of ABC. Circum circle of ABC. Okay. I'm giving you one minute for this. Everybody please try it out very easy. Okay. And just try it done on the chat box if you are done. Yes. Could anybody get the answer for this? Anybody? Okay. So let's look into this. See, we had already figured out the length of the sides of the orthic triangle. Right. Okay. And let's say this is your, this is your, you know, triangle with circumscribe this, sorry, a circle with circumscribe this particular triangle. Okay. So let's say its radius is R dash. Okay. So let's say this radius is R dash. Now you can actually apply your sign law and say, okay, let me just write down this side. Okay. And this angle which was 180 minus 2A. Okay. So you can say that 2R dash is going to be A cos A divided by a sign of 180 degree minus 2A. Yes or no? So 2R dash will be A cos A divided by now sin 180 minus 2A is as good as a sin 2A. And sin 2A we can always write it as 2 sin A cos A. So cos A cos A goes off. Right. And what is A by sin A? A by sin A is 2R. Right. So it's 2R by 2 which is R. So what does it mean? It means 2R dash is equal to R which means R dash is equal to R by 2 which clearly proves that the radius of the orthic, the radius of the circum circle of the orthic triangle is half the radius of the, half the radius of the triangle ABC, radius of the circum circle or triangle ABC. Is it fine? So all these results that we have derived over here, please keep them in mind because they can be utilized in solving any questions. Is it fine? Any questions, any concerns here? Please copy this down and if you have any concerns, please get it immediately addressed. Okay. So we'll take a quick break over here. I know I'm giving you some early break today. We never had a break as early as 5.58. But I'm going to start a new subtopic under this concept of orthic triangle which might take around 15-20 minutes. So I don't want to keep you so late for your break. So we'll have a small break of 15 minutes. Right now, the time is 5.58. We'll exactly meet at 6.13 p.m. Okay. On the other side, we'll discuss about X-central triangle, the one which was formed by connecting the centers of the S-curved circles. So we'll be talking about that in our next discussion after the break. So all of you see you on the other side. The next concept which is very much in sync with our orthic triangle concept is the concept of our X-central triangle. In fact, in the afternoon itself, we had a quick exposure of X-central triangle. So X-central triangle, again, I would like to mention, it's a triangle formed by connecting the centers of the S-curved circles. So I'll just be drawing that triangle for you. Just the X-central triangle, I'll be drawing it. Just the triangle, not the circles themselves. So let's say this was your centers of the S-curved circles opposite to vertex A. I2 is the center of the S-curved circle opposite to vertex B. And I3 is the S-curved circle opposite to vertex C. Now here, what I claim is that our original triangle ABC, our original triangle ABC is actually nothing but it's the orthic triangle of I1, I2, I3. Again, listen to my claim, what I'm making here. The original triangle ABC whose S-curved centers opposite to ABC where I1, I2, I3, the triangle ABC is the orthic triangle of I1, I2, I3. So if you draw a triangle by connecting the feet of the perpendicular within the triangle, that is called the orthic triangle. But the triangle whose orthic triangle is the triangle itself, that is actually called the X-central triangle. So I1, I2, I3, orthic triangle, I1, I2, I3's orthic triangle. Please note that I have written the apostrophe S here. Is your triangle ABC? Are you getting my point? So now what used to be inside is now your ABC, that orthic triangle is now your ABC and I1, I2, I3 is the original triangle now, thing like that. So that means if you drop perpendicular from I1 onto the opposite sides, please note that. Oh, I am so sorry. I think this is going to be something like just a second. I'll just redraw the structure once again. Small shuffling may be required here and there. So this will become your, let's say I call this as I point. So this I is your in-center of triangle ABC and the same I is your orthocenter of triangle I1, I2, I3. Are you getting this point? What I'm trying to say here? Okay. So now ABC is nothing but your triangle obtained by connecting the feet of the perpendicular. That's very easy to prove this. See, all of us know that when we were making the, let's say I talk about I1 position. When we were making I1, we internally, we externally bisected this angle. So this angle was 5 by 2. This angle was 5 by 2 minus B by 2. Correct? Yes or no? And this whole angle is 90 degree. This whole angle is 90 degree. So this automatically becomes B by 2, which is nothing but the bisector of the angle B. Here the diagram is such that it is making a big dot over here. So let me just make an arrangement for the same. Please treat this to be a dot for the purpose of understanding. So this is B by 2. And similarly, this angle, this angle was also, see, when you were making this particular line, what did we do? We bisected this angle. So this whole angle was B, this was B. So this was pi minus B. So you bisected this angle to make pi by 2 minus B by 2. And this whole thing is 90 degree, which means this angle is again B by 2. So you are actually bisecting the angle B. And again, same way you are bisecting the angle A and same way you are bisecting the angle C. And that's what makes this I as the in-center. But also it makes I1A, the altitude of I1, I2, I3. Getting my point? So I1B is perpendicular to I1, I3. I1A is perpendicular to I2, I3. And I1C is perpendicular to I1, I2. So A, B, C is nothing but the orthic triangle of I1, I2, I3. Is there any questions? Okay. Now we will do a similar exercise as what we did in our case of an orthic triangle. So there are certain things that I would like you to prove here. Number one, the angle of the x-central triangle. Let's say these are my three angles. Okay. Angle I1, angle I2 and angle I3. Prove that these angles are 90 minus A by 2, 90 minus B by 2 and 90 minus C by 2. That means if you know the angle of A, this angle is going to be, this angle is going to be 90 degree minus A by 2. How do you prove that? How do you show that? That's number one question. Okay. We'll come to the proof. We'll come to the proof. Okay. Second thing that I would like you to prove is the length of the sides of the x-central triangle. So I2, I3 length is 4r cos A by 2. Okay. That means this length, I2, I3 length, that is equal to 4r cos A by 2. I1, I3 length will be 4r cos B by 2. Okay. And I1, in fact, I can say I2, I1, I2, that will be 4r cos C by 2. Okay. Don't worry. We will prove this slot and this slot together. Okay. So let's first prove the angle part. But before that, you please make a note of this because this is very, very important. Okay. So now let's prove that angle I2, I1, I3 is 90 minus A by 2. So that proof is very easy. Actually, see, let us say this angle was x. Let's say this angle is x. Let me write it here. Let angle I2, I1, I3 be x. So this angle is x. In terms of this x, what should be this angle? What should be this angle in terms of x? Please note that A, B, C is the orthic triangle of I1, I2, I3. So if this is x, what should be this angle? As per our previous isn't. Do you want me to go to the previous isn't? Okay. So let me just go to the previous isn't. I think before this. Yeah. See here, if this is A, this angle is 180 minus 2A, right? If this is A, this angle is 180 minus 2A. So similarly, similarly, if this is x, if this is x, what should be this angle? Of course, it is angle A, but what should be that angle in terms of x? 5 minus 2x, isn't it? Or 180 degree minus 2x, which means A angle is 180 degree minus 2x, which means 2x is 180 degree minus A, which means x value is 90 degree minus A by 2. That's what we wanted to prove. Hence, okay. So this angle x will be 90 degree minus A by 2, which is your first result. And using the same logic, you can prove that this angle, this angle will be 90 degree minus B by 2. And this angle will be 90 degree minus C by 2. And of course, this is something which we just now figured out 90 degree minus A by 2. Is it fine? Any questions? Any questions? Any concerns? Okay. All good. Okay. Now let's move on towards proving the sides. Okay, sides of the orthic triangle. So I need to prove that I2, I3 side is 4R cos A by 2. Okay. Again, let's start with the fact that I2, I3 length is x. So let's say I2, I3 length is x. Okay. Now can I say the length BC will be x cos 90 minus A by 2? Now remember, again, let me go back to the previous slide. In the previous slide, we had learned that this length ML or LM is A cos A. It's written over here as well. Okay. So if this is A and this angle is A, then this length is A cos of this angle. Using the same logic, using the same logic. Can I say if this I2, I3 length is x and this is angle 90 minus A, then BC will be x cos 90 minus A by 2. Okay. So from here I can say BC length is x cos 90 degree minus A by 2 by the same logic of A cos A. So that A is your x now and small A. Small A is your x now and capital A is 90 minus A by 2. And BC is your small A. So small A is x sin sin A by 2. Now we all know that in terms of circum-circle radius and angle A, we can write small A like this. And not only that, we can break this as 2 sin A by 2 cos A by 2. So this is x sin A by 2. Okay. Let's cancel out this. And from here, you get x as 4R because there's a 2 into 2R, 4R cos A by 2. That's what we wanted to prove. Hence that. Okay. So see all these results which I am giving right now, they're not coming from sky. They're coming from the result which we already derived for the orthic track. So these are the results which are already coming from your orthic triangle concept. Okay. So in a similar way, you can prove I1, I3 length which is going to be 4R cos B by 2 and I1, I2 length which is going to be 4R cos C by 2. Okay. And by the very same logic, can I now also add one more thing? I also want to add one more thing here. Again, one more step. Can I say that the radius of the circumcircle of I1, I2, I3 triangle will be twice the radius of the circumcircle of ABC. Yes, sir. Why? Because we had learned that radius of the orthic triangle is half the radius of the original triangle. So here the original triangle, let's say you suppose it to be your I1, I2, I3. So your orthic triangle is your ABC. So its circumcircle will be half the radius of the x-central triangle. That means x-central triangle circumcircle radius is twice the radius of the circumcircle of ABC. Okay. So these are the only things that we need to know with respect to our x-central triangle, nothing more. Okay. So I will be not discussing this any further. By the way, KBPY exams which has been postponed, whenever it happens, this chapter, many of the concepts you may use it in order to solve few questions because their geometry based questions is mostly out of the concept of properties of triangles. Is it okay? Any questions, any concerns you have, please do highlight. So what did we learn? We learned that the triangle which is obtained by joining the ascribed circle centers I1, I2, I3 is called the x-central triangle of ABC. That means ABC is nothing but the orthic triangle of I1, I2, I3. Angles of I1, I2, I3 triangle is given by these three, 90 minus A by 2, 90 minus B by 2, 90 minus C by 2. And side lengths of the x-central triangle is given by these three expressions 4R cos A by 2, 4R cos B by 2, 4R cos C by 2. And we also learned that if you draw a circle circumscribing the x-central triangle, its radius will be twice the radius of, twice the radius of, yeah, twice the radius of the actual circle. Good enough. Okay. So with this, we move on to our further concept which is actually distances between critical points, distance between critical points of a triangle or important points of a triangle. What are the important points of a triangle? Of course, circum center, in-center, ortho center, etc., etc. Okay. So I'll be giving you just an idea of how to find the distances and I mean the process is actually lengthy. So first let me start with distance between the circum center, circum center. Let's call circum center by S and ortho center. Okay. Ortho center we'll call it by H. Okay. So let us derive the distance between the circum center and the ortho center which will come out to be this figure I'm already giving you but I'll be deriving it shortly for you. So note down, distance between the circum center and the ortho center is given by R times R being the radius of the circum circle under root of 1 minus 8 cos A cos V cos C. Okay. So I'll be deriving this result very shortly. So let's look into the diagram. Let's say this is our triangle. I'm purposely making a big one so that we are not finding ourselves short of space. We already know that the circum center is obtained by perpendicularly bisecting the sides of a triangle, isn't it? Okay. So please consider them to be perpendicular bisectors. That means this length is equal to this length. This length is equal to this length and this length is equal to this length and of course these are 90 degrees. Okay. So this is your circum center S. How do you obtain ortho center by dropping altitudes? By the way, I will not draw all the altitudes because it is very obvious. Okay. This is an altitude. Let me name the triangle. Now what are we looking for? We are looking for this length HS which is shown by this red line HS which is shown by this red line. Okay. Now all of you, let me connect A to S. So what I'm making here is I'm making a triangle AHS. All of you, please focus on the triangle AHS. AHS. I hope you can see this triangle, this triangle. I'm also shading it out. Okay. So all of you can see the triangle. Fine. Now I have few questions for you. What is the AS length? Write it down on the chat box. What is the length of AS? That means the distance of the vertex A from your circum center. You see capital R, sir. This is circum circle radius only, isn't it? This is capital R, isn't it? Correct. Okay. What is the H length? You see, sir, I remember that you had done H length a little while ago when you were talking about the distance of the ortho center from the vertices that was 2R cos A, right? Correct. And what is your this angle? Now this is something very important. How would you find this angle? Okay. Now all of you, please pay attention. What is the angle HAS is what we are looking for? Now all of you, please pay attention. This angle, I mean, had you completed a triangle over here and circle over here, this whole thing would have been twice of B, isn't it? So each one of them will be BB each. So this is also B and this is also B, right? See, please imagine this is the center of the circle, which is passing through ABC. So the angle subtended at the center would be double this angle. So if this is B, this is 2B, even though I've not made a circle, but I hope you can well imagine that. And if this is B and this is 90 degree, this is obviously 90 minus B, correct? And again, pay attention. If this is C and this is 90 degree, this whole angle was 90 minus C. So from 90 minus C, if you subtract, okay, so this angle was C, okay, right? So this is 90 minus C and this part is 90 minus B. So from 90 minus B, sorry, 90 minus C, if you subtract 90 minus B, what are you going to get? You're going to get B minus C. So this angle that I'm asking you is actually B minus C. So HAS is going to be B minus C. Now most of you would be wondering yourself, what are you trying to do here? Why are you focusing on AHS? The reason for focusing on AHS is now, all of you please pay attention. Let me draw that AHS triangle once again. I'll be using the same font color. Okay, just to not confuse you. So this was your AH and I think this was HS. I made it in red. I remember it. And this was your, yeah, so this was your AHS. So what I've done is I've just made the same triangle, AHS, again, okay. So AH was 2R cos A. This length was R and this angle was B minus C. So can I say I can apply my cosine law and get my X, isn't it? So can I say cos of B minus C is nothing but this square, this square minus X square by 2 into 2R cos A into R, yes or no? So from here, can I say X square is going to be 4R square cos square A plus R square minus 4R square cos A cos B minus C. Agreed? Yes or no? Any questions, any concerns so far? Now let's take an R square out common because each of these terms have R square, R square. So this R square, if it loses, an R square will become a 1. And from these two terms, what I will do is I'll take 4 cos A common. Okay, so 4 cos A common will leave you with cos A minus cos B minus C. And let me use a curly bracket over it just to keep it different from the square bracket. Okay, so far so good. Any questions, any concerns till this step? Any questions, any concerns till this step? Okay, now can I say cos A is negative cos B plus C? Yes or no? Cos A is negative cos B plus C because A is pi minus B plus C. So cos of A is cos 180 degree minus B plus C cos 180 minus theta is minus cos theta, isn't it? In other words, can I just pull out this negative sign outside? Because both these terms will have negative sign. So this will make it plus. This you can erase it. Okay, and put a negative sign here. Okay, now use your formula for, use your formula for cos B plus C plus cos B minus C which is going to be 2 cos B cos C. Right? And there you go. You end up getting x square as R square 1 minus 8 cos A cos B cos C. In other words, you get your x as R under root 1 minus 8 cos A cos B cos C. Okay, so of course, they will not ask you the derivation anywhere, but it is good to know the the method that was used to get this result so that in case you need it anytime, you can always derive it. So that's what proves the distance between the ortho center and the circum center. Is it fine? Any questions? So I'm waiting on this page. Okay, so that if you want to copy down anything, you can definitely do it. So it all started with our cosine rule applied to this B minus C. And after that, I just used my conditional identities and I kept on reducing it and finally stumbled upon our final result which was R under root 1 minus 8 cos A cos B cos C. Is it fine? Any questions? No questions? All right. So if there's no questions, then we'll do a small analysis on this. Okay, so let me make the triangle once again. Okay, so let's make the triangle once again. Again, let's make the perpendicular bisectors. Okay, so this is your S and let's make the altitudes. Okay, so this is your H. Okay, now what I'm going to do is very, very interesting. I'm going to draw a median. I'm going to draw a median. Okay, all of us know what is a median. Median is a line drawn from opposite vertex to the midpoint of the other side. Please note that. Let's say I call this as M. M is the midpoint of B C and this S is the circum center. Okay, H is the ortho center. Fine. And let me connect. Let me connect H and S. Now here, H and S meets at this point. Let's say I call this point as capital X point. So if I draw a median, it is cutting the join of H and S at let's say capital X. Okay, now I would like you to prove that. I would like you to prove that HX is to XS. HX is to XS is 2 is to 1. Okay, so let's prove that HX is to XS that is this is to this is 2 is to 1. Okay, how to prove it? Very simple. So all of you first focus on the triangle A, H, X. This triangle A, H, X. Okay, and focus on the triangle MSX. Okay, which is your small triangle here, which I'm shading with yellow. Now, do you see some kind of a relationship between these two triangles? Does anybody see the relationship between this white shaded triangle and this yellow shaded triangle? Are they similar to each other? Do you think so? Are they similar to each other? What do you think? Yes, no. Shalini says yes. Anybody else? Exactly, they have to be similar to each other. The reason being, see this angle triangle, sorry, angle SMX is same as triangle, sorry, same as angle HAX. The reason being, these two lines are perpendicular, so they are parallel to each other, isn't it? So let's say AM is like a transverse you are making through it. So this angle, which I'm showing here, this angle and this angle, they are going to be equal, isn't it? And these two are anyways equal. This angle and this angle is anyways equal. So angle HXA is equal to angle SXM, opposite angles. So can I say by angle angles similarity, these two triangles will be similar? Yes or no? Yes or no? So if these two triangles are similar, can I make this statement that HX by XS, HX by HXS, HX by XS will be equal to AH by SM because they are similar triangles. And what is AH? AH is 2R cos A that we had already done in our distance of the ortho center from the vertex. Remember, we had done that and that time I told you that result is going to be very, very useful and you are going to use it in so many other proofs. So AH is 2R cos A and what is SM? Now, that is also very easy. Let's connect. So this is your R length, that is your length of the radius of the circumcircle and this is your angle A. So SM is the base of the triangle SBM. So SM will be R cos A. So R cos A, R cos A goes off and this becomes 2 is 2R. Yes or no? Hands proved? Okay. Yes or no? Any questions, any concerns in this proof? You can please let me know. Good enough? Now, by the same logic, can I also say that? I mean, thanks to this proof that I also know that AX is 2XS, that is this length. Okay, sorry, AX is 2XM, my bad, my bad, XM. Yeah. That means this length is to this length. That will also be 2 is to 1. Yes or no? Just like HX is 2, XS was 2 is to 1, AH is to SM was 2 is to 1, same will be AX is to XM. Now, this tells us something very interesting about X. What is the, which point on the median divides the median in the ratio 2 is to 1? That means X divides, divides the median AM in the ratio 2 is to 1. So which is this fellow X now? Now you know that this X happens to be now the centroid. Isn't it? So centroid is the point which divides a median in the ratio 2 is to 1, 2 towards the vertex side. So this XXX, which I was saying so far, that X was actually your centroid. Okay. So here is an interesting theorem that I would like to present. The interesting theorem is note this down. The orthocenter, the centroid and the circumcenter are collinear okay. And the line on which they lie, that is called the Euler's line. So they lie on, they lie on a line which we call as the Euler's line after the famous Swiss mathematician Leonhard Euler. So HGNS, let me make it HGNS, they lie on this line and this line is what we call as the Euler's line. And not only that, this ratio is 2 is to 1. So HG is to GS, is 2 is to 1, that is what we proved. Some direct questions have been framed on this concept, okay, in past, in cognitive exams. So kindly note this down. Okay. So in any triangle, the orthocenter, the centroid and the circumcenter would be in the same line. And this ratio of HG is to GS will be 2 is to 1. Now from here, I can actually find out the distance between orthocenter and centroid. So what's the distance between orthocenter and centroid? That can be easily figured out from here. So you'll say, sir, we already know the distance between orthocenter and circumcenter which was R under root 1 minus 8 cos A cos B cos C. So distance between orthocenter and the centroid will be just two-third of that distance. That's this is another point which you can keep in your formulas. In the same way, you can find out the distance between the centroid and the circumcenter distance between circumcenter and centroid that also can be found out. So this is one-third of your distance between the circumcenter and the centroid. So we already know that distance between circumcenter and orthocenter is R under root 1 minus 8 cos A cos B cos C. So this will be one-third of that easy. Getting my point. Any questions, any concerns? Okay. Now likewise, I will give you some results which I will be requesting you to prove as a homework question. Okay. So I will not be deriving everything else it will become like a spoon feeding. So I'll give you some results for homework, for your homework to derive out. Okay. So please prove the following, prove the following for your homework. Okay. Distance between, distance between circumcenter and in-center. I hope everybody knows the distance, what is circumcenter in-center? This distance is given by under root of capital R square minus 2R small r. Okay. Capital R square minus 2R small r. Distance between, by the way, this is what we call as I. This is what we call as S. So we can call it as Si. Okay. Distance between, distance between circumcenter and center of circle or you can say scribe centers, scribe centers. Okay. So as all of you know, scribe centers are named as I1, I2, I3. So I know, we already know it's I1, I2, I3. So let me write down the results. So Si1 is under root of R square plus 2R R1. Si2 is under root of R square plus 2R R2. And Si3 is under root of R square plus 2R R3. This is something that you will prove as a homework. So these four results you'll be proving. And please send it across to me if you have been able to prove it. I'll be more than happy to receive these results from you all. Okay. So with this, we are now going to move towards our concept of circumcircle radius and in-circle radius of regular polygons. Okay. So so far, we are talking about triangles. Now we're talking about regular polygons. Okay. So let me take that, but only after you have copied this. Everybody, please copy whatever result you want to. This all supposed to be proved. Right. So this is for your homework. I'm writing it down for your homework. Apart from the homework that you will get from me after the class, you have to prove these as well. Okay. Now moving on quickly without much waste of time, the circumcircle and in-circle radii of a regular polygon. Okay. Polygon or n-gon, whatever you want to call it. Polygon is also called as n-gon. Where instead of saying a polygon, we put the side length side next to it. For example, a triangle. A triangle is a three gong. Okay. A quadrilateral is a four gong. A pentagon is a five gong. A hexagon is a six gong. A heptagon is a seven gong. Octagon, eight gong. Nonagon, nine gong. Decagon, ten gong like that. Okay. So many people, many books will call it as a regular n-gon also. Okay. So let's say we have a regular, regular n-sided polygon. Okay. I'm not drawing everything here. I'm just making a few, a few lengths over here. This is, let's say an n-sided figure. So let's say I call this as a one, a two, a three, a four. Till back to n. Okay. So this is a, let's say n, you can say regular n-gon. This is a regular n-gon. Okay. Now if I ask you that, I'm making a circum circle, which circumscribes this n-gon. That means a circle which passes through the vertices of this n-gon. I'll make it again. Sorry about that. I should have actually made a circle and then made a n-gon. Okay. Anyways, I'll redraw the n-gon once again. Doesn't take much time. Yeah. So let's say this is your n-gon. And so on. Okay. So a one, a two, a three, a four. Okay. Now let's say this is the center of this circle. Okay. Now let's say the side length of a one, a two is small a. So this length is small a. Okay. And this is an n-gon. Then prove that. In fact, I'm giving this result already to you. Prove that your capital R will be given by a by two cosec pi by m. Okay. Your circum circle, circumscribing this regular n-gon will have a radius of a by two a being the side length of this n-gon and n is your number of sites of this n-gon. Prove that your capital R is going to be this. Okay. Now I'll be doing this proof for you. This is not a rocket science. It is a very simple proof. First, let me ask you, what is this angle? Let's say I call this as C. What is this angle? What is this whole angle? Who will tell me? What is this angle a one, c a two? A one, c a two, who will tell me? Write it down on the chat box. A one, c a two. Nobody. Okay. Can I say, can I say that there will be n such triangles which will be made like this? Correct. Right. So if this whole thing was two pi and there will be n such triangles, how much will come in one of the triangle? This will be two pi by n. You will say, correct. So this will be two pi by n. Isn't it? Now, if I drop a perpendicular, let's say if I drop a perpendicular, if I drop a perpendicular from let's say c on to a one a two, do you all agree that this is going to not only bisect this, but these two angles will become pi by n, pi by n each. Okay. Yes or no? Let's say this perpendicular is C m. Now, all of you please, please pay attention in the triangle c a one m. So in the triangle, in the triangle c a one m, can I say, can I say sign of pi by n will be a by two divided by R that is opposite by hypotenuse. In other words, R is nothing but a by two cosec pi by n and that's what we wanted to prove. That's it done. Clear. Any questions, any concerns related to the radius of the circum, circle of a regular polygon of n sides? Now, many people ask me, said this result seems to be coming from such a simple process. So do we need to remember it? I would say no. If you're fine with the process, you can do it very quickly in your examination. No. So don't over burden your memory with so many formulas because there's already so many formulas that we have already done in this chapter and there are many more to come actually by the way. Okay. So please keep your memory space available because you have other subjects also to cater to. Right. All right. Now what about the in-circle radius? So what is an in-circle? In-circle is nothing but a circle which will, which will just touch all the sides of the n-gone. So I'm just making one such example. Okay. So as you can see this gray circle is your in-circle. Okay. And let's say the radius of this in-circle. By the way, cm itself will be the radius of this in-circle that is written as small r. Okay. That will be given by a by two cot pi by n. Okay. So this capital r is the radius of your circum-circle. This small r is the radius of your in-circle. So how do you find the radius of this in-circle? Or how do you prove that it is a by two cot pi by n? Now I've actually done all the hard work already while finding capital R. So please note that this length, this length is your small r. If you want, I can draw that structure once again. Okay. Let me just, yeah, let me just draw that. Okay. So I'm just drawing this triangle C a1 a2. Okay. So this whole thing was a and if I drop a perpendicular, this becomes pi by n and this is your small r. Okay. So this is a by two, this is small r. So from triangle, from triangle, C a1 m, can I say opposite by base is tan pi by n. So if I make r the subject of the formula, it becomes a by two cot pi by n. That's the result we wanted to prove. Okay. No need to remember it, guys. No need to remember it if you can do this in the examination condition. And this is not time ticking. This will hardly take your, I would say 20 seconds, not more than that. That's a good investment. At least your memory space will be free a little bit. Is it fine? All right. So based on this, I would like to give you a small question. We haven't done a question for a long, long, long time now. So a small question I will be giving you. So please copy this down r capital r value and small r value. Okay. All right. So this is a small question from my side. Let's say a1 is the area of a circle. I'm writing it in shorthand. I'm not writing everything. A2 is the area of a regular pentagon, a regular five gone, let's say inscribed, inscribed in the circle or inscribed within the circle. Prove that a2 by a1 is 2 pi by 5 seek 18 degree. Prove that a2 by a1, or sorry, a1 by a2, a1 by a2. Hey, what did I write again? Again, I wrote the same thing. a1 by a2. a1 by a2 is 2 pi by 5 seek 18 degree. And let me know with the done on the chat box if you're done. I'm giving you one minute time for the same. It's a very, very, very, very, very easy question. So the area of this guy is a1. An area of this guy is a2. So to prove that a1 by a2 will be 2 by 5 pi seek 18 degree. Anybody who's done just can sit down on the chat box also. Nobody. I thought it was so easy. Okay. Let me make a simple triangle here. Now let's focus on this triangle. This triangle, the side lengths are rr and this angle is 72 degrees. Correct? Agreed? Yes or no? Okay. So let's say I call this as a and p and q. So cpq is a triangle. So what are the area of cpq? Area of cpq will be half rr sin 72. Please remember the area of a triangle is given by half into product of the two adjacent sides into sin of the angle included between them. So can I say area of the pentagon will be 5 times this and by the way sin 72 is as good as cos 18. And what are the area of the circle a1? A1 circle area will be pi r square. So as per your question you're asked to find out a1 by a2. So a1 by a2 will be pi r square upon 5 r square by 2 cos 18 degree which means 2 pi by 5 reciprocal of cos 18 which is nothing but seek 18 degree. That's what you wanted to prove. Easy. How much time it takes? Good enough? Any questions? Any questions? Any concerns? Okay. So next our topic of discussion will be quadrilaterals. Okay. Quadrilaterals. So from triangles we are slowly moving towards the quadrilateral. The first thing that we are going to discuss is your area of any quadrilateral. So let's say we have a quadrilateral abcd. Okay. So what is the formula for area of a quadrilateral? Okay. So no doubt and we'll be proving it in some time. Area of a quadrilateral is actually nothing but half the product of its diagonals into sign of the angle included between the diagonals. So no doubt this is a formula for area of a quadrilateral. So area of any quadrilateral, whether cyclic, non-cyclic is given by half into product of its diagonals into sign of the angle included between them. Sign of the included angle between the diagonals. Now many people ask this question sir. Included angle could be this also. So why not sign of this angle? See it doesn't matter whether you do sign alpha, sign pi minus alpha and result is sign alpha only. So you take any of the two angles whether you take alpha or the supplement of that. Your sign is going to be same value, isn't it? So it's going to be half in our case it is going to be half bd into ac sign of alpha. But what is the proof for this? How do we prove this? How do you prove that area of a quadrilateral is half the product of its diagonals into sign of the included angle? So let us derive this result. Can I say from the figure that area of the quadrilateral abcd will be sum of area of the two triangles adc which is this triangle plus area of this triangle? Yes or no? I'm just removing that off because I don't want to write it. So area of adc and area of abc. Any doubt related to that? Obviously it is very obvious. So area of abcd is sum of these two angles. Now what's the area of adc? adc itself is made up of two more triangles. Let me call this sp. Can I say it is made up of apd and dpc. In the same way our abc is made up of two triangles which is again this triangle dpc and this triangle which is bpc. Yes or no? Right? Now area of apd will be half ap into pd into the angle included between them. So if this is alpha this also has to be alpha. So can I say this sign alpha? Similarly area of dpc. I hope you can see the figure dpc. Yeah this dpc is going to be half dp or pd you can call it. Okay into pc into sign of pi minus alpha which is again sign alpha. I need not write sign pi minus alpha for that. Right convinced? Similarly similarly what I was saying dpc adc dpc and dpc oh once again I wrote dpc twice. So sorry bpc and bp a or apb sorry about that. So I was I was actually writing the area of abc so it is apb times sorry plus area of bpc. Yeah so what is the area of apb? So here I can say half of half of ap bp sign of pi minus alpha ap bp sign of pi minus alpha which is alpha and area of bpc is half bp cp sign of alpha half bp cp sign of alpha. Okay now pay attention to the simplification. So from these two what I'm going to do I'm going to take half pd common. In fact half pd sign alpha common. So if I take half pd sign alpha common I will be left with ap plus pc okay and from these two and from these two I will take half bp sign alpha common. So that will leave me with half sorry that will leave me with ap ap and pc. Let's start write cp let's write pc for it. Okay so now look at the figure and tell me look at the figure and tell me what is ap plus pc ap plus pc is ac isn't it? So I can say this is going to be this a to ac. Similarly the other expression which we have half bp sign alpha ap plus pc will be what? Ap plus pc will be ac. Okay that will be again ac. So again I can take half ac sign alpha common that will be giving me bp plus pd and bp plus pd my dear bp plus pd will be bd. Yes that will be bd. So it's half ac bd sign alpha and ac bd happens to be your diagonals. These two are your diagonals of that given quadrilateral isn't it? Hence prove this is what we wanted to prove. So in a quadrilateral the area of a quadrilateral is half the product of its diagonals into sign of the included angle. What's the point? Any questions? Okay now is this proof important? No the result is important. Here this result is an important one. This will help you to you know solve many many questions. Note down so that we can move on to our next expression. Everybody's done with this? All right now this is one more formula for the area of a quadrilateral which is actually slightly complicated okay and I will not be proving it right now. Maybe I can give this as another assignment to you to prove okay. So let's say abcd is you know a quadrilateral. So I am going to give you another formula for area another formula for area okay. This formula is slightly interesting. Let's say I call these side lengths as abcd okay. Now just like we have it in our triangle there is something called semi perimeter of a quadrilateral. The word itself means half of the perimeter semi perimeter. So if let's say small a small b small c small d are the sides of this quadrilateral okay. Then there is a formula for area of a quadrilateral which is very lesser known. So area of your quadrilateral abcd and this is given by under root of s minus a s minus b s minus c s minus d minus abcd cos square of arithmetic mean of any two opposite angles okay. So arithmetic mean of any two opposite angle means it could be either a plus c by two or it could be b plus d by two. So any two opposite angles you just take the arithmetic mean. So either you take a plus c by two or you take b plus d by two. However the result is not going to change. It is going to be the same right because because if a plus c is 360 degree minus b plus d because we all know that in any quadrilateral the sum of all the angles is 360 degree okay. So can I say a plus c is 360 minus b plus d. So a plus c by two would be 180 degree minus b plus d by two isn't it yes or no. So if you take a cos on both the sides this is going to be cos of 180 degree minus b plus c b plus d by two which is nothing but minus cos b plus d by two okay. So if you square it so if you square it you will end up getting the same expression for both of them. So that is why it doesn't matter whether you take arithmetic mean of a and c or you take an arithmetic mean of b and d okay. By the way this formula is a complicated formula and you will be proving it as your homework okay. So prove it for homework okay. Is it fine? Any questions? Actually there is no question to be asked here because I just gave you a result and I have asked you to prove the result. Could you scroll along yeah yeah why not fine all right. Now let's talk about a cyclic quadrilateral. In fact we had a brief exposure about cyclic quadrilateral when we are talking about the angles of the orthic triangle. So let's talk about cyclic quadrilateral. So a quadrilateral which is basically inscribed within a circle that is called a cyclic quadrilateral not a new concept for you you already know it. So let's say this is a quadrilateral which is inscribed within a circle okay. So this is called a cyclic quadrilateral. In cyclic quadrilateral you are already aware that the sum of the opposite angles is 180 degree okay. So if this is the case you can actually find out the area of a cyclic quadrilateral by using the previous formula that we discussed. So in our previous formula that was true for any quadrilateral area of the quadrilateral was under root of s minus a s minus b s minus c s minus d minus a b c d d cos square either a plus c by 2 or b plus d by 2 doesn't matter. So when you use a cyclic quadrilateral when this guy becomes cyclic then what will happen this fellow will become a 90 degree isn't it right. Yes or no and because of that what will happen this whole thing will get simplified to this expression s minus a s minus b s minus c s minus d where you already know what's your s s is your semi-perimeter of this quadrilateral. So if you're able to prove this formula which I've given you as a homework question then you will be able to find out the area of a cyclic quadrilateral by using this expression okay. So please note this is a mistake which many people do and I keep highlighting this even your seniors are doing this mistake that this formula is not true for any quadrilateral this formula is not the area of any quadrilateral it is the area of a very very specific quadrilateral which is what we call as a cyclic quadrilateral got the point. So many people ask me sir can we use half the product of the diagonals into sign of the included angle still in case of cyclic quadrilateral yes why not that our old formula of half product into the of diagonals into sign of the included angle will still be valid for this case as well will still be valid for this case as well okay see at the end of the day cyclic quadrilateral is a quadrilateral only know so whatever we derived was for any quadrilateral in general so whether cyclic or non-cyclic that half into product of the diagonals into sign of the included angle between the diagonals is still going to hold true is still going to hold all right okay so the last concept of the day that we are going to talk about is your cosine law for cyclic quadrilateral that's a very you know interesting law which is valid for or cyclic quadrilaterals or cyclic quadrilaterals so again let me draw the figure i'll be you know deriving the result here quickly so let's say this is a cyclic quadrilateral okay a b c d and let me name the sites here for your reference this is small a small b small c small d okay so in a cyclic quadrilateral the following laws hold true or following cosine rule hold true so since there are four since there are four angles i'll be writing cosine of these four angles one by one so no down cos a is given by cos a is given by cos a is given by this angle is given by a square plus d square minus b square minus c square by 2 ad plus bc write this down first okay so cos a there's a way to remember it also cos a is whichever two sites enclose that particular angle write their squares add them whichever sites do not enclose that angle write them with a negative sign square so negative b square negative c square then to you write then group ad and group bc and write it like this this is the way to remember okay similarly cos b you tell me what should be cos b this angle b what is cos b you'll say sir a square b square because a and b are including these this angle minus c square minus d square upon 2 ab plus cd similarly cos c will be what b square plus c square minus a square minus d square upon 2 bc plus ad and cos d will be what cos d will be c square plus d square minus a square minus b square upon 2 cd plus ab note this down i'll just prove one of them quickly it'll not take much time but what is important for you to appreciate here that cos a is negative of cos c and cos b is negative of cos d and it is rightly so because a plus c add up to 180 degree isn't it so look here cos c cos a negatives of each other isn't it and cos b and cos d they are negatives of each other have that observation very very important very very important okay now i will quickly prove one of them maybe let's let's try to prove which one first one first one is good enough okay so for first one what i'm going to do i'm just going to connect d to b okay so i'm proving the first one okay so prove for cos a now all of you refer to this figure all of you refer to this figure maybe i'll sketch that figure once again for you all okay i'm just taking a snapshot so that i can replicate that on the right side of my screen else i will have to keep moving left right left right every time yeah so refer to this figure in this figure can i say let's say this length is x bd length is x if i use cos of a in triangle adb in triangle adb if i use cos of a formula lb a square plus d square minus x square by 280 cosine law yes sir right so from here i can say my x square is going to be a square plus d square minus 280 cos a let's call this as expression one okay in a similar way in triangle bdc let's write cos of c now remember cos of c is cos pi minus a which means you're actually writing an expression for minus cos a oh are you there's no place over it uh so let me just write it down yeah so let me write it down so in triangle bdc in triangle bdc which is your this triangle bdc okay uh cos of c is cos of pi minus a which is actually minus cos a can i write it as c square plus b square minus x square by 2 bc so b square plus c square minus x square by 2 bc so from here i get x square as correct me if i'm wrong b square plus c square plus 2 bc cos a that's your expression number two so from one and two since both of them represent the same expression of x square can i say a square plus d square minus 2 ad cos a is equal to b square plus c square plus 2 bc cos a in other words a square plus d square minus b square minus c square is twice cos a times ad plus bc so i'm just taking this result on the other side and these two i'm taking it to the left side correct so from here i get cos a value as a square plus d square minus b square minus c square by 2 ad plus bc that's what we had to prove a square plus e square minus b square minus e square by 2 ad plus bc okay hence proof right now this result could be obtained for cos b cos c cos d in a similar way so i will not be deriving them okay it is obvious that it can be obtained in a similar fashion now this result also has to be noted down please note this down because it may be helpful and it will be helpful in proving a very interesting theorem which is applicable to a quadrilateral cyclic quadrilateral called toll me's theorem which i'll be taking in the next class so our next class agenda would be toll me's theorem and we'll also talk about solution of triangle concept which is you know small part of this chapter and we'll be you know wrapping up this concept fine