 Hello and welcome to another session on Gems of Geometry. So, guys in the last session we tried to validate Brahma Gupta's formula which is formula to calculate the area of a cyclic quadrilateral. The formula k is under root s minus a, s minus b, s minus c, s minus d where s is the semi-parameter that is sum of the four sides divided by 2. That is what we evaluated in the previous video. In this session we are going to prove this Brahma Gupta's formula. Okay and how are we going to prove it? We are going to use trigonometry for it. Okay, so let us begin. So, you can clearly see this is angle alpha and this is beta. So, we can use cosine rule in triangle ADB. So, let us apply cosine rule in triangle ADB. So, what will that be? So, I can write f square is equal to how much f square will be equal to a square plus d square minus twice AD cos of alpha. And similarly in triangle if you see BCD. Okay, here also you will see f square is equal to b square plus c square minus 2 BC and we have cos of beta. So, I can equate both f square here. So, what will I get? I will get a square plus d square minus 2 AD cos of alpha is equal to b square plus c square minus 2 BC and cos of beta. Now if you look at this particular thing what that alpha plus beta is 180 degrees in a cyclic quadrature. Quadrangle it happens alpha plus beta is 180 degrees. Opposite angles are supplementary in cyclic quadrature. So, hence alpha is equal to 180 degrees minus beta hence cos of alpha will be equal to cos of 180 minus beta which I can write as minus cos beta. So, that means if you see this is a square plus d square minus b square minus c square and here I can write 2 AD cos of alpha and minus cos beta can be written as. So, this can be written as 2 BC minus cos beta which I am writing again as cos alpha. So, I hope you understood it. Okay. So, this minus sign here along with this cos beta will give you cos alpha. So, that is I am writing here. Okay. So, you can now simplify this to a square plus d square minus b square minus c square equals cos alpha common and within brackets 2 AD plus 2 BC. So, let it be equation number 1. Okay friends. Now what is the area of this cyclic quadrangle? So, if you see let us consider that to be k. So, k is equal to area of what triangle ABD plus area of triangle BCD. Okay. So, area of triangle now we can use sign using the sign rule and so area of triangle ABD can be written as half into A into D half AD sin of alpha and area of BCD will be half BC sin of beta and if you see again alpha plus beta is 180 degrees. So, sin of alpha is equal to sin of 180 degrees minus alpha which is sin alpha itself. Okay. So, both can be equated to be sin alpha right. So, hence I can write half common sin alpha common and this will be AD plus BC. I hope this is understood to all. Okay. So, from here what can we say? We can say 2 k by AD plus BC is equal to sin alpha right. This is equation number 2. Now using these two equations let us proceed. So, if you see sin alpha is this and cos alpha from here let me write it here cos of alpha will be from equation number 1. If you check it is A square plus D square minus B square minus C square divided by twice AD plus BC twice AD plus BC. Is it not? Now I can square 2 and 3 and add together. So, cos square alpha plus sin square alpha will give you 1 is it not? So, square both sides squaring 2 and 3 and adding together will give me cos square alpha plus sin square alpha is equal to 1 right. That means I can write what do I write? So, just to save space I will be writing here A square plus D square minus B square minus C square divided by twice AD plus BC. Okay. So, this whole squared okay plus sin square alpha which is 4 k square by AD plus BC whole square and that equals 1. Okay. Now if you simplify let me utilize a little you know or what I can do is I can just take this out so that I get some space. So, you know already. So, let me shift it here. Yep. So, then I get some space. Okay. Now, so from here now what do I do? So, let us now simplify this. So, this will be simply if I write it here A square plus D square minus B square minus C square whole square okay plus 16 k square. If you simplify you will get it. So, this is capital K. So, let me use capital K only. So, on the other side also I used small k no this will be capital K. Okay. So, let me write capital K here. So, hence capital K here 16 k square and in the denominator you will get what 4 AD plus BC whole square right and this is equal to 1. So, you can now write A square plus D square minus minus B square minus C square plus 16 k square again capital K square. So, this is A square plus D square minus D square minus C square and this is whole square actually here. So, square was missing plus 16 k square will be equal to 4 AD plus BC whole square is it. So, this is okay fair enough. Now, what do we do? We say that this is 16 k square right and then let us arrange this. So, if you arrange it it will be A square plus D square minus B square minus C square whole square and then minus 4 or rather I can write this as twice AD plus BC whole square correct looks good. And now I can write this as 16 k square is equal to now this if you see the right right hand side of the previous step it is A square minus B square form. So, hence I can directly write this as A square plus D square minus B square minus C square plus twice AD plus twice BC this is the first term and the second term will be A square plus B square minus B square minus C square minus 2 AD minus 2 BC I hope this is clear A square minus B square form. Okay friends now there are something interesting here. So, if you see A square plus D square and then plus 2 AD what can I write this as. So, I can write that as A plus D whole square and the second one is minus B plus C right sorry minus B minus C whole square if you check this will it be and then last the second term is A minus D whole square and the second one minus B plus C whole square is it guys okay. So, again if you see this can be factored further. So, this is 16 k square and factorize again square square difference of square. So, what will this be within curly brackets I can write A plus D minus B plus C see first term second term A plus D minus B sorry plus B and minus C very good third term A minus D minus B minus C and this is A minus D A minus D then minus B minus C plus B plus C yeah correct this is fine now. So, what do we see here. So, we see this that 16 k square can be written as 16 k square is equal to now if A plus D plus C can be written as 2S minus B and there was a minus B again pay attention I am writing here for clarity. So, if you consider A plus D or yeah A plus D or let me write it here. So, if you consider A plus D minus B plus C can be written as A plus B plus C plus D minus 2B isn't it check and A plus B plus C plus D is what 2S twice semi perimeter is right. So, I can write like that. So, I am writing 2S minus this then this one again A plus B plus B. So, this you can see this will be 2S minus 2C oh actually there is a correction here. So, if you see this will be this is fine but this will be 16 k square is equal to the signs will be opposite. So, here there will be a negative sign my dear friends negative sign. So, just bear with me. So, this is negative sign okay now it is fine. So, this is negative sign negative sign because here I made a mistake right fair enough. So, this is good 16 minus 16 k square and then here it is 2A minus 2S and then this is 2S minus 2D okay. Now, if you change take away this minus sign it will become 16 k square is equal to 2S minus 2B then 2S minus 2C then 2S minus 2A. So, I am changing the sign here because I am just eliminating this minus sign here and this is 2S minus 2D and hence proved. So, if you you know remove all the tools common tools you will get S minus B S minus A S minus C S minus D and this is what we wanted to prove. So, k is equal to under root sorry. So, sorry for the small space here. So, k is equal to let me write one part or like this I write S minus A S minus B S minus C S minus D whole to the power half okay. This is what we needed to prove and this is what is called Brahma Gupta's formula here. So, whenever a cyclic quadrilateral is given you know how to find out the area of that cyclic quadrandom okay. So, this was the proof I hope you understood the proof.