 Hello and welcome to the session. In this session we discussed the following question which says, if a upon b is equal to c upon d prove that a square plus ab plus b square whole upon a square minus ab plus b square is equal to c square plus cd plus d square whole upon c square minus cd plus d square. Before we move on to the solution, let's discuss the Componendo-Dividendo relation. According to this, if we have a is to b is equal to c is to d that is the four quantities ab, c and d are in proportion then a plus b that is sum of the first and the second quantity upon the difference of the first and second quantity that is a plus b upon a minus b is equal to the sum of the third and the fourth quantity that is c and d and the difference of third and fourth quantity that is c minus d. So when a is to b is equal to c is to d then a plus b upon a minus b is equal to c plus d upon c minus d. This is the key idea that we use for this question. Let's move on to the solution now. In the question we are given that a upon b is equal to c upon b and we are supposed to prove that a square plus ab plus b square whole upon a square minus ab plus b square is equal to c square plus cd plus d square whole upon c square minus cd plus d square. Now first of all we take let a upon b equal to c upon d be equal to k. This means that a would be equal to bk and c would be equal to dk. Let us now consider the LHS that is this a square plus ab plus b square this whole upon a square minus ab plus b square. Now to simplify this expression we apply componento and dividendo. So applying componento dividendo we get the LHS would be equal to the numerator that is a square plus ab plus b square plus the denominator that is a square minus ab plus b square this whole upon the numerator that is a square plus ab plus b square minus the denominator. So here we have minus a square plus ab minus b square. Now in the numerator ab and minus ab cancels and in the denominator a square minus a square cancels b square minus b square cancels. So this is equal to two times a square plus b square this whole upon two times ab. Now this two and two cancels and this is equal to a square plus b square upon ab. Now we know that a is equal to bk. So in place of a we put bk in the numerator as well as in the denominator. So this is equal to bk the whole square plus b square upon bk into b. So further we get b square k square plus b square upon b square k. Taking b square common in the numerator we get b square into k square plus 1 the whole this whole upon b square into k. Now this b square b square cancels and we are left with k square plus 1 the whole upon k. This is our LHS. Similarly let's consider the RHS which is c square plus cd plus d square this whole upon c square minus cd plus d square. Now again to simplify this expression we apply componento and dividendo. So applying componento dividendo we get RHS is equal to the numerator that is c square plus cd plus d square plus the denominator that is c square minus cd plus d square this whole upon the numerator c square plus cd plus d square minus the denominator. So we would get here minus c square plus cd minus d square. Now in the numerator cd and minus cd cancels and in the denominator c square minus c square cancels d square minus d square cancels and we are left with 2 times c square plus d square this whole upon 2 times cd. Now this 2 and 2 cancels and we are left with c square plus d square and this whole upon cd. We know that c is equal to dk. So in place of c we put dk so this is equal to dk whole square plus d square this whole upon dk into d. So we get this is equal to d square k square plus d square and this whole upon d square k. Now for the taking d square common from the numerator we get d square into k square plus 1 d whole upon d square k. This d square d square cancels and we are left with k square plus 1 whole upon k. This is our RHS. So we find that LHS and the RHS are equal that is k square plus 1 upon k. So we have the LHS is equal to the RHS. Hence finally we get a square plus ab plus b square and this whole upon a square minus ab plus b square is equal to c square plus cd plus d square this whole upon c square minus cd plus d square. This is what we were supposed to prove. So this come to each c session hope you have understood the solution of this question.