 Hello and how are you all today? The question that we need to discuss today says, In figure 9.34, ABC is the right triangle, right angle at A. B, C, E, D, A, C, F, T and A, B, M, N are squares on the side, B, C, C, A and A, B respectively. Line segment AX is perpendicular to D, means BC at Y. To show that, these are the six parts of the question which we need to show and we will be proceeding on one by one. The diagram that we need to refer is figure 9.34 and this is the figure which we need to refer. Now these are the squares that are on the sides of A, B, C and C respectively and we are given that they all are squares. Let us proceed on with our solution. First of all, for the first part, we need to prove that triangle M, B, C is congruent to triangle A, B, D. Right, have the figure. Now here, triangle M, B, C is congruent to A, B, D. Now these are the two triangles which we need to refer. If you carefully see, M, B and A, B are the side of the same square, hence they have to be equal and since M, B is this side, it has to be side C and this has to be the side D. Right, now in order to prove that these two triangles are congruent, we need to have three things which are common. Let us have them. We know that M, B is equal to E, B because A, B, M, N is a square and all sides of a square are equal to each other. Further, we can say that angle M, B, C is equal to angle M or A, B, D because here we know that this is a right-angled triangle and this is also right-angled. Now if you add this angle which is A, B, C common to both these 90 degrees, first of all we will get angle M, B, C and then we will get A, B, D which is the required. So we will write it as that angle M, B, A is equal to angle C, B, D because they are equal to 90 degrees and angle A, B, C is common to both of them. So hence these are equal to each other and further B, C is equal to B, D because B, C, A, D is the square. We can say that triangle M, B, C is congruent to triangle A, B, D, Y, S, A, S congruency criteria. Right, this completes our first part. Now let us proceed on with the second part. Here we need to prove that area of BYXT is equal to twice area of M, B, C. Now if you observe in this question BYXT is this and M, B, C is this triangle. We are given that Ax is perpendicular to D and also Ax is parallel to B, D. Ax is parallel to B, D. So now we can straight away say that therefore area of triangle ABD is equal to half of area of BYXT because of the reason that parallelograms because of the reason that triangles and parallelogram if triangle and parallelogram are on the same base and between the same parallel lines then area of the triangle is the area of parallelogram. This is what you learnt in the example 2 of page 158. Also we know that area of triangle M, B, C is equal to area of triangle ABD because congruent triangles have equal areas and these are the congruent triangles that we proved in the first part. So on comparing we can see that therefore if we compare these two parts and if you notice area of ABD are equal to area of M, B, C and half of this. We can say that therefore area of M, B, C is equal to half area of BYXT or taking two on the left hand side we can say twice area of M, B, C is equal to area of BYXT. This is the second part's proof. Now in third part we need to prove that area of BYXT is equal to area of ABMN. Now we can directly say that area of M, B, C is equal to half area of ABMN because of the same reason that they are having the same base or between the same panel. So area of triangle is half the area of the parallelogram that we have used above also and from second part we can say that area of M, B, C is half area of BYXT. So on comparing these two we can say that therefore half of area of BYXT is equal to half of area of ABMN half and half will get cancelled and therefore we can say that area of BYXT is equal to area of ABMN and this is our third part's proof. Let us proceed on with our fourth part. Here we need to prove that triangle FCB is congruent to triangle ACE. Now we know that in these two triangles FC is equal to AC, FC is equal to AC because they are the sides of this square and similarly FCB will be equal to ACE because they are 90 degrees and this is the common angle and CB is equal to CE. CB is equal to C again, size of a square. So because of SAS congruent rule we can say that these two triangles are congruent. Let us write it down very quickly. Now we have written down all the three criterias and therefore by SAS as we discussed in the diagram also by SAS these two triangles are congruent to each other. So this completes the fourth part. Now in the fifth part we need to prove that area of CYXC is equal to twice area of FCB. As we proceeded in our second part also similarly we will proceed on here also. We know that area of ACE is equal to half of area of CYXC because they are on the same base and between the same panel line. Also from above we can say that area of ACE is equal to area of FCB by CPCT, congruent parts of congruent or we can just mention that congruent triangles are equal in area and on comparison we can notice that half of area of CYXE is equal to area of FCB or we can write it down like area of CYXE is twice area of FCB. If you understood the second part well you can easily write down the fifth part yourself. This completes the fifth part. Let us proceed on with the sixth part. Here we need to prove that area of CYXC is equal to area of ACFG like we proceeded in the third part similarly here also we will proceed on. We can say that area of FCB is half of area of ACFG because they are between the same panel line and they have CF their common base. Also area of FCB is equal to half area of CYXC that we have taken from just the above part here and on comparison we have the left hand side same therefore the right hand side will be equal half the area of ACFG is equal to half the area of CYXE and on cancelling we can say that therefore just write this on the right left hand side CYXC is equal to area of ACFG and this proves our sixth part. Now the last part which we need to prove is area of BCED is equal to area of ABMN plus area of ACFG. Now we have area of BYXT equal to area of ABMN that we proved in part 3. Let this be the first equation. Also we know that area of CYXE is equal to area of ACFG that we just proved in the part 6. Let this be the second equation. Now by adding 1 and 2 we get area of BYXT plus area of CYXE is equal to area of ABMN plus area of further. If you notice from the figure given to you BYXT BYXT plus CYCYXE is which figure? It is BDCE only so here we have therefore area of BCED is equal to the sum that we have mentioned above that is this only that is area of ABMN plus area of ACFG and this is what we needed to prove. So we can just write down in the end that hence we have proved the required question. I hope you enjoyed the session this was a big question but if you know the first three parts well you can easily find out the remaining parts. Concentrate on your sessions, bye for now.