 Hi and how are you all today? The question says the size AB of a parallelogram ABCD is produced to any point E. Aligned through A and parallel to Cp needs to be produced at Q. And then parallelogram PbQr is completed. See figure 9.26 show that area of ABCD is equal to area of PbQr. Now this is the figure which we need to refer. We need to prove that area of ABCD is equal to the area of parallelogram PbQr. Let us proceed on with our solution. Have we given that AQ is parallel to Cp, Pcd and PbQr parallelograms to prove that the area of ABCD is equal to area of PbQr. We need to join start with our proof. Qcq is equal to area of Apq. Apq because triangle is on the same base. Now if you notice in both these triangles the area of triangle Apq is common between both these triangles. On subtracting of triangle Apq that is the shaded triangle. From both sides we have we are left with the triangle Apcb will be equal to the triangle Qpp this triangle in the first equation. Now we also know that diagonal of a parallelogram divides it into equal right. So we can say that area of parallelogram ABCD is equal to twice area of ABCD. That is area of ABCD is twice this area as the diagonal AC is divided into two equal halves. Similarly area of Pqr is equal to twice area of Qp. Now from the first equation we can see that area of ACb that can also be written as ABC is equal to Qbp that is Qbp. Since the left hand side of these two right hand side of these two equations are equal then the left hand side will also be equal. So let this be the second equation. So from the first and second equation we can say area of ABCD is equal to area of Pbqr and this is our required proof. This ends the session. Hope you have a good day. Bye for now.