 Hi and welcome to the session. I'm Kanekan. I'm going to help you to solve the following question. The question says in figure 9.28, this is the figure 9.28, AP is parallel to BQ is parallel to CR. Prove that area of AQC is equal to area of PBR. Let's first write down the given information. We are given that AP is parallel to BQ is parallel to CR and we have to prove that area of AQC is equal to area of PBR. Let's now begin with the solution since triangle ABQ triangle BVQ on the same base BQ BQ is equal to area of triangle BQ theorem 9.2 given in your book which states that it's on the same base are equal in area. Triangle BQC and BQR. Now same BQC on the same base BQ between the same parallel BQ and CR. Therefore triangle BQC is equal to area of triangle BQR. My theorem 9.2 which says that two triangles on the same base the same parallels are equal in area. Let's say this equation is equation number one and this as two adding one and two we get area of triangle ABQ plus area of triangle BQC is equal to area of triangle PBQ plus area of triangle BQR. Now triangle ABQ plus triangle BQC is equal to triangle AQC. So this means area of triangle ABQ plus area of triangle BQC is equal to area of triangle AQC. Triangle PBQ plus triangle BQR is equal to triangle PBR. So area of triangle PBQ plus area of triangle BQR is equal to area of triangle PBR. Hence we have proved that area of triangle AQC is equal to area of triangle PBR. This completes the session. Bye and take care.