 Hi and how are you all today? I'm Priyanka and let us discuss this question. In figure 9.32 that is given below, ADCD is a parallelogram and BC is produced to a point Q such that AD is equal to CQ. Now if AQ intersects BC at P, show that area of BPC that is this triangle is equal to area of DPQ that is this triangle. So let us start with our solution. First of all here we are given that ABCD is a parallelogram, AD is equal to CQ and we need to show that that area of BPC is equal to area of DPQ. Right? So here we need to join AC in order to proceed on with our solution like this. This will be under construction. What we have done here? We have joined AC. Let us start with our proof. Now this is the figure and where, what we did? We joined AC. We know that in triangle PC and triangle CPA, CPC is this triangle and CPA that is this triangle. That means we are talking about this and this triangle. If you notice they are on the same base and between the same parallel lines. How come they are between the same parallel line? We know that ABCD is a parallelogram that means opposite sides will be equal and parallel to each other. Since PC is a part of DC therefore AB will also be parallel to PC. Right? So if ABCD is a parallelogram as we discussed that means AB is parallel to CD. Right? Now we know that since P is a point on DC therefore AB is parallel to PC also. Now we can say that area of triangle ABC is equal to area of triangle BPC because of the simple reason that triangles on the same base, equal base between the same parallel lines in area. So this is the term that you learned before starting off the session. So we can say that area of ABC is equal to area of VPC. Right? And let this be the first equation. We know that ABCD is a parallelogram as it is given to us in the question. Therefore we can say that AD is equal to BC as it is the opposite side of parallelogram. Right? Further we know that AD is equal to, equal to CQ. It's given to us in the question. Therefore by these two that is we can say that AD is parallel to BC. Since BC is parallel to AD and BC is produced to Q that means this line will also be parallel to AD. Right? So since BC is produced to Q we can say that AD is parallel to CQ. And let this be the second equation and this be the third equation. So by N3DQC, ADQC is a parallelogram because opposite sides and therefore we can say that since we have proved that this is a parallelogram so we can say that AC is equal to DQ also because opposite sides of a parallelogram are equal and parallel to each other. Now we also know that is equal to PQ because it is a diagonal of the parallelogram ADQC. So diagonals of a parallelogram bisect each other. This is one of the properties of parallelogram that we know and we can say that similarly that is this is equal to D and if you notice that we have proved that CPA is congruent to triangle DPQ by SSS congruency criteria. We know that when we have proved that these two triangles are congruent to each other so we can say that area of these two triangles are also equal to each other because congruent triangles equal in let this be the fourth equation. So by third and fourth or by one and fourth we get that area of Q because PPC and DPQ are both equal to a same triangle's area that is CPA that you have proved in one and four. So this was session do take care of yourself. Bye for now.