 Hi and welcome to the session. Let us discuss the following question. The question says in figure 9.29, this is the figure 9.29. Area of triangle DRC is equal to area of triangle DPC and area of triangle BDP is equal to area of triangle ARC. Show that both the quadrilaterals A, P, C, D and D, C, P, R are trapezius. Let us now begin with the solution. We are given that area of triangle DPC is equal to area of triangle DRC and area of triangle BDP is equal to area of triangle ARC. Let us name this as equation number 1 and this as 2. On subtracting we get area of triangle BDP minus area of triangle DPC is equal to area of triangle ARC minus area of triangle DRC. Now, triangle BDP minus triangle DPC is equal to triangle BCT, right? So, area of triangle BDP minus area of triangle DPC is equal to area of triangle BCT. C minus triangle DRC is equal to triangle ACD. So, this means area of triangle ARC minus area of triangle DRC is equal to area of triangle equation as equation number 3. Now, since the BCD, triangle BCD and ACD are, from 3 we have concluded that area of triangle BCD is equal to area of triangle ACD. So, BCD triangle ACD lie between the same parallels AB and DC.3 given in your book which stable areas lie between the same parallel. It is parallel to DC. Therefore, BCD is a trapezium. Trapezium exactly its sides, we have proved that ABCD is a trapezium. The quadrilateral DCPR is a trapezium. Thus, that area of triangle DRC is equal to area of triangle DPC. Now, consider triangle DRC and DPC DRC lie between DC and RP. So, RP therefore DR is a trapezium exactly one way is a trapezium.