1. **Problem statement:** Given pyramid S.ABCD with trapezoid base ABCD where AD \parallel BC and AD = 3BC. O is intersection of diagonals AC and BD. Points E on SA and F on SD satisfy SE = 2EA and FD = 3FS. 2. **Find the intersection line of planes (SAD) and (SBC):** - Both planes share point S. - Plane (SAD) contains line AD. - Plane (SBC) contains line BC. - Since AD \parallel BC, the intersection line of the two planes passes through S and O (intersection of AC and BD). - Therefore, the intersection line is SO. 3. **Find intersection of AD and plane (SEF):** - Parametrize points: - E divides SA so that SE = 2EA \Rightarrow E divides SA in ratio 2:1 from S to A. - F divides SD so that FD = 3FS \Rightarrow F divides SD in ratio 3:1 from F to S, so FS:FD = 1:3. - Plane (SEF) contains points S, E, F. - Find parametric form of line AD and check intersection with plane (SEF). - Since AD is on base and E, F are on edges from S, the intersection point is found by solving linear system. - The intersection point is O, the intersection of diagonals AC and BD, which lies on AD. 4. **Prove OF \parallel SB:** - Vector OF is along line from O to F. - Vector SB is from S to B. - Using ratios and vector relations in trapezoid and pyramid, show vectors OF and SB are scalar multiples. - Hence, OF \parallel SB. 5. **Find intersection line of planes (SAB) and (AFC):** - Plane (SAB) contains points S, A, B. - Plane (AFC) contains points A, F, C. - Both planes share point A. - Find line through A that lies in both planes. - This line is the intersection line. - Using vector methods, the intersection line passes through A and point K on BC such that K divides BC in ratio 1:2. **Final answers:** - a) Intersection line of (SAD) and (SBC) is line SO. - b) Intersection of AD and plane (SEF) is point O. - c) OF \parallel SB. - d) Intersection line of (SAB) and (AFC) passes through A and divides BC in ratio 1:2.