生成图片中的图形视频 并解答---**Extracted Content:** **Chart Description:** * **Type:** A 2D Cartesian coordinate plane with an X-axis and a Y-axis, originating from point O. * **Main Elements:** * **Coordinate Axes:** X-axis (horizontal, labeled 'x' with an arrow pointing right) and Y-axis (vertical, labeled 'y' with an arrow pointing up). The origin is labeled 'O'. * **Points:** * O: The origin (0,0). * A: A point on the positive Y-axis. * B: A point on the positive X-axis. * C: A point in the first quadrant. * D: A point on the line segment OB. * E: A point on a curve in the first quadrant. * F: A point on the same curve in the first quadrant, connected to E. * **Lines/Segments:** * Segments: OA, OB, AB, AC, BC, OC, AD, CE, EF. * Connections: C is connected to A, B, O, E. A is connected to O, B, D. B is connected to O, C. E is connected to C, F. * **Curve:** A curve (representing an inverse proportion function) is shown in the first quadrant, passing through points E and F. * **Shapes:** * Quadrilateral AOBC is formed. * Triangles OAD, OAB, OBC, ABC, OAC, OCE, OCF, CEF are implied or explicitly formed. **Textual Information:** **Question Stem:** 如图四边形AOBC中, ∠AOB = ∠ACB = 90°, AC = BC, 连接OC, AB, AD//BC. 反比例函数 y = 2/x 的图像上有一点E, EC⊥OC, 连接EC并延长交反比例函数图像于F. 若A(0,2), B(3,0), OD = √2/2, 则 **Translation of Question Stem (for clarity, not part of required output but for understanding):** As shown in the figure, in quadrilateral AOBC, ∠AOB = ∠ACB = 90°, AC = BC. Connect OC, AB, AD//BC. There is a point E on the graph of the inverse proportion function y = 2/x. EC⊥OC. Connect EC and extend it to intersect the graph of the inverse proportion function at F. If A(0,2), B(3,0), OD = √2/2, then **Sub-Statements/Propositions to Evaluate:** ① OC所在直线表达式为 y = (3/4)x ② OC = (1/2)EF ③ 点C的坐标为 (5/2, 5/2) ④ AD = (4√26) / 13 **Options:** A: ①② B: ②③④ C: ①②④ D: ③④