利用中国初中数学的知识解答图片中的题目，并将解题思路和步骤制作成小视频，视频时长控制在三分钟以内，如果有需要仅显示关键步骤---**Question Stem:** In △ABC, ∠ACB = 90°, ∠ABC = α. Point D is on ray BC. Connect AD. Rotate line segment AD counter-clockwise around point A by 180° - 2α to obtain line segment AE (point E is not on line AB). Draw a line from point E, EF // AB, intersecting line BC at point F. **(1) Sub-question 1:** As shown in Figure 1, if α = 45° and point D coincides with point C, prove: BF = AC. **(2) Sub-question 2:** As shown in Figure 2, points D and F are both on the extension of BC. Express the quantitative relationship between DF and BC using an equation, and prove it. --- **Chart/Diagram Description:** **Figure 1 (图1):** * **Type:** Geometric figure illustrating a specific configuration for the problem. * **Main Elements:** * **Points:** A, C(D), B, E, F. * **Lines/Segments:** * Line segment AC, which is vertical. * Line segment CBF, which is horizontal. Point C and D are at the same location. Point B is to the right of C. Point F is to the right of B. * Line segment AB, connecting A to B. * Line segment AE, connecting A to E. * Line segment EF, connecting E to F. This segment is drawn parallel to AB (EF // AB). * **Shapes:** Implied right-angled triangle △ABC (with ∠ACB = 90°). Quadrilateral AEFB is formed. * **Labels:** "图1" (Figure 1) below the diagram. * **Relative Position:** Point A is above point C. Points C, B, F are collinear horizontally, with C being the leftmost, then B, then F. Point E is positioned such that A, E, F form a quadrilateral AEFB with AB and EF being parallel sides. **Figure 2 (图2):** * **Type:** Geometric figure illustrating another configuration for the problem. * **Main Elements:** * **Points:** A, D, F, C, B, E. * **Lines/Segments:** * Horizontal line segment D-F-C-B, representing the line BC and its extension. Point D is the leftmost, followed by F, then C, and then B. * Line segment AC, which is vertical, connecting A to C. This implies ∠ACB is a right angle. * Line segment AB, connecting A to B. * Line segment AD, connecting A to D. * Line segment AE, connecting A to E. Point E is below the line D-F-C-B. * Line segment EF, connecting E to F. * **Shapes:** Triangles △ABC, △ADC, △AEF are formed. * **Labels:** "图2" (Figure 2) below the diagram. * **Relative Position:** Point A is above point C. Points D, F, C, B are collinear on a horizontal line in that order from left to right. Point E is below this horizontal line. Line segment EF connects E to F. Line segment AD connects A to D. Line segment AE connects A to E.