解答？---(11 分) 综合与实践
问题情境:
如图 1, E 为正方形 ABCD 内一点, ∠AEB = 90°. 将 Rt△ABE 绕点 B 按顺时针方向旋转 90°, 得 △CBE' (点 A 的对应点为点 C). 延长 AE 交 CE' 于点 F, 连接 DE. 猜想证明:
(1) 试判断四边形 BE'FE 的形状, 并说明理由;
(2) 如图 2, 若 DA = DE, 请猜想线段 CF 与 E'F 之间的数量关系, 并加以证明;
解决问题:
(3) 如图 1, 若 AB = 15, BE' = 9, 请直接写出 DE 的长.
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Chart/Diagram Description:
Figure 1:
Type: Geometric figure.
Main Elements:
- Square ABCD with vertices labeled counterclockwise.
- Point E is inside the square.
- Line segments AE, BE connect A and B to E.
- Angle AEB is marked with a right angle symbol (∠AEB = 90°).
- Triangle ABE is shown.
- Point E' is located such that triangle CBE' is the result of rotating triangle ABE 90 degrees clockwise around point B (A corresponds to C, E corresponds to E').
- Line segments BE' and CE' form triangle CBE'.
- Line segment AE is extended to intersect line segment CE' at point F.
- Line segment DE is drawn.
- Points are labeled A, B, C, D, E, F, E'.
- The figure illustrates the initial setup and the results of the rotation and intersection.
Figure 2:
Type: Geometric figure.
Main Elements:
- Similar to Figure 1, showing square ABCD, point E inside, ∠AEB = 90°.
- Triangle ABE.
- Triangle CBE' obtained by rotating △ABE 90° clockwise around B.
- Line segment AE extended intersects CE' at F.
- Line segment DE is shown.
- The figure represents the situation for question (2), where an additional condition DA = DE is given.