初中数学技巧---Here is the complete and accurate extraction of the content from the image: --- **Question Stem:** 例 如图1,在Rt △ABC 中,∠ABC=90°,AB=8,BC=6,D 是 AB 上一点,且 AD=2,过点 D 作 DE//BC 交 AC 于 E,将△ADE 绕 A 点顺时针旋转转到图2的位置,则图2中 BD/CE 的值为____. **Chart/Diagram Description:** * **图1 (Figure 1):** * **Type:** Geometric figure, a right-angled triangle ABC with an inscribed segment DE. * **Main Elements:** * Triangle ABC: A right-angled triangle with the right angle at vertex B. * Points: Vertices are A, B, C. Point D is located on side AB. Point E is located on side AC. * Lines: Line segment DE is drawn connecting D and E. * Relationship: DE is parallel to BC (DE//BC). * **图2 (Figure 2):** * **Type:** Geometric figure, showing a transformation (rotation) of a part of the original triangle from Figure 1, with additional connecting lines. * **Main Elements:** * Triangle ABC: The original right-angled triangle, vertices A, B, C. * Triangle ADE (rotated): Triangle ADE from Figure 1 has been rotated clockwise around point A. The new positions of D and E are shown. * Points: Vertices A, B, C are in their original positions. The rotated points D and E are inside the triangle ABC. * Lines: Original sides AB, BC, AC are present. The rotated segments AD and AE are present. New line segments BD and CE are drawn, connecting vertex B to the rotated D, and vertex C to the rotated E. **Solution Explanation (解析):** 在 Rt △ABC 中,∠ABC=90°,AB=8,BC=6, ∴ AC = sqrt(AB^2 + BC^2) = sqrt(8^2 + 6^2) = sqrt(64 + 36) = sqrt(100) = 10. ∴ DE//BC, ∴ ∠ADE = ∠ABC = 90°, ∠AED = ∠ACB, ∴ △ADE ~ △ABC, ∴ AD/AE = AB/AC, ∴ AD/AB = AE/AC. 将△ADE绕A点顺时针旋转 ∴ ∠BAC = ∠DAE, ∴ ∠BAC + ∠CAD = ∠DAE + ∠CAD, ∴ ∠BAD = ∠CAE, ∴ △ABD ~ △ACE, ∴ BD/CE = AB/AC = 8/10 = 4/5. 故答案为 4/5. **Final Answer (答案):** 4/5 ---