根据附件图片内容，解释”什么是手拉手模型“？---Here is the extracted content from the image: **Title:** 模型 50 “手拉手”模型 (Model 50 "Hand-in-Hand" Model) **模型展现 (Model Presentation)** * **Table Header:** 图示 (Diagram) | 条件 (Conditions) | 结论 (Conclusion) * **Table Row 1:** * 图示: AD 在 △ABC 内且拉手线无交点 (AD is inside △ABC and the hand-in-hand lines do not intersect) * 条件: 在 △ABC 中, 点 D, E 分别在边 AB, AC 上, 将 △ADE 绕点 A 旋转 (In △ABC, points D, E are on sides AB, AC respectively. Rotate △ADE around point A) * 结论: 1. △ADE ∽ △ABC, △ADB ∽ △AEC; 2. 两条拉手线 BD, CE 交于点 F, 则: (1) ∠BFC = ∠BAC = α; (2) 点 A, B, C, F 四点共圆. (1. △ADE is similar to △ABC, △ADB is similar to △AEC; 2. The two hand-in-hand lines BD, CE intersect at point F, then: (1) ∠BFC = ∠BAC = α; (2) Points A, B, C, F are concyclic.) * **Table Row 2:** * 图示: AD 在 △ABC 外且拉手线无交点 (AD is outside △ABC and the hand-in-hand lines do not intersect) * 条件: (Same as Row 1) * 结论: (Same as Row 1) * **Table Row 3:** * 图示: AD 在 △ABC 外且拉手线有交点 (AD is outside △ABC and the hand-in-hand lines intersect) * 条件: (Same as Row 1) * 结论: (Same as Row 1) **模型展现 Diagrams:** * Three geometric diagrams labeled A, B, C. * Diagram A: Triangle ABC with vertex A at the top. Point D on AB, E on AC. Segment DE. Lines BD and CE. D is inside triangle ABC. BD and CE do not intersect within the triangle. * Diagram B: Triangle ABC with vertex A at the top. Point D on extension of AB, E on extension of AC. Segment DE. Lines BD and CE. D is outside triangle ABC. BD and CE do not intersect in the shown area. * Diagram C: Triangle ABC with vertex A at the top. Point D on extension of AB, E on extension of AC. Segment DE. Lines BD and CE. D is outside triangle ABC. BD and CE intersect at point F. **【结论分析】 (Conclusion Analysis)** * 结论 1: △ADE ∽ △ABC, △ADB ∽ △AEC * 证明: ∵ DE // BC, ∴ △ADE ∽ △ABC, AD/AB = AE/AC 即 AD/AE = AB/AC 由旋转的性质得, ∠BAD = ∠CAE, ∴ △ADB ∽ △AEC. (两边成比例且夹角相等的两个三角形相似) (Proof: Since DE // BC, Therefore △ADE is similar to △ABC, AD/AB = AE/AC which means AD/AE = AB/AC. From the property of rotation, we get ∠BAD = ∠CAE. Therefore △ADB is similar to △AEC. (Two triangles are similar if two pairs of sides are proportional and the included angles are equal)) * 结论 2: 两条拉手线 BD, CE 交于点 F, 则 (1) ∠BFC = ∠BAC * 证明: 如图 ①, AC, BD 交于点 G, ∵ △ADB ∽ △AEC, ∴ ∠ABD = ∠ACE, ∴ ∠AGB = ∠CGF (对顶角相等), ∴ ∠BAC = ∠BFC. (Proof: As shown in Figure ①, AC, BD intersect at point G, Since △ADB is similar to △AEC, Therefore ∠ABD = ∠ACE, Therefore ∠AGB = ∠CGF (Vertically opposite angles are equal), Therefore ∠BAC = ∠BFC.) * (2) A, B, C, F 四点共圆 (A, B, C, F are concyclic) * 证明: 如图 ②, ∵ BC 为 △ABC 和 △BCF 的公共边, 且点 A, F 在 BC 的同侧, 由结论 2(1) 得 ∠BAC = ∠BFC = α, ∴ 点 A, B, C, F 在同一个圆上, 即 A, B, C, F 四点共圆. (Proof: As shown in Figure ②, Since BC is the common side of △ABC and △BCF, and points A, F are on the same side of BC, From conclusion 2(1) we get ∠BAC = ∠BFC = α, Therefore points A, B, C, F are on the same circle, i.e., A, B, C, F are concyclic.) **结论分析 Diagrams:** * Figure ①: Geometric diagram. Triangle ABC with vertex A at the top. Point D on AB, E on AC. Lines BD and CE intersect at F. AC and BD intersect at G. Shaded triangle ADE. Labels A, B, C, D, E, F, G are present. * Figure ②: Geometric diagram. Triangle ABC with vertex A at the top. Lines BD and CE intersect at F. A dashed circle passes through points A, B, C, F. Center O is labeled. Labels A, B, C, D, E, F, O, G are present. **思考延伸 (Thinking Extension)** * Text: 在公共角为直角的 手拉手模型中, 除了存在一般三角形手拉手模型的结论外, 以下结论同样成立吗? 让我们一起开启探索之旅吧! 结论: 1. BD ⊥ CE; 2. BE² + CD² = BC² + DE². (In the Hand-in-Hand model where the common angle is a right angle, besides the conclusions of the general triangle Hand-in-Hand model, do the following conclusions also hold? Let's embark on a journey of exploration together! Conclusions: 1. BD is perpendicular to CE; 2. BE squared + CD squared = BC squared + DE squared.) * Diagram: Geometric diagram. Triangle ABC with vertex A, ∠BAC is a right angle. Point D on AB, E on AC. Lines BD and CE are drawn and intersect at F. Labels A, B, C, D, E, F are present. **模型链接 (Model Link)** * Text: 若“手拉手”模型中, 双等腰, 共顶点, 顶角相等, 旋转得全等, 则为“手拉手”全等模型.(见 P40 模型 16) (If in the "Hand-in-Hand" model, the two triangles are isosceles, have a common vertex, equal vertex angles, and are congruent after rotation, then it is a "Hand-in-Hand" congruent model. (See P40 Model 16)) **高频图形 (High-Frequency Diagrams)** * Diagram 1: Geometric diagram. A square or rectangle with diagonals drawn, and lines connecting a vertex to midpoints of opposite sides, or lines connecting midpoints. Appears complex with multiple intersecting lines. * Diagram 2: Geometric diagram. A square or rectangle with diagonals and lines connecting vertices to some points on opposite sides, forming internal triangles and quadrilaterals. **例 1 (Example 1)** * **Question Stem:** 如图, 在等腰 △ABC 和 △ADE 中, ∠ABC = ∠ADE = 90°, 连接 BD, CE. 若 ∠CED = 75°, 则 ∠ADB 的度数为 (As shown in the figure, in isosceles △ABC and △ADE, ∠ABC = ∠ADE = 90°, connect BD, CE. If ∠CED = 75°, then the degree of ∠ADB is) * **模型猜想 (Model Conjecture):** 存在共顶点的两个等角且等角两边成比例, 故为“手拉手”模型 (There exist two angles with a common vertex that are equal and the sides containing the equal angles are proportional, thus it is a "Hand-in-Hand" model) * **Example 1 Diagram:** Geometric diagram. Triangle ABC and Triangle ADE with common vertex A at the top. Point D on AB, E on AC. Lines BD and CE are drawn. Right angle symbols at B (in △ABC) and D (in △ADE). Labels A, B, C, D, E are present. * **Options:** A. 110° B. 120° C. 130° D. 140° **解题思路 (Problem-Solving Steps)** * **Header:** 第一步: 依据特征找模型 (Step 1: Identify model based on features) | 第二步: 抽象模型 (Step 2: Abstract model) | 第三步: 用模型 (Step 3: Use model) * **第一步 (Step 1) Content:** 特征 1: 是否存在 共顶点的两个三角形 △ABC 和 △ADE (Feature 1: Are there two triangles with a common vertex? △ABC and △ADE) 特征 2: 共顶点的两个三角形 是否存在 两组成比例的线段 AB/AC = AD/AE (Feature 2: Are there two pairs of proportional segments in the two triangles with a common vertex? AB/AC = AD/AE) * **第一步 (Step 1) Diagram:** Same as Example 1 Diagram. * **第二步 (Step 2) Diagram:** Same as Example 1 Diagram. * **第三步 (Step 3) Content:** △BAD ∽ △CAE * **第三步 (Step 3) Diagram:** A rectangular box containing the text △BAD ∽ △CAE.