第一部分：文本内容需要按照我给的一字不差 标题：8字模型 文本内容：10秒破解角度难题的几何神器来了！8字模型！它能瞬间打通复杂图形中的角度关系，让你不再被‘倒角难题’卡住，从此看到‘交叉线’就笑出声！ 绘图：需绘制一个8字模型的图片，按照我上传的图片一摸一样 第二部分：文本内容需要按照我给的一字不差 标题：8字模型的定义 文本内容：8字模型的标准定义是，两条线段相交于O点（形成"×"形），连接线段端点构成两个三角形（如△AOB和△COD），则：∠A + ∠B = ∠C + ∠D，8字模型的核心特征是： 图形形似“8”或沙漏，对顶角所在三角形内角和相等。 绘图：根据我文本内容的描述，生成一个描述动画 第三部分：文本内容需要按照我给的一字不差 标题：8字模型的原理剖析 文本内容：根据对顶角相等原理， ∠AOB = ∠COD，在△AOB与△COD中；∠A + ∠B + ∠AOB = 180°，∠C + ∠D + ∠COD = 180°，则∠AOB = ∠COD ，因此验证得出∠A + ∠B = ∠C + ∠D 动画绘图：绘制两条线段相交与两条线段相交于O点（形成"×"形），连接线段端点构成两个三角形（如△AOB和△COD），在根据我文本描述的内容生成动画 第四部分：文本内容需要按照我给的一字不差 标题：8字模型的典型例题 文本内容：例题：如图，直线AB与CD交于点O，∠A=50°，∠D=30°，∠C=40°，求∠B的度数？解题步骤：第一步识别模型：AB与CD交叉形成"8字" → 锁定△AOB和△COD；第二步套用公式：∠A + ∠B = ∠C + ∠D，因此50° + ∠B = 40° + 30°；第三步求解未知角∠B = 70° - 50° = 20°；第四步验证：对顶角∠AOB=∠COD，△AOB内角和：50°+20°+∠AOB=180° → ∠AOB=110°，△COD内角和：40°+30°+∠COD=180° → ∠COD=110°，等式成立！因此∠B等于 20° 绘图：根据我文本内容的描述，生成一个描述动画 第四部分：文本内容需要按照我给的一字不差 标题：总结 文本内容： 模型本质：交叉三角形，对顶角作桥；不相邻两角，和相等！ 核心思想： 等角代换：利用对顶角实现角度关系转移 整体求和：将分散角整合为等式两端的整体 避坑指南： ✅ 必有一条交叉线（无交叉非8字！） ❌ 勿与飞镖模型混淆（飞镖有凹点，8字必交叉） 结尾金句：交叉线一出，8字显神威！记住‘上下两角和相等’，分进合击秒杀角度难题！ 绘图：需绘制一个8字模型的图片，按照我上传的图片一摸一样---**Extracted Content:** **Question Stem:** No question stem is provided in the image. **Options:** No options (A, B, C, D) are provided in the image. **Other Relevant Text:** No other relevant text, explanations, hints, or data sources are present in the image. --- **Chart/Diagram Description:** * **Type:** Geometric Figure. The diagram displays a configuration of points and line segments, specifically two intersecting line segments forming two triangles. * **Main Elements:** * **Points:** * Point A: Located at the top-left. * Point B: Located at the top-right. * Point C: Located at the bottom-left. * Point D: Located at the bottom-right. * Point O: The central intersection point of two line segments. * **Lines/Segments:** * A straight line segment connects A and D, passing through point O. * A straight line segment connects B and C, passing through point O. * A straight line segment connects A and B. * A straight line segment connects C and D. * **Shapes:** * The figure consists of two triangles: * Triangle AOB, with vertices A, O, and B. * Triangle DOC (or COD), with vertices D, O, and C. * These two triangles share the common vertex O and are positioned such that they are "vertically opposite" to each other at the intersection point O, forming a shape resembling an hourglass or bow-tie. * **Relative Position and Direction:** * Points A, O, and D are collinear. * Points B, O, and C are collinear. * Point O is the intersection of the line segments AD and BC. * Triangle AOB is positioned above and to the left of O (relative to triangle DOC). * Triangle DOC is positioned below and to the right of O (relative to triangle AOB). * The side AB of triangle AOB is opposite to the side CD of triangle DOC, with O lying between them.