八年级数学下册期末易错点专题复习教学设计

八年级数学下册期末易错点专题复习教学设计一、设计理念与复习定位本节课是基于“教学评一体化”理念指导下的一节八年级下册期末综合复习课，内容锁定为人教版初中数学八年级下册全册核心内容的易错点辨析与矫正3。作为期末冲刺阶段的复习课，本设计摒弃了简单的知识罗列和题海战术，转而聚焦于学生在学习“二次根式”、“勾股定理”、“平行四边形”以及“一次函数”这四大知识板块中暴露出的共性、顽固性错误。本节课的核心目标是通过精准的诊断、深刻的归因以及变式的训练，帮助学生建立正确的思维模型，完善认知结构，从而在期末考试中实现从“会做”到“做对”的质的飞跃。本设计强调以学生为中心，通过暴露错误、合作辨析、总结规律，最终达到提升数学核心素养的目的。二、教材与学情分析【基础】人教版八年级下册数学内容抽象性增强，逻辑推理要求显著提高。二次根式侧重运算的严谨性，勾股定理强调数形结合，平行四边形是培养逻辑推理能力的重要载体，一次函数则是函数的入门，对变量对应思想和数形结合思想要求极高。【非常重要】从学情来看，学生经过近一年的初中学习，已经具备了一定的运算能力和逻辑推理基础，但在思维的深刻性和严谨性上仍有待发展。期末复习阶段，学生的易错点往往不是简单的“不会”，而是“会而不对，对而不全”。具体表现为：对二次根式双非负性的忽视、对勾股定理应用场景的混淆、平行四边形判定定理与性质定理的混用、以及在处理函数问题时对自变量取值范围和数形结合思想的麻木。这些问题具有普遍性和顽固性，需要在复习课上进行专题化的集中诊治。三、教学目标1.（基础）通过典型错题的再现与分析，学生能精准识别二次根式运算、勾股定理应用、平行四边形判定与性质、一次函数解析式确定及图像应用中的常见陷阱。2.（重要）经历“错在哪里—为何出错—如何修正—变式巩固”的探究过程，学生能深刻理解易错点的本质，掌握规避错误的策略，完善解题规范。3.（高频考点）通过对核心易错题的变式训练，学生能灵活运用分类讨论思想、方程思想、数形结合思想解决问题，提升思维的批判性和深刻性。四、教学重难点1.教学重点：二次根式化简中的隐含条件、平行四边形判定与性质的综合应用、一次函数实际应用中的自变量取值范围。2.教学难点：勾股定理应用中的分类讨论（如锐角与钝角三角形的高）、动态几何问题中函数关系式的建立、复杂图形中全等三角形或平行四边形的构造。五、教学过程设计（核心环节）（一）导入：聚焦错因，诊断先行【基础】展示一组从学生近期模拟考试和作业中摘录的典型错解，不公布答案，而是直接提问：“这些解法看似都有道理，为什么会被扣分？问题究竟出在哪里？”以此激发学生的认知冲突，快速将注意力集中到本节课的核心任务上。教师明确本节课的目的：不是为了做题，而是为了给我们的思维“排雷”，确保在期末考试中颗粒归仓。（二）模块一：二次根式——“埋伏”在看似简单处的陷阱【高频考点】【难点】二次根式的核心陷阱集中在概念的“双非负性”和化简法则的运用条件上。1.典型错误1：无视被开方数的非负性1.2.原题呈现：若(a−3)2=3−a\sqrt{(a3)^2}=3a(a−3)2<pathd="M263,681c0.7,0,18,39.7,52,119c34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120c340,704.7,510.7,1060.3,512,1067l00c4.7,7.3,11,11,19,11H40000v40H1012.3s271.3,567,271.3,567c38.7,80.7,84,175,136,283c52,108,89.167,185.3,111.5,232c22.3,46.7,33.8,70.3,34.5,71c4.7,4.7,12.3,7,23,7s12,1,12,1s109,253,109,253c72.7,168,109.3,252,110,252c10.7,8,22,16.7,34,26c22,17.3,33.3,26,34,26s26,26,26,26s76,59,76,59s76,60,76,60zMHv40hz">​=3−a成立，则实数a的取值范围是_______。2.3.错解展示：学生往往直接得出a=3。3.4.【非常重要】归因分析：错解源于对公式a2=∣a∣\sqrt{a^2}=|a|a2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​=∣a∣的理解仅停留在机械记忆，忽视了绝对值化简时需考虑a的正负。教师应引导学生从“算术平方根的结果为非负数”这一根本性质出发，逆向推导：因为(a−3)2=3−a\sqrt{(a3)^2}=3a(a−3)2<pathd="M263,681c0.7,0,18,39.7,52,119c34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120c340,704.7,510.7,1060.3,512,1067l00c4.7,7.3,11,11,19,11H40000v40H1012.3s271.3,567,271.3,567c38.7,80.7,84,175,136,283c52,108,89.167,185.3,111.5,232c22.3,46.7,33.8,70.3,34.5,71c4.7,4.7,12.3,7,23,7s12,1,12,1s109,253,109,253c72.7,168,109.3,252,110,252c10.7,8,22,16.7,34,26c22,17.3,33.3,26,34,26s26,26,26,26s76,59,76,59s76,60,76,60zMHv40hz">​=3−a，而(a−3)2≥0\sqrt{(a3)^2}\ge0(a−3)2<pathd="M263,681c0.7,0,18,39.7,52,119c34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120c340,704.7,510.7,1060.3,512,1067l00c4.7,7.3,11,11,19,11H40000v40H1012.3s271.3,567,271.3,567c38.7,80.7,84,175,136,283c52,108,89.167,185.3,111.5,232c22.3,46.7,33.8,70.3,34.5,71c4.7,4.7,12.3,7,23,7s12,1,12,1s109,253,109,253c72.7,168,109.3,252,110,252c10.7,8,22,16.7,34,26c22,17.3,33.3,26,34,26s26,26,26,26s76,59,76,59s76,60,76,60zMHv40hz">​≥0，所以3−a≥03a\ge03−a≥0，即a≤3a\le3a≤3。这才是本质解法。4.5.矫正策略：回归定义，强化算术平方根的非负性。任何关于二次根式的化简问题，首先要关注被开方数的取值范围。6.典型错误2：分母有理化中的运算错误1.7.原题呈现：计算13−2\frac{1}{\sqrt{3}\sqrt{2}}3<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​−2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​1​。2.8.错解展示：13−2=3+2\frac{1}{\sqrt{3}\sqrt{2}}=\sqrt{3}+\sqrt{2}3<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​−2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​1​=3<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​+2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​（部分学生直接记结论，过程乱写）或分母有理化时符号出错。3.9.【重要】归因分析：对平方差公式(a+b)(a−b)=a2−b2(a+b)(ab)=a^2b^2(a+b)(a−b)=a2−b2在根式运算中的迁移不熟练。4.10.矫正策略：慢下来，写清楚过程。强调分子分母同乘的是分母的有理化因式3+2\sqrt{3}+\sqrt{2}3<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​+2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​。同时，展示逆用：若已知3+2\sqrt{3}+\sqrt{2}3<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​+2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​，能否反推回原式？加深对互逆关系的理解。11.变式训练：1.12.若x−1+1−x=y+4\sqrt{x1}+\sqrt{1x}=y+4x−1<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​+1−x<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​=y+4，则xyx^yxy的值为多少？（考察被开方数的非负性，得出x=1，进而求解）（三）模块二：勾股定理——当“形”不唯一时【难点】【热点】勾股定理的应用中，当图形不确定时，分类讨论思想是学生最易遗漏的。1.典型错误1：直角三角形中，直角边与斜边不分1.2.原题呈现：已知直角三角形的两边长分别为3和4，求第三边的长。2.3.错解展示：绝大多数学生脱口而出：5。3.4.【非常重要】归因分析：思维定势，将3、4默认为两条直角边，忽视了4可能是斜边的情况。4.5.矫正策略：构建“陷阱题”模型。教师在黑板上画出两种情况，并板书：1.5.6.当3和4为直角边时，斜边=32+42=5\sqrt{3^2+4^2}=532+42<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​=5。2.6.7.当4为斜边，3为直角边时，另一直角边=42−32=7\sqrt{4^23^2}=\sqrt{7}42−32<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​=7<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​。3.7.8.总结：遇直角三角形两边求第三边，必分类讨论（谁是斜边）。9.典型错误2：锐角与钝角三角形的混淆1.10.原题呈现：在△ABC\triangleABC△ABC中，AB=15，AC=13，BC边上的高AD=12，求△ABC\triangleABC△ABC的面积。2.11.错解展示：直接画出高在三角形内部的情况，求得BC=9+5=14，面积=84。3.12.【难点】归因分析：学生潜意识里认为高一定在三角形内部，缺乏对钝角三角形高的认知。这是典型的数感缺失和空间想象能力不足。4.13.矫正策略：利用几何画板动态演示，当△ABC\triangleABC△ABC是钝角三角形时，高AD会落在BC的延长线上。此时，图形变了，计算逻辑也变了。引导学生画出两种图形：1.5.14.锐角三角形：BC=BD+DC=152−122+132−122=9+5=14\sqrt{15^212^2}+\sqrt{13^212^2}=9+5=14152−122<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​+132−122<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​=9+5=14。2.6.15.钝角三角形（角B为钝角）：此时D在CB延长线上，BC=BDDC=95=4。3.7.16.最终面积为84或24。8.17.变式训练：已知等腰三角形一腰上的高与另一腰的夹角为40°，求这个等腰三角形的顶角度数。（同样涉及高在三角形内外的讨论，答案50°或130°）（四）模块三：平行四边形——判定与性质的“迷魂阵”【高频考点】【基础】平行四边形是几何证明的核心，学生极易将判定定理和性质定理混淆，导致逻辑链条断裂。1.典型错误1：判定定理运用条件模糊1.2.原题呈现：如图，在四边形ABCD中，AB∥CD，∠A=∠C。求证：四边形ABCD是平行四边形。2.3.错解展示：有的学生直接由AB∥CD和AD=BC来证，或者由∠A=∠C和∠B=∠D来证（但题目未给出后两个条件）。3.4.【非常重要】归因分析：对平行四边形的五种判定方法理解不透彻，死记硬背，不会根据已知条件灵活选择。4.5.矫正策略：引导分析法。要证平行四边形，目前有AB∥CD，即一组对边平行。思路一：再证另一组对边平行（AD∥BC）。思路二：证这组对边相等（AB=CD）。但题目给的是角的关系，所以应选择思路一。如何由AB∥CD和∠A=∠C推出AD∥BC？利用同旁内角互补。因为AB∥CD，所以∠A+∠D=180°，又因为∠A=∠C，所以∠C+∠D=180°，所以AD∥BC（同旁内角互补，两直线平行）。从而得证。6.典型错误2：特殊平行四边形（矩形、菱形、正方形）性质混淆1.7.原题呈现：下列说法中，正确的是（）A.对角线相等的四边形是矩形B.对角线互相垂直的四边形是菱形C.对角线互相垂直且相等的四边形是正方形D.对角线互相平分的四边形是平行四边形2.8.错解展示：大量学生选A、B或C。3.9.【基础】归因分析：将特殊图形的性质当成了判定，或者忽略了判定中的前提条件（如平行四边形）。4.10.矫正策略：思维导图对比。在黑板画出包含关系图，并强调判定时的核心条件：1.5.11.矩形：平行四边形+对角线相等（或一个直角）。2.6.12.菱形：平行四边形+对角线垂直（或一组邻边相等）。3.7.13.正方形：平行四边形+对角线垂直且相等（或矩形+一组邻边相等，或菱形+一个直角）。4.8.14.而对于D，对角线互相平分是判定平行四边形的直接定理，正确。15.变式训练：在□ABCD\squareABCD□ABCD中，对角线AC、BD交于点O，AB=6，AC=8，BD=10，则□ABCD\squareABCD□ABCD的面积为多少？（考察平行四边形对角线互相平分，利用勾股定理逆定理证得AC⊥BD，从而面积=对角线乘积的一半，即40）