两“线”定“形”——对一道课本例题的探究

• 作 者: 张妹燕
• 机 构: 江苏省常熟实验中学,
• 出 处: 初中生世界
• 发表日期: 2015年38期
• 关键词: 等腰三角形 苏科版 等角 等边

苏科版八(上)64页例2已知:如图1,∠EAC是△ABC的外角,AD平分∠EAC,AD∥BC.求证:AB=AC.【分析】要判断一个三角形是等腰三角形一是根据定义,二是根据“等角对等边”.本题根据已知条件“角平分线”和“平行”都能转化成角来说明,所以选择方法二.【反思】(1)将条件“AD平分∠EAC”和结论“AB=AC”互换,命题是否仍然成立?(2)将条件“AD∥BC”和结论“AB=AC”互换,命题是否仍然成立?

As shown in Figure 1, ∠EAC is the outer corner of △ ABC, AD bisects ∠EAC, AD∥BC. Proof: AB = AC. [Analysis] To determine whether a triangle is equal The first is based on the definition of the waist triangle, the second is based on the “equal angle peer.” This question is based on known conditions “angle bisector ” and “parallel ” can be transformed into angle to illustrate, so choose method two . [Reflections] (1) Will the condition “AD bisect ∠ EAC ” and the conclusion “AB = AC ” be interchanged and the proposition still holds? (2) The condition “AD // BC AB = AC ”exchange, proposition still holds?