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我们在处理课本习题时要根据题目的特征,让其中的元素“动”起来,让习题“活”起来,通过一题多变,深入挖掘其隐藏的价值,扩展我们的思路,起到举一反三的作用.题目一(人教版义务教育教科书八年级《数学》下第62页第15题)如图1,四边形ABCD是正方形,G是BC上的任意一点,DE⊥AG于点E,BF∥DE,且交AG于点F.求证:AF-BF=EF.分析要证AF-BF=EF,观察图形可知AF-AE=EF,所以只需证明AE=BF即
When we deal with textbooks exercises according to the characteristics of the topic, let the elements “move ” up, exercises “living ” up, through a question changeable, dig deeper its hidden value, expand our train of thought, Title one (eighth edition of the textbook of the people’s education compulsory education under the sixth title of the first 15) as shown in Figure 1, the quadrilateral ABCD is a square, G is any point on the BC, DE ⊥ AG at the point E, BF ∥ DE, and pay AG at point F. Verification: AF-BF = EF. Analysis to prove AF-BF = EF, the observed graph shows AF-AE = EF, so only prove that AE = BF