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学习数学,特别是解题时,对习题进行变式,举一反三,常常收到事半功倍的效果.下面给同学们介绍一下数学习题的几种简单变式:一、把题中的部分题设与结论交换位置例:已知:如图1,AB∥CD,请说明∠BED=∠B+∠D成立的理由.解:过点E作EF∥AB,因为AB∥CD,所以AB∥CD∥EF,所以∠BEF=∠B,∠FED=∠D,所以∠BEF+∠FED=∠B+∠D,即∠BED=∠B+∠D.本题也可以延长BE交CD于点G,再根据平行线的性质及三角形的性质,也可以证出.变式一已知:如图1,∠BED=∠B+∠D,请说明AB∥CD成立的理由.本题是将例题中的题设与结论交换位置,解法如下.过点E作EF∥AB,所以∠BEF=∠B,因为∠BED=∠B+∠D,所以∠BEF+∠FED=∠B+∠D,所以
Learning mathematics, especially when solving problems, is a variant of the exercises. By analogy, it often results in a multiplier effect. Here are a few simple variants of the mathematics problems introduced to the students: First, some of the questions in the questions are designed and concluded. Example of exchange position: Known: As shown in Figure 1, AB∥CD, please explain why ∠BED=∠B+∠D is established. Solution: Exceed E for EF∥AB, because AB∥CD, then AB∥CD∥EF, So ∠BEF=∠B, ∠FED=∠D, so ∠BEF+∠FED=∠B+∠D, that is, ∠BED=∠B+∠D. This question can also extend BE to CD at point G, and then according to the nature of parallel lines And the nature of the triangle, you can also prove that the variable is known as: Figure 1, BED = ∠ B + ∠ D, please explain the reasons for the establishment of AB ∥ CD. This question is to set the position of the example and the conclusion exchange position, Solution is as follows. Over point E for EF∥AB, so ∠BEF=∠B, because ∠BED=∠B+∠D, so ∠BEF+∠FED=∠B+∠D, so