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1题目呈现如图1,直线EF将矩形纸片ABCD分成面积相等的两部分,E、F分别与BC交于点E,与AD交于点F(E,F不与顶点重合).(1)求证:AF=CE.(2)(3)略.2解答概况近75%的学生采用如下方式证明:连结AC交EF于点O,如图2.因为矩形是中心对称图形,而,所以EF,AC都过对称中心,故有OA=OC.然后用全等证得AF=CE.显然,“EF,AC都过对称中心,故有OA=OC”的得出缺乏依据.本题解答的正确率只有15%左右.
1 questions appear as shown in Figure 1, a straight line EF rectangular paper ABCD will be divided into two equal parts, E, F and BC were at the point E, and AD at the point F (E, F does not coincide with the vertex. ) Proof: AF = CE. (2) (3) Slightly .2 Solution Summary Nearly 75% of students use the following method to prove: Link AC AC EF at point O, as shown in Figure 2. Because the rectangle is a centrosymmetric graph, EF, AC are symmetric center, so there OA = OC. And then use the congruent AF = CE. Obviously, “EF, AC are symmetric center, so there OA = OC ” lack of basis. The correct answer rate is only about 15%.