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今年的全国初中数学竞赛二试的第二题,笔者认为这是一道源于教材,高于教材,内涵丰富,不落俗套的好题。题如图1,在△ABC中,AB=AC,D是底边BC上一点,E是线段AD上一点,且∠BED=2∠CED=∠A。求证:BD=2CD。等腰三角形极为常见,但给出连等式∠BED=2∠CED=∠A,使其内涵丰富,可用元素多,证题方法灵活,可用知识点几乎涉及整个平几知识,本文给出标准答案以外的若干证法,大家就不难体会这一点。
This year’s second national junior high school math contest second test, the author believes that this is a good source from the teaching material, higher than the teaching material, rich in content, unconventional. The problem is shown in Figure 1. In △ABC, AB=AC, D is a point on the bottom BC, E is a point on line AD, and ∠BED=2∠CED=∠A. Proof: BD=2CD. Isosceles triangles are very common, but they give the equivalent equation BED=2∠CED=∠A, which makes it rich in connotation, more elements are available, and the method of proofs is flexible. The available knowledge covers almost the entire level of knowledge, and the standard is given in this paper. It is not difficult for everyone to understand this point of evidence.