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一、争论焦点例1如图1所示,在平行四边形ABCD中,E是DC的中点,AE交BD于M,试用向量的方法证明:M是BD的一个三等分点.文[1]作者在上述源自苏教版《数学4》的例1及依据平面向量基本定理的原证基础上,给出了如下简证.简证:AM+MB=DC=2DE=2DM+2ME,由平面向量基本定理得AM=2ME,MB=2DM,M是BD的一个三等分点.
First, the focus of controversy Example 1 As shown in Figure 1, in the parallelogram ABCD, E is the midpoint of DC, AE AC BD in M, the trial vector method proved: M is a BD third grade points. The author gives the following brief proofs on the basis of the above example 1 derived from the Sujiao edition of “Mathematics 4” and the original evidences based on the basic theorem of plane vectors: The proofs are as follows: AM + MB = DC = 2DE = 2DM + 2ME, According to the basic theorem of plane vector, AM = 2ME, MB = 2DM, M is a third-degree point of BD.