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这是一道向量的典型题目,几乎在所有的练习册中都要出现,如果不观察点的位置,盲目列式,就得不到有效的方程组,使问题得以解决,本人用这道题将这个问题进行说明。已知:在△ABC中,点D是边AB的三等分点,点E是边AC的中点,BE与CD相交于点F。求:(1)用AB和AC表示AF(2)用BC和BA表示BF解:(1)方法一:【分析】利用1平面向量基本定理2A、B、C三点共线OA=xOB+yOC且x+y=1列方程组求解。∵点D是边AB的三等分点,点E是边AC的中点
This is a typical example of a vector that appears in almost all exercise books. If you do not observe the position of a point or a blind list, you will not have a valid system of equations that will solve the problem. I will use this question This question is explained. It is known that in Δ ABC, point D is the trisection of edge AB, point E is the midpoint of edge AC, and BE and CD intersect at point F. (1) AB and AC for AF (2) BC and BA for BF solution: (1) Method one: [Analysis] Using the basic theorem of 1 plane vector 2A, B, C Three points collinear OA = xOB + yOC and x + y = 1 column equations. ∵ Point D is the third bisection of edge AB, and point E is the midpoint of edge AC