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It is known that the vertex of triangle ABC is a (0,1) B (8,0) C (4,10) if vector BD = vector DC, and vector CE = 2, vector EA, a The intersection of AD and be is f, and the vector AF is obtained.
∵ vector BD = vector DC, ∵ D is the midpoint of BC, and the coordinates of ∵ D are (6,5)
∵ vector CE = 2, vector EA, ∵ CE = 2EA, ∵ CE / EA = 2,
The coordinate of E is (4 / 3,4)
Let the midpoint of CE be g, then the coordinates of G are (8 / 3,7)
The vector DG = (- 10 / 3,2)
From CE = 2EA, CG = Ge, Ge = EA, BD = DC, ∥ DG ∥ BG, ∥ DF = FA,
The coordinates of F are (3,3)
The vector AF = (3,2)
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