Hình vuông ABCD, A(-1,2), C(3,-2), E là trung điểm AD, đường thẳng qua B vuông góc với CE tại M có phương trình: 2x-y-4=0, N là trung điểm MB. Tìm tọa độ giao điểm của AN và DM 
CE: $x+2y+1=0; E\in CE\Rightarrow E(-2t-1;t)$
$\Rightarrow D(-4t-3;2t+2) \Rightarrow B (5+4t;-2t-2)$
mà $B\in BM \Rightarrow t=\frac{-4}{5}$
$\Rightarrow B(\frac{9}{5};\frac{-2}{5});D(\frac{1}{5};\frac{2}{5})$
$M=BM cắt CE \Rightarrow M(\frac{7}{5};\frac{-6}{5} ) \Rightarrow  N(\frac{8}{5};\frac{-4}{5})$
$\Rightarrow pt AN:...;; pt DM:..$ từ đó suy ra tđ

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