Bài 1: Ta có: \(\left\{{}\begin{matrix}\frac{AM}{MB}=\frac{2}{4}=\frac{1}{2}\\\frac{AN}{NC}=\frac{3}{6}=\frac{1}{2}\end{matrix}\right.\)\(\Rightarrow\frac{AM}{MB}=\frac{AN}{NC}\)
Theo định lý Thales đảo suy ra ĐPCM
Bài 2: Theo định lý Thales, ta có:
a/ \(\frac{AN}{NC}=\frac{AM}{MB}\Leftrightarrow\frac{1,2}{4,8}=\frac{1}{x}\Leftrightarrow x=\frac{4,8}{1,2}=4\left(cm\right)\)
b/ \(\frac{MN}{BC}=\frac{AN}{AC}\Leftrightarrow\frac{0,9}{BC}=\frac{1,2}{1,2+4,8}\Leftrightarrow BC=\frac{0,9.6}{1,2}=4,5\left(cm\right)\)