a) -Xét △AMN và △ABC có:
\(\widehat{BAC}\) là góc chung.
\(\widehat{AMN}=\widehat{ABC}\) (MN//BC và đồng vị)
\(\Rightarrow\)△AMN ∼ △ABC (g-g)
\(\Rightarrow\dfrac{AM}{AB}=k\)
\(\Rightarrow\dfrac{\dfrac{2}{3}AB}{AB}=k\)
\(\Rightarrow k=\dfrac{2}{3}\)
b) -Có: △AMN ∼ △ABC (cmt)
\(\Rightarrow\dfrac{AM}{AB}=\dfrac{AN}{AC}=\dfrac{MN}{BC}=k=\dfrac{2}{3}\)
\(\Rightarrow AM=\dfrac{2}{3}AB=\dfrac{2}{3}.3=2\left(cm\right)\)
\(AN=\dfrac{2}{3}AC=\dfrac{2}{3}.10=\dfrac{20}{3}\left(cm\right)\)
\(MN=\dfrac{2}{3}BC=\dfrac{2}{3}.8=\dfrac{16}{3}\left(cm\right)\)
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