a) Ta có: \(\dfrac{AB}{AN}=\dfrac{15}{6}=\dfrac{5}{2}\)
\(\dfrac{AC}{AM}=\dfrac{20}{8}=\dfrac{5}{2}\)
=> \(\dfrac{AB}{AN}=\dfrac{AC}{AM}\)
Xét \(\Delta ABC\) và \(\Delta ANM\) có:
\(\dfrac{AB}{AN}=\dfrac{AC}{AM}\) (cmt)
\(\widehat{A}\): chung
=> \(\Delta ABC\) ~ \(\Delta ANM\) (c.g.c)
b) Vì \(\Delta ABC\) ~ \(\Delta ANM\) (cmt)
=> \(\dfrac{AB}{AN}=\dfrac{BC}{NM}\)
\(\Rightarrow NM=\dfrac{AN\cdot BC}{AB}=\dfrac{6\cdot25}{15}=3,6\left(cm\right)\)
\(2P_{AMN}=AN+AM+MN=6+8+3,6=17,6\left(cm\right)\)