a. -△ABC có AD là phân giác \(\Rightarrow\dfrac{DB}{DC}=\dfrac{AB}{AC}=\dfrac{16}{12}=\dfrac{4}{3}\)
b. -△ABC có DH//AC \(\Rightarrow\dfrac{DH}{AC}=\dfrac{BD}{BC}=\dfrac{BD}{BD+CD}\)
\(\Rightarrow\dfrac{DH}{12}=\dfrac{4}{4+3}\Rightarrow DH=\dfrac{12.4}{4+3}=\dfrac{48}{7}\left(cm\right)\)