-△ABH∼△CBA (g-g) \(\Rightarrow\dfrac{AB}{CB}=\dfrac{BH}{BA}\Rightarrow BH=\dfrac{AB^2}{CB}\)
-△CAH∼△CBA (g-g) \(\Rightarrow\dfrac{CA}{CB}=\dfrac{CH}{CA}\Rightarrow CH=\dfrac{AC^2}{CB}\)
\(\dfrac{S_{HAB}}{S_{HAC}}=\dfrac{HB}{HC}=\dfrac{\dfrac{AB^2}{CB}}{\dfrac{AC^2}{CB}}=\left(\dfrac{AB}{AC}\right)^2=\left(\dfrac{4}{9}\right)^2=\dfrac{36}{81}\)