a)
Theo tính chất đường phân giác, ta có:
\(\dfrac{AB}{AC}=\dfrac{BD}{DC}=\dfrac{36}{60}=\dfrac{3}{5}\)
Mà \(\left\{{}\begin{matrix}AB^2=BH.BC\\AC^2=CH.BC\end{matrix}\right.\Rightarrow\dfrac{BH}{CH}=\dfrac{AB^2}{AC^2}=\dfrac{3^2}{5^2}=\dfrac{9}{25}\)