\(AB=\dfrac{AC}{\cot C}=12\left(cm\right)\\ \Rightarrow BC=\sqrt{AB^2+AC^2}=15\left(cm\right)\\ \Rightarrow\dfrac{AD}{CD}=\dfrac{AB}{BC}=\dfrac{12}{15}=\dfrac{4}{5}\Rightarrow AD=\dfrac{4}{5}CD\\ AD+CD=AC\\ \Rightarrow\dfrac{9}{5}CD=9\Rightarrow CD=5\left(cm\right)\)