\(AC=\sqrt{BC^2-AB^2}=16\left(cm\right)\left(pytago\right)\)
Áp dụng HTL tam giác
\(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot BC\\AH^2=CH\cdot BH\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}HB=\dfrac{AB^2}{BC}=3,24\left(cm\right)\\HC=\dfrac{AC^2}{BC}=10,24\left(cm\right)\\AH=\sqrt{3,24\cdot10,24}=5,76\left(cm\right)\end{matrix}\right.\)