a, \(AC=\sqrt{BC^2-AB^2}=20\left(cm\right)\left(pytago\right)\)
Áp dụng HTL: \(\left\{{}\begin{matrix}HB=\dfrac{AB^2}{BC}=9\left(cm\right)\\HC=\dfrac{AC^2}{BC}=16\left(cm\right)\\HA=\dfrac{AB\cdot AC}{BC}=12\left(cm\right)\end{matrix}\right.\)
b, Vì AIHK là hcn nên \(IK=AH\Leftrightarrow IK^2=AH^2\)
Áp dụng HTL: \(HB\cdot HC=HA^2\)
Do đó \(IK^2=HB\cdot HC\)