Vì $∆ABC$ vuông tại $A$ nên
$BC=\sqrt{AB^2+AC^2}=\sqrt{6^2+4^2}=2\sqrt{13}$
$AH=\dfrac{AB.AC}{BC}=\dfrac{6.4}{2\sqrt{13}}=\dfrac{12}{\sqrt{13}}$
$BH=\dfrac{AB^2}{BC}=\dfrac{6^2}{2\sqrt{13}}=\dfrac{18}{\sqrt{13}}$
$CH=BC-BH=2\sqrt{13}-\dfrac{18}{\sqrt{13}}=\dfrac{8}{\sqrt{13}}$