Ta có: $\dfrac{AB}{AC}=\dfrac{3}{4}\Rightarrow AC=\dfrac{4AB}{3}$
Áp dụng định lý Pitago vào $\Delta$ vuông $ABC$ ta có:
$\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}=\dfrac{1}{AB^2}+\dfrac{9}{16AB^2}=\dfrac{25}{16AB^2}$
$\Rightarrow AH=\dfrac{4AB}{5}$
$\Rightarrow AB=\dfrac{5AH}{4}=\dfrac{5.9}{4}=11,25$
$\Rightarrow AC=4AB=15$
$\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{11,25^2+15^2}=\dfrac{75}{4}$
$AC^2=CH.CB\Rightarrow CH=\dfrac{AC^2}{CB}=12$.
