a, Xét $ΔABC(A=90^o)$ $⇒BC^2=AB^2+AC^2$ $⇒BC=\sqrt{AB^2+AC^2}=\sqrt{3^2+4^2}=5(cm)$ $AB^2=BH.BC$ $⇒BH=\dfrac{AB^2}{BC}=\dfrac{3^2}{5}=1,8(cm)$ $⇒HC=BC-BH=5-1,8=3,2(cm)$
$AH^2=BH.HC$ $⇒AH=\sqrt{BH.HC}=\sqrt{1,8.3,2}=2,4(cm)$ Ta có $AD$ là phân giác $A$ $⇒\dfrac{AB}{AC}=\dfrac{BD}{DC}$ $⇒\dfrac{AB}{AC}=\dfrac{BC-DC}{DC}$ $⇒\dfrac{3}{4}=\dfrac{5-DC}{DC}$ Giải pt ta được $DC=\dfrac{20}{7}(cm)$
