Xét $ΔABC$ vuông tại $\widehat{A}$ có:
$\widehat{B}+\widehat{C}=90^o$
$⇒\widehat{B}=90^o-\widehat{C}=90^o-30^o=60^o$
$cos_C=\dfrac{AC}{BC}$
$⇒BC=\dfrac{AC}{cos_C}=\dfrac{8}{cos30^o}=\dfrac{16\sqrt{3}}{3}(cm)$
$BC^2=AB^2+AC^2$
$⇒AB=\sqrt{BC^2-AC^2}=\sqrt{(\dfrac{16\sqrt{3}}{3})^2-8^2}=\dfrac{8\sqrt{3}}{3}(cm)$
Ta có $AD$ phân giác góc $\widehat{A}$
$⇒\dfrac{AB}{AC}=\dfrac{BD}{DC}=\dfrac{\dfrac{8\sqrt{3}}{3}}{8}=\dfrac{\sqrt{3}}{3}(cm)$
$⇒\dfrac{BD}{\sqrt{3}}=\dfrac{DC}{3}=\dfrac{BD+DC}{\sqrt{3}+3}=\dfrac{\dfrac{16\sqrt{3}}{3}}{3+\sqrt{3}}=\dfrac{8\sqrt{3}-8}{3}(cm)$
$\dfrac{BD}{\sqrt{3}}=\dfrac{8\sqrt{3}-8}{3}⇒BD=\dfrac{8\sqrt{3}-8}{3}.\sqrt{3}=\dfrac{24-8\sqrt{3}}{3}(cm)$
$\dfrac{DC}{3}=\dfrac{8\sqrt{3}-8}{3}⇒DC=\dfrac{8\sqrt{3}-8}{3}.3=8\sqrt{3}-8(cm)$
