a.△ABC có AD là phân giác \( \widehat{BAC}\) (gt)
⇒\(\frac{BD}{CD}=\frac{AB}{AC}\)
⇒\(\frac{BD}{CD+BC}=\frac{AB}{AB+AC}\)
hay \(\frac{BD}{CD}=\frac{3}{3+4,5}\)
\(\frac{BD}{5}=\frac{3}{7,5}\)
⇒\(BD=\frac{3.5}{7,5}=\frac{15}{7,5}=2\)cm
Có: BC=BD+DC
hay 5=2+DC
⇒DC=3cm