\(\frac{BD}{DC}=\frac{AB}{AC}=\frac{6}{9}=\frac{2}{3}\\ \Leftrightarrow\frac{BD}{BD+DC}=\frac{2}{2+3}=\frac{2}{5}\\ \Leftrightarrow\frac{BD}{BC}=\frac{2}{5}\\ \Rightarrow BD=BC\cdot\frac{2}{5}=4\left(cm\right)\) Theo tính chât phân giác ngoài:
\(\frac{EB}{EC}=\frac{AB}{AC}=\frac{6}{9}=\frac{2}{3}\\ \Leftrightarrow\frac{EB}{EB+EC}=\frac{2}{3}\\ \Leftrightarrow\frac{EB}{EB+10}=\frac{2}{3}\\ \Leftrightarrow3EB=2\cdot\left(EB+10\right)\\ \Rightarrow EB=20\left(cm\right)\)