Kẻ CD//IA'
Ta có: $\frac{DC}{IA'}=$ $\frac{AC}{AA'}=$ $\frac{1}{3}$
$\frac{IB}{DC}=$ $\frac{B'B}{B'C}=$ $\frac{1}{2}$
⇒$IB=\frac{1}{6}IA'=\frac{1}{5}BA'$
⇒$\vec{BI}=\frac{1}{5}\vec{A'B}=\frac{1}{5}(\vec{A'C}+\vec{CB})=\frac{1}{5}(2\vec{a}+\vec{b})$
$\vec{CI}=\vec{CB}+\vec{BI}=\vec{b}+\frac{1}{5}(2\vec{a}+\vec{b})=\frac{2}{5}\vec{a}+\frac{6}{5}\vec{b}4
⇒m/n=1/3
⇒ A