Đáp án:
$\begin{array}{l}
Do:\dfrac{{BD}}{{DC}} = \dfrac{2}{3} \Rightarrow \dfrac{{BD}}{{BC}} = \dfrac{2}{5}\\
Trong\,\Delta BCI:DN//BI\\
\Rightarrow \dfrac{{IN}}{{CI}} = \dfrac{{BD}}{{BC}} = \dfrac{2}{5}\left( {theo\,Talet} \right)\\
\Rightarrow IN = \dfrac{2}{5}CI\\
Trong:\Delta ADN:EI//DN\\
\dfrac{{IN}}{{AI}} = \dfrac{{DE}}{{AE}} = \dfrac{1}{2}\left( {theo\,Talet} \right)\\
\Rightarrow IN = \dfrac{1}{2}AI\\
\Rightarrow \dfrac{2}{5}CI = \dfrac{1}{2}AI\\
\Rightarrow \dfrac{{AI}}{{CI}} = \dfrac{4}{5}\\
hay\,\dfrac{{AI}}{{IC}} = \dfrac{4}{5}
\end{array}$