Từ $M$ kẻ $MN//AB \quad (N\in DC)$
$\Rightarrow ΔDAI \sim ΔNMI \, (g.g)$
$\Rightarrow \dfrac{MN}{AD} = \dfrac{MI}{ID}$
Ta cũng có: $ΔMNC \sim ΔBDC \, (g.g)$
$\Rightarrow \dfrac{MN}{BD} = \dfrac{NC}{DC} = \dfrac{MC}{BC}\dfrac{1}{2}$
$\Rightarrow MN = \dfrac{BD}{2}$
$\Rightarrow MN = \dfrac{\dfrac{2}{3}AB}{2} = \dfrac{1}{3}AB$
mà $AD = \dfrac{1}{3}AB$
nên $MN = AD$
$\Rightarrow \dfrac{MN}{AD} = 1$
$\Rightarrow \dfrac{MI}{ID} = 1$
$\Rightarrow MI = ID = \dfrac{1}{2}DN$
Ta lại có: $\dfrac{NC}{DC} = \dfrac{1}{2}$
$\Rightarrow NC = DN$
$\Rightarrow \dfrac{DN}{DC} = \dfrac{1}{2}$
$\Rightarrow \dfrac{DI}{DC} = \dfrac{1}{4}$
hay $DI = \dfrac{1}{4}DC$