#breaking-web2.0calc's-ice.

Connect AK and extend it to meet BC at F.

By Ceva's theorem,

$\dfrac{\text{AE}}{\text{EB}} \cdot \dfrac{\text{BF}}{\text{FC}} \cdot \dfrac{\text{CD}}{\text{DA}} = 1\\ \text{Let }\dfrac{\text{BF}}{\text{FC}} = k,\\ \dfrac{2}{3} \cdot k \cdot \dfrac{1}{2} = 1\\ k = 3\\ \text{BF}:\text{FC} = 3:1$

By Menelaus' theorem,

$\dfrac{\text{BF}}{\text{FC}} \cdot \dfrac{\text{CA}}{\text{AD}}\cdot \dfrac{\text{DK}}{\text{KB}} = 1\\ 3\cdot \dfrac{3}{2} \cdot \dfrac{\text{DK}}{\text{KB}} = 1\\ \boxed{\dfrac{\text{DK}}{\text{KB}} = \color{red}{\dfrac{2}{9}}}$