Let $AK=t$, $AB=x, BC=2x$. Notice that $\omega$ is inscribed in $AKCB$. Therefore, $AK+BC=AB+KC \Rightarrow KC=t+x$. Applying Pythagorean theorem in $\triangle DKC$ gives that $KC^2=DC^2+DK^2 \Rightarrow t=\frac{2x}{3}$. Now you've to finish it!