$\frac{1}{2}=\sin(ABC_1)=\sin(KC_1D)=\frac{DK}{KC_1}=\frac{DK}{AB-DK}$ $\to \frac{DK}{AB}=1/3$

As $BC=BC_1$ we have that $\frac{AC_1}{BC_1}=\frac{0.5AD}{BC}=\frac{0.5BC}{BC}=\frac{1}{2}$. Hence $\angle ABC_1=30^\circ$. Having that $\angle ABC_1=\angle DC_1K=30^\circ$ we get $sin 30^\circ =\frac{KD}{KC_1}=\frac{KD}{KC}=\frac{KD}{DC-KD}=\frac{KD}{AB-KD}$ Thus $AB-DK=2DK$, $AB=3DK$ and $\frac{DK}{AB}=\frac{DK}{3DK}=\frac{1}{3}$.

It's clear that, $\Delta BC_1C$ is equilateral, If $\angle KC_1D =x \implies \angle C_1CD=45^{\circ}-\frac {x}{2}=30^{\circ} \implies \boxed {x=30^{\circ}}$ Hence, $$\frac {DK}{KC_1}=\frac {DK}{CK}=\frac {1}{2} \implies \frac {CK}{DK}=2 \implies DK:AB=1:3$$

A similar kind of problem appeared in Indian RMO 1990 P3, here, https://artofproblemsolving.com/community/c6h56050p347212