Let the radii of $\omega_1$ and $\omega_2$ be $r_1$ and $r_2$ respectively. Let $E$ be the second intersection of $AK$ and $\omega_2$ (Fig. 40). As $AK$ and $KE$ are diameters of $\omega_1$ and $\omega_2$ respectively, the angles $ABK$ and $KDE$ must be right angles. In triangle $AKC$, the segment $KB$ is both a median and an altitude, thus $KA = KC$ and $\angle KAC = \angle ACK$. As $CDEK$ is cyclic, we have
$$\angle EAD = \angle KAC = \angle ACK = \angle DEA.$$
Thus $ADE$ is isosceles, also $KBA$ and $KDE$ are similar. Therefore
$$\frac{r_1}{r_2} = \frac{AK}{KE} = \frac{AB}{DE} = \frac{AB}{AD} = \frac{1}{3}.$$

Fig. 40

Let the radii of $\omega_1$ and $\omega_2$ be $r_1$ and $r_2$ respectively. Let $P$ be the midpoint of $CD$ and $Q$ the center of $\omega_2$ (Fig. 41). Then $PQ$ is perpendicular to $CD$, as it connects the midpoint of the chord $CD$ of $\omega_2$ and the center of $\omega_2$. Thus $\angle APQ = \angle CPQ = 90^\circ$. As $AK$ is a diameter of $\omega_1$, we have $\angle ABK = 90^\circ$. Therefore $BK \parallel PQ$. Together with $AB = BC = CD$, we get $\frac{AQ}{AK} = \frac{AP}{AB} = 2.5$ or $2AQ = 5AK$. From here $2(2r_1 + r_2) = 5 \cdot 2r_1$, from which $\frac{r_1}{r_2} = \frac{1}{3}$.

Fig. 41