(10.) Circles $\omega_{1}$ and $\omega_{2}$ intersect at points $A$ and $B$. Let $P Q$ and $R S$ be the external common tangents of these two circles (points $P$ and $R$ are on $\omega_{1}$, points $Q$ and $S$ are on $\omega_{2}$). Given that $R B / / P Q$, the ray $R B$ intersects circle $\omega_{2}$ again at point $W$. Find the ratio $\frac{R B}{B W}$.
Let the intersection point of line $AB$ and $PQ$ be $X$. Then, $XP^2 = AX \cdot XB = XQ^2$, which means $X$ is the midpoint of segment $PQ$. It is easy to see that $AB \parallel PR$, in fact, they are both perpendicular to the line connecting the centers of circles $\omega_1$ and $\omega_2$. From the problem, we know that $PXB R$ is a parallelogram, so $BR = XP = \frac{1}{2} PQ = \frac{1}{2} RS$ (the last equality comes from the symmetry of $PQ$ and $RS$). Consequently, since $RS$ is a tangent to circle $\omega_2$, we have $RB \cdot RW = RS^2 = (2RB)^2$, thus $RW = 4RB$. This indicates that $\frac{RB}{BW} = \frac{1}{3}$ (refer to Figure 8).