10.6. Circles $\omega_{1}$ and $\omega_{2}$ intersect at points $A$ and $B$. Let $PQ$ and $RS$ be the external common tangents of these two circles (points $P$ and $R$ are on circle $\omega_{1}$, points $Q$ and $S$ are on circle $\omega_{2}$). Given that $RB \parallel PQ$, the ray $RB$ intersects circle $\omega_{2}$ again at point $W$. Find the ratio $\frac{RB}{BW}$.
10.6. As shown in Figure 7, let the intersection point of line $AB$ and $PQ$ be $X$. Then
$$
\begin{array}{l}
X P^{2}=A X \cdot X B \\
=X Q^{2},
\end{array}
$$

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 quadrilateral $PXBR$ is a parallelogram, so
$$
B R=X P=\frac{1}{2} P Q=\frac{1}{2} R S,
$$

the last equality comes from the symmetry of $PQ$ and $RS$.
Since $RS$ is a tangent to circle $\omega_{2}$, we have
$$
R B \cdot R W=R S^{2}=(2 R B)^{2} \text {. }
$$

Therefore, $R W=4 R B$, which means $\frac{R B}{B W}=\frac{1}{3}$.