$\odot O_{2}$ is externally tangent to $\odot O_{1}$ at point $A$, and an external common tangent of the two circles is tangent to $\odot O$ at point $B$. If $AB$ is parallel to another external common tangent of the two circles, then the ratio of the radii of $\odot O_{1}$ and $\odot O_{2}$ is ( ).
(A) $2: 5$
(B) $1: 2$
(C) $1: 3$
(D) $2: 3$
5.(C).

As shown in Figure 7, let another external common tangent line touch $\odot O_{1}$ at point $C$ and $\odot O_{2}$ at point $D$. Draw a perpendicular from $O_{1}$ to $O_{2} D$, with the foot of the perpendicular being $E$. The radii of the two circles are $r_{1}$ and $r_{2}$, respectively.
By symmetry, we have
$$
\begin{array}{l}
\angle C O_{1} B \\
=\angle C O_{1} A \\
=\angle A O_{1} B=120^{\circ} . \\
\text { Therefore, } \angle O_{2} O_{1} E=30^{\circ} .
\end{array}
$$

Thus, $r_{1}+r_{2}$
$$
=2\left(r_{2}-r_{1}\right) \text {. }
$$

Hence, $r_{1}: r_{2}=1: 3$.