16. The radius of $\odot O$ is $R$, and the radius of $\odot O_{1}$ is $r (R > r)$. The two circles are externally tangent at $A$. $OD$ is tangent to $\odot O_{1}$ at $D$, and $O_{1}E$ is tangent to $\odot O$ at $E$. $B$ and $C$ are the midpoints of $OE$ and $O_{1}D$, respectively. (1) Prove that points $B$, $A$, and $C$ are not collinear;
(2) If $R=6$ and $r=3$, connect $BA$ and extend it to intersect $O_{1}D$ at $C_{1}$. Find the length of $O_{1}C_{1}$.
16.6 (1) Let $\angle B A O=\alpha, \angle C A O_{1}=\beta$. To prove that $B, A, C$ are not collinear is equivalent to proving $\sin \alpha \neq \sin \beta$.

We can calculate $\sin \alpha=\sqrt{\frac{r(r+2 R)}{(R+5 r)(R+r)}}, \sin \beta=\sqrt{\frac{R(R+2 r)}{(5 R+r)(R+r)}}$. The following proof by contradiction shows that $\sin \alpha \neq \sin \beta \Leftrightarrow R \neq r$.
(2) Extend $B A$ to intersect $O_{1} D$ at $C_{1}$, and let $O_{1} C_{1}=x$. We can calculate $\sin \alpha=\sqrt{\frac{5}{21}}$. By the Law of Sines, $\sin \angle O_{1} A C_{1}=\frac{2 \sqrt{2}}{3}$.
$$
\frac{x}{A C} \text {. }
$$

Using the Law of Cosines to express $A C^{2}$ in terms of $x$, we obtain an equation and solve for $x$: $x=\frac{24 \sqrt{10}-15}{41}$.