(9) (20 points) Let $A+B+C=180^{\circ}$, and satisfy: $\frac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}=$
1, find the value of $\frac{\cos 2 A+\cos 2 B+\cos 2 C}{\cos A+\cos B+\cos C}$.
9 From
$$
\frac{\sin A+\sin B+\sin C}{\cos A+\cos B+\cos C}=1,
$$

that is
$$
\sin A+\sin B+\sin C=\cos A+\cos B+\cos C,
$$

squaring both sides gives
$$
\begin{aligned}
& \sin ^{2} A+\sin ^{2} B+\sin ^{2} C+2(\sin A \sin B+\sin B \sin C+\sin C \sin A) \\
= & \cos ^{2} A+\cos ^{2} B+\cos ^{2} C+2(\cos A \cos B+\cos B \cos C+\cos C \cos A)
\end{aligned}
$$

Therefore,
$$
\begin{aligned}
& \left(\cos ^{2} A-\sin ^{2} A\right)+\left(\cos ^{2} B-\sin ^{2} B\right)+\left(\cos ^{2} C-\sin ^{2} C\right) \\
= & -2[\cos (A+B)+\cos (B+C)+\cos (C+A)],
\end{aligned}
$$

which means
$$
\cos 2 A+\cos 2 B+\cos 2 C=2(\cos A+\cos B+\cos C),
$$

so
$$
\frac{\cos 2 A+\cos 2 B+\cos 2 C}{\cos A+\cos B+\cos C}=2 \text {. }
$$