(50 points) In $\triangle ABC$, $I$ is the incenter, and lines $BI$, $CI$ intersect sides $AC$, $AB$ at points $D$, $E$ respectively.
(1) Find the necessary and sufficient condition for $\frac{IC}{IB} = \frac{CD}{BE}$ to hold;
(2) If $\frac{IC}{IB} = \frac{CD}{BE} \neq 1$, find the range of $k = \frac{S_{\triangle BIC}}{S_{\triangle ABC}}$.
(1) From $\frac{I C}{C D}=\frac{\sin \left(A+\frac{B}{2}\right)}{\sin \frac{B+C}{2}}$,
$$
\begin{array}{l}
\frac{I B}{B E}=\frac{\sin \left(A+\frac{C}{2}\right)}{\sin \frac{B+C}{2}}, \\
\text { then } \frac{I C}{I B}=\frac{C D}{B E} \Leftrightarrow \sin \left(A+\frac{B}{2}\right)=\sin \left(A+\frac{C}{2}\right) \\
\Leftrightarrow 2 \sin \frac{B-C}{4} \cdot \cos \frac{4 A+B+C}{4}=0 \\
\Leftrightarrow 2 \sin \frac{B-C}{4} \cdot \cos \frac{3 A+\pi}{4}=0 \\
\Leftrightarrow \frac{\angle B-\angle C}{4}=0 \text { or } \frac{3 \angle A+\pi}{4}=\frac{\pi}{2} \\
\Leftrightarrow \angle B=\angle C \text { or } \angle A=\frac{\pi}{3} .
\end{array}
$$
(2) From condition (1) we know
$$
\frac{I C}{I B}=\frac{C D}{B E} \neq 1 \Leftrightarrow \angle A=\frac{\pi}{3}, \angle B \neq \angle C .
$$

By the properties of the incenter of a triangle, it is easy to know
$$
k=\frac{S_{\triangle B B C}}{S_{\triangle A B C}}=\frac{a}{a+b+c}=\frac{\sin A}{\sin A+\sin B+\sin C} .
$$

Given $\angle A=\frac{\pi}{3}$, let $\angle B=\frac{\pi}{3}+\alpha, \angle C=\frac{\pi}{3}-\alpha, \alpha \in$ $\left(-\frac{\pi}{3}, 0\right) \cup\left(0, \frac{\pi}{3}\right)$, then
$$
\begin{array}{l}
k=\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}+\sin \left(\frac{\pi}{3}+\alpha\right)+\sin \left(\frac{\pi}{3}-\alpha\right)} \\
=\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}+2 \sin \frac{\pi}{3} \cdot \cos \alpha}=\frac{1}{1+2 \cos \alpha} .
\end{array}
$$

Since $\alpha \in\left(-\frac{\pi}{3}, 0\right) \cup\left(0, \frac{\pi}{3}\right)$,
thus, $\cos \alpha \in\left(\frac{1}{2}, 1\right), k \in\left(\frac{1}{3}, \frac{1}{2}\right)$.