In $\triangle A B C$, $\angle B>\angle A>\angle C$, and $\angle B, \angle A, \angle C$ form an arithmetic sequence. $I$ is the incenter of $\triangle A B C$, the extension of $B I$ intersects $A C$ at point $D$, and the extension of $C I$ intersects $A B$ at point $E$.
(1) Prove that $B C^{2}=B E \cdot B A+C D \cdot C A$;
(2) Find the value of $D E\left(\frac{1}{B I}+\frac{1}{C I}\right)$.
Proof: (1) As shown in Figure 5, since $\angle B, \angle A, \angle C$ form an arithmetic sequence, thus, $2 \angle A = \angle B + \angle C$.

Also, $\angle A + \angle B + \angle C = 180^{\circ}$, then
$$
\begin{array}{l}
\angle A = 60^{\circ}, \\
\frac{\angle B + \angle C}{2} = \angle A = 60^{\circ}.
\end{array}
$$

Since $I$ is the incenter of $\triangle ABC$, $BI$ and $CI$ bisect $\angle ABC$ and $\angle ACB$ respectively.
Thus, $\angle IBC = \frac{1}{2} \angle ABC, \angle ICB = \frac{1}{2} \angle ACB$.
Therefore, $\angle IBC + \angle ICB$
$$
\begin{array}{l}
= \frac{1}{2} (\angle ABC + \angle ACB) \\
= \frac{\angle B + \angle C}{2} = 60^{\circ}.
\end{array}
$$

Thus, $\angle BIC = \angle EID = 120^{\circ}$.
Noting that $\angle A + \angle EID = 180^{\circ}$, therefore, $A, E, I, D$ are concyclic.

Taking a point $M$ on $BC$ such that $\angle IMC = \angle IDA$, thus, $I, M, C, D$ are concyclic.

Noting that $\angle BEI = \angle IDA = \angle IMC$, thus, $I, M, B, E$ are concyclic. Therefore,
$$
\begin{array}{l}
CM \cdot CB = CI \cdot CE = CD \cdot CA, \\
BM \cdot BC = BI \cdot BD = BE \cdot BA.
\end{array}
$$

Hence $BC^2 = CM \cdot CB + BM \cdot BC = CD \cdot CA + BE \cdot BA$.
Solution: (2) Connecting $AI$, then $AI$ bisects $\angle BAC$. Since $A, E, I, D$ are concyclic, then $IE = ID$, and $\angle EID = 120^{\circ}$.
Thus, $IE = ID = \frac{\sqrt{3}}{3} ED$.
In $\triangle BIC$, by the cosine rule we have
$$
\begin{array}{l}
BC^2 = BI^2 + CI^2 - 2 BI \cdot CI \cos 120^{\circ} \\
= BI^2 + CI^2 + BI \cdot CI.
\end{array}
$$

From (1) we know
$$
\begin{array}{l}
BC^2 = BI \cdot BD + CI \cdot CE \\
= BI(BI + ID) + CI(CI + IE) \\
= BI^2 + BI \cdot ID + CI^2 + CI \cdot IE.
\end{array}
$$

From equations (1) and (2) we know
$$
BI \cdot CI = BI \cdot ID + CI \cdot IE.
$$

Since $ID = IE$, we have
$$
IE \left( \frac{1}{BI} + \frac{1}{CI} \right) = 1.
$$

Thus, $DE \left( \frac{1}{BI} + \frac{1}{CI} \right) = \sqrt{3}$.