Example 4 As shown in Figure 7, in $\triangle A B C$, $B C: C A: A B$ $=5: 4: 3$. The three excenters $A^{\prime}$, $B^{\prime}$, and $C^{\prime}$ are located inside $\angle A$, $\angle B$, and $\angle C$, respectively. Find the ratio of the sides of $\triangle A^{\prime} B^{\prime} C^{\prime}$.
For simplicity, let
$BC=5, CA=4, AB=3$, then $\triangle ABC$ is a right triangle, $\angle BAC=90^{\circ}$.

By property 3 of the excenter, $B^{\prime}, A, C^{\prime}$, $C^{\prime}, B, A^{\prime}$, and $A^{\prime}, C, B^{\prime}$ are collinear, and $A^{\prime}A, B^{\prime}B, C^{\prime}C$ are the three altitudes of $\triangle A^{\prime}B^{\prime}C^{\prime}$. Thus, their intersection point $I$ is both the orthocenter of $\triangle A^{\prime}B^{\prime}C^{\prime}$ and the incenter of $\triangle ABC$, and
$$
\angle BAI=\angle CAI=45^{\circ}.
$$

It is easy to prove that $A, I, B, C^{\prime}$ and $A, I, C, B^{\prime}$ are concyclic, so
$$
\begin{array}{l}
\angle BC^{\prime}I=\angle BAI=45^{\circ}, \\
\angle CB^{\prime}I=\angle CAI=45^{\circ},
\end{array}
$$

which implies
$$
\angle A^{\prime}C^{\prime}C=45^{\circ}=\angle A^{\prime}B^{\prime}B.
$$

Thus, $\triangle A^{\prime}B^{\prime}B$ and $\triangle A^{\prime}C^{\prime}C$ are both isosceles right triangles.
Let $A^{\prime}B=x, A^{\prime}C=y$, then
$$
A^{\prime}B^{\prime}=\sqrt{2}x, A^{\prime}C^{\prime}=\sqrt{2}y.
$$

It is easy to prove that $B, C, B^{\prime}, C^{\prime}$ are concyclic, hence
$$
\angle BCA^{\prime}=\angle BC^{\prime}A.
$$

Also, $\angle A^{\prime}BC=\angle ABC^{\prime}$, so,
$\triangle A^{\prime}BC \sim \triangle ABC^{\prime}$
$$
\begin{array}{l}
\Rightarrow \frac{A^{\prime}B}{AB}=\frac{BC}{BC^{\prime}} \\
\Rightarrow A^{\prime}B \cdot BC^{\prime}=AB \cdot BC,
\end{array}
$$

which means $x(\sqrt{2}y-x)=3 \times 5=15$.
Similarly, $A^{\prime}C \cdot CB^{\prime}=AC \cdot CB$, which means
$$
y(\sqrt{2}x-y)=4 \times 5=20.
$$

In $\triangle A^{\prime}BC$, noting that $\angle BA^{\prime}C=45^{\circ}$, we have
$$
BC^{2}=A^{\prime}B^{2}+A^{\prime}C^{2}-2A^{\prime}B \cdot A^{\prime}C \cos \angle BA^{\prime}C,
$$

which is
$$
\begin{array}{l}
5^{2}=x^{2}+y^{2}-2xy \cos 45^{\circ} \\
=x^{2}+y^{2}-\sqrt{2}xy.
\end{array}
$$

Thus, $25=x^{2}-y(\sqrt{2}x-y)$
$$
=y^{2}-x(\sqrt{2}y-x).
$$

Substituting equations (1) and (2) into equation (3), we get
$y=\sqrt{40}$ (negative root discarded), $x=\sqrt{45}$ (negative root discarded).
Therefore, $A^{\prime}C^{\prime}=\sqrt{2}y=\sqrt{80}$,
$$
A^{\prime}B^{\prime}=\sqrt{2}x=\sqrt{90}.
$$

Since $B, C, B^{\prime}, C^{\prime}$ are concyclic, $B^{\prime}C^{\prime}$ is the diameter of the circle, thus,
$$
B^{\prime}C^{\prime}=\frac{BC}{\sin \angle BC^{\prime}C}=\frac{5}{\sin 45^{\circ}}=\sqrt{50}.
$$

Hence, $A^{\prime}B^{\prime}: A^{\prime}C^{\prime}: B^{\prime}C^{\prime}=\sqrt{9}: \sqrt{8}: \sqrt{5}$.