Example 4 As shown in Figure 4, in the acute triangle $\triangle ABC$, side $BC$ is the shortest side. Points $O, G, I, H$ are the circumcenter, centroid, incenter, and orthocenter of the triangle, respectively. When $OG = GI = IH$, then $AB : AC : BC =$
Explanation: From the definition of the Euler line, the circumcenter $O$, centroid $G$, and orthocenter $H$ are collinear. By the properties of the Euler line, we have
$$
O G: G H=1: 2 \text {. }
$$

Given $O G=G I=I H$, it follows that $I$ is the midpoint of $G H$, meaning point $I$ lies on the Euler line $O H$.

As shown in Figure 4, connect $O B$ and $O C$, and draw $O M \perp A B$, with the foot of the perpendicular at $M$.
It is easy to prove that $\angle B O M=\angle A C B$, thus
$\angle O B M=\angle C B H$.
Since $B I$ bisects $\angle A B C$, it follows that $B I$ bisects $\angle O B H$.
Similarly, $C I$ bisects $\angle O C H$.
Because $\frac{O B}{H B}=\frac{O I}{I H}=\frac{O C}{H C}$, we have
$H B=H C$.
Therefore, $\angle H C B=\angle H B C$, and consequently,
$$
90^{\circ}-\angle A B C=90^{\circ}-\angle A C B \text {, }
$$

which implies $\angle A B C=\angle A C B$.
Thus, $A C=A B$.
Hence, $\triangle A B C$ is an isosceles triangle with $B C$ as the base.
Since $\triangle O B H \cong \triangle O C H$, it follows that $O H$ bisects $\angle B O C$. Therefore, $O H \perp B C$ at $D$.
It is easy to prove that Rt $\triangle B O M \backsim$ Rt $\triangle B H D$.
Thus, $\frac{B M}{B D}=\frac{O B}{H B}=\frac{O I}{I H}=2$.
Therefore, $\frac{A B}{B C}=\frac{B M}{B D}=2$.
Hence, $A B: A C: B C=2: 2: 1$.