Example 6 Given that the bases of two congruent regular triangular pyramids are glued together, exactly forming a hexahedron with all dihedral angles equal, and the length of the shortest edge of this hexahedron is 2. Then the distance between the farthest two vertices is $\qquad$
(1996, National High School League)
Analysis: There are two difficulties in this problem: one is to determine which of $AC$ and $CD$ is 2 in length; the other is to determine which of the line segments $AC$, $CD$, and $AB$ represents the distance between the farthest two vertices.

Solution: As shown in Figure 6, construct $CE \perp AD$, and connect $EF$. It is easy to prove that $EF \perp AD$, so $\angle CEF$ is the plane angle of the dihedral angle formed by plane $ADF$ and plane $ACD$.
Let point $G$ be the midpoint of $CD$. Similarly, $\angle AGB$ is the plane angle of the dihedral angle formed by plane $ACD$ and plane $BCD$. Given that $\angle CEF = \angle AGB$. Let the side length of the base $\triangle CDF$ be $2a$, and the length of the side edge $AC$ be $b$. In $\triangle ACD$, $CE \cdot b = AG \cdot 2a$, thus
$$
CE = \frac{AG \cdot 2a}{b} = \frac{\sqrt{b^2 - a^2} \cdot 2a}{b}.
$$

In $\triangle ABC$, it is easy to find that
$$
\begin{aligned}
AB & = 2 \sqrt{b^2 - \left(\frac{2}{3} \sqrt{3} a\right)^2} \\
& = 2 \sqrt{b^2 - \frac{4}{3} a^2}.
\end{aligned}
$$

From $\triangle CEF \sim \triangle AGB$, we get
$$
\frac{AB}{CF} = \frac{AG}{CE},
$$

which means $\frac{2 \sqrt{b^2 - \frac{4}{3} a^2}}{2a} = \frac{\sqrt{b^2 - a^2}}{\frac{\sqrt{b^2 - a^2} \cdot 2a}{b}}$.
Thus, $b = \frac{4}{3} a = \frac{2}{3} \times 2a$.
Therefore, $AC < CD$.
Given that $b = 2$, we have $2a = 3$, and $AB = 2$.
Thus, $AC = AB < CD$.
Hence, the distance between the farthest two vertices is 3.