Example 2 In the tetrahedron $A-B C D$, it is known that
$$
\begin{array}{l}
\angle A C B=\angle C B D \\
\angle A C D=\angle A D C=\angle B C D=\angle B D C=\theta,
\end{array}
$$

and $\cos \theta=\frac{\sqrt{10}}{10}$.
If the length of edge $A B$ is $6 \sqrt{2}$, then the volume of this pyramid is
$\qquad$
Solve As shown in Figure 2, from the problem, we know $\triangle A C D \cong \triangle B C D$,
and $\square$
$$
\begin{array}{l}
A C=A D \\
=B C=B D .
\end{array}
$$

Then $\triangle A C D$
$\cong \triangle B C D$
$\cong \triangle C A B$
$\cong \triangle D A B$.
Therefore, $\angle A C B$
$$
\begin{array}{l}
\because \angle C B D=\angle C A D=\angle A D B, \\
A B=C D .
\end{array}
$$

Take the midpoints $E$ and $F$ of $C D$ and $A B$ respectively, and connect $A E$, $B E$, and $E F$.
Then $A E \perp C D, B E \perp C D$.
Thus, $C D \perp$ plane $A B E$.
Therefore, plane $A B E$ is the perpendicular plane of $C D$.
It is calculated that $A E=B E=9 \sqrt{2}, E F=12$. $=\frac{1}{3} \times \frac{1}{2} \times 6 \sqrt{2} \times 12 \times 6 \sqrt{2}=144$.