Example 8 The edge length of the regular tetrahedron $ABCD$ is $a$. Take points $A_{1}, B_{1}, C_{1}, D_{1}$ inside the faces $BCD, CDA, DAB, ABC$ respectively, such that $A_{1}B_{1}, B_{1}C_{1}, C_{1}D_{1}, D_{1}A_{1}$ are perpendicular to the faces $BCD, CDA, DAB, ABC$ respectively. Find the volume of the tetrahedron $A_{1}B_{1}C_{1}D_{1}$.
Solve: As shown in Figure $17-13, A_{1} B_{1} \perp C D$, construct a plane perpendicular to $C D$ through $A_{1} B_{1}$, intersecting $C D$ at $M$. Connect $A_{1} M, B_{1} M$. $\angle B_{1} A_{1} M=90^{\circ}$.
It is easy to see that $\angle A_{1} M B_{1}$ is the plane angle of the dihedral angle $B-C D-A$.
Since the cosine value of the dihedral angle $B-C D-A$ is $\frac{\sqrt{3}}{6} \div \frac{\sqrt{3}}{2}=\frac{1}{3}$,
we have $\cos \angle A_{1} M B_{1}=\frac{1}{3}$, thus $\frac{A_{1} M}{M B_{1}}=\frac{1}{3}$.
Similarly, construct a plane perpendicular to $A D$ through $B_{1} C_{1}$, intersecting $A D$ at $N$, and connect $B_{1} N, C_{1} N$.
Then $\frac{B_{1} N}{N C_{1}}=\frac{1}{3}$.
Similarly, we can find points $K, L$ on $A B, B C$ and their respective ratios.
When the surface of the tetrahedron $A B C D$ is unfolded, $\triangle A B C, \triangle B C D, \triangle C D A, \triangle D A B$ are sequentially laid out on a plane, as shown in Figure $17-14$.
In the figure, $A D \parallel B C$, and the distance between them is $\frac{\sqrt{3}}{2} a$.
Let $A_{1} M=x, B_{1} N=y, C_{1} K=z \cdot D_{1} L=u$.
Then $3 u+y+4 x \cos 60^{\circ}=\frac{\sqrt{3}}{2} a$,
which simplifies to $3 u+2 x+y=\frac{\sqrt{3}}{2} a$.
Similarly, $3 x+2 y+z=\frac{\sqrt{3}}{2} a, 3 y+2 z+u=\frac{\sqrt{3}}{2} a, 3 z+2 u+x=\frac{\sqrt{3}}{2} a$.
Therefore, $x=y=z=u=\frac{\sqrt{3}}{12} a$.
Thus, $A_{1} B_{1}=B_{1} C_{1}=C_{1} D_{1}=D_{1} A_{1}=\sqrt{\left(B_{1} M\right)^{2}-\left(A_{1} M\right)^{2}}$
$$
=\sqrt{(3 x)^{2}-x^{2}}=\frac{\sqrt{6}}{6} a
$$

Since $A_{1} B_{1}, B_{1} C_{1}, C_{1} D_{1}, D_{1} A_{1}$ are perpendicular to the faces $B C D, C D A, D A B, A B C$ respectively, $\angle B_{1} A_{1} D_{1}$ and $\angle A_{1} B_{1} C_{1}$ are complementary to the plane angles of the dihedral angles $A-B C-D$ and $B-C D-A$. Both are equal to $\arccos \frac{1}{3}$.

As shown in Figure 17-15, construct $A_{1} H \perp B_{1} D_{1}$ at $H$. Then $H$ is the midpoint of $B_{1} D_{1}$, and connect $C_{1} H$. Construct $H G \perp A_{1} C_{1}$ at $G$, then $G$ is the midpoint of $A_{1} C_{1}$.
Therefore, $B_{1} D_{1} \perp$ plane $A_{1} H C_{1}$, thus $V_{A_{1} B C_{1} D_{1}}=\frac{1}{3} \cdot S_{\triangle A H} \cdot B_{1} D_{1}$
In the isosceles $\triangle B_{1} A_{1} D_{1}$,
$$
\begin{aligned}
\sin \angle A_{1} B_{1} D_{1} & =\sin \frac{\pi-\angle \frac{B_{1}}{A_{1} D_{1}}}{2}=\cos \frac{\angle B_{1} A_{1} D_{1}}{2} \\
& =\sqrt{\frac{1+\frac{1}{3}}{2}}=\frac{\sqrt{6}}{3}
\end{aligned}
$$

So, $A_{1} H=A_{1} B_{1} \cdot \sin \angle A_{1} B_{1} D_{1}=\frac{\sqrt{6}}{6} a \cdot \frac{\sqrt{6}}{3}=\frac{a}{3}$
$$
B_{1} D_{1}=2 A_{1} B_{1} \cdot \cos \angle A_{1} B_{1} D_{1}=2 \cdot \frac{\sqrt{6}}{6} a \cdot \frac{\sqrt{3}}{3}=\frac{\sqrt{2} a}{3}
$$

By symmetry, $A_{1} C_{1}=B_{1} D_{1}=\frac{\sqrt{2} a}{3}$
$$
\begin{array}{c}
H G=\sqrt{\left(A_{1} H\right)^{2}-\left(A_{1} G\right)^{2}}=\sqrt{\left(\frac{\sqrt{6} a}{6}\right)^{2}-\left(\frac{a}{3}\right)^{2}}=\frac{\sqrt{2} a}{6} \\
S_{\triangle A, H C}=\frac{1}{2} \cdot A_{1} C_{1} \cdot H G=\frac{1}{2} \cdot \frac{\sqrt{2}}{3} a \cdot \frac{\sqrt{2}}{6} a=\frac{a^{2}}{18} \\
V_{A, B, C, D}=\frac{1}{3} \cdot S_{\triangle A, H C} \cdot B_{1} D_{1}=\frac{1}{3} \cdot \frac{a^{2}}{18} \cdot \frac{\sqrt{2} a}{3}=\frac{\sqrt{2} a^{3}}{162}
\end{array}
$$