Example 16 Given that the area of $\triangle ABC$ is $S_{\triangle}$, and the circumradius is $R$. Draw perpendiculars from $A, B, C$ to the plane $ABC$ and take points $A_{1}, B_{1}, C_{1}$ on the same side of the plane $ABC$ such that $A A_{1}=h_{a}, B B_{1}=h_{b}, C C_{1}=h_{c}$, where $h_{a}, h_{b}, h_{c}$ are the altitudes from vertices $A, B, C$ to the opposite sides $BC, CA, AB$ respectively. Find the volume of the tetrahedron formed by the four planes $A_{1} B_{1} C, B_{1} C_{1} A, C_{1} A_{1} B$, and $ABC$.
The key to solving this problem is to determine the positions of the four vertices of the tetrahedron.
Let the plane $A_{1} B_{1} C$ intersect the line $A B$ at point $K$, then point $K$ lies on $A_{1} B_{1}$. From $A A_{1} / / B B_{1}$, we know $\frac{K A}{K B} = \frac{A A_{1}}{B B_{1}} = \frac{h_{a}}{h_{b}} = \frac{b}{a} = \frac{A C}{B C}$. Therefore, $K$ is the intersection of the external angle bisector of $\angle B C A$ with $B A$, and thus the intersection line of the plane $A_{1} B_{1} C$ with $A B C$ is the external angle bisector of $\angle B C A$.
Similarly, we can deduce that the intersection line of the plane $B_{1} C_{1} A$ with $A B C$ is the external angle bisector of $\angle C A B$.
The intersection of these two external angle bisectors is the excenter $I_{B}$ of $\triangle A B C$, so $I_{B}$ is the common point of the planes $A_{1} B_{1} C$, $B_{1} C_{1} A$, and $A B C$, which is one of the vertices of the tetrahedron.

Thus, the excenters $I_{A}$, $I_{B}$, and $I_{C}$ are three vertices of the tetrahedron. Let the fourth vertex be $P$, then $P$ is the common point of the planes $A_{1} B_{1} C$ and $B_{1} C_{1} A$, and thus lies on the line $B_{1} I_{B}$. The projection of $P$ onto the plane $A B C$ lies on $B I_{B}$ and also on $A I_{A}$, so the projection of $P$ is the incenter $I$ of $\triangle A B C$.

By similar triangles, $P I / / A A_{1}$, and since $A_{1} P$ intersects $A I$ at $I_{A}$, we have $\frac{P I}{A A_{1}} = \frac{P I}{h_{a}} = \frac{I I_{A}}{A I_{A}} = \frac{r_{a} - r}{r_{a}}$, where $r$ is the inradius of $\triangle A B C$ and $r_{a}$ is the exradius.
Let the perimeter of $\triangle A B C$ be $2 l$, then $P E = h_{a} \cdot \frac{r_{a} - r}{r_{a}} = \frac{2 S_{\triangle}}{a} \cdot \frac{\frac{1}{l-a} - \frac{1}{l}}{\frac{1}{l-a}} = \frac{2 S_{\triangle}}{l}$.
By plane geometry, it is easy to prove that the area of $\triangle I_{A} I_{B} I_{C}$ is
$$
\begin{aligned}
S_{\Delta}^{\prime} & = S_{\Delta} + \frac{1}{2}\left(a r_{a} + b r_{b} + c r_{c}\right) = \frac{S_{\triangle}}{2}\left(2 + \frac{a}{l-a} + \frac{b}{l-b} + \frac{c}{l-c}\right) \\
& = \frac{a b c \cdot S_{\Delta}}{2(l-a)(l-b)(l-c)} .
\end{aligned}
$$

Therefore, the volume $V_{P I_{A} I_{B} I_{C}} = \frac{1}{3} S^{\prime}_{\Delta} \cdot P I = \frac{1}{3} \cdot \frac{2 S_{\Delta}}{l} \cdot \frac{a b c \cdot S_{\triangle}}{2(l-a)(l-b)(l-c)}$
$$
= \frac{1}{3} a b c = \frac{4}{3} S_{\triangle} \cdot R .
$$