Example 2 The two lateral edges $SB$ and $SC$ of the tetrahedron $S-ABC$ are equal, and the projection of vertex $S$ on the base is the orthocenter $H$ of $\triangle ABC$. The base edge $BC=a$, and the dihedral angle between the lateral face $SBC$ and the base is $\alpha$. Find the volume of this tetrahedron.
Solution: Connect $S H, B H, A H, C H$, and extend $A H$ to intersect $B C$ at $M$, connect $S M$, as shown in Figure 16-2.

From $S B=S C$, we get $H B=H C$, and since $H$ is the orthocenter, then $A M \perp B C$.
Thus, $M B=M C=\frac{1}{2} B C=\frac{a}{2}$, which implies $A B=A C$.

Since $H$ is the orthocenter, we have $M B^{2}=A M \cdot H M$, so $\left(\frac{a}{2}\right)^{2}=A M \cdot H M$.

From $S B=S C, B M=C M$, we get $S M \perp B C$, and since $A M \perp B C$, then $\angle S M A=\alpha$. Therefore, from $\angle S H M=90^{\circ}$, we get $S H=\operatorname{tg} \alpha \cdot H M$.
Thus, $S_{\triangle A B C}=\frac{1}{2} A M \cdot B C=\frac{a}{2} A M$. Hence,
$$
\begin{aligned}
V & =\frac{1}{3} S H \cdot S_{\triangle A B C}=\frac{1}{3} \operatorname{tg} \alpha \cdot H M \cdot \frac{1}{2} A M \cdot a \\
& =\frac{a}{6} \operatorname{tg} \alpha \cdot H M \cdot A M=\frac{a}{6} \operatorname{tg} \alpha \cdot \frac{a^{2}}{4}=\frac{a^{3}}{24} \operatorname{tg} \alpha .
\end{aligned}
$$