8. Given a tetrahedron $S-ABC$ with the base being an equilateral triangle, the projection of point $A$ on the face $SBC$ is $H$, which is the orthocenter of $\triangle SBC$. $SA=a$, find the maximum volume of this tetrahedron.
8. Solution From the given, we know that $A H \perp$ plane $S B C$, and let $B H \perp S C$ at $E$.

By the theorem of three perpendiculars, we know that $S C \perp A E, S C \perp A B$, hence $S C \perp$ plane $A B E$.
Construct $S O \perp$ plane $A B C$ at $O$, then $O C$ is the projection of $S C$ in plane $A B C$,
thus $C O \perp A B$, with the foot of the perpendicular being $F$.
Similarly, $B O \perp A C$, so point $O$ is the orthocenter of $\triangle A B C$.
Since $\triangle A B C$ is an equilateral triangle, point $O$ is also its centroid.
Therefore, $S A=S B=S C=a$,
As shown in the figure, let $A B=x$, then $S_{\triangle A B C}=\frac{\sqrt{3}}{4} x^{2}, O C=\frac{2}{3} F C=\frac{\sqrt{3}}{3} x, S O=\sqrt{a^{2}-\frac{1}{3} x^{2}}$.
Thus, $V_{S-A B C}=\frac{1}{3} S_{\triangle A B C} \cdot S O=\frac{1}{12} \sqrt{x^{4}\left(3 a^{2}-x^{2}\right)}=\frac{1}{6} \sqrt{\frac{x^{2}}{2} \cdot \frac{x^{2}}{2} \cdot\left(3 a^{2}-x^{2}\right)}$
$$
\leqslant \frac{1}{6} \sqrt{\left(\frac{\frac{1}{2} x^{2}+\frac{1}{2} x^{2}+3 a^{2}-x^{2}}{3}\right)^{3}}=\frac{1}{6} a^{3} .
$$

Equality holds if and only if $x=\sqrt{2} a$.
Therefore, the maximum volume of this tetrahedron is $\frac{1}{6} a^{3}$.