Example 9 Let points $E$, $F$, $G$ be the midpoints of the edges $AB$, $BC$, $CD$ of the regular tetrahedron $ABCD$. Then the size of the dihedral angle $C-FG-E$ is ( ).
(A) $\arcsin \frac{\sqrt{6}}{3}$
(B) $\frac{\pi}{2}+\arccos \frac{\sqrt{3}}{3}$
(C) $\frac{\pi}{2}-\arctan \sqrt{2}$
(1) $\pi-\operatorname{arccot} \frac{\sqrt{2}}{2}$
Analysis: It is easy to know that $AC$ is parallel to plane $EFG$ and $AC \perp FG$ (the intersection line of plane $EFG$ and plane $BCD$). Therefore, by Basic Conclusion 12, the required dihedral angle is equal to the supplementary angle of the angle formed by $AC$ and plane $BCD$. Draw $AO \perp$ plane $BCD$, with the foot of the perpendicular being $O$, then $O$ is the center of $\triangle BCD$, and we get $\angle ACO$ $=\operatorname{arccot} \frac{\sqrt{2}}{2}$, so the required dihedral angle is $\pi-\operatorname{arccot} \frac{\sqrt{2}}{2}$. Therefore, the correct choice is (D).

Note: Since the required dihedral angle is greater than $\frac{\pi}{2}$, after obtaining $\cos \angle ACO$ $=\frac{\sqrt{3}}{3}$, we can directly determine that the correct choice is (D).