$23 \cdot 10$ Let $E$, $F$, and $G$ be the midpoints of the edges $AB$, $BC$, and $CD$ of a regular tetrahedron $ABCD$, respectively. Then the dihedral angle $C-FG-E$ is 
(A) $\arcsin \frac{\sqrt{6}}{3}$.
(B) $\frac{\pi}{2}+\arccos \frac{\sqrt{3}}{3}$.
(C) $\frac{\pi}{2}-\operatorname{arctg} \sqrt{2}$.
(D) $\pi-\operatorname{arcctg} \frac{\sqrt{2}}{2}$.
［Solution］As shown in the figure, draw $E E_{1} \perp$ plane $B C D$, then $E E_{1} \perp$ $B G$.
$\because A B C D$ is a regular tetrahedron,
$\therefore A C \perp B D$
Also, $\because E F / / A C, F G / / B D$,
$\therefore \quad E F \perp F G$.
Hence $E_{1} F \perp F G$.
Therefore, $\angle E F E_{1}$ is the plane angle of the dihedral angle $E-F G-B$, denoted as $\alpha$.
Assume the edge length of the regular tetrahedron $A B C D$ is 1, then its height
$$
A H=\sqrt{A B^{2}-B H^{2}}=\sqrt{1^{2}-\left(\frac{\sqrt{3}}{3}\right)^{2}}=\frac{\sqrt{6}}{3} \text {. }
$$
$\therefore \quad E E_{1}=\frac{1}{2} A H=\frac{\sqrt{6}}{6}$.
Also,
$E F=\frac{1}{2} A C=\frac{1}{2}$,
$\therefore \quad \alpha=\arcsin \frac{\frac{\sqrt{6}}{6}}{\frac{1}{2}}=\arcsin \frac{\sqrt{6}}{3}$.
Thus, the size of the dihedral angle $C-F G-E$ is
$$
\pi-\arcsin \frac{\sqrt{6}}{3}=\pi-\operatorname{arcctg} \frac{\sqrt{2}}{2} \text {. }
$$

Therefore, the answer is $(D)$.