2. Let point $O$ be inside a regular tetrahedron $A B C D$, and $G$ be the center of the regular tetrahedron. The line $O G$ intersects the faces of the tetrahedron at points $A_{1} \cdot B_{1}, C_{1} \cdot D_{1}$. Then the value of $\frac{A_{1} O}{A_{1} G}+\frac{B_{1} O}{B_{1} G}+\frac{C_{1} O}{C_{1} G}+\frac{D_{1} O}{D_{1} G}$ is
A. $\frac{4}{3}$
B. 2
C. $\frac{8}{3}$
D. 4
2. D Let the distance from point $G$ to each face of the tetrahedron be $r$, and the distances from point $O$ to the planes $BCD$, $CDA$, $DAB$, $ABC$ be $h_{1}$,
$$
\begin{array}{c}
h_{2}, h_{3}, h_{4} \text {, then } \frac{V_{O-BCD}}{V_{G-BCD}}=\frac{h_{1}}{r}=\frac{A_{1} O}{A_{1} G}, \frac{V_{O}-CDA}{V_{G}-CDA}=\frac{h_{2}}{r}=\frac{B_{1} O}{B_{1} G}, \\
\frac{V_{O-DAB}}{V_{G}-DAB}=\frac{h_{3}}{r}=\frac{C_{1} O}{C_{1} G}, \frac{V_{O-ABC}}{V_{G-ABC}}=\frac{h_{4}}{r}=\frac{D_{1} O}{D_{1} G} .
\end{array}
$$

Since $V_{O-BCD}+V_{O-CDA}+V_{O-DAB}+V_{O-ABC}=V_{D-ABC}$,
and $V_{G-BCD}=V_{G-CDA}=V_{G-DAB}=V_{G-ABC}=\frac{1}{4} V_{D-ABC}$,
thus $\frac{A_{1} O}{A_{1} G}+\frac{B_{1} O}{B_{1} G}+\frac{C_{1} O}{C_{1} G}+\frac{D_{1} O}{D_{1} G}=4$.