1. For tetrahedron $ABCD$, all dihedral angles are acute, and opposite edges are pairwise equal. The six dihedral angles of the tetrahedron are $\alpha_{i}(i=1,2, \cdots, 6)$. Then $\sum_{i=1}^{6} \cos \alpha_{i}=(\quad$.
(A) 1
(B) 2
(C) 4
(D) not a constant value
Suddenly, the four triangles of the tetrahedron are denoted as $S$.

Consider the projection of $A$ onto the plane $BCD$ as $O$, and the dihedral angle between the plane $ABC$ and the plane $BCD$ as $\alpha_{1}$, then
$$
S_{\triangle OBC}=S \cdot \cos \alpha_{1} .
$$

Thus, $S \cos \alpha_{1}+S \cos \alpha_{2}+S \cos \alpha_{3}=S$, which implies $\cos \alpha_{1}+\cos \alpha_{2}+\cos \alpha_{3}=1$.
$$
\text { Hence } \sum_{i=1}^{6} \cos \alpha_{i}=1 \text {. }
$$