1. In tetrahedron $ABCD$, $AB \perp CD$, $AB=a$, $AB$ forms an angle $\alpha$ with the plane $BCD$, and the dihedral angle $A-CD-B$ is $\beta$. Then the minimum value of the perimeter of the triangular section of the tetrahedron passing through edge $AB$ is $\qquad$
1. $\frac{a}{\sin \beta}[\sin \alpha+\sin \beta+\sin (\alpha+\beta)]$.

Draw a section $A B E$ through $A B$ such that $C D \perp$ plane $A B E$, then $C D \perp A E, C D \perp B E$.
Therefore, the perimeter of $\triangle A B E$ is minimized.
At this point, $\angle A B E$ is the angle between $A B$ and plane $B C D$, so $\angle A B E=\alpha ; \angle A E B$ is the plane angle of the dihedral angle $A-C D-B$, so $\angle A E B=\beta$.
By the Law of Sines, we have $\frac{A B}{\sin \beta}=\frac{A E}{\sin \alpha}=\frac{B E}{\sin (\alpha+\beta)}$.
By the properties of proportions, we have
$$
\begin{array}{l}
\frac{A B+A E+B E}{\sin \beta+\sin \alpha+\sin (\alpha+\beta)}=\frac{A B}{\sin \beta}=\frac{a}{\sin \beta} . \\
\text { Therefore } A B+A E+B E \\
\quad=\frac{a}{\sin \beta}[\sin \alpha+\sin \beta+\sin (\alpha+\beta)] .
\end{array}
$$