4. As shown in the figure, for the regular triangular prism $A B C-A_{1} B_{1} C_{1}$, all edges are of length 1. The section $B C_{1} D$ intersects the edge $A A_{1}$ at point $D$. Let the dihedral angles formed by this section with the base $A B C$ and the three side faces $A B B_{1} A_{1}$, $B C C_{1} B_{1}$, and $C A A_{1} C_{1}$ be $\alpha_{1}$, $\alpha_{2}$, $\alpha_{3}$, and $\alpha_{4}$, respectively. If $\cos \alpha_{1}+\cos \alpha_{4}=\cos \alpha_{3}+\cos \alpha_{4}$, then the area of the section $B C_{1} D$ is . $\qquad$
4. $\frac{3}{8} \sqrt{3}$.

Let $A D=x$, then $S_{\triangle B C_{1} D}=\frac{1}{4} \sqrt{4 x^{2}-4 x+7}$.
Thus, the projection areas of the section $B C_{1} D$ on the base $A B C$ and the three side faces $A B B_{1} A_{1}, B C C_{1} B_{1}, C A A_{1} C_{1}$ are $\frac{\sqrt{3}}{4}, \frac{1}{4}(2-x), \frac{1}{4}(2 x-1), \frac{1}{4}(x+1)$, respectively.
By the projection area theorem, we get
$$
\begin{array}{l}
\cos \alpha_{1}=\frac{\sqrt{3}}{\sqrt{4 x^{2}-4 x+7}}, \cos \alpha_{2}=\frac{2-x}{\sqrt{4 x^{2}-4 x+7}}, \\
\cos \alpha_{3}=\frac{2 x-1}{\sqrt{4 x^{2}-4 x+7}}, \cos \alpha_{4}=\frac{x+1}{\sqrt{4 x^{2}-4 x+7}} .
\end{array}
$$

From $\cos \alpha_{1}+\cos \alpha_{2}=\cos \alpha_{3}+\cos \alpha_{4}$, we get
$$
\sqrt{3}+(2-x)=(2 x-1)+(x+1) \text {, }
$$

Solving for $x$ gives $x=\frac{2}{4}-\frac{1}{3}$.
Thus, $S_{\triangle B C_{1} D}=\frac{1}{4} \sqrt{4 x^{2}-4 x+7}=\frac{3 \sqrt{3}}{8}$.