8. On the edge $AS$ of the tetrahedron $S-ABC$, mark points $M, N$ such that $AM=MN=NS$. If the areas of $\triangle ABC$, $\triangle MBC$, and $\triangle SBC$ are $1$, $2$, and $\sqrt{37}$, respectively, find the area of $\triangle NBC$.
8. Let the areas of $\triangle A B C$, $\triangle M B C$, $\triangle N B C$, $\triangle S B C$ be $S_{1}$, $S_{2}$, $S_{3}$, $S_{4}$, and let $h_{1}$, $h_{2}$, $h_{3}$, $h_{4}$ be the heights from these triangles to the common base $B C$, as shown in Figure 2.

Points $A^{\prime}$, $B^{\prime}$, $C^{\prime}$, $S^{\prime}$, $M^{\prime}$, $N^{\prime}$ represent the orthogonal projections of points $A$, $B$, $C$, $S$, $M$, $N$ onto a plane perpendicular to the edge $B C$, as shown in Figure 3. Points $B^{\prime}$ and $C^{\prime}$ coincide, and
$$
\begin{array}{l}
A^{\prime} M^{\prime}=M^{\prime} N^{\prime}=N^{\prime} S^{\prime}=a, A^{\prime} B^{\prime}=h_{1}, \\
M^{\prime} B^{\prime}=h_{2}, N^{\prime} B^{\prime}=h_{3}, S^{\prime} B^{\prime}=h_{4}.
\end{array}
$$

Considering that $M^{\prime} B^{\prime}$ and $N^{\prime} B^{\prime}$ are the medians of $\triangle A^{\prime} B^{\prime} N^{\prime}$ and $\triangle M^{\prime} B^{\prime} S^{\prime}$, respectively, we have
$$
\begin{array}{l}
h_{1}^{2}+h_{3}^{2}-2 h_{2}^{2}=2 a^{2}=h_{2}^{2}+h_{4}^{2}-2 h_{3}^{2} \\
\Leftrightarrow 3 h_{3}^{2}=3 h_{2}^{2}+h_{4}^{2}-h_{1}^{2} \\
\Rightarrow h_{3}=\sqrt{h_{2}^{2}+\frac{h_{4}^{2}-h_{1}^{2}}{3}} \\
\Rightarrow S_{3}=\sqrt{S_{2}^{2}+\frac{S_{4}^{2}-S_{1}^{2}}{3}}=4.
\end{array}
$$