In the pentagram $A B C D E$, the intersection points of each line segment are shown in Figure 3, $A G=G F=F D, B G=$ $3 E G, F C=2 F E$. If the area of pentagon $F G H I J$ is 91, try to find the area of the pentagram $A B C D E$.
Solution: As shown in Figure 3, connect $F H, F I, G J, H J, A E,$ and $B C$.
From $\frac{S_{\triangle A E C}}{S_{\triangle A C F}} \cdot \frac{S_{\triangle C F A}}{S_{\triangle C A G}}$.
$\frac{S_{\triangle C A G}}{S_{\triangle C A E}}=1$, we get
$$
\frac{E C}{C F} \cdot \frac{F A}{A G} \cdot \frac{G H}{H E}=1,
$$

which means $\frac{3}{2} \cdot \frac{2}{1} \cdot \frac{G H}{H E}=1$.
Thus, $\frac{G H}{H E}=\frac{1}{3}$.
Also, $B G=3 E G$, so
$$
E G: G H: H B=2: 1: 5.
$$

Similarly, $E F: F J: J C=5: 3: 7$.
Notice that
$\frac{S_{\triangle A E F}}{S_{\triangle A F C}} \cdot \frac{S_{\triangle F C A}}{S_{\triangle F A H}} \cdot \frac{S_{\triangle F A H}}{S_{\triangle F A E}}=1$,
$\frac{S_{\triangle B C J}}{S_{\triangle B E E}} \cdot \frac{S_{\triangle \triangle E B}}{S_{\triangle B H}} \cdot \frac{S_{\triangle B B}}{S_{\triangle A B C}}=1$,
so $\frac{E F}{F C} \cdot \frac{C A}{A H} \cdot \frac{H G}{G E}=1, \frac{C J}{J E} \cdot \frac{E B}{B H} \cdot \frac{H I}{I C}=1$,
which means $\frac{1}{2} \cdot \frac{C A}{A H} \cdot \frac{1}{2}=1, \frac{7}{8} \cdot \frac{8}{5} \cdot \frac{H I}{I C}=1$.
Therefore, $A H: H I: I C=4: 5: 7$.
Similarly, $D J: J I: I B=8: 7: 25$.
Let $S_{\triangle A G H}=x$. Noting that "the ratio of the areas of triangles with the same height is equal to the ratio of the corresponding bases," and using the line segment ratios obtained earlier, we have
$$
\begin{array}{c}
S_{\triangle F G H}=x, S_{\triangle E F G}=2 x, S_{\triangle E F H}=3 x, \\
S_{\triangle F H J}=\frac{3}{5} S_{\triangle E F H}=\frac{9}{5} x, \\
S_{\triangle F G J}=\frac{3}{5} S_{\triangle E F G}=\frac{6}{5} x, \\
S_{\triangle D F J}=S_{\triangle F G J}=\frac{6}{5} x, \\
S_{\triangle F I J}=\frac{7}{8} S_{\triangle D F J}=\frac{21}{20} x, \\
S_{\triangle C I J}=\frac{7}{3} S_{\triangle F I J}=\frac{49}{20} x, \\
S_{\triangle H H J}=\frac{5}{7} S_{\triangle C I J}=\frac{7}{4} x, \\
S_{\triangle B H I}=\frac{25}{7} S_{\triangle H I J}=\frac{25}{4} x .
\end{array}
$$

Notice that the area of pentagon $F G H I J$ is the sum of the areas of $\triangle F G H$, $\triangle F H J$, and $\triangle H I J$, so
$$
x+\frac{9}{5} x+\frac{7}{4} x=91 .
$$

Solving for $x$, we get $x=20$.
Therefore, $S_{\text {pentagram } A B C D E}$
$$
\begin{array}{l}
=S_{\triangle A G H}+S_{\triangle E F G}+S_{\triangle D F J}+S_{\triangle C I J}+S_{\triangle B H I}+91 \\
=x+2 x+\frac{6}{5} x+\frac{49}{20} x+\frac{25}{4} x+91=349 .
\end{array}
$$