Example 3.4.1 (All-Soviet MO 17) The four shaded triangles in the figure have equal areas. If the area of one of these triangles is $1 \mathrm{~cm}^{2}$, what is the area of the unshaded quadrilateral?
Connect $A C_{1}$ and $D B_{1}$, as shown in the figure.
Let the areas of the four shaded small triangles be $S_{1}=S_{\triangle A D A_{1}}$,
$S_{2}=S_{\triangle B E B_{1}}, S_{3}=S_{\triangle C F C_{1}}, S_{0}=$ the middle shaded small triangle.
From $S_{1}=S_{0}$, we get $S_{\triangle A_{1} D}=S_{\triangle A_{1} B_{1}}, \therefore B_{1} D \parallel A C_{1}$,
Therefore, $\triangle A_{1} B_{1} D \sim \triangle A_{1} A C_{1}$, and
$$
\triangle A B C_{1} \sim \triangle D B B_{1} \cdot \frac{A A_{1}}{A_{1} B_{1}}=\frac{A C_{1}}{B_{1} D}=\frac{B C_{1}}{B B_{1}}=1+\frac{B_{1} C_{1}}{B B_{1}} .
$$

Let $\frac{A A_{1}}{A_{1} B_{1}}=a, \frac{B B_{1}}{B_{1} C_{1}}=b, \frac{C C_{1}}{C_{1} A_{1}}=c$,
Then we have $a=1+\frac{1}{b}, b=1+\frac{1}{c}, c=1+\frac{1}{a}$. Eliminating $b$ and $c$, we get $a^{2}-a-1=0$. Similarly, $b^{2}-b-1=0, c^{2}-c-1=0$.
Thus, $a=b=c=\frac{\sqrt{5}+1}{2}$. From $\triangle A_{1} B_{1} D \sim \triangle A_{1} A C_{1}$, we get
$$
a=\frac{A A_{1}}{A_{1} B_{1}}=\frac{C_{1} A_{1}}{A_{1} D}=\frac{\sqrt{5}+1}{2}, A_{1} D=\frac{2 A_{1} C_{1}}{\sqrt{5}+1}=\frac{\sqrt{5}-1}{2} \cdot A_{1} C_{1} .
$$

Therefore, $\frac{C_{1} D}{C C_{1}}=\frac{A_{1} D+A_{1} C_{1}}{C C_{1}}=\left(\frac{\sqrt{5}-1}{2}+1\right) \cdot \frac{A_{1} C_{1}}{C C_{1}}=\frac{\sqrt{5}+1}{2} \cdot \frac{1}{c}=1$.
That is, $C_{1} D=C C_{1}$.
$B C_{1}$ is the median of $\triangle B C D$, $S_{\triangle B C C_{1}}=S_{\triangle B C_{1} D}$, thus the two unshaded quadrilaterals are equal in area. Similarly, we can prove that the other two quadrilaterals are also equal in area, i.e., the three unshaded quadrilaterals are all equal in area.

From $\frac{B D}{D A}=\frac{B B_{1}}{B_{1} C_{1}}=\frac{\sqrt{5}+1}{2}$, we get $S_{\triangle B D C}=\frac{\sqrt{5}+1}{2} \cdot S_{\triangle A D C}$.
Let the area of one quadrilateral be $x$, then we have $(x+2)(\sqrt{5}+1) / 2=2 x+2$, solving for $x$ gives $x=\sqrt{5}+1$, which is the area of the unshaded quadrilateral.