One, (50 points) As shown in Figure $1, \triangle A B C$ has an area of 1. Points $E, F, G$ are on sides $B C, C A, A B$ respectively, and $A E$ is bisected by $B F$ at point $R$, $B F$ is bisected by $C G$ at point $S$, and $C G$ is bisected by $A E$ at point $T$. Find the area of $\triangle R S T$.
$$
\text { I. Let } \frac{C F}{F A}=x, \frac{A G}{G B}=y, \frac{B E}{E C}=z(x, y, z>0) \text {. }
$$

By $B F$ intersecting $\triangle C A G$, according to Menelaus' theorem, we get
$$
\frac{C F}{F A} \cdot \frac{A B}{B G}=1 \text {. }
$$

Similarly, by $C G$ and $A E$ intersecting $\triangle A B E$ and $\triangle B C F$ respectively, we get
$$
\frac{A G}{G B} \cdot \frac{B C}{C E}=1, \frac{B E}{E C} \cdot \frac{C A}{A F}=1 .
$$

Thus, $x(1+y)=y(1+z)=z(1+x)=1$.
Then $x=\frac{1}{1+y}=\frac{1}{1+\frac{1}{1+z}}=\frac{1}{1+\frac{1}{1+\frac{1}{1+x}}}$
$$
=\frac{x+2}{2 x+3} \text {. }
$$

Rearranging gives $x^{2}+x-1=0$.
Similarly, $y^{2}+y-1=0, z^{2}+z-1=0$.
But these three equations have only one positive root, hence
$$
x=y=z=\frac{\sqrt{5}-1}{2} \Rightarrow \frac{F C}{C A}=\frac{3-\sqrt{5}}{2} \text {. }
$$

By $C G$ intersecting $\triangle A B F$, we get
$$
\begin{array}{l}
\frac{A G}{G B} \cdot \frac{B S}{S F} \cdot \frac{F C}{C A}=1 \\
\Rightarrow \frac{\sqrt{5}-1}{2} \cdot \frac{B S}{S F} \cdot \frac{3-\sqrt{5}}{2}=1 \\
\Rightarrow \frac{B S}{S F}=\sqrt{5}+2 \Rightarrow \frac{B S}{B F}=\frac{2+\sqrt{5}}{3+\sqrt{5}}=\frac{\sqrt{5}+1}{4} .
\end{array}
$$

Given $S_{\triangle A B C}=1$ and $\frac{F C}{C A}=\frac{3-\sqrt{5}}{2}, \frac{B S}{B F}=\frac{1+\sqrt{5}}{4}$, we get
$$
S_{\triangle C B S}=\frac{3-\sqrt{5}}{2} \cdot \frac{\sqrt{5}+1}{4}=\frac{\sqrt{5}-1}{4} \text {. }
$$

Thus, $S_{\triangle R S T}=1-3 \times \frac{\sqrt{5}-1}{4}=\frac{7-3 \sqrt{5}}{4}$.