As shown in Figure 1, given a square $A B C D$ with a side length of 1, the diagonals $A C$ and $B D$ intersect at point $O, E$ is a point on the extension of side $B C$, connect $A E$, which intersects $B D$ and $C D$ at points $P$ and $F$ respectively, connect $B F$ and extend it to intersect segment $D E$ at point $G$, connect $O E$ and intersect $C D$ at point $S$. If $P S / / A C$, find the length of $B G$.

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Given $P S / / A C$, we have
$\triangle D P S \backsim \triangle D O C \Rightarrow \frac{D S}{D C}=\frac{P S}{O C}$.
Similarly, $\frac{E S}{E O}=\frac{P S}{O A}$.
Since quadrilateral $A B C D$ is a square, we know $O A=O C$.
Thus, $\frac{D S}{D C}=\frac{E S}{E O} \Rightarrow D E / / A C$.
Also, $A D / / C E$, so quadrilateral $A C E D$ is a parallelogram.
Therefore, $C E=A D=B C, D F=C F$.
Taking the midpoint $H$ of $E G$ and connecting $C H$, $C H$ is the midline of $\triangle E B G$. Hence,
$$
C H=\frac{1}{2} B G, C H / / B G .
$$

Given $D F=C F, C H / / B G$, we know $F G$ is the midline of $\triangle D C H$. Thus, $F G=\frac{1}{2} C H$.
Therefore, $F G=\frac{1}{4} B G \Rightarrow B F=B G-F G=\frac{3}{4} B G$.
In the right triangle $\triangle B C F$, it is easy to see that
$B F=\sqrt{B C^{2}+C F^{2}}=\sqrt{1^{2}+\left(\frac{1}{2}\right)^{2}}=\frac{\sqrt{5}}{2}$.
Thus, $B G=\frac{4}{3} B F=\frac{2 \sqrt{5}}{3}$.