2. As shown in Figure 1, in the square $A_{1} A_{2} A_{3} A_{4}$ with side length 1, points $A_{5}$ and $A_{6}$ are taken on the sides $A_{2} A_{3}$ and $A_{3} A_{4}$ respectively, such that $A_{3} A_{5}=A_{3} A_{6}=p$, and the lines $A_{1} A_{2}$ and $A_{4} A_{5}$, $A_{1} A_{4}$ and $A_{2} A_{6}$, $A_{1} A_{5}$ and $A_{4} A_{3}$, $A_{1} A_{6}$ and $A_{2} A_{3}$ intersect at points $B_{1}$, $B_{2}$, $B_{3}$, $B_{4}$ which are collinear. Then the value of $p$ is $(\quad)$.
(A) $\frac{\sqrt{5}+2}{3}$
(B) $\frac{\sqrt{5}-2}{2}$
(C) $\frac{3+\sqrt{5}}{3}$
(D) $\frac{3-\sqrt{5}}{2}$
2. D

Let $A_{3} B_{3}=A_{3} B_{4}=q, A_{4} B_{2}=r$.
Since $\triangle B_{3} A_{3} A_{5} \sim \triangle B_{3} A_{4} A_{1}$, we have $\frac{q}{p}=\frac{q+1}{1}$.
Thus, $q=\frac{p}{1-p}$.
Also, since $\triangle B_{2} A_{4} A_{6} \sim \triangle B_{2} A_{1} A_{2}$, we have $\frac{r}{1-p}=\frac{r+1}{1}$.
Thus, $r=\frac{1-p}{p}$.
It is easy to see that $\triangle B_{3} A_{3} B_{4} \sim \triangle B_{3} A_{4} B_{2}$, so,
$$
\frac{q}{r}=\frac{q}{1+q},
$$

which means $r=1+q$.
Therefore, $\frac{1-p}{p}=1+\frac{p}{1-p}$.
Simplifying and rearranging, we get $p^{2}-3 p+1=0$. Since $0<p<1$, we have $p=\frac{3-\sqrt{5}}{2}$.