9. As shown in Figure 3, given that pentagon $A B C D E$ is a regular pentagon, and the area of the pentagram $A C E B D$ (shaded part) is 1. Let $A C$ intersect $B E$ at point $P$, and $B D$ intersect $C E$ at point $Q$. Then the area of quadrilateral $A P Q D$ is $\qquad$
9. $\frac{1}{2}$.

As shown in Figure 6, connect
$Q R$.
It is easy to see that quadrilateral
$A P Q R$ is a rhombus,
$\triangle Q P R \cong \triangle A R P$,
$\triangle S P Q \cong \triangle T R Q$.
Let $S_{\triangle A P R}=a$,
$S_{\triangle P S Q}=b$.
Then $1=S_{\text {figure}}=6a+2b$,
$S_{\text {quadrilateral } A P Q D}=3a+b=\frac{1}{2}$.