Example 7 (2003 Belarusian Mathematical Olympiad) Given a convex pentagon $A B C D E$ satisfying $A B=B C, C D=$ $D E, \angle A B C=150^{\circ}, \angle C D E=30^{\circ}, B D=2$. Find the area of pentagon $A B C D E$.
Let $K$ be the point symmetric to $C$ with respect to the line $B D$, then
$$
\begin{array}{l}
B K=B C=B A \\
D K=D C=D E
\end{array}
$$

Construct the angle bisectors of $\angle A B K$ and $\angle E D K$, and let them intersect at point $M$. Thus, $B M$ is the perpendicular bisector of $A K$, and $D M$ is the perpendicular bisector of $E K$. In particular, $M A=M K=M E$, which means $M$ is the circumcenter of $\triangle A K E$.

Since $\angle M B D=\angle M B K+\angle K B D$
$$
\begin{aligned}
& =\frac{1}{2} \angle A B K+\frac{1}{2} \angle B K C=\frac{1}{2} \angle A B C \\
\angle M D B & =\frac{1}{2} \angle C D E
\end{aligned}
$$

Therefore, $\angle M B D+\angle M D B=\frac{1}{2}(\angle A B C+\angle C D E)$
$$
=\frac{1}{2} \times 180^{\circ}=90^{\circ}
$$

Thus, $\angle B M D=90^{\circ}$, which means $B M \perp D M$.
Since $A K \perp B M, D M \perp K E$,
it follows that $A K \perp K E$.
Hence, $A E$ is the hypotenuse of the right triangle $\triangle A K E$, which means $M$ is the midpoint of $A E$.
Since $S_{\triangle A B M}=S_{\triangle K B M}, S_{\triangle E D M}=S_{\triangle K D M}, S_{\triangle B C D}=S_{\triangle B K D}$,
it follows that $S_{\text {pentagon } A B C D E}=2 S_{\triangle B M D}=B M \cdot M D$
$$
\begin{array}{l}
=B D \cos \angle M B D \cdot B D \cdot \sin \angle M B D \\
=\frac{1}{2} B D^{2} \cdot \sin 2 \angle M B D \\
=\frac{1}{2} B D^{2} \cdot \sin \angle A B C \\
=\frac{1}{2} \times 2^{2} \times \frac{1}{2}=1
\end{array}
$$