In rhombus $A B C D$, $\angle A=60^{\circ}, E$ is a point on segment $C D$, extend $A E$, and intersect the extension of $B C$ at point $F$, connect $F D$, extend $B E$, and intersect $D F$ at point $G$. If $B G=a$, try to express the area of quadrilateral $B C G D$ in terms of $a$.
Solve As shown in Figure 2, draw $D H \perp B E$ at point $H$, $C K \perp$ $B G$ at point $K$, and take point $M$ on $B G$ such that $D M=D G$.
It is easy to know that $\triangle A B D$ is an equilateral triangle.
From $A D / / B F \Rightarrow \triangle A D E \backsim \triangle F C E$ $\Rightarrow \frac{F C}{C E}=\frac{A D}{D E}=\frac{B D}{D E}$.
Then, from $\triangle F E C \backsim \triangle F A B$
$$
\Rightarrow \frac{F C}{C E}=\frac{F B}{B A}=\frac{F B}{B D} \text {. }
$$

Therefore, $\triangle D B E \backsim \triangle D F B$
$$
\begin{array}{l}
\Rightarrow \angle D F B=\angle D B E \\
\Rightarrow \angle C D G=\angle C B G . \\
\Rightarrow B, C, G, D \text { are concyclic } \\
\Rightarrow \angle D C G=\angle D B .
\end{array}
$$

Combining $B C=B D, D M=D G$, we get
$$
\begin{array}{l} 
\triangle D B M \backsim \triangle D C G \\
\Rightarrow B M=C G \\
\Rightarrow B G=G M+M B=D G+C G \\
\Rightarrow S_{\text {of pentagon BBCCD }}=\frac{1}{2} B G(D H+C K) \\
=\frac{1}{2} B G(C G+D G) \sin 60^{\circ} \\
=\frac{\sqrt{3}}{4} B G^{2}=\frac{\sqrt{3}}{4} a^{2} .
\end{array}
$$