3. As shown in Figure 1, quadrilateral DEFG is inscribed in $\triangle A B C$, and it is known that the areas of $\triangle A D E$, $\triangle E F C$, and $\triangle D B G$ are $1$, $3$, and $1$, respectively. Therefore, the area of $\square D E F G$ is ( ).
(A) $2 \sqrt{3}$
(B) 2
(C) 3
(D) 4
3. D.

As shown in Figure 4, draw $AO \parallel DG$ intersecting $BC$ at point $O$, then
$$
\begin{array}{l}
\frac{S_{\triangle DBC}}{S_{\triangle OOB}}=\frac{S_{\triangle EFC}}{S_{\triangle AOC}} \\
=\frac{S_{\triangle EFC}+S_{\triangle DOCC}}{S_{\triangle BBC}} .
\end{array}
$$

Therefore, $S_{\triangle \triangle OB}=\frac{1}{4} S_{\triangle \triangle BC}$,
$S_{\triangle \triangle OC}=\frac{3}{4} S_{\triangle BB C}$.
$$
\begin{array}{c}
\text { Also, } \frac{\sqrt{S_{\triangle ADE}}}{\sqrt{S_{\triangle BBC}}}=\frac{DE}{BC}, \\
\frac{\sqrt{S_{\triangle EFC}}+\sqrt{S_{\triangle DBC}}}{\sqrt{S_{\triangle AOC}}+\sqrt{S_{\triangle AOB}}}=\frac{CF+BG}{BC},
\end{array}
$$

Thus, $\frac{\sqrt{S_{\triangle ADE}}}{\sqrt{S_{\triangle BBC}}}+\frac{\sqrt{S_{\triangle SFC}}+\sqrt{S_{\triangle DBC}}}{\sqrt{\frac{3}{4} S_{\triangle ABC}}+\sqrt{\frac{1}{4} S_{\triangle BBC}}}=1$.
Therefore, $S_{\triangle \triangle BC}=9$.
Hence, $S_{\square DEFC}=9-(1+3+1)=4$.