4. Given in $\triangle A B D$, $A B=m, A D=n$, and an equilateral $\triangle B D C$ is constructed with $D B$ as a side (not overlapping with $\triangle A B D$). When the area of quadrilateral $A B C D$ is maximized, $\angle B A D$ equals ( ).
(A) $60^{\circ}$
(B) $90^{\circ}$
(C) $120^{\circ}$
(D) $150^{\circ}$
4.D.
$S_{\triangle A B D}=\frac{1}{2} m n \sin \theta$, where $\theta=\angle B A D$.
In $\triangle A B D$, by the cosine rule we have
$$
\begin{array}{l}
B D^{2}=A B^{2}+A D^{2}-2 A B \cdot A D \cos \theta \\
=m^{2}+n^{2}-2 m n \cos \theta .
\end{array}
$$

Also, $S_{\triangle B D C}=\frac{\sqrt{3}}{4} B D^{2}$, from (1) we get
$$
S_{\triangle B B C}=\frac{\sqrt{3}}{4}\left(m^{2}+n^{2}-2 m n \cos \theta\right) \text {. }
$$

Since $S_{\text {quadrilateral } A B C D}=S_{\triangle A B D}+S_{\triangle B D C}$
$$
\begin{array}{l}
=\frac{\sqrt{3}}{4}\left(m^{2}+n^{2}\right)+m n\left(\frac{1}{2} \sin \theta-\frac{\sqrt{3}}{2} \cos \theta\right) \\
=\frac{\sqrt{3}}{4}\left(m^{2}+n^{2}\right)+m n \sin \left(\theta-60^{\circ}\right),
\end{array}
$$

thus when $\theta=150^{\circ}$, the area of quadrilateral $A B C D$ is maximized.