3. On the side $P C$ of $\triangle A B C$, there is a point $D$, $\angle A D B$ is an acute angle, $P 、 Q$ are the circumcenters of $\triangle A B D 、 \triangle A C D$ respectively, and the quadrilateral $A P D Q$ has the same area as $\triangle A B C$. Then $\angle A D B=(\quad)$.
(A) $60^{\circ}$
(B) $75^{\circ}$
(C) $30^{\circ}$
(D) $45^{\circ}$
3. (D).

As shown in Figure 4, let $\angle A D B^{\circ} = a$, then $\angle A P B = 2 \alpha$. By the properties of the circumcenter, it is easy to see that $\angle A P Q = \angle A B C$, $\angle A Q P = \angle A C B \Rightarrow \triangle A P Q \backsim \triangle A B C$. Clearly, $\triangle A P Q \cong \triangle D P Q$. Therefore,
$$
\begin{array}{l}
\frac{1}{2}=\frac{S_{\triangle A P Q}}{S_{\triangle P D Q}}=\frac{S_{\triangle A P Q}}{S_{\triangle A B C}}=\left(\frac{A P}{A B}\right)^{2} \\
\Rightarrow A B=\sqrt{2} \cdot A P \Rightarrow 2 \alpha=\angle A P B=90^{\circ} .
\end{array}
$$

Thus, $\angle A D B = \alpha = 45^{\circ}$.