3. Given that $D$ is a point inside the acute-angled $\triangle A B C$, satisfying $A D=D C, \angle A D C$ $=2 \angle D B C, A B=$ $12, B C=10$. As shown in Figure 2. Then the area of $\triangle B D C$ is $\qquad$
3. $5 \sqrt{11}$.

Let $\angle D B C=\theta$. Then $\angle A D C=2 \theta$.
With $D$ as the center of rotation, rotate $\triangle B D C$ to the position of $\triangle A D P$, as shown in Figure 4.
Then $A D=D C, D P=D B, A P=C B=10$, $\angle D P A=\angle D B C=\theta, \angle P D A=\angle B D C$, i.e., $\angle P D C+\angle C D A=\angle B D P+\angle P D C$.
Thus, $\angle B D P=\angle A D C=2 \theta$.
Draw $D K \perp B P$ at point $K$. Then
$$
\begin{array}{l}
\angle B P D=\frac{1}{2}\left(180^{\circ}-2 \theta\right)=90^{\circ}-\theta \\
\Rightarrow \angle A P B=\angle A P D+\angle D P B=90^{\circ} .
\end{array}
$$

In the right triangle $\triangle A B P$,
$$
B P=\sqrt{A B^{2}-A P^{2}}=2 \sqrt{11} \text {. }
$$

Draw $D H \perp A P$ at point $H$. Then the quadrilateral $P H D K$ is a rectangle, so
$$
D H=K P=\frac{1}{2} B P=\sqrt{11} .
$$

Therefore, $S_{\triangle B D C}=S_{\triangle P D A}=\frac{1}{2} \times 10 \times \sqrt{11}=5 \sqrt{11}$.