3. Given that $A, B, C, D$ are points on a circle with radius 1, $\overrightarrow{A B}+2 \overrightarrow{A C}=\overrightarrow{A D}$, and $|A C|=1$. Then the area of quadrilateral $A B D C$ is $\qquad$.
3. $\frac{3 \sqrt{3}}{4}$.
$$
\begin{array}{l}
\text { Given } \overrightarrow{A B}+2 \overrightarrow{A C}=\overrightarrow{A D} \Rightarrow 2 \overrightarrow{A C}=\overrightarrow{B D} \\
\Rightarrow|B D|=2 .
\end{array}
$$

Then $B D$ is the diameter of the circumcircle $\odot O$ of quadrilateral $A B D C$, and quadrilateral $A B D C$ is an isosceles trapezoid.
Thus, $S_{\text {quadrilateral } A B D C}=3 S_{\text {equilateral } \triangle A O C}$
$$
=\frac{\sqrt{3}}{4} \times 3=\frac{3 \sqrt{3}}{4} \text {. }
$$