3. As shown in Figure 2, quadrilateral $ABCD$ is inscribed in $\odot O$, with $BD$ being the diameter of $\odot O$, and $\overparen{AB}=\overparen{AD}$. If $BC + CD = 4$, then the area of quadrilateral $ABCD$ is $\qquad$ .
3.4 .

From $\overparen{A B}=\overparen{A D}$, we know $A B=A D$.
Also, $B D$ is the diameter of $\odot 0$, so $\angle A=\angle C=90^{\circ}$.
Let $A B=A D=x, B C=y, C D=z$.
By the Pythagorean theorem, we have
$$
x^{2}+x^{2}=y^{2}+z^{2} \text {. }
$$

Thus, $2 x^{2}=y^{2}+z^{2}=(y+z)^{2}-2 y z$.
Also, $S_{\text {quadrilateral } A B C D}=\frac{1}{2} A B \cdot A D+\frac{1}{2} B C \cdot C D$ $=\frac{1}{2} x^{2}+\frac{1}{2} y z$,

From equation (1), we know $S_{\text {quadrilateral } A B C D}=\frac{1}{4}(y+z)^{2}=4$.