3. The maximum value of the shortest side of a cyclic quadrilateral inscribed in a circle with radius 1 is ( ).
(A) $\frac{\sqrt{2}}{2}$
(B) 1
(C) $\sqrt{2}$
(D) 2
3. (C).

As shown in the figure, let $AB$ be the shortest side, then $\overparen{AB} \leqslant 90^{\circ}, \angle ADB = \alpha \leqslant 45^{\circ}$.
$$
\begin{array}{c}
\therefore AB = 2R \sin ADB \leqslant \\
2 \sin 45^{\circ} = \sqrt{2},
\end{array}
$$

Thus, the maximum value of $AB$ is $\sqrt{2}$.