5. In quadrilateral $A B C D$, it is known that $A B=B C=1$, $\angle A B C=100^{\circ}, \angle C D A=130^{\circ}$. Then $B D=$ ( ).
(A) $\frac{\sqrt{2}}{2}$
(B) 1
(C) $\sqrt{2}$
(D) $\sqrt{3}$
5. B.

As shown in Figure 2, extend $A B$ to point $E$ such that $B E = B C$.
Connect $C E$. Then
$$
\begin{array}{l}
\angle A E C = \angle B C E \\
= \frac{1}{2} \angle A B C = 50^{\circ}.
\end{array}
$$

Thus, $\angle A E C + \angle A D C = 180^{\circ}$. Therefore, points $A, E, C, D$ are concyclic, and the center of the circle is $B$. Hence, $B D = B A = 1$.