1. Given that $a, b, c, d$ are four distinct real numbers, and $(b+d)(b+a)=1,(c+d)(c+a)=1$. Then the value of $(b+d)(c+d)$ is $(\quad)$.
(A) 1
(B) -1
(C) 2
(D) 0
$$
-、 1 . B
$$

From the conditions, we have
$$
b^{2}+a b+b d+a d=1, c^{2}+a c+c d+a d=1 \text {. }
$$

Subtracting the two equations yields $b^{2}-c^{2}+a(b-c)+d(b-c)=0$.
Thus, $(b-c)(b+c+a+d)=0$.
Since $b \neq c$, i.e., $b-c \neq 0$, then
$$
\begin{array}{l}
b+c+a+d=0, c+d=-(b+a) . \\
\text { Therefore, }(b+d)(c+d)=-(b+d)(b+a)=-1 .
\end{array}
$$