5. Given real numbers $a, b, c, d$ satisfy
$$
\frac{a c-b^{2}}{a-2 b+c}=\frac{b d-c^{2}}{b-2 c+d} \text {. }
$$

Then $\frac{(a-b)(c-d)}{(b-c)^{2}}=(\quad)$.
(A) 1
(B) $\pm 1$
(C) 0
(D) Cannot be determined
5.A.

Let $\frac{a c-b^{2}}{a-2 b+c}=\frac{b d-c^{2}}{b-2 c+d}=k$. Then
$$
\begin{array}{l}
a c-k(a+c)=b^{2}-2 b k, \\
b d-k(b+d)=c^{2}-2 c k . \\
\text { Hence }(a-k)(c-k)=(b-k)^{2}, \\
(b-k)(d-k)=(c-k)^{2} .
\end{array}
$$

Thus, $(a-k)(d-k)=(b-k)(c-k)$.
Simplifying, we get $k=\frac{a d-b c}{a-b-c+d}$.
Also, $k=\frac{a c-b^{2}+b d-c^{2}}{a-b-c+d}$, so,
$$
a d-b c=a c-b^{2}+b d-c^{2},
$$

i.e., $\square$
$$
\begin{array}{l}
(a-b)(c-d)=(b-c)^{2} . \\
\text { Hence } \frac{(a-b)(c-d)}{(b-c)^{2}}=1 .
\end{array}
$$