1. If real numbers $a, b, c, d$ satisfy $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}$, then the value of $\frac{a b+b c+c d+d a}{a^{2}+b^{2}+c^{2}+d^{2}}$ is ( ).
(A) 1 or 0
(B) -1 or 0
(C) 1 or -2
(D) 1 or -1
-.1.D.
Let $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=k$, we can get $k= \pm 1$.
When $k=-1$, $a=-b=c=-d$, the original expression $=-1$;
When $k=1$, the original expression $=1$.