1.52 If $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, and $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0$, then the value of the algebraic expression $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}$ is
(A) 1.
(B) 2.
(C) 3.
(D) 4.
(Anhui Province Wuhu City Junior High School Mathematics Competition, 1991)
[Solution] Given $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, then we have $\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^{2}=1$,
which means

and
$$
\begin{array}{cc} 
& \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}+2\left(\frac{x y}{a b}+\frac{y z}{b c}+\frac{z x}{c a}\right)=1, \\
& \frac{x y}{a b}+\frac{y z}{b c}+\frac{z x}{c a}=\frac{x y z}{a b c}\left(\frac{c}{z}+\frac{a}{x}+\frac{b}{y}\right)=0, \\
\therefore \quad & \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1 .
\end{array}
$$

Therefore, the answer is $(A)$.