Example 8 If $\frac{a^{2}}{b+c-a}+\frac{b^{2}}{c+a-b}+\frac{c^{2}}{a+b-c}=0$, find the value of the algebraic expression $\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}$.

Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
$$
\begin{array}{l}
\text { Let } \left\{\begin{array} { l } 
{ b + c - a = 2 x , } \\
{ c + a - b = 2 y , } \\
{ a + b - c = 2 z . }
\end{array} \text { Then } \left\{\begin{array}{l}
a=y+z, \\
b=z+x, \\
c=x+y .
\end{array}\right.\right. \\
\text { Therefore } \frac{(y+z)^{2}}{2 x}+\frac{(z+x)^{2}}{2 y}+\frac{(x+y)^{2}}{2 z}=0 \\
\Rightarrow \frac{(y+z)^{2}}{x}+\frac{(z+x)^{2}}{y}+\frac{(x+y)^{2}}{z}=0 \\
\Rightarrow\left[\frac{(y+z)^{2}}{x}-x\right]+\left[\frac{(z+x)^{2}}{y}-y\right]+\left[\frac{(x+y)^{2}}{z}-z\right] \\
=-(x+y+z) \\
\Rightarrow(x+y+z)\left(\frac{y+z-x}{x}+\frac{z+x-y}{y}+\frac{x+y-z}{z}+1\right) \\
=0 .
\end{array}
$$
(1) If $\frac{y+z-x}{x}+\frac{z+x-y}{y}+\frac{x+y-z}{z}+1=0$, then
$$
\begin{array}{l}
\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}=2 \\
\Rightarrow \frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}=1 .
\end{array}
$$
(2) If $x+y+z=0$, then
$$
\begin{array}{l}
\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \\
=\frac{y+z}{2 x}+\frac{z+x}{2 y}+\frac{x+y}{2 z}=-\frac{3}{2} .
\end{array}
$$

In summary, the value of the algebraic expression is 1 or $-\frac{3}{2}$.
$$