Question 4 If real numbers $a, b, c$ satisfy
$$
\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1 \text{, }
$$

find the value of $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}$.
(1999, Changsha Junior High School Mathematics Competition)
Solution: Construct the identity
$$
\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1 \text {. }
$$

Subtract the above identity from the given equation to get
$$
\begin{array}{l}
\left(\frac{a}{b+c}-\frac{a}{a+b+c}\right)+\left(\frac{b}{a+c}-\frac{b}{a+b+c}\right)+ \\
\left(\frac{c}{a+b}-\frac{c}{a+b+c}\right)=0 .
\end{array}
$$

Simplifying, we get
$$
\begin{array}{l}
\frac{a^{2}}{(b+c)(a+b+c)}+\frac{b^{2}}{(a+c)(a+b+c)}+ \\
\frac{c^{2}}{(a+b)(a+c+b)}=0,
\end{array}
$$

which means $\square$
$$
\begin{array}{l}
\frac{1}{a+b+c}\left(\frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{a+b}\right)=0 . \\
\text { Hence } \frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{a+b}=0 .
\end{array}
$$