Example 1 (Fill in the blank Question 1) Given $\frac{1}{4}(b-c)^{2}=$ $(a-b)(c-a)$ and $a \neq 0$. Then $\frac{b+c}{a}=$ $\qquad$ .
Solution 1: From the given, we have
$$
\begin{aligned}
0 & =\frac{1}{4}[-(a-b)-(c-a)]^{2}-(a-b)(c-a) \\
& =\frac{1}{4}\left[(a-b)^{2}+(c-a)^{2}-2(a-b)(c-a)\right] \\
& =\frac{1}{4}[(a-b)-(c-a)]^{2} \\
& =\frac{1}{4}(2 a-b-c)^{2} .
\end{aligned}
$$

Thus, $\frac{b+c}{a}=2$.
Solution 2: The given condition indicates that the quadratic equation with roots $\frac{a-b}{2}, \frac{c-a}{2}$ is
$$
\begin{array}{l}
\left(x-\frac{a-b}{2}\right)\left(x-\frac{c-a}{2}\right)=0 \\
\Leftrightarrow x^{2}+\frac{1}{2}(b-c) x+\frac{1}{4}(a-b)(c-a)=0 .
\end{array}
$$

We have $\Delta=\frac{1}{4}(b-c)^{2}-(a-b)(c-a)=0$.
Therefore, the two roots are equal, i.e., $\frac{a-b}{2}=\frac{c-a}{2}$.

Thus, $\frac{b+c}{a}=2$.
Analysis: Let $x=a-b, y=c-a$. The background knowledge of this problem is: in the identity
$$
\frac{1}{4}(x+y)^{2}-x y=\frac{1}{4}(x-y)^{2}
$$

if the left side is 0, then the right side is also 0.