Example 3 Given $\frac{1}{4}(b-c)^{2}=(a-b)(c-a)$, and $a \neq 0$. Then $\frac{b+c}{a}=$ $\qquad$ .
(1999, National Junior High School Mathematics League) This problem has a resemblance to a 1979 college entrance exam question.
Example $\mathbf{3}^{\prime}$ If $(z-x)^{2}-4(x-y)(y-z)=0$, prove that $x, y, z$ form an arithmetic sequence.

In Example 3, taking $a=b=c$ can guess the answer to be 2. But this poses a risk of "root reduction," because the condition is a quadratic expression, while the conclusion
$$
\frac{b+c}{a}=2 \Leftrightarrow b+c=2 a
$$

is a linear expression in the letters. Transitioning from quadratic to linear, could it result in two solutions? There are many safer methods (see [4] P60), one of which is a method of constructing equations: from the given condition, we transform to get
$$
(b-c)^{2}-4(a-b)(c-a)=0 \text {. }
$$

This transformation suggests a discriminant of 0; from the discriminant, we construct the equation
$$
(a-b) x^{2}+(b-c) x+(c-a)=0,
$$
which has equal roots $x_{1}=x_{2}$.
Since the sum of the coefficients of equation (10) is 0, we get
$$
x_{1}=x_{2}=1 \text {. }
$$

Substituting into the relationship between roots and coefficients, we get
$$
1=x_{1} x_{2}=\frac{c-a}{a-b} \text {, }
$$

which means $a-b=c-a$.
Therefore, $\frac{b+c}{a}=2$.
This idea is creative, but a good idea was wasted due to a closed mindset.
(1) Equation (10) is a quadratic equation, which requires $a-b \neq 0$ to ensure, and it is necessary to supplement the discussion when $a=b$ to make the proposition complete.
(2) The entire process uses the discriminant of a quadratic equation, the method of finding roots by observation, and the relationship between roots and coefficients, making the process somewhat convoluted.

The reason is that subconsciously, it is assumed that the discriminant (2) corresponds to a unique equation (1). Eliminating this mental obstacle, a more sensible equation can be constructed. The idea is: from equation (9), think of the discriminant of a quadratic equation being 0, and from the equation, think of equal roots, thus constructing an equation with roots $x_{1}=a-b, x_{2}=c-a$.

Solution: The given condition indicates that the quadratic equation with roots $a-b, c-a$
$$
\begin{array}{l}
{[x-(a-b)][x-(c-a)]=0} \\
\Leftrightarrow x^{2}+(b-c) x+(a-b)(c-a)=0
\end{array}
$$

has a discriminant of 0, i.e.,
$$
\begin{array}{l}
\Delta=(b-c)^{2}-4(a-b)(c-a) \\
=4\left[\frac{1}{4}(b-c)^{2}-(a-b)(c-a)\right]=0 .
\end{array}
$$

Thus, the two roots of the equation are equal, i.e.,
$$
a-b=c-a \text {. }
$$

Therefore, $\frac{b+c}{a}=2$.
Note: If we let $\alpha=a-b, \beta=c-a$, then
$$
b-c=-(\alpha+\beta) \text {. }
$$

The problem can be transformed into
$$
[-(\alpha+\beta)]^{2}-4 \alpha \beta=0 \Rightarrow \alpha=\beta \text {. }
$$

This can be directly derived from $(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta$. And this is exactly the equation background in equation (6).