Example 5 Given
$\frac{1}{4}(b-c)^{2}=(a-b)(c-a)$, and $a \neq 0$.
Then $\frac{b+c}{a}=$ $\qquad$
From the given equation, we have
$$
(b-c)^{2}-4(a-b)(c-a)=0 \text {. }
$$

When $a \neq b$, by the discriminant of a quadratic equation, we know that the quadratic equation in $x$
$$
(a-b) x^{2}+(b-c) x+(c-a)=0
$$

has two equal real roots.
$$
\text { Also, }(a-b)+(b-c)+(c-a)=0 \text {, thus, }
$$
$x=1$ is a root of equation (1).
Therefore, both roots of equation (1) are $x=1$.
So, $1 \times 1=\frac{c-a}{a-b} \Rightarrow \frac{b+c}{a}=2$.
When $a=b$, we have
$$
(b-c)^{2}=0 \Rightarrow b=c \text {. }
$$

In this case, $\frac{b+c}{a}=\frac{b+b}{b}=2$.
In summary, $\frac{b+c}{a}=2$.