Example 5 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 Competition)
Solution: According to the characteristics of the required expression, the given equation can be regarded as a quadratic equation in $a$ or $b+c$. Therefore, there are two methods of solution.

Solution One: Transform the given equation into a quadratic equation in $a$
$$
a^{2}-(b+c) a+\frac{1}{4}(b+c)^{2}=0 .
$$

Its discriminant: $\Delta=(b+c)^{2}-4 \times \frac{1}{4}(b+c)^{2}=0$. Therefore, equation (1) has two equal roots $a=\frac{b+c}{2}$. Thus, $\frac{b+c}{a}=2$.

Solution Two: Transform the given equation into a quadratic equation in $(b+c)$
$$
(b+c)^{2}-4 a(b+c)+4 a^{2}=0 .
$$

Its discriminant $\Delta=16 a^{2}-4 \times 4 a^{2}=0$. Therefore, equation (2) has two equal roots $b+c=2 a$. Thus, $\frac{b+c}{a}=2$.