$2 \cdot 40$ If the quadratic equation $(b-c) x^{2}+(a-b) x+c-a=0$ has two equal real roots, and $b \neq c$, then the relationship between $a, b, c$ is
(A) $b=\frac{a-c}{2}$.
(B) $b=\frac{a+c}{2}$.
(C) $a=\frac{b+c}{2}$.
(D) $c=\frac{a+b}{2}$.
(China Sichuan Province Junior High School Mathematics Competition, 1988)
[Solution] By observation, we know that $x=1$ is a root of the equation, and according to the problem, $x=1$ is a repeated root. By Vieta's formulas, $\quad \frac{c-a}{b-c}=1$,
which means $c-a=b-c$, or $c=\frac{a+b}{2}$. Therefore, the answer is $(D)$.