1. Given the numbers $a, b, c$ satisfy
$$
\frac{c-a}{2(a-b)}=\frac{2(b-c)}{c-a} \text {. }
$$

then the value of $a+b-2c$ is $\qquad$
Solution 1: From the given, we have
$$
\frac{(c-a)+2(a-b)}{2(a-b)}=\frac{2(b-c)+(c-a)}{c-a},
$$

which simplifies to $\frac{a+c-2 b}{2(a-b)}=-\frac{a+c-2 b}{c-a}$.
Thus, $0=(a+c-2 b)\left[\frac{1}{2(a-b)}+\frac{1}{c-a}\right]$
$$
=\frac{(a+c-2 b)^{2}}{2(a-b)(c-a)} \text {. }
$$

This gives $a+c-2 b=0$.
Solution 2: From the given,
$$
\begin{aligned}
0 & =(c-a)^{2}-4(a-b)(b-c) \\
& =[(a-b)+(b-c)]^{2}-4(a-b)(b-c) \\
& =[(a-b)-(b-c)]^{2}=(a+c-2 b)^{2} .
\end{aligned}
$$

This gives $a+c-2 b=0$.
Note: If we directly apply the proportion to the original equation, we have
$$
\begin{aligned}
\frac{c-a}{2(a-b)} & =\frac{2(b-c)}{c-a}=\frac{(c-a)+2(b-c)}{2(a-b)+(c-a)} \\
& =\frac{-a-c+2 b}{a+c-2 b}=-1 .
\end{aligned}
$$

Thus, $\frac{c-a}{2(a-b)}=-1$, which means $c-a=-2(a-b)$.
This gives $a+c-2 b=0$.
Although the conclusion is coincidentally correct, the principle is wrong, as the denominator (*) is 0, which does not satisfy the conditions of the proportion theorem.