Let A, B, C, D be 4 an ordered 4 tuple of different points on a line, assume cross ratio [A, B, C, D]=[B, A, C, D], then [A, B, C, D, ]=-1.
This seems to be an easy question but I cannot figure out. By definition of cross ratio: (AC/AD)(BD/BC)=(BC/BD)(AD/AC), then $(AC/AD)^2(BD/BC)^2=1$, $(AC/AD)(BD/BC)= 1$, or $-1$, but I don't know why 1 is not possible.
The Wikipedia article Cross ratio states
If $\,A,B,C,\,$ and $\,D\,$ are four points on an oriented affine line, their cross ratio is: $$ (A,B;C,D) = \frac{AC:BC}{AD:BD}, $$ with the notation $\,WX:YZ\,$ defined to mean the signed ratio of the dispacement from $W$ to $X$ to the dispacement from $Y$ to $Z$.
Later, it states
The first set of fixed points is $\{0, 1, \infty\}.$ However, the cross-ratio can never take on these values if the points $A, B, C,$ and $D$ are all distinct. These values are limit values as one pair of coordinates approach each other
The specific reason that the cross ratio can not be equal to $1$ is that
$$ \frac{AC:BC}{AD:BD} = \frac{AC\cdot BD}{BC\cdot AD} = (AC/AD)(BD/BC) = 1 $$
implies $\, AC\,BD = BC\, AD.\,$ Introduce coordinates to get
$$(C-A)(D-B) = (B-C)(A-D).$$
Use the "Special Algebraic Identity" ($\texttt{id4_3_1_2a}\,$ from my collection of such identities)
$$(a-b)(c-d) - (a-c)(b-d) + (a-d)(b-c) = 0$$
to show that $\,(A-B)(C-D) = 0.\,$ This implies that either $\,A-B=0\,$ or else $\,C-D=0.\,$ The first is not possible if $A$ and $B$ are distinct and the second is not possible if $C$ and $D$ are distinct.
