On ideal quadrilaterals in hyperbolic geometry 
Consider the ideal quadrilateral $Q_x$ comprising geodesics $S_1$ between $0$ and
$\infty$, $S_2$ between $x$ and $\infty$, $T_1$ between $1$ and $x$, and $T_2$ between $1$ and $0$.
We have hyperbolic isometries $A$ and $B$ that send $S_1$ to $T_1$ and $S_2$ to $T_2$ respectively.
Call the domain bounded by the quadrilateral $D$. Then prove that both $A$ and $B$ move $D$ disjoint from itself, that is $A.D\cap D = \emptyset$ and $B.D\cap D=\emptyset$.
I have managed to find out the isometries $A$ and $B$. But I don't know how to proceed with showing that they move $D$ disjoint from itself.
Any help is kindly appreciated. Thank you.
Edit:
$D$ is the domain bounded by the ideal quadrilateral $Q_x$. The sides are not included in $D$. The isometries $A$ and $B$ have to be hyperbolic and such that the commutator $\left[A,B\right]$ is a parabolic isometry fixing $\infty$.
In my attempt, I have found possible isometries $A$ and $B$.
$A=\frac{1}{\sqrt{x-1}}\left(\begin{matrix} x & 1 \\
1 & 1 \end{matrix}\right)$
$B=\left(\begin{matrix} 0 & 1 \\
-1 & x+1 \end{matrix}\right)$
$\left[A,B\right]= A^{-1}B^{-1}AB=\left(\begin{matrix} -1 & 2x+2 \\
0 & -1 \end{matrix}\right)$, which is parabolic and fixes $\infty$.
I cannot proceed further.
 A: Let's start with the isometry $A$, which takes the line $S_1$ to the line $T_1$. 
Let's examine how $A$ behaves with respect to various different types of orientations. 
I observe that your $A$ preserves orientation on the hyperbolic plane because the determinant is positive. I also observe that $A$ takes $0 \to 1$ and $\infty \to x$, and therefore $A$ takes the "upward" orientation on $S_1$ to the "rightward" orientation on $T_1$. Putting these together, and using the right-hand-rule, it follows that $A$ takes the rightward transverse orientation on $S_1$ to the downward transverse orientation on $T_1$, and therefore $A$ takes the rightward half plane of $S_1$ to the downward half plane of $T_1$. Since $D$ is contained in the rightward half plane of $S_1$ it follows that $A \cdot D$ is contained in the downward half plane of $T_1$. But also, $D$ is contained in the upward half plane of $T_1$ and thereore $A \cdot D \cap D = \emptyset$.
A similar orientation analysis of $B$ shows that $B$ takes the leftward half plane of $S_2$ to the downward half plane of $T_2$, and by the same argument we get $B \cdot D \cap D = \emptyset$.
