*This is a follow-up question to a question from math.stackexchange: https://math.stackexchange.com/q/1436253/67563
Had it not been for the exchange there between myself and @Lee_Mosher in the comments I would not have thought the question that follows research-level, but apparently the details I'm concerned with have not been worked out in the literature.*

Consider the Fuchsian group $\Gamma:=\Big\langle\begin{pmatrix}1&1\\1&2\end{pmatrix}, \begin{pmatrix}1&-1\\-1&2\end{pmatrix}\Big\rangle$. In the literature it is stated (with varying justifications) that $T:=\mathbb{H}^2/\Gamma$ is a punctured torus and that the Dirichlet domain $D$ for $T$ centered at $i\in\mathbb{H^2}$ is the region colored in yellow in the figure below. I am interested in verifying this explicitly via the following method of constructing $D$.

A Dirichlet domain can be formed by taking the images under each $\gamma\in\Gamma$ of its center $c$, then looking at the geodesic $g_{\gamma}$ from $c$ to $\gamma(c)$, then taking the perpendicular bisector $b_{\gamma}$ of $g_{\gamma}$, then taking the intersection of all the half-spaces that are on the same side of $b_{\gamma}$ as $c$ (varying over all $\gamma\in\Gamma$).

Let $c=i$ and let $\gamma=\begin{pmatrix}1&1\\1&2\end{pmatrix}$. $\gamma(i)=\dfrac{i+1}{i+2}=\frac{3}{5}+\frac{1}{5}i$. The geodesic $g_{\gamma}$ from $i$ to $\gamma(i)$ is a segment of the circle centered at $-\frac{1}{2}$ of radius $\frac{5}{4}$, shown in blue in the figure. This can be verified by calculating that the Euclidean distance between $-\frac{1}{2}$ and $i$, as well as the Euclidean distance between $-\frac{1}{2}$ and $\frac{3}{5}+\frac{1}{5}i$, is $\frac{5}{4}$.

The problem arises when we consider $b_{\gamma}$. The distance between $-\frac{1}{2}$ and $p:=\frac{1}{2}+\frac{1}{2}i$ is also $\frac{5}{4}$, therefore $g_{\gamma}$ intersects the edge of the Dirichlet domain that connects $0$ and $1$ at $p$, as shown in the figure. But that edge is not $b_{\gamma}$ because the geodesic perpendicular to it at $p$ is the vertical line rising out of $\frac{1}{2}$ (the dotted line in the figure). Even worse, the hyperbolic distance between $i$ and $\frac{1}{2}+\frac{1}{2}i$ is equal to the hyperbolic distance between $\frac{1}2+\frac{1}2i$ and $\frac{3}{5}+\frac{1}{5}i$, as follows. \begin{align} d_{\mathbb{H}^2}(x_1+iy_1,x_2+iy_2)&=\mathrm{arcosh}\Big(1+\frac{(x_2-x_1)^2+(y_2-y_1)^2}{2y_1y_2}\Big)\\ \Longrightarrow d_{\mathbb{H}^2}(i,\tfrac{1}{2}+\tfrac{1}{2}i)&= \mathrm{arcosh}\Big(1+\frac{(\tfrac{1}{2}-0)^2+(\tfrac{1}{2}-1)^2}{2\cdot 1\cdot\tfrac{1}{2}}\Big)=\mathrm{arcosh}(\tfrac{3}{2}),\\ \text{and }d_{\mathbb{H}^2}(\tfrac{1}{2}+\tfrac{1}{2}i,\tfrac{3}{5}+\tfrac{1}{5}i)&= \mathrm{arcosh}\Big(1+\frac{(\tfrac{3}{5}-\tfrac{1}{2})^2+(\tfrac{1}{5}-\tfrac{1}{2})^2}{2\cdot\tfrac{1}{2}\cdot\tfrac{1}{5}}\Big)=\mathrm{arcosh}(\tfrac{3}{2}). \end{align} And so $b_{\gamma}$ must be a segment of a half-circle which traverses the Dirichlet domain like the red curve in the figure, but this is not possible.

Where is the mistake?

Low-Dimensional Geometry, but looking for the exact statement now I'm not finding it. He discusses this fundamental domain in sections 5.5 and 8.2. Exercise 5.13 is to show $D$ has the symmetry group of the square, from which I thought this followed. I could have sworn I saw some online references stating this too but I'm not finding those either! I would be very happy to learn that the assumption is false and that my computations are correct. $\endgroup$ – j0equ1nn Sep 24 '15 at 22:403more comments