Consider the following mapping:
In a square grid on a unit disk we shift the angles between intersecting segments, every shift in general different at different $(x,y)$ points. Take the continuum limit. The image of the gridlines (i.e. the blue lines below) will be curves with the same lengths, but will be contained in a different domain $D$ (shaded blue below right). Moreover, each family (i.e. verticals and horizontals in the disk), is composed of congruent curves (meaning that they differ just by a translation and, or, a rotation; equivalently, that they have the same curvature). The question is
Show that Perimeter$\left(D\right)<2\pi$.
Analytically, it can be re-formulated as follows. Let's call $0<\gamma(x,y)<\pi$ the shift angle at every point. Since the image curves are congruent within each family, we have that $\gamma(x,y)=\psi(x)-\phi(y)$, being $\psi$ and $\phi$ arbitrary, not even functions (satisfying the condition on $\gamma$ for every $(x,y)\in\:$Disk). Then
\begin{align} \text{Perimeter}\left(D\right)=\int_0^{2\pi}\mathbb{d}t\sqrt{1-2\sin t \cos t \cos\gamma(t)}, \end{align}
with $\gamma(t)=\gamma(x=\cos t,y=\sin t)=\psi(\cos t)-\phi(\sin t)$. The integrand in general varies between $0$ and $\sqrt{2}$, so it is not immediate to establish that $\text{Perimeter}\left(D\right)<2\pi$. Expanding the square root one can show that the quadratic terms satisfy some inequalities, but the linear term has no definite sign for any pair $\psi,\:\phi$. So I don't think that handling this integral expression may lead to an answer, but rather some geometrical proof/analysis using the fact the curves are congruent.
This is not an answer, only a related comment. We are dealing with different geometries.
Fig 1: Sine-Gordon equation of Chebychev net $ \alpha = 2 \psi $ is
$$ \alpha^{''}= \sin \alpha/a^2~; \psi^{''}= \sin \psi \cos \psi/a^2 ~$$
where $a$ is cuspidal radius of a pseudosphere in hyperbolic geometry.
Fig 2: When $a\to \infty $ we try to get at the flat euclidean net.
$$ \alpha^{''}=0, \quad \psi^{''}=0. ~ $$
Polar plots of the net are computed and plotted below.
( Boundary conditions taken in numerical work $\psi_i=0.12, \text{radius }r_i=0.2) $
Although similar in appearance the hyperbolic Chebyshev and euclidean nets are not the same.
The integrands are quite different.
In Fig 1 situation
$ \psi^{'}= \dfrac{\sin \psi}{r} $ is true for a geodesic in hyperbolic geometry. When integrated $\dfrac{\sin \psi}{r}$ is a constant, represented by circles through origin, so that $ \psi^{'}$ is constant.
In Fig 2 situation
$ \psi^{'}\ne \dfrac{\sin \psi}{r} $ and $\dfrac{\sin \psi}{r}$ is not a constant.