Constructing a quadrilateral in hyperbolic plane Suppose we have a set of angles, $\{\alpha_1, \alpha_2, \alpha_3, \alpha_4 \}$, and we want to construct some quadrilateral at Poincare Disc Model with angle $\alpha_i$ at vertex $i$.
The question is, how to do it?
I know how solve this problem for triangles. We can place one vertex, say $A$, at the origin (center of Poincare Disc), then two sides of the triangle will be Euclidian segments (not circle arcs). Construct two arbitrary rays originating from $A$ such that the angle between them is $\alpha_1$. Since in hyperbolic plane triangle's angles define triangle's edge lengths (from the law of cosines), we now can place points $B$ and $C$ on this rays knowing their distance from $A$. I'm wondering if this technique can be extended to quadrilaterals somehow?
UPDATE
I want to 'construct' it in a sence that I want to obtain (explicitly, through some formula, or implicitly, trough some algorithm) all vertices' Euclidian coordinates or anything that can be translated into them (for example, hyperbolic distance and angles). 
 A: Well, it seems that this problem has no nice analytic solution, so I've come up with a numerical one.
So, given a set of angles, $\{\alpha, \beta,\gamma,\delta\}$, we want to cosntruct an quadrilateral $ABCD$ with appropriate angles. Construct the diagonal $AD$ and denote the angles as in the picture:

If we obtain $\xi$ and $\eta$, then we will be able to construct the ABCD since we know how to construct triangles (using a method I described in the original post). 
Consider the following function:
$$
F(x,y,z) = \dfrac{\cos(x)\cos(y) - \cos(z)}{\sin(x)\sin(y)}
$$ 
From the hyperbolic law of cosines, it tells hyperbolic cosine of side overlooking angle $z$ in triangle with angles $\{x,y,z\}$. Then for some pair ${(\xi, \eta)}$ to be valid it is neccesarry that
$$
\Phi(\xi, \eta) = F(\xi, \eta, \beta) - F(\alpha - \xi, \gamma -\eta,\delta) =0. 
$$
It can be straighforwardly checked that
$$
\dfrac{\partial\Phi(\xi, \eta)}{\partial \xi} < 0,\ \dfrac{\partial\Phi(\xi, \eta)}{\partial \eta} < 0\ \text{for all}\ (\xi, \eta)\in(0,\alpha)\times(0,\gamma).
$$
Also, it is obvious that
$$
\lim\limits_{\eta\to 0}\Phi(\xi, \eta) = +\infty,\ \lim\limits_{\eta\to\gamma }\Phi(\xi, \eta) = -\infty\text{ for all }\xi\in(0,\alpha),
$$
and
$$
\lim\limits_{\xi\to 0}\Phi(\xi, \eta) = +\infty,\ \lim\limits_{\xi\to\alpha }\Phi(\xi, \eta) = -\infty\text{ for all }\eta\in(0,\gamma),
$$
Hence, for all $\xi\in(0,\alpha)$ the equation $\Phi(\xi, \eta)=0$ has a unique solution $\eta^*$, and for all $\eta\in(0,\gamma)$ it has a unique solution $\xi^*$. Moreover, from monotonyty, these equations can be easily numerically solved using, say, bisection method.
So, the algorithm can now be formulated as follows.


*

*Select a grid of points $\{\xi_0, \xi_1,\ldots, \xi_N\}$ in segment $[\varepsilon, \alpha - \varepsilon]$. 

*For a given $\xi_i$, find the root $\eta_i$ of equation $\Phi(\xi_i, \eta) = 0$. Repeat untill $\eta_i\in(0,\gamma)$.

*Construct a triangle $ABC$ with angles $(\xi, \eta,\beta)$.

*Construct a triangle $AD'F$ with angles $(\alpha-\xi, \gamma-\eta, \delta)$ such thath $AF$ is colinear with $AC$.

*Rotate triangle $ADF$ around $A$ by $\xi$ to triangle $ADF'$. Than points $A,B,C,D$ will form the desired quadrilateral.


This method can obviously extended to an arbitrary polygon.
