Geometry question regarding existence of a quadrilateral Let $a,b,c,d>0$ be edges, in that order, of a given quadrilateral with two opposing angles $\alpha > 0$ and $\beta < \pi$. 

What conditions on do we need on $a,b,c,d,\alpha, \beta$ for such a quadrilateral to exist?
Added question: How can I show that such a quadrilateral has maximal area if it can inscribed in a cirle? I know a quadrilateral can be inscribed in a circle if  $\alpha + \beta = \pi$. So, this is my constrain. How can I express the area of the quadrilateral in terms of $\alpha$ and $\beta $ ?
thanks

Let's assume that $0 \lt \alpha,\beta \lt \pi$ was intended.$^*$ 
As the diagram nicely illustrates, what is required is for a common length $e$ to exist as a diagonal "opposite" angles $\alpha$ and $\beta$, which we may express in terms of the law of cosines:
$$ a^2 + b^2 - 2ab\cos \alpha = e^2 = c^2 + d^2 - 2cd\cos \beta $$
* We can certainly allow for one of the angles, say $\alpha \lt 2\pi$ to be reflexive, greater than $\pi$, and so obtain a nonconvex quadrilateral (and similar compatibility conditions), but I see no need to allow a negative $\beta$.
