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
 A: The first question was already answered by hardmath
using the Law of Cosines: the quadrilateral exists iff
$$
(*) \phantom{\infty\infty\infty\infty\infty\infty}
a^2 - 2ab \cos \alpha + b^2 = c^2 - 2cd \cos \beta + d^2,
\phantom{\infty\infty\infty\infty\infty\infty (*)}
$$
because both expressions must equal $e^2$.  This can be used
to answer the remaining question, proving that for given $a,b,c,d$
the area of the quadrilateral is maximized when $\alpha + \beta = \pi$,
and deriving a formula for the area of the quadrilateral
in terms of $a,b,c,d$ and $\alpha,\beta$.
We first show that such a quadrilateral exists as long as there exists
any quadrilateral with sides $a,b,c,d$, that is, provided that
each of $a,b,c,d$ is smaller than the sum of the other three sides.
Indeed $\alpha + \beta = \pi$ iff $\cos \beta = - \cos \alpha$, so
(*) yields
$$
\cos \alpha = \frac{a^2+b^2-c^2-d^2}{2ab+2cd}
$$
and the necessary condition $|\cos \alpha| < 1$ becomes
$$
-(2ab+2cd) < a^2+b^2-c^2-d^2 < 2ab+2cd.
$$
If the first inequality fails then $(c-d)^2 \geq (a+b)^2$, so
$\left|c-d\right| \geq a+b$, and likewise if the second inequality fails then
$\left|a-b\right| \geq c+d$.  In either case we find that $a,b,c,d$ cannot be
the sides of a quadrilateral.
Now let the area of the quadrilateral be $K$.  Using Mann's
pictured decomposition of the quadrliateral into two triangles,
and the formula $\frac12 ab \sin C$ for the area of a triangle, we find
$$
2K = ab \sin \alpha + cd \sin \beta.
$$
Thus we are to maximize $2K = ab \sin \alpha + cd \sin \beta$ subject to
$$
ab \cos \alpha - cd \cos \beta = -\frac12 (a^2+b^2-c^2-d^2) =: Q.
$$
This is easily done using calculus: implicit differentiation gives
$$
ab \sin\alpha = cd \sin\beta \frac{d\beta}{d\alpha},
$$
while 
$$
\frac{d(2K)}{d\alpha} = ab \cos \alpha + cd \cos \beta \frac{d\beta}{d\alpha},
$$
so if $d(2K)/d\alpha = 0$ then
$$
ab \cos\alpha = -cd \cos\beta \frac{d\beta}{d\alpha}.
$$
Dividing our two formulas for $d\beta/d\alpha$ we find
$\tan\alpha = -\tan\beta$, whence $\alpha+\beta = \pi$ as claimed.
Alternatively, we can square the formulas for $Q$ and $2K$ and add
to find
$$
Q^2 + (2K)^2
 = (ab \cos \alpha - cd \cos \beta)^2 + (ab \sin \alpha + cd \sin \beta)^2
$$ $$
 = (ab)^2 (\cos^2\alpha + \sin^2\alpha)
 - abcd (\cos\alpha\cos\beta - \sin\alpha\sin\beta)
 + (cd)^2 (\cos^2\beta + \sin^2\beta)
$$ $$
 = (ab)^2 + (cd)^2 - abcd \cos(\alpha+\beta).
$$
Thus $(2K)^2 = (ab)^2 + (cd)^2 - Q^2 - abcd \cos(\alpha+\beta)$,
which is equivalent with
Bretschneider's
formula for the area of a quadrilateral (and indeed this derivation
is equivalent to the proof recited on that Wikipedia page).
Therefore if $a,b,c,d$ are fixed then $K$ is maximized when
$\cos(\alpha+\beta) = -1$, which is to say $\alpha+\beta = \pi$.
The resulting formula for the area of a quadrilateral inscribed in a circle
is the special case of Bretschneider's formula already obtained by
Brahmagupta:
$$
K = \sqrt{(s-a)(s-b)(s-c)(s-d)},
$$
where $s = \frac12(a+b+c+d)$ is the semiperimeter.  This in turn is
a generalization of Heron's formula, which is the limiting case
as one of the sides tends to zero.
A: 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 $\beta$.
A: (a) e can be generated, by cosine law, from (a, b, α).
Once e is known, the triangle involving (e, c, d) can be solved, using a combination of sine law and cosine law, without the existence of β.
As a further generalization, any one of unknowns from (c, d, β) can be relaxed.
Thus, only 5 of the givens will be sufficient to determine a quadrilateral.
From the given a, b, c, d > 0 only, please refer to 1.
(b) [quad] = (1/2)ab sin α + (1/2)cd sin β
= (1/2)(ab sin α + cd sin (π – α).....The quadrilateral is cyclic.
= (1/2)(ab sin α + cd sin α)
= (1/2) (sin α)(ab + cd)
∴ Max[quad] =  (1/2) (ab + cd) Max [sin α]..... Assuming that a, b, c, d are given constants
= (1/2) (ab + cd)
This occurs when α = 90 degrees. (i.e. when e is the diameter of that circle)
A: It takes $5$ parameters to specify a quadrilateral in the plane.  
The $6$ numbers $a,b,c,d,\alpha,\beta$ satisfy the relation

"$e$ computed from $(a,b,\alpha)$ = $e$ computed from $(c,d,\beta)$". 

(The computation is done using the law of cosines.)  
This reduces the parameter space to $5$ dimensions, therefore any additional conditions are inequalities, such as $a+b+c > d$ and its cyclic permutations, and $a+b > e$ and $c+d > e$, with $e$ computed from the other data.  However, all of those inequalities are implied by the equation relating the $6$ parameters. The equation says that the two triangles (which exist for any positive $a,b,c,d$ and arbitrary $\alpha, \beta$ as long as the equation holds) can be glued along the $e$ side, and the gluing constructs a quadrilateral, which by its very existence must satisfy any additional inequality requirements.
Thus the equation from computing $e$ in two ways is necessary and sufficient.
