What is the Maximum Area of a Quadrilateral with sides of length a,b,c,d (in sequence). I am given the lengths of 4 sides of a convex quadrilateral as $a$, $b$, $c$, $d$.
I am also given the sequence of the sides (i.e. it is $abcd$, not $abdc$, or $acbd$, or $acdb$,... etc).
I know that Bretschneider's_formula
$$
K=\sqrt{(s-a)(s-b)(s-c)(s-d)-a b c d \cdot \cos ^{2}\left(\frac{\alpha+\gamma}{2}\right)}
$$
(where $s=\frac{a+b+c+d}2$ is the semi-perimeter)
can give the area if the sum of either pair of opposite angles are known.
But I do not know the angles of any of the vertices (interior or exterior).
I can generate two quadrilateral equations by generating diagonal lines $e$, $f$ and using the cosine rule of the triangle...
$$  a^2 + b^2 - 2.a.b.\cos(\alpha)  (= e^2) =  c^2 + d^2 - 2.c.d.\cos(\gamma) $$
$$  b^2 + c^2 - 2.b.c.\cos(\beta)  (= f^2) =  a^2 + d^2 - 2.a.d.\cos(\delta) $$
These two equations have four unknowns (i.e. the interior angles: $\alpha$, $\beta$, $\gamma$, $\delta$).  $\delta$ is given by:
$$ \delta = 360 - \alpha-\beta-\gamma $$
so this leave three unknowns but only two equations.
Where do I go from here?
ANSWER
I have accepted Ted Shifrin's answer, if opposite angles sum to $\pi$ then we can use a simplified form of Bretschneider's formula to get the (maximum) area K.
$$ K^2 = [(s-a)(s-b)(s-c)(s-d)] $$
For finding the maximum area we dont need to know what any of the four individual vertex angles actually are.
NOTES
Ted indicates that it can be proved that a quadrilateral with opposite angles summing to $\pi$ must be concyclic and vice-versa.  It is straightforward to prove that, in a concyclic quadrilateral, the opposite angles are supplementary, by drawing diagonals and using the rule that the angle subtended by a chord at the centre is twice the angle subtended at the perimeter.
Related question:- geometric-argument-as-to-why-the-cyclic-quadrilateral-has-the-maximal-area
This paper (PDF) by Thomas Peter (2003):- Maximising the area of a quadrilateral gives a proof that:- "For any quadrilateral with given edge lengths, there is a cyclic quadrilateral with the same edge lengths".  It goes on to prove that:- "The cyclic quadrilateral Q has the largest area of all quadrilaterals with sides of the same length as those of Q".
Also I deduce that any angle of a cyclic quadrilateral can be obtained from:-
$$ cos(\alpha) = (a^2 + b^2 - c^2 - d^2)/(2(a.b +c.d)) $$
where $\alpha$ is the interior angle between sides a and b.  Which is nice. :)
 A: You will find that the optimal solution is for the opposite angles to add to $\pi$ and, thus, for the quadrilateral to be concyclic (inscribed in a circle).
A: Notice that on the right hand side of Bretschneider's formula, everything is constant except $\alpha$ and $\gamma$. So the area is maximized if and only if $\cos^2 (\frac{\alpha+\gamma}{2})$ is minimized. But $\alpha$ and $\gamma$ are not independent. From the figure one can see that $\alpha$ and $\gamma$ are constrained by applying the law of cosines to two of the triangles in the picture:
$$(*) \qquad b^2 + c^2 - 2bc \cos(\gamma) = a^2 + d^2 - 2ad \cos(\alpha)
$$
So now you have a constrained optimization problem, and you can try to solve it using Lagrange's method: find the minimum of $\cos^2 (\frac{\alpha+\gamma}{2})$ subject to the constraint $(*)$ (and restricted to $\alpha,\gamma \in [0,\pi]$). 
You might be able to simplify the problem even further using some symmetry. For example: if you restrict to $\alpha,\gamma \in [0,\pi/2]$, so that $\cos(\alpha+\gamma) \ge 0$, then $\cos^2(\frac{\alpha+\gamma}{2})$ is minimized if and only if $\cos(\frac{\alpha+\gamma}{2})$ is also minimized. 
