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 http://en.wikipedia.org/wiki/Bretschneider%27s_formula
(where $ s = (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?
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 the angles actually are.
I dont think it matters either that the maximum area quadrilateral is concyclic although Ted indicates that it can be proved that a quadrilateral with opposite angles summing to $\pi$ must be concyclic.
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. :)