Take any quadrilateral ABCD. Write AB=a, BC=b, CD=c, DA=a; σ = ^{1}⁄_{2}(a+b+c+d); area of ABCD = S.
Note that a+b+cd = 2(σd) and similarly for the other sides.
The diagonals of ABCD are AC and BD. Let the angle between them be θ. Then S = ^{1}⁄_{2}AC.BD.sin(θ).
If ABCD is convex and the diagonals intersect at P, this is easily proved by considering the four triangles ABP, BCP, CDP, DAP since S is the sum of the areas of these four triangles. IF ABCD is not convex, then one of the vertices, say C, must lie inside the triangle ABD. We then find S as the area of ABD less the area of BCD.
Let angle A+C = 2α. To find S in terms of the sides and alpha;.
We can find BD^{2} by applying the cosine theorem to either of the triangles BAD, BCD. This means that
 a^{2}+d^{2}2ad.cos(A) = b^{2}+c^{2}2bc.cos(C)
so
 a^{2}+d^{2}b^{2}c^{2} = 2ad.cos(A)2bc.cos(C) ... (i)
Also
 S = area(BAD) + area(BCD) = ^{1}⁄_{2}ad.sin(A) + ^{1}⁄_{2}bc.sin(C)
so
 4S = 2ad.sin(A) + 2bc.sin(C) ... (ii)
Taking (ii)^{2} + (i)^{2},
 16S^{2} + (a^{2}+d^{2}b^{2}c^{2})^{2} = 4a^{2}d^{2} + 4b^{2}c^{2}  8abcd.cos(A+C)
But
 cos(A+C) = cos(2α) = 2cos^{2}(α)1
so
 16S^{2} = 4(ad+bc)^{2}  (a^{2}+d^{2}b^{2}c^{2})^{2}  16abcd.cos^{2}(α).
Simplifying,
 S^{2} = (σa)(σb)(σc)(σd)  abcd.cos^{2}(α).
This expression becomes even simpler for a cyclic quadrilateral, because then cos(α) = 0 so the last term disappears.
The diagonals and circumradius of a cyclic quadrilateralEdit
In the expression (i) above, for a cyclic quadrilateral cos(C) = cos(A), so
 .
From the cosine theorem,
 .
Similarly,
 .
Thus AC.BD = ac+bd (as we already knew) and
 .
The circumcircle of ABCD is also the circumcircle of triangle ABD, so
 .
 .
Further resultsEdit
If a quadrilateral is circumcyclic, i.e. such that a circle can be inscribed in it touching all four sides, then a+c=b+d. This is easily proved by noting that the lengths of the two tangents from a point to a circle are equal in length.
The radius of the inscribed circle is called the inradius and equals S/σ.
Theorem: If a quadrilateral ABCD is both cyclic and circumcyclic, then
 .
Proof: Since ABCD is cyclic,
 .
Since a+c=b+d, ad=bc, i.e. a^{2}+d^{2}b^{2}c^{2} = 2(adbc). Thus
 .
If a quadrilateral ABCD is both cyclic and circumcyclic, then its area is √(abcd) and inradius is 2√(abcd)/(a+b+c+d).

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