Trigonometry/Area of a quadrilateral

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+c-d = 2(σ-d) and similarly for the other sides. The diagonals of ABCD are AC and BD. Let the angle between them be θ. Then S = 1 2AC.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 BD2 by applying the cosine theorem to either of the triangles BAD, BCD. This means that a2+d2-2ad.cos(A) = b2+c2-2bc.cos(C) so a2+d2-b2-c2 = 2ad.cos(A)-2bc.cos(C) ... (i) Also S = area(BAD) + area(BCD) = 1 2ad.sin(A) + 1 2bc.sin(C) so 4S = 2ad.sin(A) + 2bc.sin(C) ... (ii) Taking (ii)2 + (i)2, 16S2 + (a2+d2-b2-c2)2 = 4a2d2 + 4b2c2 - 8abcd.cos(A+C) But cos(A+C) = cos(2α) = 2cos2(α)-1 so 16S2 = 4(ad+bc)2 - (a2+d2-b2-c2)2 - 16abcd.cos2(α). Simplifying, S2 = (σ-a)(σ-b)(σ-c)(σ-d) - abcd.cos2(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 quadrilateral 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 results 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, a-d=b-c, i.e. a2+d2-b2-c2 = 2(ad-bc). Thus . If a quadrilateral ABCD is both cyclic and circumcyclic, then its area is √(abcd) and inradius is 2√(abcd)/(a+b+c+d).