Take any quadrilateral ABCD. Write AB=a, BC=b, CD=c, DA=a; σ = 12(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 = 12AC.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

so

Also

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.

In the expression (i) above, for a cyclic quadrilateral cos(C) = -cos(A), so

$2(ad+bc)\cos(A)=a^{2}+d^{2}-b^{2}-c^{2}$ .

From the cosine theorem,

$BD^{2}=a^{2}+d^{2}-2ad.\cos(A)=a^{2}+d^{2}-{\frac {ad(a^{2}+d^{2}-b^{2}-c^{2})}{ad+bc}}={\frac {(ab+cd)(ac+bd)}{ad+bc}}$ .

Similarly,

$AC^{2}={\frac {(ab+cd)(ad+bc)}{ab+cd}}$ .

Thus AC.BD = ac+bd (as we already knew) and

${\frac {AC}{BD}}={\frac {ad+bc}{ab+cd)}}$ .

The circumcircle of ABCD is also the circumcircle of triangle ABD, so

$R={\frac {BD}{2\sin(A)}}={\frac {(ad+bc)BD}{2(ad+bc)\sin(A)}}={\frac {(ad+bc)BD}{4S}}={\frac {1}{4S}}{\sqrt {(ab+cd)(ac+bd)(ad+bc)}}$ .
$\tan ^{2}\left({\frac {B}{2}}\right)={\frac {(\sigma -a)(\sigma -b)}{(\sigma -c)(\sigma -d)}}$ .

## 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

$\tan ^{2}\left({\frac {A}{2}}\right)={\frac {bc}{ad}}$ .

Proof: Since ABCD is cyclic,

$\cos(A)={\frac {a^{2}+d^{2}-b^{2}-c^{2}}{2(ad+bc)}}$ .

Since a+c=b+d, a-d=b-c, i.e. a2+d2-b2-c2 = 2(ad-bc). Thus

$\cos(A)={\frac {ad-bc}{ad+bc}}$
$\tan ^{2}\left({\frac {A}{2}}\right)={\frac {1-\cos(A)}{1+\cos(A)}}={\frac {bc}{ad}}$ .

If a quadrilateral ABCD is both cyclic and circumcyclic, then its area is √(abcd) and inradius is 2√(abcd)/(a+b+c+d).