Bretschneider's formula

A quadrilateral

In geometry, Bretschneider's formula is the following expression for the area of a quadrilateral,

$\text{area} = \sqrt {(T-p)(T-q)(T-r)(T-s) - pqrs \cos^2 \frac{A+C}{2}}.$

Here, p, q, r and s are the sides of the quadrilateral, T is half the perimeter, and A and C are two opposite angles.

Bretschneider's formula works on any quadrilateral regardless of whether it is cyclic or not.

[edit] Proof of Bretschneider's formula

Denote the area of the quadrilateral by S. Then we have

$\begin{align} S &= \text{area of } \triangle ADB + \text{area of } \triangle BDC \\ &= \tfrac{1}{2}pq\sin A + \tfrac{1}{2}rs\sin C \end{align}$

Therefore

$4S^2 = (pq)^2\sin^2 A + (rs)^2\sin^2 C + 2pqrs\sin A\sin C. \,$

The cosine law implies that

$p^2 + q^2 -2pq\cos A = r^2 + s^2 -2rs\cos C, \,$

because both sides equal the square of the length of the diagonal BD. This can be rewritten as

$\tfrac14 (r^2 + s^2 - p^2 - q^2)^2 = (pq)^2\cos^2 A +(rs)^2\cos^2 C -2 pqrs\cos A\cos C. \,$

Substituting this in the above formula for $4 S 2$ yields

$4S^2 + \tfrac14 (r^2 + s^2 - p^2 - q^2)^2 = (pq)^2 + (rs)^2 - 2pqrs\cos (A+C). \,$

This can be written as

$16S^2 = (r+s+p-q)(r+s+q-p)(r+p+q-s)(s+p+q-r) - 16pqrs \cos^2 \frac{A+C}2.$

Introducing the semiperimeter

$T = \frac{p+q+r+s}{2},$

the above becomes

$16S^2 = 16(T-p)(T-q)(T-r)(T-s) - 16pqrs \cos^2 \frac{A+C}2$

and Bretschneider's formula follows.

Bretschneider's formula is a more generalized case of Brahmagupta's formula for the area of a quadrilateral, which is in turn a more generalized case of Heron's Formula for the area of a triangle.