# Bretschneider's formula

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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.

## Proof of Bretschneider's formula

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

$S = \text{area of } \triangle ADB + \text{area of } \triangle BDC = \tfrac{1}{2}ps\sin A + \tfrac{1}{2}qr\sin C$

Therefore

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

The cosine law implies that

$p^2 + s^2 -2ps\cos A = q^2 + r^2 -2qr\cos C, \,$

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

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

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

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

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.

## Related formulas

Bretschneider's formula generalizes Brahmagupta's formula for the area of a cyclic quadrilateral, which in turn generalizes Heron's formula for the area of a triangle.