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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.