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

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

Here, $p,q,r,s$ are the sides of the quadrilateral, $T$ is half the perimeter, and $A,C$ 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

$S=\text{area}(\triangle ADB)+\text{area}(\triangle BDC)=\tfrac12ps\sin(A)+\tfrac12qr\sin(C)$

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

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

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

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

Introducing the semiperimeter

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

the above becomes

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

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.