Bretschneider's formula

Suppose we have a quadrilateral with edges of length $a,b,c,d$ (in that order) and diagonals of length $p, q$ . Bretschneider's formula states that the area $[ABCD]=\frac{1}{4}*\sqrt{4p^2q^2-(b^2+d^2-a^2-c^2)^2}$ .

It can be derived with vector geometry.

The Proof

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}$ (Error compiling LaTeX. Unknown error_msg)

Therefore

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

The law of cosines 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 $4S^2$ 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.

NOTE TO ALL: this proof was taken from Wikipedia on December the 1st, 2006.

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Bretschneider's formula

The Proof

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