Bretschneider's Formula

Theorem
Let $ABCD$ be a general quadrilateral.

Then the area $\AA$ of $ABCD$ is given by:


 * $\AA = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} - a b c d \map {\cos^2} {\dfrac {\alpha + \gamma} 2} }$

where:
 * $a, b, c, d$ are the lengths of the sides of the quadrilateral
 * $s = \dfrac {a + b + c + d} 2$ is the semiperimeter
 * $\alpha$ and $\gamma$ are opposite angles.

Proof

 * Bretschneider's Formula.png

Let the area of $\triangle DAB$ and $\triangle BCD$ be $\AA_1$ and $\AA_2$.

From Area of Triangle in Terms of Two Sides and Angle:
 * $\AA_1 = \dfrac {a b \sin \alpha} 2$ and $\AA_2 = \dfrac {c d \sin \gamma} 2$

From to the second axiom of area, $\AA = \AA_1 + \AA_2$, so:

The diagonal $p$ can be written in 2 ways using the Law of Cosines:
 * $p^2 = a^2 + b^2 - 2 a b \cos \alpha$
 * $p^2 = c^2 + d^2 - 2 c d \cos \gamma$

Equality is transitive, so:

Now add this equation to $(1)$. Then trigonometric identities can be used, as follows:

By expanding the square $\paren {a^2 + b^2 - c^2 - d^2}^2$:

Adding and subtracting $8 a b c d$ to and from the numerator of the first term of $(2)$:

allows the product $\paren {-a + b + c + d} \paren {a - b + c + d} \paren {a + b - c + d} \paren {a + b + c - d}$ to be formed:

Hence the result.

Also see

 * Brahmagupta's Formula is a specific version of Bretschneider's Formula for a cyclic quadrilateral.

In this case, from Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, $\alpha + \gamma = 180^\circ$ and the formula becomes:
 * $\AA = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} }$


 * Heron's Formula is Brahmagupta's Formula for triangles, so $d = 0$ and the formula becomes:
 * $\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$

He published a proof in 1842.