Bretschneider's Formula

Theorem
Let $ABCD$ be a general quadrilateral.

Then the area $\mathcal{A}$ of $ABCD$ is given by:


 * $\mathcal{A}=\sqrt{(s-a) (s-b) (s-c) (s-d) - abcd \cdot \cos^2 \left( \dfrac {\alpha + \gamma} {2} \right)}$

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

Proof

 * Bretschneider's Formula.png

Let the area of $\triangle DAB$ and $\triangle BCD$ be $\mathcal{A}_1$ and $\mathcal{A}_2$.

Then, using the corollary of the half base times height formula:
 * $\mathcal{A}_1 = \dfrac {ab \sin \alpha} {2}$ and $\mathcal{A}_2 = \dfrac {cd \sin \gamma} {2}$

From to the second axiom of area, $\mathcal{A} = \mathcal{A}_1 + \mathcal{A}_2$, so:

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

Equality is transitive, so:

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

The term $8abcd$ was added and substracted from the numerator of the first term of the equation, as then the product $(-a+b+c+d) \cdot (a-b+c+d) \cdot (a+b-c+d) \cdot (a+b+c-d)$ can be formed.

Thus:

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, $\alpha + \gamma = 180^\circ$ and the formula becomes:
 * $\mathcal{A} = \sqrt{(s-a) (s-b) (s-c) (s-d)}$


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

He published a proof in 1842.