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
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:
| \(\text {(1)}: \quad\) | \(\ds \AA^2\) | \(=\) | \(\ds \frac 1 4 \paren {a^2 b^2 \sin^2 \alpha + 2 a b c d \sin \alpha \sin \gamma + c^2 d^2 \sin^2 \gamma}\) |
The diagonal $p$ can be written in 2 ways using the Law of Cosines:
- $p^2 = a^2 + b^2 - 2 a b \cos \alpha$
- $p^2 = c^2 + d^2 - 2 c d \cos \gamma$
| \(\ds a^2 + b^2 - 2 a b \cos \alpha\) | \(=\) | \(\ds c^2 + d^2 - 2 c d \cos \gamma\) | ||||||||||||
| \(\ds a^2 + b^2 - c^2 - d^2\) | \(=\) | \(\ds 2 a b \cos \alpha - 2 c d \cos \gamma\) | adding $2 a b \cos \alpha - c^2 - d^2$ to both sides | |||||||||||
| \(\ds \paren {a^2 + b^2 - c^2 - d^2}^2\) | \(=\) | \(\ds 4 a^2 b^2 \cos^2 \alpha - 8 a b c d \cos \alpha \cos \gamma + 4 c^2 d^2 \cos^2 \gamma\) | squaring both sides | |||||||||||
| \(\ds 0\) | \(=\) | \(\ds \frac 1 4 \paren {a^2 b^2 \cos^2 \alpha - 2 a b c d \cos \alpha \cos \gamma + c^2 d^2 \cos^2 \gamma}\) | algebraic manipulation | |||||||||||
| \(\ds \) | \(\) | \(\, \ds - \, \) | \(\ds \frac 1 {16} \paren {a^2 + b^2 - c^2 - d^2}^2\) |
Now add this equation to $(1)$. Then trigonometric identities can be used, as follows:
| \(\ds \AA^2\) | \(=\) | \(\ds \frac 1 4 \paren {a^2 b^2 + c^2 d^2 - 2 a b c d \map \cos {\alpha + \gamma} } - \frac 1 {16} \paren {a^2 + b^2 - c^2 - d^2}^2\) | Sum of Squares of Sine and Cosine and Cosine of Sum | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 {16} \paren {4 a^2 b^2 + 4 c^2 d^2 - \paren {a^2 + b^2 - c^2 - d^2}^2} - \frac 1 2 a b c d \cdot \map \cos {\alpha + \gamma}\) |
By expanding the square $\paren {a^2 + b^2 - c^2 - d^2}^2$:
| \(\text {(2)}: \quad\) | \(\ds \AA^2\) | \(=\) | \(\ds \frac 1 {16} \paren {-a^4 - b^4 - c^4 - d^4 + 2 a^2 b^2 + 2 a^2 c^2 + 2 a^2 d^2 + 2 b^2 c^2 + 2 b^2 d^2 + 2 c^2 d^2} - \frac 1 2 a b c d \map \cos {\alpha + \gamma}\) |
Adding and subtracting $8 a b c d$ to and from the numerator of the first term of $(2)$:
| \(\ds \AA^2\) | \(=\) | \(\ds \frac 1 {16} \paren {-a^4 - b^4 - c^4 - d^4 + 2 a^2 b^2 + 2 a^2 c^2 + 2 a^2 d^2 + 2 b^2 c^2 + 2 b^2 d^2 + 2 c^2 d^2 + 8 a b c d - 8 a b c d} - \frac 1 2 a b c d \map \cos {\alpha + \gamma}\) |
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:
| \(\ds \AA^2\) | \(=\) | \(\ds \frac 1 {16} \paren {-a + b + c + d} \paren {a - b + c + d} \paren {a + b - c + d} \paren {a + b + c - d}\) | ||||||||||||
| \(\ds \) | \(\) | \(\, \ds - \, \) | \(\ds \frac 1 2 a b c d - \frac 1 2 a b c d \map \cos {\alpha + \gamma}\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} - \frac 1 2 a b c d - \frac 1 2 a b c d \map \cos {\alpha + \gamma}\) | as $s = \dfrac {a + b + c + d} 2$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} - \frac 1 2 a b c d \paren {1 + \map \cos {\alpha + \gamma} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} - a b c d \map {\cos^2} {\dfrac {\alpha + \gamma} 2}\) | Half Angle Formula for Cosine |
Hence the result.
$\blacksquare$
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:
- $\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
Source of Name
This entry was named for Carl Anton Bretschneider.
He published a proof in 1842.
Sources
- Weisstein, Eric W. "Bretschneider's Formula." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/BretschneidersFormula.html