Bretschneider's Formula

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Theorem

Let $ABCD$ be a general quadrilateral.

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

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

where:

$a, b, c, d$ are the lengths of the sides of the quadrilateral
$s = \dfrac {a + b + c + d} 2$ is the 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 $\mathcal A_2$.

From Area of Triangle in Terms of Two Sides and Angle:

$\mathcal A_1 = \dfrac {a b \sin \alpha} 2$ and $\mathcal A_2 = \dfrac {c d \sin \gamma} 2$

From to the second axiom of area, $\mathcal A = \mathcal A_1 + \mathcal A_2$, so:

\((1):\)      \(\displaystyle \mathcal A^2\) \(=\) \(\displaystyle \frac 1 4 \left({a^2 b^2 \sin^2 \alpha + 2 a b c d \sin \alpha \sin \gamma + c^2 d^2 \sin^2 \gamma}\right)\)                    

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:

\(\displaystyle a^2 + b^2 - 2 a b \cos \alpha\) \(=\) \(\displaystyle c^2 + d^2 - 2 c d \cos \gamma\)                    
\(\displaystyle a^2 + b^2 - c^2 - d^2\) \(=\) \(\displaystyle 2 a b \cos \alpha - 2 c d \cos \gamma\)          adding $2 a b \cos \alpha - c^2 - d^2$ to both sides          
\(\displaystyle \left({a^2 + b^2 - c^2 - d^2}\right)^2\) \(=\) \(\displaystyle 4 a^2 b^2 \cos^2 \alpha - 8 a b c d \cos \alpha \cos \gamma + 4 c^2 d^2 \cos^2 \gamma\)          squaring both sides          
\(\displaystyle 0\) \(=\) \(\displaystyle \frac 1 4 \left({a^2 b^2 \cos^2 \alpha - 2 a b c d \cos \alpha \cos \gamma + c^2 d^2 \cos^2 \gamma}\right)\)          algebraic manipulation          
\(\displaystyle \) \(\) \(\displaystyle -\) \(\displaystyle \frac 1 {16} \left({a^2 + b^2 - c^2 - d^2}\right)^2\)                    

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

\(\displaystyle \mathcal A^2\) \(=\) \(\displaystyle \frac 1 4 \left({a^2 b^2 + c^2 d^2 - 2 a b c d \cos \left({\alpha + \gamma}\right)}\right) - \frac 1 {16} \left({a^2 + b^2 - c^2 - d^2}\right)^2\)          Sum of Squares of Sine and Cosine and Cosine of Sum          
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {16} \left({4 a^2 b^2 + 4 c^2 d^2 - \left({a^2 + b^2 - c^2 - d^2}\right)^2}\right) - \frac 1 2 a b c d \cdot \cos \left({\alpha + \gamma}\right)\)                    


By expanding the square $\left({a^2 + b^2 - c^2 - d^2}\right)^2$:

\((2):\)      \(\displaystyle \mathcal A^2\) \(=\) \(\displaystyle \frac 1 {16} \left({-a^4 - b^4 - c^4 - d^4 + 2 a^2 b^2 + 2 a^2 c^2 + 2 a^2 d^2 + 2 b^2 c^2 + 2 b^2 d^2 + 2 c^2 d^2}\right) - \frac 1 2 abcd \cos \left({\alpha + \gamma}\right)\)                    


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

\(\displaystyle \mathcal A^2\) \(=\) \(\displaystyle \frac 1 {16} \left({-a^4 - b^4 - c^4 - d^4 + 2 a^2 b^2 + 2 a^2 c^2 + 2 a^2 d^2 + 2 b^2 c^2 + 2 b^2 d^2 + 2 c^2 d^2 + 8 a b c d - 8 a b c d}\right) - \frac 1 2 a b c d \cos \left({\alpha + \gamma}\right)\)                    

allows the product $\left({-a + b + c + d}\right) \left({a - b + c + d}\right) \left({a + b - c + d}\right) \left({a + b + c - d}\right)$ to be formed:

\(\displaystyle \mathcal A^2\) \(=\) \(\displaystyle \frac 1 {16} \left({-a + b + c + d}\right) \left({a - b + c + d}\right) \left({a + b - c + d}\right) \left({a + b + c - d}\right)\)                    
\(\displaystyle \) \(\) \(\displaystyle -\) \(\displaystyle \frac 1 2 a b c d - \frac 1 2 a b c d \cos \left({\alpha + \gamma}\right)\)                    
\(\displaystyle \) \(=\) \(\displaystyle \left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right) - \frac 1 2 a b c d - \frac 1 2 a b c d \cos \left({\alpha + \gamma}\right)\)          as $s = \dfrac {a + b + c + d} 2$          
\(\displaystyle \) \(=\) \(\displaystyle \left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right) - \frac 1 2 a b c d \left({1 + \cos \left({\alpha + \gamma}\right)}\right)\)                    
\(\displaystyle \) \(=\) \(\displaystyle \left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right) - a b c d \cos^2 \left({\dfrac {\alpha + \gamma} 2} \right)\)          Half Angle Formula for Cosine          

Hence the result.

$\blacksquare$


Also see

In this case, from Opposite Angles of Cyclic Quadrilateral, $\alpha + \gamma = 180^\circ$ and the formula becomes:

$\mathcal A = \sqrt{\left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right)}$
$\mathcal A = \sqrt{s \left({s - a}\right) \left({s - b}\right) \left({s - c}\right)}$


Source of Name

This entry was named for Carl Anton Bretschneider.

He published a proof in 1842.


Sources