Bretschneider's Formula

From ProofWiki
Jump to: navigation, search

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 $\mathcal{A}_2$.

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

$\mathcal{A}_1 = \dfrac {ab \sin \alpha} {2}$ 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:

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

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:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle a^2 + b^2 - 2ab \cos \alpha\) \(=\) \(\displaystyle \) \(\displaystyle c^2 + d^2 - 2cd \cos \gamma\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle a^2 +b^2 - c^2 - d^2\) \(=\) \(\displaystyle \) \(\displaystyle 2ab \cos \alpha - 2cd \cos \gamma\) \(\displaystyle \) \(\displaystyle \)          by adding $2ab \cos \alpha - c^2 -d^2$ to both sides          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \left( a^2 +b^2 - c^2 - d^2 \right)^2\) \(=\) \(\displaystyle \) \(\displaystyle 4 a^2 b^2 \cos^2 \alpha - 8 abcd \cdot \cos \alpha \cos \gamma + 4 c^2 d^2 \cos^2 \gamma\) \(\displaystyle \) \(\displaystyle \)          by squaring both sides          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle 0\) \(=\) \(\displaystyle \) \(\displaystyle \frac 1 4 \left( a^2 b^2 \cos^2 \alpha - 2 abcd \cdot \cos \alpha \cos \gamma + c^2 d^2 \cos^2 \gamma \right) - \frac 1 {16} \left( a^2 +b^2 - c^2 - d^2 \right)^2\) \(\displaystyle \) \(\displaystyle \)          by dividing by $16$ and substracting $\left( a^2 +b^2 - c^2 - d^2 \right)^2$          

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

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \mathcal A^2\) \(=\) \(\displaystyle \) \(\displaystyle \frac 1 4 \left( a^2 b^2 + c^2 d^2 - 2 abcd \cdot \cos \left( \alpha + \gamma\right) \right) - \frac 1 {16} \left( a^2 +b^2 - c^2 - d^2 \right)^2\) \(\displaystyle \) \(\displaystyle \)          $\sin^2 x + \cos^2 x = 1$ and $\cos \left({x - y}\right) = \cos x \cos y + \sin x \sin y$          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \mathcal A^2\) \(=\) \(\displaystyle \) \(\displaystyle \frac 1 {16} \left( 4 a^2 b^2 + 4 c^2 d^2 - \left( a^2 +b^2 - c^2 - d^2 \right)^2 \right) - \frac 1 2 abcd \cdot \cos \left( \alpha + \gamma\right)\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \mathcal A^2\) \(=\) \(\displaystyle \) \(\displaystyle \frac 1 {16} \left(-a^4 - b^4 - c^4 - d^4 + 2 a^2 b^2 + 2 a^2 c^2 + 2 a^2 d^2 + 2 b^2 c^2 + 2 b^2 d^2 + 2 c^2 d^2 \right) - \frac 1 2 abcd \cdot \cos \left( \alpha + \gamma\right)\) \(\displaystyle \) \(\displaystyle \)          by expanding $\left( a^2 +b^2 - c^2 - d^2 \right)^2$          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \mathcal A^2\) \(=\) \(\displaystyle \) \(\displaystyle \frac 1 {16} \left(-a^4 - b^4 - c^4 - d^4 + 2 a^2 b^2 + 2 a^2 c^2 + 2 a^2 d^2 + 2 b^2 c^2 + 2 b^2 d^2 + 2 c^2 d^2 + 8abcd - 8abcd\right) - \frac 1 2 abcd \cdot \cos \left( \alpha + \gamma\right)\) \(\displaystyle \) \(\displaystyle \)                    

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:

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

Hence the result.

$\blacksquare$


Also see

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

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


Source of Name

This entry was named for Carl Anton Bretschneider.

He published a proof in 1842.


Sources