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 $\AA_2$.

$\AA_1 = \dfrac {a b \sin \alpha} 2$ 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

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

$\AA = \sqrt {\paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} }$
$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$

Source of Name

This entry was named for Carl Anton Bretschneider.

He published a proof in 1842.