Theorem: Bretschneider's Formula

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:
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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 the second axiom of area:
:$\AA = \AA_1 + \AA_2$
so:
{{begin-eqn}}
{{eqn | l = \AA^2
      | r = \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}
      | n = 1
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = a^2 + b^2 - 2 a b \cos \alpha
      | r = c^2 + d^2 - 2 c d \cos \gamma
}}
{{eqn | l = a^2 + b^2 - c^2 - d^2
      | r = 2 a b \cos \alpha - 2 c d \cos \gamma
      | c = adding $2 a b \cos \alpha - c^2 - d^2$ to both sides
}}
{{eqn | l = \paren {a^2 + b^2 - c^2 - d^2}^2
      | r = 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
      | c = squaring both sides
}}
{{eqn | l = 0
      | r = \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}
      | c = algebraic manipulation
}}
{{eqn | ro= -
      | r = \frac 1 {16} \paren {a^2 + b^2 - c^2 - d^2}^2
      | o = 
}}
{{end-eqn}}
Now add this equation to $(1)$.
Then trigonometric identities can be used, as follows:
{{begin-eqn}}
{{eqn | l = \AA^2
      | r = \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
      | c = Sum of Squares of Sine and Cosine and Cosine of Sum
}}
{{eqn | r = \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}
}}
{{end-eqn}}
By expanding the square $\paren {a^2 + b^2 - c^2 - d^2}^2$: 
{{begin-eqn}}
{{eqn | n = 2
      | l = \AA^2
      | r = \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}
}}
{{end-eqn}}
Adding and subtracting $8 a b c d$ to and from the numerator of the first term of $(2)$:
{{begin-eqn}}
{{eqn | l = \AA^2
      | r = \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}
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \AA^2
      | r = \frac 1 {16} \paren {-a + b + c + d} \paren {a - b + c + d} \paren {a + b - c + d} \paren {a + b + c - d}
      | c = 
}}
{{eqn | ro= -
      | r = \frac 1 2 a b c d - \frac 1 2 a b c d \map \cos {\alpha + \gamma}
      | o = 
}}
{{eqn | r = \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}
      | c = as $s = \dfrac {a + b + c + d} 2$
}}
{{eqn | r = \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} }
      | c = 
}}
{{eqn | r = \paren {s - a} \paren {s - b} \paren {s - c} \paren {s - d} - a b c d \map {\cos^2} {\dfrac {\alpha + \gamma} 2}
      | c = Half Angle Formula for Cosine
}}
{{end-eqn}}
Hence the result.
{{qed}}