# Brahmagupta's Formula/Corollary

## Corollary to Brahmagupta's Formula

The area of a cyclic quadrilateral with sides of lengths $a, b, c, d$ is:

$\dfrac {\sqrt {\paren {a^2 + b^2 + c^2 + d^2}^2 + 8 a b c d - 2 \paren {a^4 + b^4 + c^4 + d^4} } } 4$

## Proof

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

where $s$ is the semiperimeter:

$s = \dfrac {a + b + c + d} 2$

Making the substitutions:

 $\ds s - a$ $=$ $\ds \frac {-a + b + c + d} 2$ $\ds s - b$ $=$ $\ds \frac {a - b + c + d} 2$ $\ds s - c$ $=$ $\ds \frac {a + b - c + d} 2$ $\ds s - d$ $=$ $\ds \frac {a + b + c - d} 2$

results in:

$\AA = \dfrac {\sqrt {\paren {a^2 + b^2 + c^2 + d^2}^2 + 8 a b c d - 2 \paren {a^4 + b^4 + c^4 + d^4} } } 4$

$\blacksquare$