Brahmagupta's Formula/Corollary

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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

Brahmagupta's Formula:

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

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:

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

$\blacksquare$