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