布雷特施奈德公式

${\displaystyle A={\sqrt {(s-a)(s-b)(s-c)(s-d)-abcd\cdot \cos ^{2}\left({\frac {\alpha +\gamma }{2}}\right)}}}$

布雷特施奈德公式的证明

{\displaystyle {\begin{aligned}A&=\triangle ADB+\triangle BDC\\&={\frac {1}{2}}ad\sin \alpha +{\frac {1}{2}}bc\sin \gamma \end{aligned}}}

${\displaystyle 4A^{2}=(ad)^{2}\sin ^{2}\alpha +(bc)^{2}\sin ^{2}\gamma +2abcd\sin \alpha \sin \gamma }$

${\displaystyle a^{2}+d^{2}-2ad\cos \alpha =b^{2}+c^{2}-2bc\cos \gamma }$

${\displaystyle {\frac {(a^{2}+d^{2}-b^{2}-c^{2})^{2}}{4}}=(ad)^{2}\cos ^{2}\alpha +(bc)^{2}\cos ^{2}\gamma -2abcd\cos \alpha \cos \gamma }$

${\displaystyle 4A^{2}+{\frac {(b^{2}+c^{2}-a^{2}-d^{2})^{2}}{4}}=(ad)^{2}+(bc)^{2}-2abcd\cdot \cos(\alpha +\gamma )}$

${\displaystyle 16A^{2}=(a+b+c-d)(a+b+d-c)(a+c+d-b)(b+c+d-a)-16abcd\cdot \cos ^{2}\left({\frac {\alpha +\gamma }{2}}\right)}$

${\displaystyle 16A^{2}=16(s-a)(s-b)(s-c)(s-d)-16abcd\cdot \cos ^{2}\left({\frac {\alpha +\gamma }{2}}\right)}$
${\displaystyle \therefore A={\sqrt {(s-a)(s-b)(s-c)(s-d)-abcd\cdot \cos ^{2}\left({\frac {\alpha +\gamma }{2}}\right)}}}$

相關公式

${\displaystyle K={\sqrt {(s-a)(s-b)(s-c)(s-d)-{\tfrac {1}{4}}(ac+bd+pq)(ac+bd-pq)}}}$

相關連接

1. ^ J. L. Coolidge, "A historically interesting formula for the area of a quadrilateral", American Mathematical Monthly, 46 (1939) 345–347.