婆羅摩笈多公式

基本形式

$\sqrt{(p-a)(p-b)(p-c)(p-d)}$

$p=\frac{a+b+c+d}{2}$

证明

$= \frac{1}{2}pq\sin A + \frac{1}{2}rs\sin C.$

$\mbox{Area} = \frac{1}{2}pq\sin A + \frac{1}{2}rs\sin A$
$(\mbox{Area})^2 = \frac{1}{4}\sin^2 A (pq + rs)^2$
$4(\mbox{Area})^2 = (1 - \cos^2 A)(pq + rs)^2 \,$
$4(\mbox{Area})^2 = (pq + rs)^2 - cos^2 A (pq + rs)^2. \,$

$\triangle ADB$$\triangle BDC$利用余弦定理，我们有：

$DB^2 = p^2 + q^2 - 2pq\cos A = r^2 + s^2 - 2rs\cos C. \,$

$2\cos A (pq + rs) = p^2 + q^2 - r^2 - s^2. \,$

$4(\mbox{Area})^2 = (pq + rs)^2 - \frac{1}{4}(p^2 + q^2 - r^2 - s^2)^2$
$16(\mbox{Area})^2 = 4(pq + rs)^2 - (p^2 + q^2 - r^2 - s^2)^2, \,$

$(2(pq + rs) + p^2 + q^2 -r^2 - s^2)(2(pq + rs) - p^2 - q^2 + r^2 +s^2) \,$
$= [ (p+q)^2 - (r-s)^2 ][ (r+s)^2 - (p-q)^2 ] \,$
$= (p+q+r-s)(p+q+s-r)(p+r+s-q)(q+r+s-p). \,$

$16(\mbox{Area})^2 = 16(T-p)(T-q)(T-r)(T-s). \,$

$\mbox{Area} = \sqrt{(T-p)(T-q)(T-r)(T-s)}.$

更特殊情況

$\sqrt{abcd}$

证明

$S=\sqrt{(p-a)(p-b)(p-c)(p-d)}$

$p=\frac{a+b+c+d}{2}$

$a+c=b+d$

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

$p-b=d$$p-c=a$$p-d=b$

$S=\sqrt{abcd}$

一般情況

$\sqrt{(p-a)(p-b)(p-c)(p-d)-a b c d \cos^2\theta}$