Brahmagupta's Formula
Brahmagupta's formula is a special case of Bretschneider's formula as applied to cyclic quadrilaterals.
Example of a Cyclic Quadrilateral
Bretschneider's formula states that the area of a quadrilateral is given by
\[\Delta^{2} = (s-a)(s-b)(s-c)(s-d) - abcd\cos^{2}\left(\frac{B+D}{2}\right),\]
where \(\Delta\) is the area of the quadrilateral, \(B\) and \(D\) are the angles, \(a, b, c,\) and \(d\) are the sides of the quadrilateral, and \(s\) is the semiperimeter of the quadrilateral, given by \(s=\frac{a+b+c+d}{2}\).
For a cyclic quadrilateral, we know that the sum of the opposite angles is \(180^{\circ}\). Hence we have \(B+D=180^{\circ}\). So the simplified version known as Brahmagupta's formula is given as follows:
Brahmagupta's Formula
\[\Delta^{2} = (s-a)(s-b)(s-c)(s-d)\]
Given a cyclic quadrilateral with side lengths \(a,b,c,d\), the area \(\Delta\) can be found as
\[ \Delta =\sqrt{(s-a)(s-b)(s-c)(s-d)}.\]
Alternatively, without using the semiperimeter, we can use
\[ \Delta = \frac{1}{4} \sqrt{(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)}. \]
Proof
Assume \(A,B,C,D\) to be the vertices of the cyclic quadrilateral, and \(p,q,r,s\) the lengths of sides \(AD,AB,BC,CD,\) respectively. Then
\[\begin{align} (\text{Area of } ABCD) \equiv K &=(\text{Area of } \triangle ADB)+(\text{Area of } \triangle BDC)\\\\ &=\frac { pq \sin A }{ 2 } +\frac { rs \sin C }{ 2 }. \end{align}\]
Since the quadrilateral is cyclic, \(\sin A=\sin C,\) so
\[\begin{align} K&=\frac { pq\sin A }{ 2 } +\frac { rs\sin A }{ 2 } \\ K^2&=\frac { 1 }{ 4 } { ( pq+rs ) }^{ 2 }\sin ^{ 2 }{ A } \\ 4K^{ 2 }&={ \left( pq+rs \right) }^{ 2 }\big( 1-\cos ^{ 2 }{ A } \big) \\ &={ \left( pq+rs \right) }^{ 2 }-{ \left( pq+rs \right) }^{ 2 }\big( \cos ^{ 2 }{ A } \big). \qquad (1) \end{align} \]
Solving for common side \(DB\) in triangles \(ADB\) and \(BDC\) and using laws of cosines, we get
\[{ p }^{ 2 }+{ q }^{ 2 }-2pq\cos A={ r }^{ 2 }+{ s }^{ 2 }-2rs\cos C.\]
As angles \(A\) and \(C\) are supplementary, \(\cos A=-\cos C\) and
\[2(pq+rs)\cos A={ p }^{ 2 }+{ q }^{ 2 }-{ r }^{ 2 }-{ s }^{ 2 }.\]
Substituting this into the equation of area \((1),\) we get
\[16{ K }^{ 2 }={ 4(pq+rs) }^{ 2 }-{ \big({ p }^{ 2 }+{ q }^{ 2 }-{ r }^{ 2 }-{ s }^{ 2 }\big) }^{ 2 }.\]
The RHS of the equation is of the form \({ a }^{ 2 }-{ b }^{ 2 }\), so it can be written as
\[\begin{align} \Big[2(pq+rs)-\big({ p }^{ 2 }+{ q }^{ 2 }-{ r }^{ 2 }-{ s }^{ 2 }\big)\Big]\Big[2(pq+rs)+\big({ p }^{ 2 }+{ q }^{ 2 }-{ r }^{ 2 }-{ s }^{ 2 }\big)\Big] &=\big[ { \left( r+s \right) }^{ 2 }-{ \left( p-q \right) }^{ 2 } \big] \big[ { \left( p+q \right) }^{ 2 }-{ \left( r-s \right) }^{ 2 } \big] \\ \\ &=(p+q+r-s)(p+q-r+s)(p-q+r+s)(-p+q+r+s). \end{align}\]
So, the area of the cyclic quadrilateral with sides \(a,b,c,d\) can be written as
\[K=\frac { 1 }{ 4 } \sqrt { (a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d) }.\ _\square\]