If we draw , we find that . Since , . Hence, . Multiplying by 2 and squaring, we get: Substituting results in By the Law of Cosines, . , so a little rearranging gives
Brahmagupta's formula reduces to Heron's formula by setting the side length .
A similar formula which Brahmagupta derived for the area of a general quadrilateral is where is the semiperimeter of the quadrilateral. What happens when the quadrilateral is cyclic?
- is a cyclic quadrilateral that has an inscribed circle. The diagonals of intersect at . If and then the area of the inscribed circle of can be expressed as , where and are relatively prime positive integers. Determine . (Source)