Three months ago I post a complex number proof of Heron’s formula, and today I am gonna show you a proof of Brahmagupta’s formula. Brahmagupta’s formula is basically a generalization of Heron’s formula. Brahmagupta was a great Indian mathematician (597–668 AD).

**Brahmagupta’s Formula**

Given a cyclic quadrilateral ABCD with side length of a, b, c, and d. And let semi-perimeter be s such that,

then the area of the cyclic quadrilateral is

Note that this formula works only if the quadrilateral is cyclic, which means that the opposite angles of the quadrilateral are supplementary.

and

or in other words, the quadrilateral ABCD is cyclic if it can be inscribed inside a circle.

*Proof:*

Join the line segment BD with length x, then apply the cosine law on triangle ABD and triangle BCD.

hence we have

Since A + C = 180 degrees, we have

so,

Now the area of the quadrilateral ABCD is the sum of the areas of the triangle ABD and triangle BCD, we have

Since A + C = 180 degrees, we have

hence,

let’s square both sides for convenient and apply the pythagoras trig identity,

Now, factor the polynomial in the bracket,

Now, since , so

Therefore,

*Special case: The Heron’s formula*

Now if one of the side length of the cyclic quadrilateral approaches to zero, then it becomes the Heron’s formula

Since all triangles are cyclic,