# Cyclic quadrilaterals & Brahmagupta’s formula

I suppose every reader of this ‘ere blog will know Heron’s formula for the area $K$ of a triangle with sides $a,b,c$:

$K = \sqrt{s(s-a)(s-b)(s-c)}$

where $s$ is the “semi-perimeter”:

$\displaystyle{s=\frac{a+b+c}{2}.}$

The formula is not at all hard to prove: see the Wikipedia page for two elementary proofs.

However, I have only recently become aware of Brahmagupta’s formula for the area of a cyclic quadrilateral. A cyclic quadrilateral, if you didn’t know, is a (convex) quadrilateral all of whose points lie on a circle:

And if the edges have lengths $a,b,c,d$ as shown, then the formula states that the area is given by

$K = \sqrt{(s-a)(s-b)(s-c)(s-d)}$

where as above $s$ is the semi-perimeter:

$\displaystyle{s=\frac{a+b+c+d}{2}.}$

This can be seen to be a generalization of Heron’s formula. Although the formula is named for Brahmagupta (598 – 670), who does indeed seem to have been the first to state it, there is no evidence that he had a proof, or even recognized that the formula was valid only for cyclic quadrilaterals.

There are of course innumerable proofs of Brahmagupta’s formula; one elegant way starts by noting that any formula for the area of a cyclic quadrilateral in terms of edge lengths must be symmetric in its variables. We can see this by considering the four triangles from the center of the circle: they can be arranged in any order, and the total area of course is unchanged. If we assume that the area $K^2$ is a symmetric 4-degree polynomial in the variables, then it’s very straightforward to obtain the result.

Cyclic quadrilaterals have a host of wonderful theorems and results. Here’s one I particularly like – it’s a Sangaku problem:

Suppose that $A,B,C,D$ are the vertices of a cyclic quadrilateral. Then the centres of the incircles of the four triangles $ABC, ABD, ACD, BCD$ form a rectangle.

Here’s a picture (created with geogebra):

You can find an angle-chasing proof here.