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:

cyclic_quad

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):

cyclic_quad_incircles

You can find an angle-chasing proof here.

2 thoughts on “Cyclic quadrilaterals & Brahmagupta's formula

  1. “If we assume that the area is a symmetric 4-degree polynomial in the variables, then it’s very straightforward to obtain the result.”

    You mean the squared area. Anyway, proving that it’s a polynomial is no easier than proving the complete formula. In fact, for a general cyclic polygon this was an open problem up to some years ago (“Robbin’s conjecture”).

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