## Brahmagupta's Formula

### Stefan Gössner

Department of Mechanical Engineering, University of Applied Sciences, Dortmund, Germany.

Keywords: Brahmagupta's formula; vectorial proof; cyclic quadrilateral; geometry; maximum area;

### Introduction

In Euclidean geometry, Brahmagupta's formula calculates the aera $A$ enclosed by a cyclic quadrilateral (a quadrilateral whose vertices lie on a common circle). The quadrilateral can be described by a loop closure of side vectors $\\bold a$, $\\bold b$, $\\bold c$, $\\bold d$ running counter-clockwise (Fig.1) [3].

Brahmagupta's formula is used to determine the aera $A$ of a cyclic quadrilateral given by its side lengths via [1]

$A = \\sqrt{(s-a)(s-b)(s-c)(s-d)}$(1)

with the semiperimeter

$s = \\dfrac{a+b+c+d}{2}\\,.$(2)

The area of a cyclic quadrilateral is the maximum possible for any quadrilateral with the given side lengths.

### Vectorial Proof

We start with the sum of the area of the two triangles in Fig.1 using half area product each [6].

$A = \\frac{1}{2}(\\tilde\\bold a \\bold b + \\tilde\\bold c \\bold d)$(3)

Both angles in the triangles opposite to diagonal $\\bold p$ sum up to $\\alpha + \\gamma = \\pi$ and thus equals $\\sin\\alpha = \\sin\\gamma$. So we can write in vectorial notation [6]

$\\frac{\\tilde\\bold a \\bold b}{ab} = \\frac{\\tilde\\bold c \\bold d}{cd}$

Substituting $\\tilde\\bold c \\bold d$ in equation (3) using this relation and multiplying by 2 gives

$2A = (ab + cd)\\frac{\\tilde\\bold a \\bold b}{ab}$

Squaring both sides yields

$4A^2 = (ab + cd)^2\\,(\\frac{\\tilde\\bold a \\bold b}{ab})^2$

and reusing Lagrange's Identity [6] $(\\bold a \\bold b)^2 + (\\tilde\\bold a \\bold b)^2 = (ab)^2$ results in

$4A^2 = (ab + cd)^2(1 - (\\frac{\\bold a \\bold b}{ab})^2)$(4)

Expressing the diagonal vector by its two triangle side vectors $\\bold p = \\bold a + \\bold b = -(\\bold c + \\bold d)$ and squaring leads us to

$a^2 + 2\\bold a\\bold b + b^2 = c^2 + 2\\bold c\\bold d + d^2$(5)

The dot products $\\bold a\\bold b$ and $\\bold c\\bold d$ correspond to cosine's of their enclosed angles by $\\cos\\alpha = -\\cos\\gamma$. In vectorial notation this now reads

$\\frac{\\bold a \\bold b}{ab} = -\\frac{\\bold c \\bold d}{cd}$

Substituting $\\bold c \\bold d$ in equation (5) resolves to

$\\frac{\\bold a \\bold b}{ab} = \\frac{-a^2 - b^2 + c^2 + d^2}{2(ab + cd)}$

After squaring this expression and bringing it into this shape

$1 - (\\frac{\\bold a \\bold b}{ab})^2 = \\frac{4(ab + cd)^2 - (-a^2 - b^2 + c^2 + d^2)^2}{4(ab + cd)^2}\\,,$

we can introduce that comfortable into equation (4)

$16A^2 = 4(ab + cd)^2 - (-a^2 - b^2 + c^2 + d^2)^2\\,.$

The right-hand side is now of the form $u^2 - v^2$, thus might be turned into $(u+v)(u-v)$, i.e.

$16A^2 = [2(ab + cd) - a^2 - b^2 + c^2 + d^2][2(ab + cd) + a^2 + b^2 - c^2 - d^2]\\,.$

Rearranging terms in the brackets leads us to

$16A^2 = [(c+d)^2 - (a-b)^2][(a+b)^2 - (c-d)^2]\\,.$

Now applying the $u^2 - v^2$ trick twice again results in

$16A^2 = (-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)\\,,$

which finally transforms to the shape of equation (1) by making use of semiperimeter (2)

$A = \\sqrt{(s-a)(s-b)(s-c)(s-d)}.$(1)

Expressing the area of the triangle with given sides $a, b, p$ and its circumradius $R$ is $\\frac{abp}{4R}$ [5]. So we can rewrite the quadrilateral's area as the sum of two triangles according to Fig.1

$A = \\frac{(ab + cd)p}{4R}\\,.$

Now with reuse of equation (4) for $p$ from Ptolemy's proof we get

$A = \\frac{1}{4R}\\sqrt{(ab+cd)(ac+bd)(ad+bc)}\\,,$

and with given area $A$ from (1) we finally yield the radius $R$ of the circle on which all vertices of the cyclic quadrilateral are lying on

$R = \\frac{1}{4A}\\sqrt{(ab+cd)(ac+bd)(ad+bc)}\\,.$(6)