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;


Fig.1: Cyclic Quadrilateral

In Euclidean geometry, Brahmagupta's formula calculates the aera AA 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 a\\bold a, b\\bold b, c\\bold c, d\\bold d running counter-clockwise (Fig.1) [3].

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

A=(sa)(sb)(sc)(sd)A = \\sqrt{(s-a)(s-b)(s-c)(s-d)} (1)

with the semiperimeter

s=a+b+c+d2.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=12(a~b+c~d) A = \\frac{1}{2}(\\tilde\\bold a \\bold b + \\tilde\\bold c \\bold d)(3)

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

a~bab=c~dcd\\frac{\\tilde\\bold a \\bold b}{ab} = \\frac{\\tilde\\bold c \\bold d}{cd}

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

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

Squaring both sides yields

4A2=(ab+cd)2(a~bab)2 4A^2 = (ab + cd)^2\\,(\\frac{\\tilde\\bold a \\bold b}{ab})^2

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

4A2=(ab+cd)2(1(abab)2) 4A^2 = (ab + cd)^2(1 - (\\frac{\\bold a \\bold b}{ab})^2)(4)

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

a2+2ab+b2=c2+2cd+d2a^2 + 2\\bold a\\bold b + b^2 = c^2 + 2\\bold c\\bold d + d^2(5)

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

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

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

abab=a2b2+c2+d22(ab+cd)\\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(abab)2=4(ab+cd)2(a2b2+c2+d2)24(ab+cd)2,1 - (\\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)

16A2=4(ab+cd)2(a2b2+c2+d2)2.16A^2 = 4(ab + cd)^2 - (-a^2 - b^2 + c^2 + d^2)^2\\,.

The right-hand side is now of the form u2v2u^2 - v^2, thus might be turned into (u+v)(uv)(u+v)(u-v), i.e.

16A2=[2(ab+cd)a2b2+c2+d2][2(ab+cd)+a2+b2c2d2].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

16A2=[(c+d)2(ab)2][(a+b)2(cd)2].16A^2 = [(c+d)^2 - (a-b)^2][(a+b)^2 - (c-d)^2]\\,.

Now applying the u2v2u^2 - v^2 trick twice again results in

16A2=(a+b+c+d)(ab+c+d)(a+bc+d)(a+b+cd),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=(sa)(sb)(sc)(sd).A = \\sqrt{(s-a)(s-b)(s-c)(s-d)}. (1)

Circle Radius

Expressing the area of the triangle with given sides a,b,pa, b, p and its circumradius RR is abp4R\\frac{abp}{4R} [5]. So we can rewrite the quadrilateral's area as the sum of two triangles according to Fig.1

A=(ab+cd)p4R. A = \\frac{(ab + cd)p}{4R}\\,.

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

A=14R(ab+cd)(ac+bd)(ad+bc), A = \\frac{1}{4R}\\sqrt{(ab+cd)(ac+bd)(ad+bc)}\\,,

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

R=14A(ab+cd)(ac+bd)(ad+bc). R = \\frac{1}{4A}\\sqrt{(ab+cd)(ac+bd)(ad+bc)}\\,. (6)