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 Math Help > Geometry > Polygons and Triangles > Cyclic Quadrilateral > Brahmagupta's Formula

Brahmagupta's formula provides the area A of a cyclic quadrilateral (i.e., a simple quadrilateral that is inscribed in a circle) with sides of length a, b, c, and d as

A = sqrt((s-a)(s-b)(s-c)(s-d)),

where s is the semiperimeter (a+b+c+d)/2

If the quadrilateral ABCD is a rectangle, then s=a+b, so A=sqrt((b)(a)(b)(a))=ab, so the formula is true in that special case.  If one of the sides goes to zero, then ABCD is a triangle, and the formula become's Heron's Formula, so it's true in that special case as well.  Here on this page, I will demonstrate that the formula is true of any other cyclic quadrilateral.  Brahmagupta first published this result in 628 in a book that included many other groundbreaking mathematical results.

You can see that opposite angles are supplementary by drawing a diagonal, AC.

Angle D subtends the arc ABC, so the measure of angle D is half the measure of arc ABC;
Angle B subtends the arc ADC, so the measure of angle B is half the measure of arc ADC;
The sum of the measures of arcs ABC and ADC is 360�, so the sum of the measures of angles B and D is 180�.

Having shown Brahmagupta's formula true for rectangles, we assume this cyclic quadrilateral is not a rectangle, so WLOG we assume AB and CD are not parallel.  Extend AB and DC until they meet at P:

So triangles PBC and PDA are similar.  The ratio of their sides is b/d, so the ratio of their areas is b²/d².

Let A be the area of the quadrilateral, and let T be the area of triangle PBC.

(1)   A = T - (d²/b²)T

(2)   A = (1-d²/b²)T

(3)   A = ((b²-d²)/b²)T

By Heron's Formula, T = sqrt(s(s-e-a)(s-f-c)(s-b)), where s is the semiperimeter of triangle PBC.

(4)   A = ((b²-d²)/b²)sqrt(s(s-e-a)(s-f-c)(s-b))

Now, let's get e and f in terms of a, b, c, and d.  By similar triangles, we have these two equations

be - df = cd

Treating these as two linear equations in e and f, the solution is

f = (abd+cd²)/(b²-d²)

Substituting these values of e and f into equation 4, we get:

And s = (a+b+c+e+f)/2
s = (1/2)b(a+b+c-d)(b+d)/(b²-d²)
s = (1/2)b(a+b+c-d)/(b-d)

Now substituting this value of s into equation 5,

(6)   A = ((b²-d²)/b²)sqrt(
((1/2)b(a+b+c-d)/(b-d))
((1/2)b(a+b+c-d)/(b-d)-(abd+cd²)/(b²-d²)-c)
((1/2)b(a+b+c-d)/(b-d)-b))

Now pulling 4 factors of (1/2) out of the sqrt,

(7)   A = (1/4)((b²-d²)/b²)sqrt(
(b(a+b+c-d)/(b-d))
(b(a+b+c-d)/(b-d)-2(abd+cd²)/(b²-d²)-2c)
(b(a+b+c-d)/(b-d)-2b))

Now pulling factors of 1/(b-d), 1/(b+d), 1/(b+d), and 1/(b-d) out of the four factors in the sqrt, respectively, canceling b²-d²,

(8)   A = (1/4)(1/b²)sqrt(
(b(a+b+c-d))
(b(a+b+c-d)(b+d)/(b-d)-2(abd+cd²)/(b-d)-2c(b+d))
(b(a+b+c-d)-2b(b-d)))

(9)   A = (1/4)(1/b²)sqrt(
(b(a+b+c-d))
(b(a-b+c+d))

(10)   A = (1/4)(1/b²)sqrt(
(b(a+b+c-d))
((b²+ab+cb+db+2cd)-2c(b+d))
(b(a-b+c+d))

(11)   A = (1/4)(1/b²)sqrt(
(b(a+b+c-d))
(b²-ab+cb+db)
(b²+ab-cb+db)
(b(a-b+c+d))

Now pull four factors of b from the sqrt, canceling b² in the denominator,

(12)   A = (1/4)sqrt(
(a+b+c-d)
(b-a+c+d)
(b+a-c+d)
(a-b+c+d))

Noting that -a+b+c+d = 2(s-a), and a-b+c+d = 2(s-b), etc., where s is now the semiperimeter of the quadrilateral,

(13)   A = sqrt((s-a)(s-b)(s-c)(s-d))

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Circumscribed Triangle