View source for Brahmagupta's formula

{{Short description|Formula relating the area of a cyclic quadrilateral to its side lengths}}
In [[Euclidean geometry]], '''Brahmagupta's formula''', named after the 7th century [[Brahmagupta|Indian mathematician]], is used to find the [[area]] of any convex [[cyclic quadrilateral]] (one that can be inscribed in a circle) given the lengths of the sides. Its generalized version, ''[[Bretschneider's formula]]'', can be used with non-cyclic quadrilateral. ''[[Heron's formula]]'' can be thought as a special case of the Brahmagupta's formula for triangles.

== Formulation ==
Brahmagupta's formula gives the area {{math|''K''}} of a convex [[cyclic quadrilateral]] whose sides have lengths {{math|''a''}}, {{math|''b''}}, {{math|''c''}}, {{math|''d''}} as

: <math>K=\sqrt{(s-a)(s-b)(s-c)(s-d)}</math>

where {{math|''s''}}, the [[semiperimeter]], is defined to be

: <math>s=\frac{a+b+c+d}{2}.</math>

This formula generalizes [[Heron's formula]] for the area of a [[triangle]]. A triangle may be regarded as a quadrilateral with one side of length zero. From this perspective, as {{math|''d''}} (or any one side) approaches zero, a cyclic quadrilateral converges into a cyclic triangle (all triangles are cyclic), and Brahmagupta's formula simplifies to Heron's formula.

If the semiperimeter is not used, Brahmagupta's formula is

: <math>K=\frac{1}{4}\sqrt{(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)}.</math>

Another equivalent version is

: <math>K=\frac{\sqrt{(a^2+b^2+c^2+d^2)^2+8abcd-2(a^4+b^4+c^4+d^4)}}{4}\cdot</math>

== Proof ==
[[File:Brahmagupta's formula Sketch.png|400x400px|Diagram for reference|thumb]]

===Trigonometric proof===
Here the notations in the figure to the right are used. The area {{math|''K''}} of the convex cyclic quadrilateral equals the sum of the areas of {{math|△''ADB''}} and {{math|△''BDC''}}:

:<math>K = \frac{1}{2}pq\sin A + \frac{1}{2}rs\sin C.</math>

But since {{math|''□ABCD''}} is a cyclic quadrilateral, {{math|∠''DAB'' {{=}} 180° − ∠''DCB''}}. Hence {{math|sin ''A'' {{=}} sin ''C''}}. Therefore,

:<math>K = \frac{1}{2}pq\sin A + \frac{1}{2}rs\sin A</math>

:<math>K^2 = \frac{1}{4} (pq + rs)^2 \sin^2 A</math>

:<math>4K^2 = (pq + rs)^2 (1 - \cos^2 A) = (pq + rs)^2 - ((pq + rs)\cos A)^2</math>

(using the [[trigonometric identity]]).

Solving for common side {{math|''DB''}}, in {{math|△''ADB''}} and {{math|△''BDC''}}, the [[law of cosines]] gives

:<math>p^2 + q^2 - 2pq\cos A = r^2 + s^2 - 2rs\cos C.</math>

Substituting {{math|cos ''C'' {{=}} −cos ''A''}} (since angles {{math|''A''}} and {{math|''C''}} are [[Supplementary angles|supplementary]]) and rearranging, we have

:<math>(pq + rs) \cos A = \frac{1}{2}(p^2 + q^2 - r^2 - s^2).</math>

Substituting this in the equation for the area,

:<math>4K^2 = (pq + rs)^2 - \frac{1}{4}(p^2 + q^2 - r^2 - s^2)^2</math>

:<math>16K^2 = 4(pq + rs)^2 - (p^2 + q^2 - r^2 - s^2)^2.</math>

The right-hand side is of the form {{math|''a''{{sup|2}} − ''b''{{sup|2}} {{=}} (''a'' − ''b'')(''a'' + ''b'')}} and hence can be written as

:<math>[2(pq + rs)) - p^2 - q^2 + r^2 +s^2][2(pq + rs) + p^2 + q^2 -r^2 - s^2] </math>

which, upon rearranging the terms in the square brackets, yields

:<math>16K^2= [ (r+s)^2 - (p-q)^2 ][ (p+q)^2 - (r-s)^2 ] </math>

that can be factored again into

:<math>16K^2=(q+r+s-p)(p+r+s-q)(p+q+s-r)(p+q+r-s). </math>

Introducing the [[semiperimeter]] {{math|''S'' {{=}} {{sfrac|''p'' + ''q'' + ''r'' + ''s''|2}}}} yields

:<math>16K^2 = 16(S-p)(S-q)(S-r)(S-s). </math>

Taking the square root, we get

:<math>K = \sqrt{(S-p)(S-q)(S-r)(S-s)}.</math>

===Non-trigonometric proof===
An alternative, non-trigonometric proof utilizes two applications of Heron's triangle area formula on similar triangles.<ref>Hess, Albrecht, "A highway from Heron to Brahmagupta", ''Forum Geometricorum'' 12 (2012), 191–192.</ref>

== Extension to non-cyclic quadrilaterals ==
In the case of non-cyclic quadrilaterals, Brahmagupta's formula can be extended by considering the measures of two opposite angles of the quadrilateral:

: <math>K=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\theta}</math>

where {{math|''θ''}} is half the sum of any two opposite angles. (The choice of which pair of opposite angles is irrelevant: if the other two angles are taken, half their sum is {{math|180° − ''θ''}}. Since {{math|cos(180° − ''θ'') {{=}} −cos ''θ''}}, we have {{math|cos<sup>2</sup>(180° − ''θ'') {{=}} cos<sup>2</sup> ''θ''}}.) This more general formula is known as [[Bretschneider's formula]].

It is a property of [[cyclic quadrilateral]]s (and ultimately of [[inscribed angle]]s) that opposite angles of a quadrilateral sum to 180°.  Consequently, in the case of an inscribed quadrilateral, {{math|''θ''}} is 90°, whence the term

:<math>abcd\cos^2\theta=abcd\cos^2 \left(90^\circ\right)=abcd\cdot0=0, </math>

giving the basic form of Brahmagupta's formula. It follows from the latter equation that the area of a cyclic quadrilateral is the maximum possible area for any quadrilateral with the given side lengths.

A related formula, which was proved by [[Julian Coolidge|Coolidge]], also gives the area of a general convex quadrilateral. It is<ref>J. L. Coolidge, "A Historically Interesting Formula for the Area of a Quadrilateral", ''American Mathematical Monthly'', '''46''' (1939) pp. 345-347.</ref>

: <math>K=\sqrt{(s-a)(s-b)(s-c)(s-d)-\textstyle{1\over4}(ac+bd+pq)(ac+bd-pq)}</math>

where {{math|''p''}} and {{math|''q''}} are the lengths of the diagonals of the quadrilateral. In a [[cyclic quadrilateral]], {{math|''pq'' {{=}} ''ac'' + ''bd''}} according to [[Ptolemy's theorem]], and the formula of Coolidge reduces to Brahmagupta's formula.

== Related theorems ==
* [[Heron's formula]] for the area of a [[triangle]] is the special case obtained by taking {{math|''d'' {{=}} 0}}.
* The relationship between the general and extended form of Brahmagupta's formula is similar to how the [[law of cosines]] extends the [[Pythagorean theorem]].
* Increasingly complicated closed-form formulas exist for the area of general polygons on circles, as described by Maley et al.<ref>{{cite journal |last1=Maley |first1=F. Miller |last2=Robbins |first2=David P. |last3=Roskies |first3=Julie |title=On the areas of cyclic and semicyclic polygons |journal=Advances in Applied Mathematics |date=2005 |volume=34 |issue=4 |pages=669–689 |doi=10.1016/j.aam.2004.09.008 |arxiv=math/0407300 |s2cid=119565975}}</ref>

== References ==
{{PlanetMath attribution|id=3594|title=proof of Brahmagupta's formula}}
{{reflist}}

==External links==
* [https://www.youtube.com/watch?v=NXtKjxVgYeM A geometric proof] from [[Sam Vandervelde]].
* {{mathworld|urlname=BrahmaguptasFormula|title=Brahmagupta's Formula}}

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[[Category:Brahmagupta]]
[[Category:Theorems about quadrilaterals and circles]]
[[Category:Area]]