Brahmagupta's formula#Trigonometric proof

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

: $K=\backslash sqrt\{(s-a)(s-b)(s-c)(s-d)\}$

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

: $s=\backslash frac\{a+b+c+d\}\{2\}.$

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

: $K=\backslash frac\{1\}\{4\}\backslash sqrt\{(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)\}.$

Another equivalent version is

: $K=\backslash frac\{\backslash sqrt\{(a^2+b^2+c^2+d^2)^2+8abcd-2(a^4+b^4+c^4+d^4)\}\}\{4\}\backslash cdot$

File:Brahmagupta's formula Sketch.png

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}}:

:$K\; =\; \backslash frac\{1\}\{2\}pq\backslash sin\; A\; +\; \backslash frac\{1\}\{2\}rs\backslash sin\; C.$

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

:$K\; =\; \backslash frac\{1\}\{2\}pq\backslash sin\; A\; +\; \backslash frac\{1\}\{2\}rs\backslash sin\; A$

:$K^2\; =\; \backslash frac\{1\}\{4\}\; (pq\; +\; rs)^2\; \backslash sin^2\; A$

:$4K^2\; =\; (pq\; +\; rs)^2\; (1\; -\; \backslash cos^2\; A)\; =\; (pq\; +\; rs)^2\; -\; ((pq\; +\; rs)\backslash cos\; A)^2$

(using the trigonometric identity).

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

:$p^2\; +\; q^2\; -\; 2pq\backslash cos\; A\; =\; r^2\; +\; s^2\; -\; 2rs\backslash cos\; C.$

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

:$(pq\; +\; rs)\; \backslash cos\; A\; =\; \backslash frac\{1\}\{2\}(p^2\; +\; q^2\; -\; r^2\; -\; s^2).$

Substituting this in the equation for the area,

:$4K^2\; =\; (pq\; +\; rs)^2\; -\; \backslash frac\{1\}\{4\}(p^2\; +\; q^2\; -\; r^2\; -\; s^2)^2$

:$16K^2\; =\; 4(pq\; +\; rs)^2\; -\; (p^2\; +\; q^2\; -\; r^2\; -\; s^2)^2.$

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

:$[2(pq\; +\; rs))\; -\; p^2\; -\; q^2\; +\; r^2\; +s^2][2(pq\; +\; rs)\; +\; p^2\; +\; q^2\; -r^2\; -\; s^2]$

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

:$16K^2=\; [\; (r+s)^2\; -\; (p-q)^2\; ][\; (p+q)^2\; -\; (r-s)^2\; ]$

that can be factored again into

:$16K^2=(q+r+s-p)(p+r+s-q)(p+q+s-r)(p+q+r-s).$

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

:$16K^2\; =\; 16(S-p)(S-q)(S-r)(S-s).$

Taking the square root, we get

:$K\; =\; \backslash sqrt\{(S-p)(S-q)(S-r)(S-s)\}.$

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

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

: $K=\backslash sqrt\{(s-a)(s-b)(s-c)(s-d)-abcd\backslash cos^2\backslash theta\}$

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|cos²(180° − θ) {{=}} cos² θ}}.) This more general formula is known as Bretschneider's formula.

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

:$abcd\backslash cos^2\backslash theta=abcd\backslash cos^2\; \backslash left(90^\backslash circ\backslash right)=abcd\backslash cdot0=0,$

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 Coolidge, also gives the area of a general convex quadrilateral. It isJ. L. Coolidge, "A Historically Interesting Formula for the Area of a Quadrilateral", American Mathematical Monthly, 46 (1939) pp. 345-347.

: $K=\backslash sqrt\{(s-a)(s-b)(s-c)(s-d)-\backslash textstyle\{1\backslash over4\}(ac+bd+pq)(ac+bd-pq)\}$

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.

• 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.{{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}}

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

{{reflist}}

• [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

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