Brahmagupta’s formula

Posted: April 11, 2013 in Mathematics

Three months ago I post a complex number proof of Heron’s formula, and today I am gonna show you a proof of Brahmagupta’s formula.  Brahmagupta’s formula is basically a generalization of Heron’s formula.  Brahmagupta was a great Indian mathematician (597–668 AD).

Brahmagupta’s Formula

brahmagupta

Given a cyclic quadrilateral ABCD with side length of a, b, c, and d.  And let semi-perimeter be s such that,

s=\dfrac{a+b+c+d}{2}

then the area of the cyclic quadrilateral is

Area=\sqrt{(s-a)(s-b)(s-c)(s-d)}

Note that this formula works only if the quadrilateral is cyclic, which means that the opposite angles of the quadrilateral are supplementary.

A+C=180^\circ and B+D=180^\circ

or in other words, the quadrilateral ABCD is cyclic if it can be inscribed inside a circle.

Proof:

Join the line segment BD with length x, then apply the cosine law on triangle ABD and triangle BCD.

brahmagupta2

x^2=a^2+d^2-2ad\cos A

x^2=b^2+c^2-2bc\cos C

hence we have

a^2+d^2-2ad\cos A = b^2+c^2-2bc\cos C

Since A + C = 180 degrees, we have

\cos A = \cos (180^\circ - C) = -\cos C

so,

a^2+d^2-2ad\cos A = b^2+c^2+2bc\cos A

\cos A = \dfrac{a^2+d^2-b^2-c^2}{2ad+2bc}

Now the area of the quadrilateral ABCD is the sum of the areas of the triangle ABD and triangle BCD, we have

Area = \dfrac{1}{2}ad\sin A + \dfrac{1}{2}bc\sin C

Since A + C = 180 degrees, we have

\sin A = \sin C

hence,

Area = \dfrac{1}{2}ad\sin A + \dfrac{1}{2}bc\sin A = \dfrac{ad+bc}{2}\sin A

let’s square both sides for convenient and apply the pythagoras trig identity,

Area^2 = \dfrac{(ad+bc)^2}{4}\sin^2 A

Area^2=\dfrac{(ad+bc)^2}{4}(1-\cos^2 A)

Area^2=\dfrac{(ad+bc)^2}{4}\left(1-\dfrac{(a^2+d^2-b^2-c^2)^2}{4(ad+bc)^2}\right)

Area^2=\dfrac{1}{16}\bigg(4(ad+bc)^2-(a^2+d^2-b^2-c^2)^2\bigg)

Now, factor the polynomial in the bracket,

Area^2=\dfrac{1}{16}(2ad+2bc-a^2-d^2+b^2+c^2)(2ad+2bc+a^2+d^2-b^2-c^2)

Area^2=\dfrac{1}{16}(b^2+c^2+2bc-a^2-d^2+2ad)(a^2+d^2+2ad-b^2-c^2+2bc)

Area^2=\dfrac{1}{16}\bigg((b+c)^2-(a-d)^2\bigg)\bigg((a+d)^2-(b-c)^2\bigg)

Area^2=\dfrac{1}{16}(b+c-a+d)(b+c+a-d)(a+d-b+c)(a+d+b-c)

Area^2=\dfrac{b+c+d-a}{2}\cdot\dfrac{a+c+d-b}{2}\cdot\dfrac{a+b+d-c}{2}\cdot\dfrac{a+b+c-d}{2}

Now, since s=\dfrac{a+b+c+d}{2}, so

s-a = \dfrac{a+b+c+d}{2} - \dfrac{2a}{2} = \dfrac{b+c+d-a}{2}

Therefore,

Area^2=(s-a)(s-b)(s-c)(s-d)

Area = \sqrt{(s-a)(s-b)(s-c)(s-d)} \qquad \blacksquare

Special case: The Heron’s formula

heron

Now if one of the side length of the cyclic quadrilateral approaches to zero, then it becomes the Heron’s formula

Since all triangles are cyclic,

s=\dfrac{a+b+c+d}{2} \stackrel{(d \to 0)}{=} \dfrac{a+b+c}{2}

Area = \sqrt{(s-a)(s-b)(s-c)(s-d)} \stackrel{(d \to 0)}{=} \sqrt{s(s-a)(s-b)(s-c)}

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