## 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

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.

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

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

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

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,

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

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