Brahmagupta's Formula

The area $K$ of a cyclic quadrilateral with side lengths $a, b, c, d$ and semiperimeter $s = \dfrac{a + b + c + d}{2}$ is given by

$$K = =\sqrt{(s — a)(s — b)(s — c)(s — d)}$$

(i) The familiar formula for triangular area (1/2 x base x height) was known to Greek mathematicians for at least three hundred years before Euclid catalogued all of known geometry in his Elements, including (Book IV, Proposition 5) the construction of the unique circumcircle which passes through the vertices of a triangle, by intersecting the perpedicular bisectors of its sides.

(ii) Three hundred years after that came the famous Heron's formula: the area $K$ of a triangle with sides $A, B, C$ and semiperimeter$s = \dfrac{A + B + C + D}{2}$ is given by 

$$K=\sqrt{(s — A)(s — B)(s — C)(s — D)}$$

(iii) Although Greek mathematics was apparently unknown to medieval Indian (as opposed to Islamic) scholars, Brahmagupta effectively put Heron's triangle back into the circle: take two non-overlapping circumscribed triangles sharing a common edge ($C$ in the picture); the result is a cyclic quadrilateral, one whose vertices all lie on a circle.

And now the area of the quadrilateral replaces the final s in Heron's formula by $s - d$. If point $P$ is allowed to approach point $Q$ then d becomes zero and c becomes $C$, recovering Heron.

(iv) Another thirteen hundred years pass and the circumscribing circle is removed once more in American mathematician Julian Lowell Coolidge's quadrilateral area formula: the area of an arbitrary convex quadrilateral with sides $a, b, c, d$, a opposite to $d$, with diagonals $e, f $, and with semiperimeter s, is given by 

$K=\sqrt{(s — a)(s — b)(s — c)(s — d)- \dfrac{1}{4}(ad+bc+ef)(ad+bc-ef)}$

This generalises Brahmagupta by virtue of another classic of antiquity, Ptolemy’s Theorem: quadrilateral $a, b, c, d$, a opposite to $d$, with diagonals $e, f$ , is cyclic if and only if $ad +bc = ef$ .

Πηγή: theoremoftheday