Now reveal the fourth vertex, D, on the circle, and draw chords from D to each
of the other points. Now ABCD is the cyclic quadrilateral, and AC and BD
are its diagonals. The lengths of the sides are a, b, c, and d, and the
lengths of the diagonals are p and q.
Now that I've labeled the figure this way, the circumradius of the
quadrilateral is the
circumradius of the triangle ABC,
R = abp/(4K), where K is the area of triangle ABC. (proof)
R = cdp/(4L), where L is the area of triangle CDA.
Let A be the area of the quadrilateral. A = K+L
R = (abp+cdp)/(4A) = p(ab+cd)/(4A)
Similarly, R = q(bc + ad)/(4A), so
R2 = pq(ab+cd)(bc + ad)/(16A2)
Ptolemy's Theorem says
ac+bd = pq, so R2 = (ac+bd)(ab+cd)(bc + ad)/(16A2)
Brahmagupta's
Theorem says the area, A, of the quadrilateral is sqrt((s-a)(s-b)(s-c)(s-d)),
so
R2 = (ac+bd)(ab+cd)(bc + ad)/(16(s-a)(s-b)(s-c)(s-d)), then by
taking the square root of each side, the result follows.
