I found a partial proof in
Howard Eves, Great Moments in Mathematics Before 1650, 10.5 p.108

Let ABCD be a cyclic quadrilateral of diameter t.
Let BD=m and AC=n.
Let theta = the angle between either diagonal and the perpendicular upon the other.

Then, (using triangle's formula ab=2hR applied to DAB and DCB), we get
mt cos theta = ab+cd
nt cos theta = ad+bc

So m/n = (ab+cd)/(ad+bc)

Also mn = ac+bd (Ptolemy relation)

Multiplying those last 2 equations, we get:
m^2=(ab+cd)(ac+bd)/(ad+bc)

Dividing instead, we get:
n^2=(ac+bd)(ad+bc)/(ab+cd)

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I also found some info there:
http://www.pballew.net/cycquad.html
http://www.vias.org/comp_geometry/geom_quad_cyclic.html
http://www.answers.com/topic/mahavira-mathematician