Does anyone know the proof of Mahavira's formulae for the calculation of the inner diagonals of the cyclic quadrilaterals? Presumably, they must be a continuation of Ptolemy's theorem. They are:

x=Sq.Root and symmetrically for y.

Thanks.

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

I do not understand it unfortunately. What formula are you referring to? What quantity is m.t.cosČ ? Could you please explain?

Have a look at

http://www.cut-the-knot.org/proofs/ptolemy.shtml#Mahavira

It's the same as Pierre's remark but perhas a little more detailed. What I want you to do is to make a diagram where you mark the angle theta and think of what cos(theta) signifies, OK?

Dear Alex,

Sorry, still nothing! I do not understand anything! Are you sure there is not a flaw in the argument? I have a feeling there is. Could you please send me a diagram and an explanation? You may draw it by hand and fax it through at +30-210-7219225 if this is easier for you.

The formula is proven very easily by using the cosine rule.

m^2=b^2+c^2-2.b.c.cosC
=a^2+d^2-2.a.d.cosA=a^2+d^2-2.a.d.cos(180-C)=a^2+d^2+2.a.d.cosC
Multiply the first with ad and the second with bc, add them up and you will arrive at m^2=(ab+cd).(ac+bd)/(ad+bc)

I am really surprised. I do not understand what cosč signifies. There might be an error in the proof.

Have a bash at the probability question I posed. I think I have got an answer, it depends on what one is really asking. I look forward to see the replies.

ab = hR - can you verify that? It says that in a triangle the product of two sides equals the product of the diameter circumscribed circle and the altitude to the third side.

In a quadrilateral, you have two pairs of triangles sharing a base (the third side in 1). Bases are the diagonals of the quadrilateral. Consider one pair at a time, say, ab = ht and cd = gt, where h is the altitude to from the ab vertex to the diagonal m, I belive. g is the altitude from the cd vertex to the sam diagonal.

Each of these triangles contains a piece of the diagonal n which plays the hypotenuse, one in a triangle with side h, the other in a triangle with side g. The triangles are similar. You can write

h = cos(theta)×(one piece) and
g = cos(theta)×(the other piece).

On summing up you get

h + g = cos(theta)·n

Thus adding ab = ht and cd = gt gives you

nt·cos(theta) = ab + cd.

Sorry if I confused m and n.

sorry for my late reply,
the angle theta is the angle between either diagonal and the perpendicular upon the other.
Four of them are marked is the included image.