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melefthe
guest

Apr1107, 07:21 PM (EST) 

"Mahavira's formula for cyclic quadrilaterals"

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<(ad+bc)/(ab+cd)x(ac+bd)> and symmetrically for y. Thanks. 

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melefthe
guest

May1907, 03:49 PM (EST) 

2. "RE: Mahavira's formula for cyclic quadrilaterals"
In response to message #1

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

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melefthe
guest

May2007, 10:28 AM (EST) 

4. "RE: Mahavira's formula for cyclic quadrilaterals"
In response to message #3

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 +302107219225 if this is easier for you. The formula is proven very easily by using the cosine rule. m^2=b^2+c^22.b.c.cosC =a^2+d^22.a.d.cosA=a^2+d^22.a.d.cos(180C)=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 cosè 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. 

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alexb
Charter Member
2013 posts 
May2007, 04:59 PM (EST) 

5. "RE: Mahavira's formula for cyclic quadrilaterals"
In response to message #4

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. 

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