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 Subject: "Mahavira's formula for cyclic quadrilaterals" Previous Topic | Next Topic
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melefthe guest
Apr-11-07, 07:21 PM (EST)

 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.

Subject     Author     Message Date     ID Mahavira's formula for cyclic quadrilaterals melefthe Apr-11-07 TOP RE: Mahavira's formula for cyclic quadrilaterals Pierre Charland May-11-07 1 RE: Mahavira's formula for cyclic quadrilaterals melefthe May-19-07 2 RE: Mahavira's formula for cyclic quadrilaterals alexb May-19-07 3 RE: Mahavira's formula for cyclic quadrilaterals melefthe May-20-07 4 RE: Mahavira's formula for cyclic quadrilaterals alexb May-20-07 5 RE: Mahavira's formula for cyclic quadrilaterals Pierre Charland May-27-07 6

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Pierre Charland
Member since Dec-22-05
May-11-07, 06:40 AM (EST)    1. "RE: Mahavira's formula for cyclic quadrilaterals"
In response to message #0

 I found a partial proof inHoward Eves, Great Moments in Mathematics Before 1650, 10.5 p.108Let 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 getmt cos theta = ab+cdnt cos theta = ad+bcSo 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)===========================I also found some info there:https://www.pballew.net/cycquad.htmlhttps://www.vias.org/comp_geometry/geom_quad_cyclic.htmlhttps://www.answers.com/topic/mahavira-mathematicianAlphaChapMtl melefthe guest
May-19-07, 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.cosČ ? Could you please explain? alexb
Charter Member
2013 posts
May-19-07, 03:56 PM (EST)    3. "RE: Mahavira's formula for cyclic quadrilaterals"
In response to message #2

 Have a look athttps://www.cut-the-knot.org/proofs/ptolemy.shtml#MahaviraIt'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? melefthe guest
May-20-07, 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 +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.cosCMultiply 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. alexb
Charter Member
2013 posts
May-20-07, 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 writeh = cos(theta)×(one piece) andg = cos(theta)×(the other piece).On summing up you geth + g = cos(theta)·nThus adding ab = ht and cd = gt gives yount·cos(theta) = ab + cd.Sorry if I confused m and n. Pierre Charland
Member since Dec-22-05
May-27-07, 06:43 PM (EST)    6. "RE: Mahavira's formula for cyclic quadrilaterals"
In response to message #2

 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.AlphaChapMtl

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