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Math Help - Ptolemy's Theorem

  1. #1
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    Ptolemy's Theorem

    In a cycic quadrilateral ABCD, let the sides AB, BC, CD, DA be of lengths a, b, c, d, respectively.

    If the diagonal AC and BD have lengths m and n, respectively, prove that ac + bd = mn
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  2. #2
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    So you need a proof of Ptolemy's Theorem , right ?

    Here's what i proved not long ago :


    There are fours arcs , choose the biggest one , say  arc~AB Let  P be a point on the arc such that  AP = BC note that  APBC is an isosceles trapzeium so we have  AB = CP and  S_{ABC} = S_{ACP} . Then consider the area of the quadrilateral  ABCD we find that it is equal to :

     \frac{1}{2} (AM \cdot BM + BM \cdot CM + CM \cdot DM + DM \cdot AM ) ~\sin(\theta) where  M is the intersection of the diagonals and  \theta is the included angle (acute one ) .

     = \frac{1}{2} \sin(\theta) ( AM+CM)(BM+DM)

     = \frac{mn}{2} \sin(\theta)


    Since  S_{ABC} = S_{ACP} , we have

     S_{ABC} + S_{ACD} = S_{ACP} + S_{ACD}

     S_{ABCD} = S_{ACP} + S_{ACD}

    Also  S_{ACP} + S_{ACD} = S_{APCD} = S_{ADP} + S_{CDP}

    S_{ADP} = \frac{1}{2} AD \cdot AP \sin(\angle DAP)

     = \frac{1}{2} AD \cdot BC \sin(\angle DAP) ~,~ (AP = BC) ~,<br />
 = \frac{1}{2} bd \sin(\angle DAP)


     S_{CDP} = \frac{1}{2} CD \cdot CP \sin(\angle DCP)

     = \frac{1}{2} CD \cdot AB \sin(\angle DAP) ~ ,~AB=CP ~ ,~\angle DCP = 180^o - \angle DAP ~,  = \frac{1}{2} ac \sin(\angle DAP)

    But  \angle DAP = \angle CAD + \angle PAC = \angle CAD + \angle ACB = \angle CAD + \angle ADB = 180^o - \theta


    Therefore,  S_{ADP} + S_{CDP} = \frac{1}{2} (ac+bd) \sin(\theta)


    By equating S_{ABCD} and  S_{ADP} + S_{CDP} we have  mn = ac + bd


    Here is another method ,

    Let  \alpha = \angle ACB ~,~ \beta = \angle ACD ~,~ x = \angle BAC ~,~ y = \angle CAD


     ac+bd = (2R\sin(\alpha))(2R \sin(y)) + (2R\sin(\beta))(2R\sin(x))


     = 4R^2 [ \sin(\alpha) \sin(y) + \sin(\beta)\sin(x)]

     = 2R^2 [ \cos(\alpha-y)- \cos(\alpha+y) + \cos(\beta - x)- \cos(\beta + x ) ]

     = 2R^2 [ (\cos(\alpha-y) + \cos(\beta - x) ) - ( \cos(\alpha+y) + \cos(\beta + x ) )]

     = 4R^2 [ \cos\left( \frac{\alpha-y+\beta - x}{2} \right)\cos\left( \frac{\alpha-y-\beta + x}{2} \right)  - \cos\left( \frac{\alpha+y+\beta + x}{2} \right) \cos\left( \frac{\alpha+y-\beta - x}{2} \right) ]

    Since  \alpha+\beta +x+y = 180^o

     = 4R^2 [\cos(90^o - (\alpha+\beta)) \cos( 90^o - (\alpha + x)) - \cos(90^o) \cos\left( \frac{\alpha+y-\beta - x}{2} \right) ]

     = 4R^2 [ sin(\alpha+\beta) \sin(\alpha + x ) -0 ]

     = (2R sin(\alpha+\beta) ) (2R \sin(180^0 -(\alpha + x )))

     = (2R \sin(\angle BCD))(2R \sin(\angle ABC) ) = mn

     ac+bd= mn
    Last edited by simplependulum; June 18th 2010 at 08:38 PM.
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  3. #3
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    LOL, uhmm, wheres all the numbers gone after this website upgrade?? can you rewrite this again?? sorry! thanks!
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