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Thread: 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 $\displaystyle arc~AB$ Let $\displaystyle P $ be a point on the arc such that $\displaystyle AP = BC $ note that $\displaystyle APBC$ is an isosceles trapzeium so we have $\displaystyle AB = CP $ and $\displaystyle S_{ABC} = S_{ACP}$ . Then consider the area of the quadrilateral $\displaystyle ABCD $ we find that it is equal to :

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

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

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


    Since $\displaystyle S_{ABC} = S_{ACP}$ , we have

    $\displaystyle S_{ABC} + S_{ACD} = S_{ACP} + S_{ACD} $

    $\displaystyle S_{ABCD} = S_{ACP} + S_{ACD} $

    Also $\displaystyle S_{ACP} + S_{ACD} = S_{APCD} = S_{ADP} + S_{CDP}$

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

    $\displaystyle = \frac{1}{2} AD \cdot BC \sin(\angle DAP) ~,~ (AP = BC) ~,
    $ $\displaystyle = \frac{1}{2} bd \sin(\angle DAP) $


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

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

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


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


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


    Here is another method ,

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


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


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

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

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

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

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

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

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

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

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

    $\displaystyle ac+bd= mn $
    Last edited by simplependulum; Jun 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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