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Math Help - Point P inside parallelogram to give certain properties

  1. #1
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    Point P inside parallelogram to give certain properties

    Let a point
    P inside a parallelogram ABCD be given such that <APB +<CPD = 180.
    Prove that

    AB
    AD = BP DP + AP CP

    (here AB, AD etc. refer to the lengths of the corresponding segments).

    I think that P could be in the very centre, but am unsure how to go about the question, and initially unsure about a possible position for 'p'.
    Help would be great
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  2. #2
    Junior Member BERMES39's Avatar
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    Re: Point P inside parallelogram to give certain properties

    <APB = <CPD so parlallelogram ABCD is a square.
    AB*AD = area of the square
    BP*PD = AP*CD = area of either small square PBFC or AEBP
    Attached Thumbnails Attached Thumbnails Point P inside parallelogram to give certain properties-scan0005.jpg  
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  3. #3
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    Re: Point P inside parallelogram to give certain properties

    So do i just use pythagoras' theorem and prove that square AEBP and PBFC are both say 2x^2, so 4x^2 = ABCD which is 4x^2?
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    Re: Point P inside parallelogram to give certain properties

    Quote Originally Posted by BERMES39 View Post
    <APB = <CPD so parlallelogram ABCD is a square.
    AB*AD = area of the square
    BP*PD = AP*CD = area of either small square PBFC or AEBP

    What about the cases in which <APB  \neq <CPD ?
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  5. #5
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    Re: Point P inside parallelogram to give certain properties

    I dont know, wouldn't that mean that <APB could be anything so long as its not degenerate? And same with the other? This prospect has confused me, a lot.
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  6. #6
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    Re: Point P inside parallelogram to give certain properties

    Quote Originally Posted by pikachu26134 View Post
    Let a point P inside a parallelogram ABCD be given such that <APB +<CPD = 180. Prove that

    AB.AD = BP.DP + AP.CP

    (here AB, AD etc. refer to the lengths of the corresponding segments).

    I think that P could be in the very centre, but am unsure how to go about the question, and initially unsure about a possible position for 'p'.
    Help would be great
    With the parallelogram ABCD and the point P as in the diagram below, draw an identical parallelogram on top of the given one, with a point P' in the same relative position as P. Then PP'= AD. The angles CPD and AP'B are equal. Thus the angles APB and AP'B are supplementary. Hence PAP'B is a cyclic quadrilateral. The result then follows from Ptolemy's theorem.
    Attached Thumbnails Attached Thumbnails Point P inside parallelogram to give certain properties-ptolemy.png  
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  7. #7
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    Re: Point P inside parallelogram to give certain properties

    Okay, ive got the quadrilateral, PP'xAB = PA x P'B + AP' x PB using ptolemy's theorem. What do i do from there?
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  8. #8
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    Re: Point P inside parallelogram to give certain properties

    Quote Originally Posted by pikachu26134 View Post
    Okay, ive got the quadrilateral, PP'xAB = PA x P'B + AP' x PB using ptolemy's theorem. What do i do from there?
    Well I already mentioned that PP' = AD (that's because DAP'P is a parallelogram). Also, P'B = PC and P'A = PD (that's because the upper parallelogram is an exact copy of the lower one).
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