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Math Help - cyclic quadrilateral question 2

  1. #1
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    cyclic quadrilateral question 2

    ABCD is a cyclic quadrilateral. Diagonals of ABCD (ie.,AC & BD) meet at right angles at P.PL is a perpendicular bisector to chord AB. If PL is extended to meet chord DC at M, show that PM bisects DC.
    Please help me on the steps to solve the above
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  2. #2
    MHF Contributor red_dog's Avatar
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    AL=LB, \ PL\perp AB\Rightarrow\Delta APB isosceles \Rightarrow\widehat{PAB}=\widehat{PBA}

    But \widehat{PAB}=\widehat{PDC}, \ \widehat{PBA}=\widehat{PCD}\Rightarrow\widehat{PDC  }=\widehat{PCD}\Rightarrow\Delta PCD isosceles.

    We have \widehat{ACD}=\widehat{CAB}\Rightarrow AB\parallel DC\Rightarrow PM\perp CD\Rightarrow DM=MC
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  3. #3
    Super Member fardeen_gen's Avatar
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    1)angle APB and DPC are vertically opposite,hence equal
    2)opposite sides are similar in ABCD because the diagonals are perpendicular to each other, hence have equal length
    3)angle PLA and angle PMD are alternate angles, hence equal

    Therefore, by ASA(angle-side-angle) postulate, triangles ABP and DCP are congruent.

    Also,
    PL=PM
    Therefore, if PL bisects AB then, PM bisects DC.
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