Friday, December 25, 2009

Problem 408: Cyclic quadrilateral, Perpendicular, Parallelogram, Congruence

Proposed Problem
Click the figure below to see the complete problem 408 about Cyclic quadrilateral, Perpendicular, Parallelogram, Congruence.

Problem 408: Cyclic quadrilateral, Perpendicular, Parallelogram, Congruence.
See also:
Complete Problem 408
Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 1:52 PM
Labels: , , , ,

3 comments:

  1. APOSTOLIS MANOLOUDISJune 17, 2016 at 7:36 AM

    Problem 408
    The points M, N are orthocenter of triangles ACD, ABD respectively. Bring OP perpendicular AD.Then BN=//2OP and CM=//2OP. So BN=//CM, therefore BCMN is parallelogram.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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  2. c.t.e.oFebruary 6, 2019 at 2:12 AM

    ang CMN + BNM = 180 => BN // CM
    Extend BN to N', CM to M' ( M', N' on circle ) => MNN'M' isosceles trapezoid
    => MN = M'N' But M'N' = BC => BC = MN

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  3. GregMay 24, 2022 at 3:57 AM

    Here is a pure synthetic solution

    See diagram here

    We see 3 pairs of similar triangles : ΔADE - ΔCME, ΔDAH - ΔBNH, ΔBDH - ΔCAE
    Therefore CM/CE = AD/AE, DA/DH = BN/BH and CE/AE = BH/DH
    Multiplying the 3 equations => CM.DA.CE/CE.DH.AE = AD.BN.BH/AE.BH.DH
    Simplifying => CM = BN
    So CM = BN and since CM and BN are both perpendicular to AD, they are also // therefore BCMN is a parallelogram
    QED

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