Wednesday, August 5, 2009

Problem 335. Cyclic Quadrilateral, Perpendicular to sides, Concyclic points

Proposed Problem
Click the figure below to see the complete problem 335 about Cyclic Quadrilateral, Perpendicular to sides, Concyclic points.

 Problem 335. Cyclic Quadrilateral, Perpendicular to sides, Concyclic points.
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Complete Problem 335
Level: High School, SAT Prep, College geometry

2 comments:

  1. Many triangles are similar. Especially
    PAE ~ PCG, PBF ~ PDH.
    Corresponding sides are in proportion,
    PA:PC = AE:CG =PE:PG,
    PB:PD = BF:DH = PF:PH.
    Product of extremes and the product of means are equal,
    PA CG = PC AE, PB DH = PD BF
    PA PG = PC PE, PB PH = PD PF.
    Multiplying the first two equations and the last two,
    PA CG PB DH = PC AD PD BF
    PA PG PB PH = PC PE PD PF.
    From the power theorem on circles, PA PB = PC PD, so by canceling,
    CG DH = AD BF
    PG PH = PE PF.
    By the converse of power theorem, four points E, F, G, and H are cyclic.

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  2. Problem 335
    Si AB CD intersect to K.Triangles KAE,KDH are similar so AE/DH= KA/KD,triangles KAD,KCB are similar so KA/KD=KC/KB and triangles KCF, KBF are similar so KC/KB=CG/BF then AE/DH=CG/BF.
    Then AE.BF=DH.CG.Is AHDE , CGBF concyclic so <FGB=FCB=<DAH=<DEH. Therefore E,H,E,F are concyclic.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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