Monday, August 9, 2010

Problem 496: Triangle, Two Squares, 90 Degrees, Concurrency

Geometry Problem
Click the figure below to see the complete problem 496 about Triangle, Two Squares, 90 Degrees, Concurrency, Angle Bisectors.

Problem 496. Triangle, Two Squares, 90 Degrees, Concurrency.


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Complete Problem 496

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 9:03 AM
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4 comments:

  1. c .t . e. oAugust 9, 2010 at 9:33 AM

    1) a) ▲DBC = ▲ABG ( side, 90+B,side) => DC = AG
    b) DPA = 45 - CDB, DAP 45 + BAC => DPA = 90°
    2) EPA = EDA = 45, (EDPA cyclic)
    FPC = PGC = 45, (PGFC cyclic)
    APC = 90
    3) see 2)

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  2. c .t . e. oAugust 9, 2010 at 10:00 AM

    a correction
    second row of 2) have to be FPC = FBC

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  3. Peter TranAugust 9, 2010 at 10:02 AM

    Note that triangles ABG congruence with Tri. DBC ( case SAS) so DC=AG
    Tri. DBC is the image of tri. ABG in the rotational transformation with center at B and rotational angle =90 so DC perpendicular to AG
    Since angle GPC=90 so quadrilateral FGFC is cyclic with diameter GC and arc GF=arc FC=90
    Angle GPF=angle FPC = 45 ( face 90 degrees arc)
    Similarly quadrilateral EDPA is cyclic and angle DPE=angle EPA=45
    E, P and F are collinear ( 45+90+45=180) and EF is angle bisector of angle GPC

    Peter Tran

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  4. Sumith PeirisNovember 18, 2017 at 2:36 AM

    Tr. DBC is congruent to Tr. ABG, SAS
    Hence CD = AG

    Also because of the congruency,< BDP = < BAP
    Hence ADBP is concyclic and so < BPD = < BPG = 45
    Also since ADBP is concyclic, < APD = 90 so APDE is concyclic
    Hence <APE = < DPE = 45
    Similarly <FPG = FPC = 45
    It follows that EPC is a straight line and so AG,CD,EF are concurrent

    Further EP or EF bisects < GPC

    Sumith Peiris
    Moratuwa
    Sri Lanka

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