Wednesday, July 28, 2010

Problem 484: Square, Angle, 90 degrees, Triangle, Measurement, Proportion

Geometry Problem
Click the figure below to see the complete problem 484 about Square, Angle, 90 degrees, Triangle, Measurement, Proportion.

Problem 484 Square, Angle, 90 degrees, Triangle, Measurement, Proportion
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Complete Problem 484

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 2:56 PM
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3 comments:

  1. Denote (XYZ) =angle XYZ
    We have ( ABC)=(AEC)=(ADC)=90
    So polygon ABECD is cyclic with AC as a diameter of circumcircle of the polygon
    In this circumcircle , arc BA=arc AD=arc DC= ¼ of full circle = 90
    (BEA)=(AFD)=(DEC)= 45 ( These angles face 90 degrees arcs )
    Consider triangle BEG and angle BEG, EF is internal angle bisector of ( BEG)
    So FB/FG=EB/EG ( property of internal angle bisector)

    Since EC perpen. To EF , EC will be external angle bisector of (BEG)
    And CB/CG= EB/EG ( property of external angle bisector)

    So FB/FG= CB/CG

    Peter Tran

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  2. Problem 484
    Is ABECD cyclic (<AEC=90=<ABC).Then <BEA=45=<AED=<DEC.So EG is internal bisector and the EB is external bisector of triangle FEC.Then FG/GC=EF/EC and EF/EC=BF/BC.
    So FG/GC=BF/BC or BF/FG=BC/CG.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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  3. Triangles BEG and CDG are similar so

    BE/EG = CD/CG.....(1)
    EF bisects < BEG hence

    BE/EG = BF/FG...(2)

    From (1) and (2) BC/CG = BF/FG

    Sumith Peiris
    Moratuwa
    Sri Lanka

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