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
See also:
Complete Problem 484

Level: High School, SAT Prep, College geometry

4 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

    ReplyDelete
  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

    ReplyDelete
  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

    ReplyDelete
  4. From similarity of ABF and CEF we get

    BF/AB=EF/EC ........(1.)

    From similarity of EFG and AED we get

    EF/FG=AE/AD ........(2.)

    From similarity of AEC and DCG we get

    GC/CD=EC/AE ........(3.)

    Multiplying equation 1 and 2 with AB=AD gives :

    BF/FG=AE/EC

    And from equation 3, with CD =BC we get:

    BF/FG=CD/GC
    BF/FG=BC/GC

    ReplyDelete