Tuesday, November 24, 2009

Problem 394: Square, 90 Degree Arc, Diagonal, Congruence

Proposed Problem
Click the figure below to see the complete problem 394 about Square, 90 Degree Arc, Diagonal, Congruence.

Problem 394: Square, 90 Degree Arc, Diagonal, Congruence.
See also:
Complete Problem 394

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 8:29 AM
Labels: , , , , ,

11 comments:

  1. AnonymousNovember 24, 2009 at 1:37 PM

    http://geometri-problemleri.blogspot.com/2009/11/problem-51-ve-cozumu.html

    ReplyDelete
  2. AjitNovember 26, 2009 at 7:57 AM

    EF = a-a/V2 and FC=V2a-a where V=square root. So Another way to prove this w/o any construction would be: a=square side. EF/FC=a(1-1/V2)/a(V2-1)=1/V2 and ED/DC=(a/V2)/a =1/V2. Thus,ED/DC=EF/FC or DG bisects angle BDC. Therefore. BG/GC=DB/DC= aV2/a=V2/1 or BG/(BG+GC)=V2*/(V2+1)or rationalizing, BG =V2*a(V2-1) = 2a-aV2.
    Now, EF=a-a/V2 or 2EF=2a - aV2. Hence, BG=2*EF.
    Ajit

    ReplyDelete
  3. vijay9290009015December 4, 2009 at 1:37 AM

    let us find the angles of angleBDG,BGD,DFE,EFD,DGC,GDC. then we can observe that DG is the angular bisector of angle BDC,and tr DEF, tr DGC are simalar triangles using these two concepts we can prove the required

    ReplyDelete
  4. c.t.e.oDecember 8, 2009 at 12:48 PM

    suggest to others

    (i did not understand anything from third comment)

    a easy way: use EF as middle line for any triangle

    ReplyDelete
  5. Peter TranApril 20, 2015 at 6:18 PM

    From E draw EH// BC ( H on GH)
    Note that H is the midpoint of GD
    And triangle EFH is isoceles ( EFH=FDA and FHE=FDA)
    So EF=EH but EH=1/2. BG
    So EF= ½. BG

    ReplyDelete
    Replies
    1. Rogerio de SouzaDecember 6, 2022 at 7:29 PM

      Excellent approach to the solution. Mine was rather cumbersome trying to express EF and BG in terms of square side.

      Delete
    2. Peter TranDecember 7, 2022 at 9:35 PM

      Thank You

      Delete
  6. Sumith PeirisAugust 26, 2015 at 3:28 AM

    Proved this by writing EF and BG in terms of a the side of the square and noting that Tr. CGF is isoceles but Peter's proof is far more easy and elegant

    ReplyDelete
  7. APOSTOLIS MANOLOUDISAugust 15, 2016 at 5:07 AM

    Problem 394
    Sulotion 1
    Is <FDC=<FBC=<FAD/2=22.5=<FDB.If DF intersect AB at point P then <BPG=<FDC=22.5=<BDP so BP=BD=2BE. But triangle BPG and triangle EBF are
    similar.Then BG/EF=BP/BE=2.Therefore PB=2BE.
    Sulotion 2
    Is <CGF=67.5=<AFD=<GFC so CG=CF .In triangle EDC DF is bisector so DE/DC=EF/FC (1)
    and in triangle BDC DG is bisector so DC/DB=CG/BG (2). Μultiply by members of (1),(2)
    σο ½=DE/DB=EF/GB or BG=2EF.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  8. Sumith PeirisDecember 20, 2018 at 9:28 AM

    Since GD bisects < BDC,
    EP/PC = 1/sqrt2 and BG/GC = sqrt2

    Dividing and since PC = CG, the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  9. MarcoSeptember 10, 2023 at 2:31 AM

    AE=(sqrt2)*a/2
    AF=AD=a (radii)
    EF=AF-AE
    =a-(sqrt2)*a/2
    =(2-sqrt2)*a/2

    In triangle EFD, <FDE=90
    tan<EDF=EF/ED
    =(2-sqrt2)/sqrt2
    =sqrt2-1
    <EDF=22.5
    <GDC=45-22.5
    =22.5

    In triangle GDC
    tan<GDC=GC/CD
    GC=(sqrt2-1)*a

    BG=BC-CG
    =a-(sqrt2-1)*a
    =(2-sqrt2)*a
    =2EF

    ReplyDelete

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