Saturday, September 10, 2016

Geometry Problem 1258: Cyclic Quadrilateral, Concyclic Points, Measurement

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1258.

Geometry Problem 1258: Cyclic Quadrilateral, Concyclic Points, Measurement.

Posted by Antonio Gutierrez at 4:38 PM
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3 comments:

  1. Sumith PeirisSeptember 10, 2016 at 5:26 PM

    Let GE = p, EF = q and FG = r
    Let AG = AF = BD = y

    Tr.s GEF and BCD are similar so,

    p/4 = q/x = r/y .........(1)

    Applying Ptolemy to cyclic quad AGEF,
    py + qy = 5r and hence from (1).

    4r + xr = 5r which yields
    x = 1

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  2. Peter TranSeptember 10, 2016 at 9:16 PM

    https://goo.gl/photos/ffCv5yAwZMzaTRv7A

    Connect FG, EF
    Extend BC to H such that CH=CD=x ( see sketch)
    We have AG=AF => ∠ (AGF)= ∠ (AFG)= u
    Let ∠ (GAE)= w => ∠ (BDC)= ∠ (BAC)= ∠ (GFE)= w
    Let ∠ (CAD)= ∠ (DBC)=v
    Since ∠ (DCH)= ∠ (BAD) => Isoceles triangles GAF simillar to triangle HCD ( case AAA)=> ∠ (CDH)= u
    Triangle AFE congruent to BDH ( case ASA)
    So AE= BH=> 5=4+x => x=1

    ReplyDelete
  3. APOSTOLIS MANOLOUDISSeptember 11, 2016 at 10:59 PM

    Problem 1258
    Let the point K on the side AE (no extension) such that AK=BC=4.So KE=1,triangle AFK=triangle BDK(AF=BD,AK=BC,<FAK=<DBC).So CD=KF then <AKF=<BCD=180-<FAG=180-<FKE. So <FAG=<FKE .But <FGA=<GFA=<FEA=<FEK,so <KFE=<KEF(<KFE=180-<FKE-<FEK=
    180-<FAG-<FGA=<AFG).So KF=KE=CD=1.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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