Sunday, July 12, 2015

Geometry Problem 1133: Circle, Chord, Tangent, Circumcircle, Similarity, Equal Product

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

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1133: Circle, Chord, Tangent, Circumcircle, Similarity, Equal Product.

Posted by Antonio Gutierrez at 12:10 PM
Labels: , , , ,

4 comments:

  1. Peter TranJuly 12, 2015 at 8:17 PM

    http://s17.postimg.org/5cijbp9e7/Pro_1133.png

    Draw line BMN// GH ( M on EC and N on GC)
    Since GH tangent to circle EFC so
    ∠ (EDF)= ∠ (GEF)= ∠ (BEH)=∠ (MBE)
    Triangle FED similar to MEB ( case AA)
    So EM/EF= EB/ED ……..(1)
    Triangle AED similar to CEB ( case AA)
    So EC/EA=EB/ED…… (2)
    From (1) and (2) we have EM/EF=EC/EA => MF//CA
    And FA/FB= MN/MB …..(3)
    Since NB//GH => EG/EH=MN/MB…..(4)
    From (3) and (4) we have FA/FB= EG/EH

    ReplyDelete
  2. Stan FulgerAugust 13, 2015 at 1:15 PM

    <BEH=<GEA=<FDE, so <CHE=<CBE-<BEH=<ADE-<FDE=<AFD ( 1 ).
    Let the circle (CGH) intersect CE at P and, easily, triangles GPH and ADB are similar and <GPE=<EHC(=<ADF as per (1)), consequently DF and PE are homologous lines within the 2 similar triangles, thus solving our problem.

    ReplyDelete
  3. Sumith PeirisOctober 24, 2018 at 6:29 AM

    Tr.s CEG & BFD are similar so DF / FB = CE / EG……(1)

    Tr.s AFD & CEH are similar so DF / AF = CE / EH …..(2)

    (1)/(2) gives AF/FB = EG/EH

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Sailendra TOctober 26, 2018 at 1:44 PM

    Connect AD and DB and let m(GEA)=m(EDF)=w, m(CAB)=m(CDB)=x, m(ACD)=m(ABD)=y and m(DCB0=m(DAB)=z
    Simple angle chasing will yield to the following
    Triangles GEC and DFB are similar -----(1)
    Triangles CEH and AFD are similar -----(2)

    From(1) GE.FB=DF.EC ------(3)
    From(2) CE.FD=AF.EH ------(4)
    From(3)&(4) the result follows

    ReplyDelete

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