Sunday, February 4, 2018

Geometry Problem 1354: Circle, Tangent Lines, Secant, Chord, Center, Parallel Lines

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

Details: Click on the figure below.

Geometry Problem 1354: Circle, Tangent Lines, Secant, Chord, Center, Parallel Lines.

5 comments:

  1. Is AE*AF=AC^2=AD*AO=>OFED=cyclic=> DF=DG.But triangleOFD=triangleOGD => <ODG=<ODF.So OD=perpendicular in FG.Therefore FG//BC.

    ReplyDelete
    Replies
    1. It is not clear to me how a cyclic quadrilateral OFED will give DF=DG in your solution.
      Please explain

      Peter

      Delete
    2. is DGF=DFG=>DG=DF (not from quadrlateral OFED)

      Delete
  2. https://photos.app.goo.gl/bNpiPWF2vo2hD4S63
    connect OE, OF, DE and DF
    since AC and AB tangent to circle O
    so OC^2=OE^2=OF^2=OD.OA
    or OE/OA=OD/OE and OF/OA=OD/OF
    triangle ODE similar to OEA ( case SAS)
    and EA/DE=OE/OD => EA=OE/OD x DE… (1)
    triangle ODF similar to OFA… ( case SAS)
    and FA/DF=OF/OD => FA= OF/OD x DF… (2)
    Divide (1) to (2) and note that OF=OE we get
    EA/FA=DE/DF => ODA is the internal bisector of angle GDF
    So DO ⊥ GF => GF//BC

    ReplyDelete
  3. Since OC2 = OD.OA = OE2,
    < OED = < OAE
    Similarly < OFD = OAE,
    So OFED is concyclic.
     
    Hence < GDO = < OFE = < OEF = < ODF = λ say
    Therefore < FOE = 180 - 2λ and < FGE = 90 – λ = < GDB

    So FG // BC
     
    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete