Sunday, December 11, 2011

Problem 702: Triangle, Circumcircle, Parallel, Concyclic Points

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 702.

Online Problem 702: Triangle, Circumcircle, Parallel, Concyclic Points.

Posted by Antonio Gutierrez at 4:00 PM
Labels: , , ,

5 comments:

  1. AjitDecember 11, 2011 at 7:55 PM

    Join G to F. /_EDF=/_EGF & /_DGE=/_DFE so /_DGF=/_DGE+/_EGF=/_EDF+/_DFE=/_DEB=/_ACB.
    Hence etc.

    ReplyDelete
  2. Peter TranDecember 11, 2011 at 10:24 PM

    http://img809.imageshack.us/img809/2501/problem702.png
    Connect GF ( see picture)
    Since quadrilateral GDEF cyclic so m(GDE)=m(GFC)
    Note that m(GDE) supplement to angle(GAC) ( DE//AC)
    So angle(GFC) supplement to angle (GAC)
    And A,G,E,F concyclic
    Peter Tran

    ReplyDelete
  3. AnonymousDecember 11, 2011 at 10:32 PM

    DE//AC ---> angle (BED)=(BCA)
    cyclic quadrilateral DEFG ---> (BED)=(BGF)
    Therefore: angle(BGF)=(BCA).
    Then the quadrilateral AGFC is cyclic.

    ReplyDelete
  4. PravinDecember 11, 2011 at 10:46 PM

    Join GF
    Angle DGF = angle BED = angle BCA
    So A,G,F,C are concyclic

    ReplyDelete
  5. PravinDecember 12, 2011 at 2:23 AM

    Triangle BGF ~ Triangle BED ~ Triangle BCA
    BG:BC = BF:BA
    BG.BA = BF.BC
    A, G, F, C are concyclic

    ReplyDelete

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