Thursday, December 19, 2013

Geometry Problem 947: Triangle, Circumcircle, Angle Bisector, Chord, Arc, Parallel

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to enlarge the problem 947.

Online Geometry Problem 947: Triangle, Circumcircle, Angle Bisector, Chord, Arc, Parallel

Posted by Antonio Gutierrez at 4:34 PM
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4 comments:

  1. nakaDecember 19, 2013 at 5:33 PM

    Angle ABE = angle AFE = angle GBD
    So BDFG concylic
    Angle FAC = angle FBG = angle FDG
    QED

    ReplyDelete
  2. Jacob HA (EMK2000)December 19, 2013 at 5:50 PM

    Let AF and BC intersect at P.

    Since ABFC concyclic, (AD+DP)×FP = (CG+GP)×BP ... (1)
    Since BDGF concyclic, DP×FP = GP×BP ... (2)
    (1) - (2): AD×FP = CG×BP ... (3)
    (2) / (3): DP/AD = GP/CG

    Hence, DG//AC.

    ReplyDelete
  3. ion raduDecember 24, 2013 at 9:14 AM

    BDGFconcyclic=>< BFD= DG||AC (formeazacu secanta BC unghiuri corespondente congruente)

    ReplyDelete
    Replies
    1. ion raduDecember 25, 2013 at 2:45 AM

      BFD=BGD,BFA=BCA, BGD=BCA,=> DG||AC

      Delete

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