Tuesday, July 9, 2013

Problem 900: Intersecting Circles, Common External Tangent, Secant, Angle, Congruence

Geometry Problem. GeoGebra, HTML5 Animation for iPad and Nexus.
Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the dynamic geometry demonstration of problem 900.

Online Geometry Problem 900: Intersecting Circles, Common External Tangent, Secant, Angle, Congruence. GeoGebra, HTML5 Animation for iPad and more tablets

Posted by Antonio Gutierrez at 12:17 PM
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4 comments:

  1. nakaJuly 9, 2013 at 3:58 PM

    By problem 899, F,B,D,G concylic,
    ∠GBD=∠GFD=∠CFA=∠CBA

    ReplyDelete
  2. Peter TranJuly 9, 2013 at 4:20 PM

    Quadri. FCAB is cyclic => ∠ (CBA)= ∠(CFA)
    Quadr. FGDB is cyclic per result of problem 899=> ∠(DBG)= ∠(GFD)= ∠(CFA)
    So ∠(ABC)= ∠(DBG)

    ReplyDelete
  3. Jacob HA (EMK2000)July 9, 2013 at 4:44 PM

    Since ABFC concyclic, and also BDGF (Problem 899).

    Thus
    ∠ABC = ∠AFC = ∠DFG = ∠DBG.

    ReplyDelete
  4. APOSTOLIS MANOLOUDISDecember 2, 2016 at 1:24 AM

    Problem 900
    In the convex quadrilateral GCADFE by Miquel’s theorem the point B is point Miquel.So the F,G,D and B are concyclic.So <CBA=<CFA=<GFD=<GBD.
    MANOLOUDIS APOSTOLIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS-GREECE

    ReplyDelete

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