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

4 comments:

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

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

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

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

    ReplyDelete
  4. 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