Tuesday, July 9, 2013

Problem 899: Intersecting Circles, Common External Tangent, Secant, Concyclic Points, Cyclic Quadrilateral

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 899.

Online Geometry Problem 899: Intersecting Circles, Common External Tangent, Secant, Concyclic Points, Cyclic Quadrilateral. GeoGebra, HTML5 Animation for iPad and more tablets

Posted by Antonio Gutierrez at 6:56 AM
Labels: , , , , ,

2 comments:

  1. nakaJuly 9, 2013 at 8:19 AM

    ∠FBD
    = ∠FBA + ∠ABD
    = (180-∠FCA) + ∠AED
    = 180-(∠CGD + ∠AED) + ∠AED
    = 180 - ∠CGD
    = 180 - ∠FGD
    So F,B,G,D are concylic,
    symmetrically, B,G,C,E are concylic.

    ReplyDelete
  2. UnknownJuly 13, 2013 at 1:00 AM

    Angle BED=angle FAB=angle FCB=180-angle GCB. Hence GCBE is cyclic. Similarly, angle BFG=angle BAE=angle BDE=180-angle BDG. Hence, BFGD is cyclic.

    ReplyDelete

Subscribe to: Post Comments (Atom)