Wednesday, February 18, 2015

Geometry Problem 1085: Intersecting Circles, Concyclic Points, Cyclic Quadrilateral

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1085: Intersecting Circles, Concyclic Points, Cyclic Quadrilateral.

3 comments:

  1. ∠GAH
    = ∠CAD
    = ∠ADF − ∠ACE
    = ∠ABF − ∠ABE
    = ∠EBF
    = ∠GBH

    Hence, ABHG are concyclic.

    ***
    Corollary

    Since ∠GHB = ∠DAB = ∠DFB
    Thus, GH // CF.

    ReplyDelete
  2. Denote ∠DAB = ∠DFB = x and ∠ABE = ∠ACE = y
    Clearly ∠G = x - y = ∠H and so A,G,H,B are concyclic

    ReplyDelete
  3. < ABF = <ADF......(1)
    <ABE = <ACF .....,(2)

    (1)-(2) < EBF = < CAD = < GAH

    so ABHG is concylic

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete