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.

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

Posted by Antonio Gutierrez at 7:12 AM
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4 comments:

  1. Jacob HA (EMK2000)February 19, 2015 at 9:06 AM

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

    Hence, ABHG are concyclic.

    ***
    Corollary

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

    ReplyDelete
  2. PravinFebruary 19, 2015 at 7:50 PM

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

    ReplyDelete
  3. Sumith PeirisNovember 24, 2015 at 1:44 AM

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

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

    so ABHG is concylic

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Sumith PeirisAugust 15, 2017 at 3:01 AM

    It also follows that CF and GH are parallel

    ReplyDelete

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