Friday, December 27, 2013

Geometry Problem 951: Intersecting Circles, Chord, Diameter, Parallel, Cyclic Quadrilateral, Concyclic Points

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to enlarge the problem 951.

Online Geometry Problem 950: Intersecting Circles, Secant, Circumcenter, Cyclic Quadrilateral, Concyclic Points

Posted by Antonio Gutierrez at 9:01 AM
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3 comments:

  1. Jacob HA (EMK2000)December 27, 2013 at 5:00 PM

    2(180°−∠EAD)=2∠AFE=∠AOE=∠OAC=90°−∠CAD
    =∠QAD=∠AQG=2∠AHG=2∠DAH

    Thus, E,A,H are collinear.
    Similarly, F,A,G are collinear.

    By the above argument,
    ∠EFG=∠EFA=∠AHG=∠EHG

    Hence, F,E,H,G concyclic.

    ReplyDelete
  2. Ivan BazarovJanuary 11, 2014 at 8:26 AM

    <CEA=<DGA=<CAD. <DAH=45-<CAD/2, and <EAD=90+<CAD+(45-<CAD/2), so <EAD+<DAH=180 and EAH are collinear. Same reasoning for FAG to be collinear. Then <EFG=<EHG=45-<CAD/2.

    ReplyDelete
  3. c.t.e.oMarch 19, 2019 at 2:07 PM

    From the give ͡ ABC = ͡ ABD => ˂FEH = ˂FGH

    ReplyDelete

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