Friday, September 30, 2016

Geometry Problem 1268:Triangle, Exterior Angle Bisector, Circumcircle, Circle, Chord, Parallel, Concyclic Points, Cyclic Quadrilateral

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

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click on the figure below to view more details of problem 1268.

Geometry Problem 1268: Triangle, Exterior Angle Bisector, Circumcircle, Circle, Chord, Parallel, Concyclic Points, Cyclic Quadrilateral
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2 comments:

  1. AnonymousSeptember 30, 2016 at 1:40 PM

    Let the angles of the triangle ABC be A,B and C respectively.
    Let Angle BFD = @ => BFG = 180-@ -----(1)

    Since ED is the ext.Angle bisector of B ,Angle CBD = 90-B/2
    => Angle BDF = 90+B/2-@

    Angle ABE = 90-B/2 = AGE = ACE => ECB = C+B/2-90 = XDB = EGB = XGB
    Hence BPGD is concyclic

    Since BXG and BDG are supplimentary, Angle EXB = 90+B/2-@ = HXG
    Hence in triangle HXG, Angle XHG = BHG = @ ---------(2)

    From (1) & (2) BHGF are concyclic --------(3)

    Since Angle ACB = C = AGB = HGB = C => HFB = C (From (3) ) Hence HF // AC // AD

    ReplyDelete
  2. Peter TranSeptember 30, 2016 at 5:48 PM

    https://goo.gl/photos/8vPumpKGSCYHa6DE9

    Connect EC and BG
    Let u= ∠CBD=∠EBA=∠ACE=∠AGE
    And v= ∠ECB= ∠EGB ( see sketch)
    In triangle BDC , ∠BDC=∠BCA-∠CBD= v
    Since ∠BDX=∠XGB= v => quadrilateral BDGX is cyclic
    And ∠ XDG=∠XBG= w
    In triangle BFD, ∠BFD= 180-u-v-w
    In triangle HGB , ∠BHG= 180-u-v-w
    So ∠BHG=∠ BFD and quadrilateral HGFB is cyclic
    And ∠HFG=∠XBG=w=∠CDF => HF//AD

    ReplyDelete

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