Monday, August 1, 2016

Geometry Problem 1241: Quadrilateral, Four Exterior Angle Bisectors, Cyclic Quadrilateral, Circle

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

Click the figure below to view more details of problem 1241.

Geometry Problem 1241: Quadrilateral, Four Exterior Angle Bisectors, Cyclic Quadrilateral, Circle. Mobile Apps

Posted by Antonio Gutierrez at 5:43 PM
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5 comments:

  1. Sumith PeirisAugust 1, 2016 at 9:57 PM

    < A1AB = 90 - A/2 and <A1BA = 90 - B/2
    Hence <A1 = A/2+B/2….(1)

    Similarly we can show that
    <C = C/2+D/2….(2)

    So <A1+<C1 = (A/2+B/2)+(C/2+D/2) = (A+B+C+D)/2 = 360/2 =180

    Hence A1B1C1D1 is a cyclic quadrilateral.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
    Replies
    1. Sumith PeirisAugust 2, 2016 at 7:43 AM

      It can be proved that the diagonals of this cyclic quadrilateral are perpendicular to each other.

      Delete
    2. Sumith PeirisAugust 3, 2016 at 5:03 PM

      Correction - this will be only if ABCD is cyclic

      Delete
  2. Jacob HA (EMK2000)August 1, 2016 at 10:59 PM

    Let ext∠A=2a, similarly for B, C, D.

    a+b+∠A₁=180°
    c+d+∠C₁=180°
    a+b+c+d+∠A₁+∠C₁=360°

    But a+b+c+d=180°
    Thus ∠A₁+∠C₁=180°
    Hence A₁B₁C₁D₁ is concyclic.

    ReplyDelete
  3. APOSTOLIS MANOLOUDISAugust 2, 2016 at 12:46 AM

    Problem 1241
    Set <A1AB=<D1AD=x,<A1BA=<B1BC=y,<B1CB=<DCC1=z and <CDC1=<ADD1=φ.Then
    2x+2y+2z+2φ=180.But <AA1B+<DC1C=180-(x+y)+180-(z+φ)=180.Therefore A1, B1, C1
    and D1 are concyclic.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete

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