Thursday, June 21, 2012

Problem 770: Parallelogram, Angle Bisector, Triangle, Circumcenter, Circle, Circumcircle, Angle Measurement

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 770.

Online Geometry Problem 770: Parallelogram, Angle Bisector, Triangle, Circumcenter, Circle, Circumcircle, Angle Measurement.

Posted by Antonio Gutierrez at 9:32 AM
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4 comments:

  1. Peter TranJune 21, 2012 at 3:47 PM

    http://img542.imageshack.us/img542/8173/problem770.png

    Draw lines per attached sketch
    Note that ADE, CEF and ABF are isosceles triangles.
    And BO⊥CE , ∠ECO=∠CEO= 90- α
    AD=BC=DE
    ∆ OCB congruence to ∆ODE … ( Case SAS)
    So OB=OD and triangle BOD is isosceles
    X= (180-72)/2=54

    ReplyDelete
  2. PravinJune 21, 2012 at 7:23 PM

    To Peter Tran:

    Please explain why
    BO ⊥ CE
    Where it is used?

    Pravin

    ReplyDelete
    Replies
    1. Peter TranJune 21, 2012 at 9:45 PM

      http://img62.imageshack.us/img62/9149/problem7701.png
      Pravin
      Thank you for your comment. The correct statement should be CO ⊥EF instead of BO⊥ CE . The solution and the result is valid with this correction. See attached sketch above for details

      Delete
  3. Μιχάλης ΝάννοςOctober 10, 2013 at 6:49 AM

    http://www.mathematica.gr/forum/viewtopic.php?f=20&t=29034&p=139447

    ReplyDelete

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