Friday, April 27, 2012

Problem 743: Trapezoid, Triangle, Perpendicular, Parallel, Angle

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

Click the figure below to see the problem 743 details.

Online Geometry Problem 743: Trapezoid, Triangle, Perpendicular, Parallel, Angle.

Posted by Antonio Gutierrez at 6:33 PM
Labels: , , , ,

6 comments:

  1. W FungApril 27, 2012 at 7:51 PM

    A,B,F,C,E Concyclic
    E,F,D,G Concyclic
    Hence AngleEBC= AngleEFC = AngleEGD =28
    x = 180 -90 - 28 = 62

    ReplyDelete
  2. Peter TranApril 27, 2012 at 10:18 PM

    http://img259.imageshack.us/img259/43/problem743.png

    Let ∠CDE= α
    Note that ∠AGE= ∠CDE= α
    We have CE/ED=AE/GE= tan( α)
    Triangle BAE similar to triangle DEG… ( Case SAS)
    So ∠EDG= ∠ABE= 90-28= 62

    ReplyDelete
  3. UnknownApril 30, 2012 at 7:22 AM

    The solution is uploaded to the foowing link:

    https://docs.google.com/open?id=0B6XXCq92fLJJUk5oRWdKSTJJbEU

    ReplyDelete
  4. AnonymousAugust 16, 2012 at 6:34 PM

    Solution by Michael Tsourakakis from Greece

    AC and DG meet at K. the triangle AGD, GE is height , DF is height .So C is orthocenter of triangle AGD so AK is height. SO the quadrangle ECKD is inscribable .so angles KDE and ACE are equal. but angle ACE is equal to angle EBA=62.So x=62

    ReplyDelete
  5. Μιχάλης ΝάννοςOctober 10, 2013 at 6:44 AM

    http://www.mathematica.gr/forum/viewtopic.php?f=22&t=27475&p=134337

    ReplyDelete
  6. Sumith PeirisDecember 5, 2016 at 3:15 AM

    ABCE is a rectangle, so < AEB = 28 = < AFB since ABFC is cyclic.

    So ABFE is cyclic and so < ABE = 62 = < AFE = x since DEFG is cyclic.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

Subscribe to: Post Comments (Atom)