Monday, June 10, 2013

Problem 887: Triangle, Altitude, Angle Bisector, Perpendicular, Midpoint, Concyclic Points. GeoGebra

Geometry Problem. GeoGebra, HTML5 Animation for iPad and Nexus.
Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the dynamic geometry demostration of problem 887.

Online Geometry Problem 887: Triangle, Altitude, Angle Bisector, Perpendicular, Midpoint, Concyclic Points. GeoGebra

Posted by Antonio Gutierrez at 6:59 PM
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3 comments:

  1. Peter TranJune 10, 2013 at 10:37 PM

    http://img853.imageshack.us/img853/4498/problem887.png
    Let CE intersect AB at F
    Triangle BFC is isosceles => FE=EC
    We have CE/EC= GC/GA => GE//AB
    Since GE//AB => ∠(DEG)= ∠(ABE)
    In cyclic quadrilateral ABDH we have ∠(DHG)= ∠(ABE)
    So ∠(DEG)=∠(DHG) => H,D,G,E are concyclic

    ReplyDelete
  2. AnonymousJune 13, 2013 at 11:42 PM

    thanks peter tran

    ReplyDelete
  3. Sumith PeirisSeptember 19, 2016 at 5:00 AM

    Key is to prove that DG//BC

    Then < EDG = < EBC = < EHC = B/2 and the result follows

    ReplyDelete

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