Thursday, June 16, 2011

Problem 623: Triangle, Incenter, Excenter, Incircle, Excircle, Concyclic Points

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

Click the figure below to see the complete problem 623.

Online Geometry Problem 623: Triangle, Incenter, Excenter, Incircle, Excircle, Concyclic Points.

Posted by Antonio Gutierrez at 1:02 PM

3 comments:

  1. PravinJune 16, 2011 at 8:00 PM

    ∠IBE = B/2 + (180°-B)/2 = 90°
    Similarly ∠ICE = 90°
    So ∠IBE + ∠ICE = 180° and I,B,E,C are concyclic

    ReplyDelete
  2. EditorJune 17, 2011 at 3:56 PM

    internal and external angle bisector are perpendicular; so the result follows

    ReplyDelete
  3. Sumith PeirisApril 27, 2016 at 6:58 AM

    < ECB = 90-C/2 = A/2 + B/2

    So < AEC = B/2 = < IBC and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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