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Incentre properties

  1. Nov 8, 2005 #1
    If I is the incentre of the triangle ABC and AI meets the circumcircle at D, then prove that DB = DC = DI. If DE and DF are perpendiculars to AB and AC then prove that AD = AE = (AB + AC)/2
  2. jcsd
  3. Nov 8, 2005 #2


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    Something seems to be wrong with the problem statement. AI meets the circumcircle at A, and in general, AB, AC, and AI can all be distinct. Also, if we have an equilateral triangle, then DB = DC > BI = CI = AI > DI as long as D is strictly between A and I. I'm assuming you wrote the problem out wrong, but it seems to be so wrong that I can't even guess what the problem was supposed to be.
  4. Nov 9, 2005 #3
    I felt so. thanks anyway
  5. Nov 14, 2008 #4
    It's a valid problem and one I need to solve now :)
    AI is a straight line and a straight line can cross a circle at two points, so it does meet the circumcircle at point D.
    DB=DC because the according arcs of the circumcircle are equal (because angles BAI and CAI are equal).
    Why is DI equal to DB and DC.
    (Sorry, if I've got bad English.)
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