Monday, June 23, 2008

Elearn Geometry Problem 126



See complete Problem 126
Incenter of Triangle, Angle Bisector, Proportions. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Posted by Antonio Gutierrez at 9:47 PM
Labels: , , , , ,

6 comments:

  1. AnonymousJanuary 17, 2009 at 10:08 AM

    in-radius=r
    e/d=[ABCI]/[AIC]=(ar/2+cr/2)/(br/2)=(a+c)/b

    ReplyDelete
  2. Peter TranApril 26, 2012 at 1:00 AM

    http://img24.imageshack.us/img24/4625/problem126.png

    From B draw a line // to AC
    Extend AI and CI to E and F ( see attached sketch)
    Note that ∆FBC and ∆ABE are isosceles
    So FB=BC=a and AB=BE=c
    ∆ AIC similar to ∆ EIF ….( Case AA)
    So FE/AC=IE/IA=IB/ID
    And e/d= (a+c)/b

    ReplyDelete
  3. Nilton LapaJune 9, 2012 at 4:15 PM

    Solution to problem 126.
    In triangle ABC, as BD is the angle bisector of B, then
    AB/AD = BC/DC = (AB+BC)/(AD+DC) = (AB+BC)/AC.
    In triangle ABD, since AI is the angle bisector of A, then AB/AD = BI/ID.
    Hence BI/ID = (AB+BC)/AC, or e/d = (a+c)/b.

    ReplyDelete
  4. Nilton LapaJune 9, 2012 at 4:18 PM

    To Antonio:
    In the enunciate of problem 126, "the angle bisector of A" should be "the angle bisector of B". Am I right?

    ReplyDelete
  5. ArifJanuary 17, 2018 at 3:39 AM

    Connect the points I and C , then using the angle bisector theorem we get :

    a/e=DC/d
    (1.) ad/e=DC

    Connect the points I and A , then using the angle bisector theorem we get :

    c/e=AD/d
    (2.) cd/e=AD

    Adding equation 1 and 2 with AD+DC=b :

    (ad+cd)/e=AD+DC
    (ad+cd)/e=b
    (a+c)/b=e/d

    ReplyDelete

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