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.

5 comments:

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

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  2. 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

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  3. 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.

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

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