Saturday, November 15, 2014

Geometry Problem 1060: Triangle, Incircle, Incenter, Inscribed circle, Tangent, Angle

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

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Online Math: Geometry Problem 1060: Triangle, Incircle, Incenter, Inscribed circle, Tangent, Angle.
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2 comments:

  1. Peter TranNovember 15, 2014 at 1:25 PM

    Connect DE, ID and IE
    Note that ID ⊥DC and IC ⊥ DE
    ∠ (IFG)= ∠ (IDE) = alpha ( symmetric property of FBDG)
    But ∠ (IDE)= ∠ (DCI)= ∠ (ICE)= alpha
    So ∠ (DCE)= 2 alpha

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  2. Sumith PeirisAugust 28, 2015 at 9:11 AM

    IDCE is cyclic so < IDE = C/2. Hence < BGD must be A/2 considering the angles of Tr. BDG = < IAE. Hence AIGE is cyclic on which circle F must also lie since AEIF is obviously cyclic.

    So alpha = < IEG = C/2

    Sumith Peiris
    Moratuwa
    Sri Lanka

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