Wednesday, April 28, 2010

Problem 440: Triangle, Incircle, Incenter, Angle Bisector, Tangency Points, Circle, Angles

Proposed Problem
Click the figure below to see the complete problem 440 about Triangle, Incircle, Incenter, Angle Bisector, Tangency Points, Circle, Angles.

Problem 440: Triangle, Incircle, Incenter, Angle Bisector, Tangency Points, Circle, Angles
See also:
Complete Problem 440

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 9:18 PM
Labels: , , , , , ,

4 comments:

  1. c .t . e. oApril 29, 2010 at 2:15 AM

    join I to E and to F
    FIE = 180°- 44 = 136° => IFE = IEF = 44:2 = 22°
    => AFE = 90° + 22° = 112°
    A/2 + B/2 = 136:2 =68°
    =>
    AFE + AIG = 112°+ 68° = 180° => AIGF cyclic =>
    x = IFE ( have same arc IG )

    x = 22°
    ----------------------------------------------

    ReplyDelete
  2. JankonyexApril 29, 2010 at 4:45 AM

    mBIA=mEIF/2=mEFF=mEFA
    AIGF, IECF concyclic
    x=mGFI=mECI=44/2=22

    ReplyDelete
  3. Sumith PeirisSeptember 11, 2015 at 11:55 PM

    A + B = 136 and so A/2 + B/2 = 68.

    Now IECF being cyclic < IEF = 22 and considering the angles of Tr. BEG, < BGE = 180 - B/2 - 90 - 22 = A/2. Hence AIGF is cyclic and so x = < IFG = 22

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Sumith PeirisSeptember 11, 2015 at 11:56 PM

    Even if the value of C is not given we can similarly show that AG is perpendicular to BG

    ReplyDelete

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