Monday, December 10, 2018

Geometry Problem 1407: Right Triangle, Incircle, Excircle, Collinear Tangency Points, Collinearity

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

Details: Click on the figure below.

Geometry Problem 1407: Right Triangle, Incircle, Excircle, Collinear Tangency Points, Collinearity, Tutoring.

Posted by Antonio Gutierrez at 2:02 PM
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2 comments:

  1. Sumith PeirisDecember 11, 2018 at 8:46 AM

    Let BC touch in circle at X & AC at Y.
    Let AB touch excircle at Z.

    Let BD = p & FX = q

    BDIX is a square of side p & BFEZ is a square of side p+q

    Now DZ = 2p+q = YG = q + 2.CF, so CF = CG = p

    Hence Tr. s BDF & GCE are congruent SAS and hence < BFD = < CEG = < CFG since CFEG is concyclic

    It follows that D,F,G are collinear

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  2. c.t.e.oDecember 11, 2018 at 2:33 PM

    Draw DIP diameter. Ang APD = ACB = 2.CGF => PDG = CGF
    but PDG = DF'B => DF'B = CFG (CF=CG equal tangencial seg)

    ReplyDelete

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