Thursday, April 16, 2009

Problem 31: Right Triangle, Incenter, Collinear

Proposed Problem

 Problem 31: Right Triangle, Incenter, Collinear.

See also: Complete Problem 31, Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

6 comments:

  1. As proven in Problem 30, BG=2r. Similarly BH=2r or BH=BG=2r. Now if B be the origin, BA the x-axis and BC the y-axis then E is (r,r) while HG is: x + y = 2r since H is (2r,0)& G:(0,2r). Clearly E satisfies the equation of HG; hemce E lies on HG or H, E & G are collinear.
    Ajit: ajitathle@gmail.com

    ReplyDelete
  2. bae deok rak(bdr@korea.com)October 5, 2010 at 10:21 PM

    In problem 30, we see that
    ang(BEG)=ang(BEQ)+ang(QEG)=45+45=90 and
    ang(BEH)=ang(BEP)+ang(PEH)=45+45=90
    Hence H, E and G are collinear.

    ReplyDelete
  3. expand DH and GF at I.because angle (BHD) = angle (BGF) = 90 so BGHI is a QUADRILATERAL that lies on circle E.Angle (2B)=angle(E).so anle (HBG)=180 that means it is collinear

    ReplyDelete
  4. if ang(EAD)=a and ang(EAC)=45-a we can easily find that ang(EFD)=ang(EFG)=90-a which means EFid the bisector of ang(GFD) and EC the bisector of ang(GCF)so E is the excenter of FGC.Then ang(BGE)=ang(EGF)=45 and similarly ang(EHD)=45 so ang(BGE)=ang(EHD) Because BG//HD H,E,G are collinear

    ReplyDelete
  5. Kindly refer my solution to Problem 30.

    We showed that Tr. BEG is an isoceles right Tr. It can similarly be shown that Tr. BEH is also a isoceles right Tr.

    Hence < HEB + < GEB = 90 + 90 = 180 proving that H, E,G are collinear

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. If HD, GF meet at M, then HBGM is a square with side 2r and centre E
    Further it can be shown that < DEF = 45 and that E is ex centre of Tr. DMF which is a Tr. similar to the original Tr. ABC

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete