Wednesday, September 28, 2016

Geometry Problem 1266 Triangle, Excircle, Circle, Tangency Points, Perpendicular, 90 Degrees, Angle Bisector

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

Click the figure below to view more details of problem 1266.


Geometry Problem 1266 Elements: Triangle, Excircle, Circle, Tangency Points, Perpendicular, 90 Degrees, Angle Bisector.

3 comments:

  1. https://goo.gl/photos/ywDg9feFQ4TrtjkP6

    Let ED meet AC at P.
    Apply Menelaus’s theorem in secant EDP of triangle ABC
    PC/PA x EA/EB x DB/DC= 1
    Replace EB=DB in above expression and simplify it . we get PC/PA=DC/EA= FC/FA
    So (ACFP)= -1
    Apollonius circle with diameter FP will pass through G
    So GF is an angle bisector of angle AGC

    ReplyDelete
  2. If 2s = a+b+c with the usual notation,
    AE = AF = s-c and CF = CD = s-c.

    If FD = m and FE = n and if X is the midpoint of FD then Tr.s GXD and EAF are similar being isoceles and having equal angles A/2

    Hence (m/2)/GD = (s-c)/n ....(1)
    Similarly (n/2)GE = (s-a)/m....(2)

    From (1) and (2) we have (s-a)/(s-c) = GD/GE and since < BED = < BDE ( each angle = 90-B/2), Tr.s AEG and CDG are similar.

    So < AGE = < CGD, hence GF bisects < AGC

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. Problem 1266
    Is triangleFEG similar triangleCOD(<CDO=90=<FGE,<FEG=<FOD/2=<COD) so FG/CD=EG/OD.
    But triangleFGD is similar with triangle AEO then FG/AE=GD/EO. By dividing by members we have AE/CD=EG/GD(OE=OD). Is BE=BD so <AEG=<CDG, therefore triangleAEG is similar with
    triangle CDG.So<AGE=<CGD.Then <FGA=<FGC (as complementary angles ).
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

    ReplyDelete