Saturday, September 24, 2016

Geometry Problem 1265: Triangle, Incircle, 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.

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Geometry Problem 1265 Triangle, Incircle, Circle, Tangency Points, Perpendicular, 90 Degrees, Angle Bisector.

7 comments:

  1. https://goo.gl/photos/mniwJNrj8t6jnXvG7

    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

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  2. Problem 1265
    Is BE=BD then <BED=<BDE or <AEG=<CDG.But <FOC=<FOD/2=<FEG and <CFO=90=<FGE so
    Triangle EGF is similar the triangleOFC so EG/OF=GF/FC similar GD/OF=GF/AF with division by states we have EG/GD=AF/FC.So triangleAEG is similar triangleGDC then <EGA=<DGC.
    Therefore <AGF=<FGC (as equal angles supplements).
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

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  3. Trigonometry solution

    EG = 2.AE.sin(A/2).sin(C/2)
    GD = 2.CD.sin(C/2).sin(A/2)

    So EG/GD = AE/CD
    Hence Tr.s AEG and CDG are similar and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Geometry Solution

    Let M be the midpoint of EF and let EF =m, FD =n, EG = p and GD = q

    Tr.s EGM and DCM are similar (isosceles and both having <C), so
    (m/2)/p = (s-c)/n ….(1)
    where 2s = a+b+c with the usual notation.

    Similarly we can show that
    (n/2)/q = (s-a)/m…(2)

    (2)/(1) gives, p/q = (s-a)/(s-c)

    Hence Tr.s AEG and CDG are similar and the result follows.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. To prove GF is angle bisector of AngleAGC, it would suffice to prove that the triangles AEG and CDG are similar.
    Let A,B,C be the angles of the triangle ABC
    Join EF and form the isosceles triangle AEF. Similarly join FD and form the isosceles triangle CDF.

    In triangle AEF the angles are A, 90-A/2, 90-A/2
    => Angle GDF = 90-A/2 ( per Ext Segment theorm) => AngleGFD = A/2
    Similarly AngleGFE = C/2

    Join OE and consider the triangle OAE. The Angles in the triangle are 90,A/2 & 90-A/2 which is similar to the triangle GFD
    Hence AE.GD = OE.GF -------------(1)

    Similarly join OD and form the triangle ADC whose angles are 90,C/2 & 90-C/2, which is similar to the triangle GFE
    Hence EG.DC = OD.GF
    => EG.DC = OE.GF (Since OE=OD) ---------(2)

    FROM (1) & (2) AE.GD = EG.DC => Triangles AEG and CDG are similar

    Therefore AG/AE = CG/CD
    => AG/AF = CG/CF ----------- (3)

    Hence (3) is the result of Angle bisector theorem and GF is the bisector of Angle AGC

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