Sunday, January 6, 2019

Geometry Problem 1411: Right Triangle, Incircle, Excircle, Tangency Points, Isosceles Right Triangle

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 1411: Right Triangle, Incircle, Excircle, Tangency Points, Isosceles Right Triangle, Tutoring.

5 comments:

  1. Replies
    1. Full explanation
      T, P, tg points on AB, AC
      PTHG isosceles trapezoid => ang GPH = ang HTG => Tr GJH isosceles
      Tr BTF ~ Tr DJF (TFB common ang, ang GTH = GPH = PDC) => GJH = 90

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  2. https://photos.app.goo.gl/UpEAXe6BoaArDLy39

    Per the result of problem 1409 we have G, D, M are collinear
    We have ∠ (AIC)= 90+ ∠ (B)/2 = 135 => ∠ (EIC)=45
    We have GH⊥IE and DM⊥IC => ILKG is cyclic
    So ∠ (LGK)= ∠ (KIC)=45
    We have triangles FCH and MCD are isosceles
    External angle ∠ (MCD)= 2 x ∠ (CHF)= 2.u
    So ∠ (LCF)= ∠ (CFH)=u => CL//HJ => HJ⊥GJ
    So GJH is an isosceles right triangle

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  3. If the tangency point of AC is X,GDJX are collinear points

    Hence < AGX = 90 - C/2 - A = 45 - A/2 and since < BGF = 45, < JGF = A/2
    But < FGG = C/2 so < JGH =A/2 + C/2 = 45

    Now < AGH = 90 - A/2 and since < FHC = C/2, < JHG = 90 - A/2 - C/2 = 45

    So in Tr. JGH, 2 angles are 45 each, hence it is right isosceles

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  4. Considering usual triangle notations,
    m(CFH)=m(BFJ)=45-A/2 --------(1)
    A,I,J & E are collinear => m(BEJ)=45-A/2 -------(2)
    From (1)&(2) B,J,F&E are concyclic ---------(3)
    Since m(BFE)=m(BGE)=90 => G also lies on the same circle as (3)
    => m(GJF)=m(GBF)=90 Q.E.D

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