Wednesday, December 19, 2018

Geometry Problem 1410: Right Triangle, Incircle, Excircle, Tangency Points, 45 Degree Angle

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 1410: Right Triangle, Incircle, Excircle, Tangency Points, 45 Degree Angle, Tutoring.

Posted by Antonio Gutierrez at 2:56 PM
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5 comments:

  1. c.t.e.oDecember 19, 2018 at 3:33 PM

    P intersection of diagonals of trapesoid. Diagonals are perpendicular
    tr FBD similar tr FPT (T tg point on AB)
    => FPG right isosceles triangle

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  2. Peter TranDecember 19, 2018 at 6:49 PM

    https://photos.app.goo.gl/UNt6uwqEA6K9km9r6
    per the result of problem 1409 we have F, D and M are colinear
    We have ∠ (AIC)= 90+ B/2= 135 => ∠ (FIC)= 45
    We have IC⊥MD and IE⊥ FG=> quadri. INPF is cyclic
    And ∠ (DFP)= ∠ (EIN)= 45

    ReplyDelete
  3. Sumith PeirisDecember 19, 2018 at 10:12 PM

    Let X be the tangency point on AC of incircle
    Let Y be the tangency point on BC of excircle

    Considering the angles of quadrilateral CYFG, < GFY = 360 - (90-A/2) - (90 + A) - 135 = 45 - A/2 .....(1)

    Now FDX is collinear as per my proof of Problem 1409 & so < XDC = 90 - C/2 = 45 + A/2.
    Considering the angles of Tr. AFX, < BFD = 45 + A/2 - A = 45 - A/2 = < GFY from (1)

    Since < BFD + < DFY = 45, < GFY + < DFY = Alpha = 45

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Sumith PeirisDecember 19, 2018 at 10:16 PM

    Much simpler proof

    Considering Tr. XFG, < FXG = 90 - C/2 & < FGX = 90 - A/2, Alpha = A/2 + C/2 = 45

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. Sailendra TDecember 24, 2018 at 9:15 AM

    Considering usual triangle notations
    Let U & V be the tangent points of incircle I with AB and AC
    => UV//FG and UF=VG=a => UVGF is an isosceles trepezoid-----(1)
    It can be deduced that the diagonals of UVGF are congruent and say X be the point of intersection => GX=FX and FXG is isosceles ------(2)

    Let W be the tangent point of excircle E with BC
    It can be proved that W lies on the diagonal GU
    (Apply menelaus to ABC => GC/GA*CW/WB*BU/UA =1
    substituting and simplifying will give us the below
    => s(s-b)=(s-a)(s-c)
    => (a+b+c)(a+c-b)=(b+c-a)(a+b-c)
    => a^2+c^2=b^2
    => G,W&U are collinear
    )
    By symmetry, F,D&V are collinear -----------(3)

    Extend BC to meet FG at Y, Since m(B)=90 and m(BFG)=90-A/2=> m(BYF)=A/2
    Consider the triangle WYG, the external angle m(FGW)=m(YWG)+m(WYG)
    =>m(FGW)=45-A/2+A/2
    =>m(FGW)=45 ---------(4)

    From (2) & (4) => m(XFG)=45 Q.E.D

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