Saturday, February 23, 2019

Geometry Problem 1416: Right Triangle, Altitude, 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 1416: Right Triangle, Altitude, Incircle, Excircle, Tangency Points, 45 Degree Angle, Tutoring.

Posted by Antonio Gutierrez at 3:51 AM
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4 comments:

  1. c.t.e.oFebruary 23, 2019 at 6:10 AM

    Draw FT, T on AC, Extend JG to P, P on FT => ∆GPT right isosceles,
    => ang PJF = JFT = 45°

    ReplyDelete
  2. Sumith PeirisFebruary 24, 2019 at 7:51 AM

    From my proof of Problem 1415, BF = BD = BJ so B is the centre of Tr. HFG

    So Alpha = < GBF /2 = 90/2 =45

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. Peter TranFebruary 24, 2019 at 9:20 AM

    https://photos.app.goo.gl/syrJ1dYNJgpap7oi9

    See sketch for position of points L, K and angle u
    per the results of previous problems we have JBG is isosceles and IC ⊥JD and ICE is isosceles right triangle.
    Draw angle bisector of angle HBD, this line meets IN at L
    In right triangle ABC we have ∠ (BAH) = 2u= ∠ (HBC)
    So ∠ (BAI)= ∠ (LBN)= u => ∠ (BLN)=90=> AL//JG
    So ∠ (IKJ)= ∠ (KIL)= 45 and ∠ (LJK)= 45

    ReplyDelete
  4. c.t.e.oFebruary 24, 2019 at 1:09 PM

    Second way
    Draw second diagonal of trapezoid, let be P intersection
    ∆PJG right isosceles

    ReplyDelete

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