Friday, February 22, 2019

Geometry Problem 1412: 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 1412: Right Triangle, Incircle, Excircle, Tangency Points, 45 Degree Angle, Tutoring.

Posted by Antonio Gutierrez at 11:17 AM
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3 comments:

  1. c.t.e.oFebruary 22, 2019 at 11:47 AM

    Isosceles trapezoid (see P 1411) is cyclic

    ReplyDelete
  2. Peter TranFebruary 22, 2019 at 7:29 PM

    https://photos.app.goo.gl/AkZzfxhqUxSaGJcL7
    See sketch for position of points K, L, M
    We have AK⊥DG and
    ∠ (AIC)= 90+ ½* ∠ (B)= 135 => ∠ (EIC)=45
    Since ID and CF perpendicular to AB => DI//FC
    We also have CF=MB=DI => DICF is a parallelogram
    So DF//IC => ∠ (DLK)= ∠ (EIC)= 45 => ∠(GDF)= 45

    ReplyDelete
  3. Sumith PeirisFebruary 24, 2019 at 12:54 AM

    If AC meets the excircle at Z, from my Proof of Problem 1407, DFZ are collinear

    So < GDA = < AGD - < GZD
    = (90 - A/2) - C/2 = 45

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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