Saturday, February 23, 2019

Geometry Problem 1414: Right Triangle, Altitude, Incircle, Excircle, Tangency Points, Isosceles 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 1414: Right Triangle, Altitude, Incircle, Excircle, Tangency Points, Isosceles Triangle, Tutoring.

Posted by Antonio Gutierrez at 3:41 AM
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3 comments:

  1. c.t.e.oFebruary 23, 2019 at 5:54 AM

    Join E to G, draw ET ꓕ AC, => ∆DBJ~∆GET

    ReplyDelete
  2. Sumith PeirisFebruary 24, 2019 at 5:06 AM

    Let the bisector of < JBF cut DF at U.

    From my proof of Problem 1412, < DGF = C/2, hence < GDF = < BFD - < DGF = 45 - C/2 = A/2 = < JBU since < JBF is obviously = to A in right Tr. ABC

    Hence BDJU is concyclic and < BJU = < BDF = 45
    So Tr.s BJU and BFU are congruent ASA and therefore
    BJ = BF

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. Peter TranFebruary 24, 2019 at 9:52 AM

    https://photos.app.goo.gl/UvMged3MaAHNfXMy6
    Define point M and angle u per sketch
    Per the results of previous problems we have D,G,M are colinear
    In right triangle DBG, angle BDG complement to u
    In right triangle JHM, angle ∠ (HJM)= ∠ (BJD) complement to u
    So ∠ (BDJ)= ∠ (BJD) and BD=BJ=BF -> JBF is isosceles

    ReplyDelete

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