Thursday, November 25, 2010

Problem 542: Right Triangle, Altitude, Angle Bisector, Perpendicular, 90 Degrees

Geometry Problem
Click the figure below to see the complete problem 542 about Right Triangle, Altitude, Angle Bisector, Perpendicular, 90 Degrees.

Problem 542: Right Triangle, Altitude, Angle Bisector, Perpendicular, 90 Degrees.
Go to Complete Problem 542

Posted by Antonio Gutierrez at 1:42 PM
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7 comments:

  1. c .t . e. oNovember 26, 2010 at 1:09 PM

    ▲BFG isoceles from 541 => BG altitude

    ReplyDelete
  2. PravinNovember 29, 2010 at 7:24 PM

    You mean triangle BHE is isosceles (where BE, AD meet at H)?

    ReplyDelete
  3. c .t . e. oNovember 30, 2010 at 12:28 PM

    ▲BEH isoceles, AE meet BD at H

    ReplyDelete
  4. AjitSeptember 29, 2011 at 5:24 AM

    /_ABD = 90-/_A or/_DBC=/_A so/_DBG = /_A/2
    Thus,/_DAG=/_A/2=/_DBG or A,B,G & D are concyclic; hence x = /_ADB =90.
    Ajit

    ReplyDelete
  5. Sumith PeirisJuly 1, 2015 at 6:45 AM

    < DBA = A hence the bisected angles are also equal. Hence ABGD is cyclic and so x = < BDA = 90.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. Sumith PeirisJune 23, 2016 at 7:50 AM

    Another method

    < ABG = < AFG and the result follows

    ReplyDelete
  7. MarcoDecember 24, 2023 at 9:21 PM

    Let <BAE=<EAC=a
    <ACB=90-2a
    <DBC=2a
    <DBF=2a/2=a
    <ABD=90-2a
    <ABF=<ABD+<DBF=90-a
    x=180-<BAE-<ABF=180-a-(90-a)=90

    ReplyDelete

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