Wednesday, November 30, 2011

Problem 694: Triangle, Angles, 30 Degrees, Auxiliary Construction, Mind Map, Polya

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 694.

Online Problem 694: Triangle, Angles, 30 Degrees, Auxiliary Construction, Mind Map, Polya.

Posted by Antonio Gutierrez at 3:07 PM
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5 comments:

  1. AjitNovember 30, 2011 at 7:43 PM

    Draw a perpendicular BF from B to AC intersecting CD in E, join AE. Since Tr ABC is isosceles, EA =EC and therefore /_EAC=30 hence /_DAE=18 or AD is angle bisector of /_BAE. Now /_AEF=60=/AED and thus /_DEB=60 so ED bisects /_AEB which means BD bisects /_ABE. So /_x =12

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  2. EditorDecember 5, 2011 at 8:20 PM

    Here is the generalization of this problem

    http://www.fmat.cl/index.php?s=&showtopic=74752&view=findpost&p=567874

    ReplyDelete
  3. AnonymousNovember 15, 2014 at 1:52 PM

    I still don't understand...

    ReplyDelete
  4. Sumith PeirisSeptember 26, 2015 at 11:04 PM

    AB = BC, let BE be the perpendicular bisector of AC meeting DC at E. By computing angles we see that D is the in centre of Tr. ABE hence x = 12

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. APOSTOLIS MANOLOUDISMay 3, 2016 at 12:05 PM

    Problem 694
    Forming equilateral ADE (point E is lower thanBC ),then <AED=60=2.30=2<ACD so E is circumcenter of triangle ADC,so AE=AD=DE=EC. But triangle ABE= triangle BCE,then
    AB=BE and <ABE=48/2=24.But triangle ABD=triangle DBE (AB=BE,AD=DE, BD=BD),
    And <ABD=x=24/2=12.

    ReplyDelete

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