Sunday, May 18, 2008

Elearn Geometry Problem 10

Geometry Problem, Triangle, Angles

See Problem 10 details at:
www.gogeometry.com/problem/problem010.htm

Triangle, Angles, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Posted by Antonio Gutierrez at 9:37 PM
Labels: , ,

17 comments:

  1. bae deok rak(bdr@krea.com)October 3, 2010 at 9:01 PM

    Let O be the circumcenter of the triangle ABC.
    Let E be the intection of the extension of BO and CA, where O is an circumcent of ABC.
    Let F be the point on the extension of CO with triangle ABF is an equilateral triangle. Then we have ang(FBO)=10 deg=ang(ECO),
    ang(FOB)=20 deg=ang(EOC) and OB=OC(circumradius) and so two triangles OFB and OEC are ongruence.
    Hence we get CE=FB=AB, that is, E=D and hence
    x= ang(DBC)=10 degree.

    ReplyDelete
  2. Eder Contreras OrdenesAugust 2, 2011 at 11:55 AM

    Similar problem... http://www.youtube.com/watch?v=4S4RJGk6Uxs

    Greetings

    ReplyDelete
  3. AjitSeptember 4, 2011 at 9:58 PM

    For the sake those who possibly couldn’t follow Newzad’s excellent solution, I re-write it here: Draw a line CE below AC such that /_ACE=40 and CB=CE so that Tr. ECB is equilateral which makes BE=BC. Now Tr. ABE is SAS congruent with Tr. DCB for AB=CD,BE =BC and /_ABE=20=/_DCB and therefore /_x =/_AEB. Further, AC=BC=CE and thus Tr. ACE is isosceles with an apex angle=40 and thus /_AEC = x+60 = 70 which makes x=10.

    ReplyDelete
  4. UnknownDecember 11, 2012 at 2:38 AM

    I think the solution can further be simplified if we
    “draw the equilateral triangle BCE on the same side as A”.

    CBA forms an 20-80-80 isosceles triangle with CB = CA.

    angle(ABE) = 80 – 60 = 20 = angle(DCB)
    AB = DV and EB = BC
    Therefore, [BAE] is SAS-congruent to [CDB].
    Thus, angle AEB = x

    CB = CA = CE implies [CAE] is an 40-(x + 60)-(x+ 60) isosceles triangle.
    In [CAE], 40 + (x + 60) + (x+ 60) = 180
    Therefore, x = 10

    Mike from Canotta

    ReplyDelete
  5. AnonymousJanuary 5, 2013 at 11:49 PM

    There is also a solution if we draw the equilateral triangle BCE on the opposite side as A. Then by SAS triangle CED is congruent to triangle ABC, so AC=BC=BE=EC=ED, so BED is isoceles and since angle BED=40, angle BDE=70, so angle x=10.

    ReplyDelete
  6. AnonymousSeptember 24, 2014 at 8:46 PM

    AC=BC?

    ReplyDelete
  7. AnonymousJanuary 8, 2015 at 3:01 PM

    I hope that the followıng solutıon ıs not among those posted on varıous lınks shown above:

    Construct equilateral triangle ABE, inside the triangle ABC; CE is angle bisector of <ACB and easily triangles BCE and CBD are congruent, (s.a.s.), so <CBD= <BCE= 10 degs.

    Best regards,
    Stan Fulger

    ReplyDelete
    Replies
    1. Gewürtztraminer68July 7, 2019 at 5:48 AM

      please explain why CE=BD ?
      it does not seem so evident
      regards
      Daniel

      Delete
  8. Benjamin LeisJune 22, 2015 at 4:46 PM

    Align triangle BCD so CD overlays AB and call the new vertex E. This leaves an inner 60 degree angle at EAC . AC = BC since angle BAC = ABC = 80. So the new edge EA = BC since we just copied the triangle and transitively EA = BC = AC. Since EA = AC, and EAC is 60 triangle EAC is equilateral.
    Then angle ECB is 40 and the triangle its in is also isosceles. So Angle BEC = angle CBE which by angle chasing means 60 + x = 80 -x or x = 10.

    ReplyDelete
  9. geoclidOctober 15, 2016 at 8:57 AM

    https://www.youtube.com/watch?v=7ECGjyk5zb8

    ReplyDelete
  10. Web AdminJuly 17, 2017 at 8:00 PM

    Solution:
    https://drive.google.com/file/d/0B5CPfrOPKJ1TZlBJRkc5cTRwTGc/view?usp=sharing

    ReplyDelete
  11. Geek37November 1, 2017 at 9:36 AM

    Here's my solution.

    https://www.youtube.com/watch?v=bZUOe6IGPio

    ReplyDelete
  12. AnonymousJune 12, 2020 at 4:29 AM

    This solution is much clearer when illustrated, so grab a pen and a piece of paper.

    Here are some starting info:
    > Angle ABC = 180° - 20° - 80° = 80°
    > Triangle ABC is isosceles as angles ABC and BAC are equal.
    > Line AC = line BC because triangle ABC is isosceles.

    Add point E to the figure and connect it to points A, B and C, such that triangle BCE is equilateral. Do note that triangle BCE should overlap triangle ABC. Line AC = line BC = line BE = line CE.

    Firstly, focus on triangle ACE.
    > Triangle ACE is isosceles because line AC = line CE.
    > Angle ACE = 60° - 20° = 40°
    > Angle CAE = angle CEA = (180° - 40°)/2 = 70°

    Next, focus on triangle ABE.
    > Angle ABE = 80° - 60° = 20°
    > Angle AEB = 70° - 60° = 10°

    Lastly, notice that triangles ABE and BCD are congruent by SAS:
    > Line AB = line CD
    > Angle ABE = angle BCD = 20°
    > Line BE = line BC

    Angle AEB = angle CBD = x because triangles ABE and BCD are congruent.

    x = 20°

    ReplyDelete
  13. cstromwJuly 20, 2022 at 9:12 PM

    Angle ABC = 180 - 20 - 80 = 80
    Add point E such that CDE is congruent to ABC. Angle BCE = 80 - 20 = 60. Also note that angle CED is 20 due to congruence.
    We have BC congruent to CE and a 60 degree angle between, so connecting EB forms an equilateral triangle BCE. Angle BED = 60 - 20 = 40.
    Now notice that triangle BED is isosceles, meaning angle EBD is (180 - 40) / 2 = 70. Therefore x is 70 - 60 = 10.

    ReplyDelete
  14. MarcoOctober 30, 2022 at 1:10 AM

    Consider triangle BCD
    sin20/BD=sinx/CD
    BD/CD=sin20/sinx-------(1)

    Consider triangle ABD
    <ABD=20+x
    sin80/BD=sin(20+x)/AB
    BD/AB=sin80/sin(20+x)----------(2)

    By equating (1) & (2)
    sin20/sinx=sin80/sin(20+x)
    sin20/sin80=sinx/sin(20+x)
    2sin10cos10/cos10=sinx/sin(20+x)
    2sin10sin(20+x)=sinx
    cos(10+x)-cos(30+x)=sinx
    cos10cosx-sin10sinx-cos30cosx+sin30sinx=sinx
    cosx(cos10-cos30)=sinx(1+sin10-sin30)
    tanx=(cos10-cos30)/(1+sin10-sin30)
    x=10

    ReplyDelete

Subscribe to: Post Comments (Atom)