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.

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

    Greetings

    ReplyDelete
  3. 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.

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

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

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

    ReplyDelete
  10. Here's my solution.

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

    ReplyDelete
  11. 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
  12. 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
  13. 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