Wednesday, February 18, 2009

Problem 12: Triangle, Cevian, Angles, Congruence

Proposed Problem
Problem 12: Triangle, Cevian, Angles, Congruence.

See complete Problem 12 at:
http://gogeometry.com/problem/problem012.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Posted by Antonio Gutierrez at 4:11 PM
Labels: , , ,

10 comments:

  1. AjitMay 3, 2009 at 11:58 PM

    Locate a point E on AC such that BC=CE. Join B to E. Now it's easy to see that AE=CD and angle AEB = 115 = angle BDC which makes BE=BD. Hence tringles ABE & CBD are SAS congruent; thus x=50.
    Ajit: ajitathle@gmail.com

    ReplyDelete
  2. davidAugust 2, 2011 at 11:44 PM

    Denote angle ABD=α=115º-x. Suppose α>65º then AD>AB, and since AD=BC, BC>AB.Now in triangle ABC we get x>50º, that is 115º-α>50º, or α<65º in contradiction to our assumption α>65º. So α=65º and x=50º.

    ReplyDelete
  3. Eder Contreras OrdenesMarch 3, 2012 at 9:36 AM

    Solution by cristian to this problem:

    http://youtu.be/mhh1YO0LLII

    ReplyDelete
  4. UnknownApril 12, 2012 at 9:35 AM

    Solution is uploaded to the following link:

    https://docs.google.com/open?id=0B6XXCq92fLJJZWplWFJKTGMxUEk

    ReplyDelete
  5. AnonymousJanuary 9, 2015 at 12:27 AM

    Since I am not able to see all links above, I post my solution, hoping it is not among the previous ones, although it is very simple: Take E - reflection of C in BD, DE=CD and easy angle chase gives <ADE=50, i.e. triangles ADE and BCD are congruent (s.a.s.), hence AE=BD and <DAE=15, therefore triangles ABE and BAD are congruent (s.s.s.) with <BAE=<ABD. From this we get x+15=115-x, or x=50.
    Best regards,
    Stan Fulger

    ReplyDelete
  6. geoclidOctober 15, 2016 at 8:58 AM

    https://www.youtube.com/watch?v=L_AY8lZynqk

    ReplyDelete
  7. AnonymousJune 12, 2020 at 5:26 AM

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

    Angle BDC = 180° - 50° - 15° = 115°
    Angle ADB = 180° - 115° = 65°

    Add point E to the figure and connect it to points A, B and E such that a duplicate of triangle BCD is formed with angle ADE = 15°. Now, we shall prove that point E sits on line AB.

    Focus on triangle BDE. It is isosceles because line DE = line BD. Angle BDE = 65° - 15° = 50°. Angle BED = (180° - 50°)/2 = 65°. Angle AED = angle BDC = 115° because triangles ADE and BDC are identical.
    Angle AED + angle BED = 180°, which is the angle measure of a straight line. Therefore, points A, E and B are collinear.

    Since points A, E and B are collinear, angle BAC = angle EAC
    So, angle BAC = angle BCD because triangles ADE and BCD are congruent.
    Therefore, angle BAC = 50°

    Angle x = 50°

    ReplyDelete
  8. MarcoFebruary 10, 2021 at 10:12 PM

    Consider triangle ABC
    sinx/BC=sin50/AB
    BC=ABsinx/sin50

    Consider triangle ABD
    sin65/AB=sin(115-x)/AD
    AD=ABsin(115-x)/sin65

    BC=AD
    sinx/sin50=sin(115-x)/sin65
    sin65/sin50/sin(115-x)/sinx
    It is easy to see that x=50

    ReplyDelete
    Replies
    1. MarcoOctober 30, 2022 at 2:17 AM

      A better solution after 1.5 years of thinking lol
      <ABD=115-x
      By considering triangle ABD with sine law
      sin(115-x)/AD=sinx/BD
      AD/BD=sin(115-x)/sinx--------(1)
      By considering triangle BCD with sine law
      sin115/BC=sin50/BD
      BC/BD=sin115/sin50---------(2)
      Equating (1) & (2)
      sin(115-x)/sinx=sin115/sin50
      sin115sinx=sin(115-x)sin50
      cos(115-x)-cos(115+x)=cos(65-x)-cos(165-x)
      cos(115-x)+cos(65-x)=cos(65-x)-cos(165-x)
      cos(115-x)=-cos(165-x)
      cos(115-x)+cos(165-x)=0
      2cos(140-x)cos50=0
      cos(140-x)=0
      140-x=90
      x=50

      Delete
  9. Sailendra TFebruary 16, 2021 at 12:10 PM

    E on AC such that BC=CE. So AE=CD and m(AEB)=m(CDB) and BE=CD => AEB is congruent to CDB => x=50 degrees.

    Additionally E is the power of a point wrt circumcircle of BDC.

    ReplyDelete

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