Wednesday, February 18, 2009

Problem 12: Triangle, Cevian, Angles, Congruence

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

See complete Problem 12 at:

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. 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.

  2. 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.

  3. Solution by cristian to this problem:

  4. Solution is uploaded to the following link:

  5. 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