I need help with the following problem:
Given a triangle ABC. The external bisector of angle A intersects the line BC at point N. The internal bisector of angle A intersects BC at point M. Let k be the circle with diameter MN. Prove that for every point Q on the circle QB:QC=AB:AC.
Remark: The problem has to be solved using only the properties of the bisectors and Thales' theorem.