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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.

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I wouldn't be surprised if this is in C. Stanley Ogilvy's book Excursions in Geometry, where it treats harmonic division. Whether his proof satisfies the constraints in your "Remark" is another question. – Michael Hardy Feb 21 '12 at 19:58

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