Given a triangle ABC, how can you find a point P on AC and a point Q on BC, such that AP=PQ=QB?

N.B. A construction method is sought, and only compass and straightedge are allowed.

(In reply to Non-iterative solution by Federico Kereki)

Federico Kereki says: There is a non-iterative solution to this problem; I have already posted it, and will make it public in a couple of weeks.

My solution is non-iterative. Does it not make sense?

Posted by Bryan on 2005-05-03 21:45:36