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
re: Solution by Charlie)
Um, sorry. Of course X and Y are at infinity. What I meant to say was to construct a perpendicular at M, not X. I have edited my solution to correct this.
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Posted by Bryan
on 2005-04-28 15:31:36 |