Let ABCD be a simple quadrilateral.

Construct with straightedge and compass a point P
such that ΔPAB ~ ΔPCD and both triangles have the
same orientation.
  

(In reply to re: Solution by Bractals)

Ah. Let's think about it as the limiting case when E -> infinity, the radii also -> infinity and the arcs tend to diagonals of the quadrilateral.
So in the case when AB and CD are parallel, P is the point of intersection of the diagonals of ABCD.

Posted by Harry on 2013-11-24 18:31:59