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.
  

Your use of complex numbers in the official solution is much appreciated
Bractals - especially the direct way in which point F can be constructed.

The final step in the construction still seems quite difficult (constructing
a similar triangle to find P). I would need to resort to drawing circles to
achieve this. Am I missing something? Is there a direct way?

Posted by Harry on 2013-11-27 14:27:13