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(5): Argand - grand! by Bractals)

Ah, - using equal arcs on equal circles. Cunning. Presumably a parallelogram IKLJ would be needed before the third circle(I,|IJ|) could be constructed.

I think I’ll stick to my original plan with one circle, through A, F and C.

Many thanks for your patience.

Posted by Harry on 2013-12-04 14:09:41