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: Argand - grand! by Bractals)

Ah. If the triangles had opposite orientation then P would be outside the quadrilateral, and the need for angles APB and CPD to be equal would require P to be the point of intersection of BC and AD (so not possible if BC and AD are parallel).
But also, the need for angles PBA and PDC to be equal would require ABCD to be a cyclic quadrilateral.

However, back to the original problem...

I’m still not sure how simple I should find the ‘standard’ constructions that you suggest.
For example: Construct ray BP’’ such that /ABP’’ = /AFC.
The only way I can see to do this is to construct the circle through A, F and C, let G be the intersection of AB and the circle, then draw BP’’ parallel to GC. Is there a simpler way?

Posted by Harry on 2013-12-03 12:47:12