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.
  

Let E be the point of intersection of AB and CD.
The required point, P, is the point of intersection of the circumcircles
of triangles BDE and CAE.

Proof:
Angles ABP and EBP are supplementary (angles on a straight line).
Angles CDP and EBP are supplementary (opp. angles of cyclic quad EBPD).
Therefore /ABP = /CDP.              (1)
By similar reasoning..
Angles DCP and ECP are supplementary (angles on a straight line).
Angles BAP and ECP are supplementary (opp. angles of cyclic quad ECPA).
Therefore /DCP = /BAP.              (2)

It follows from (1) and (2) that triangles PAB and PCD are similar.

Posted by Harry on 2013-11-23 20:44:38