Problem 5. Let ABCD be a given convex quadrilateral with sides BC and AD equal in length and not parallel. Let E and F be interior points of the sides BC and AD respectively such that BE = DF. The lines AC and BD meet at P, the lines BD and EF meet at Q, the lines EF and AC meet at R. Consider all the triangles P QR as E and F vary. Show that the circumcircles of these triangles have a common point other than P.
Looking at the segments we know, and using generalised Menelaus we hve CR/AR = DQ/QB
We now take an inversion about P sending the general point X to X'. If we consider a point at infinity P', under inversion this point goes to P. And since cross ratios are preserved under inversion, we can say; [A,P',R,C] = [B,P',Q,D] implies the following [A',P,R',C'] = [B',P,Q',D'] we know the former is true because of Menelaus and distances from points to a point at infinity are equal.
Does this imply that since we know there exists a point where A'B' meets D'C' and a line through this point T and P, that R'Q' also goes through this point for all R'Q' and so the circles intersect at a fixed point as desired?
If this is not true can someone please explain why and if possible complete the proof with inversion?