Let ABCD be a rectangle. let P be a point on the side BC, and let Q be a point on the diagonal AC. The circle through A, Q and D intersects the circle through B, D and P at points D and R. Prove that points P, Q and R are collinear.
So far I have been able to find many different congruent angles but the problem is none of these prove the line is straight/collinear.
Thanks for any help
Let $E$ be the intersection of the circle $BFD$ with $AD$. The trapeze $BPDE$ is isosceles and thys $BD = EP = AC$. Furthermore, since $\angle DEP = \angle ADB = \angle ACB$ the quadrilateral $EACP$ is a parallelogramme and $AC || EP$.
Now we have some way to compare angles around $Q$. Let's note that in the circle $AQDR$ we have $\angle AQR = \angle ADR = \angle EDR$.
In the circle $BPD$ we have $\angle EDR = \angle EPR $. Thus we have $\angle EPR = \angle AQR$. Since $EP || AQ$ we conclude that $R,Q,P$ are collinear.