Let $ABCD$ be a rectangle. Let $P$ be a point on 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.
By a quick angle-chase we have:
$$\angle DRQ = \angle DAQ = \angle DAC = \angle DBC = \angle DBP = \angle DRP$$
Therefore $P,Q,R$ are collinear. Actually we can generalize the problem even more, by considering $ABCD$ as a cyclic quadrilateral, instead of rectangle.