I'm trying to go through the section of Newer Results in Hartshorne's Geometry: Euclid and Beyond.
This particular exercise has been bugging me for a good while:
My first approach was to show that the perpendicular bisectors of $A'B'$ and $A'D'$ intersect at the same point as the perpendicular bisectors of $A'B'$ and $B'C'$. I figured since the center of the circle circumscribed around a triangle has its center at the intersection of the three perpendicular bisectors, this would show that $A'$, $B'$, $C'$ and $D'$ would all be on the same circle. However, I didn't see any way to show this.
I also tried extending $A'D'$ and $A'C'$ down into the circle containing $D$, $D'$ and $C$ with hopes that it $\angle D'A'C'\cong\angle D'B'C'$, but this seemed like a dead end also.
I'm going a little mad, if any one sees a possible solution, I'd be quite grateful. Thanks.