Kiselev's Geometry Problem Problem 107 
Prove that if A and A' and B and B' are 2 pairs of points
  symmetric about a line XY, then the 4 points lie on the same circle.

I thought about it for a long time, until I obtained this proof:
Since A and A' and B and B' are symmetric about XY, A and A' are equidistant from XY and so are B and B'. Now, because XY is the locus of points equidistant from A, A' and B, B', there is bound to be at least one point, say R, which is equidistant from all points A, A', B, and B'. That is, AR = A'R = BR = B'R. $\therefore$, the points A, A', B, B' are the geometric locus of R, which is a circle. Thus A, A', B, B' are on the same circle.
After writing this proof, I felt a bit uncomfortable, because something didn't seem quite right with the way that I wrote it. If anyone could give suggestions or point out if I was wrong, that would be greatly appreciated.
Thanks.
 A: If you take a point $O_1$ on $XY$ on the semiplane of $AA'$ that doesn't contain to $B$ and $B'$ the distance from that point to $A$ (and $A'$) is smaller than the distance from that point to $B$ (and $B'$). Similarly if you take the point $O_2$ on $XY$ on the semiplane of $BB'$ that doesn't contain $A$ and $A'$, the the distance from that point to $B$ (and $B'$ is smaller). 
By continuity of the distance there is a point on $XY$ between $O_1$ and $O_2$ such that the distance to $A$ is the same as the distance to $B$ (and as you said, the distances to $A'$ and $B'$ are also the same amount).
A constructive way is just to draw the perpendicular bisector of $AB$ and take its intersection with $XY$. The point of intersection is your center.
You have to allow for the degenerate case in which $A,A',B,B'$ are colinear.
A: Connect AA' and BB'. XY is the perpendicular bisector of AA' and BB'. Also AB=A'B'.Thus ABB'A' is an isosceles trapezium. Since $\angle AA'B'+\angle ABB'=180º$, AA'B'B is cyclic.
