Draw a pentagon ABCDE and extend its sides to form the pentagram A*B*C*D*E*. The circumcircles of the triangles of the pentagram: A*CD, B*DE, ... etc. have second intersections A', B', C', ... etc. lying on a circle.
The key key-fact is that some other pentagons, e.g. D*B'CD'B* are cylic. This follows from another theorem of Miquel on four intersecting lines ( look at Miquel_Point.html ). This enables to show that A'B'D'E' is cyclic. Then by repeating the argument one shows that A'E'D'C' is cyclic too.