Let $ABCDEF$ be a cyclic hexagon, such that $AF,BE,CD$ concur. Prove that $(F,D;E,C)=(A,C;B,D)$.
I'm relatively new to projective geometry. This problem would be solved by perspectivity through the point $AF\cap BE\cap CD$, but I'm not sure you can do that, since the intersection point doesn't necessarily lie on the circle?
