Come on, man! Why do we need all that stuff? Obviously, we need $CD\bot AD$, i.e. $\angle ADC=90^\circ = \angle ABE$! I think it would have been more interesting to ask something using the fact that $CD=BD$, resulting from the given relation! Best regards, sunken rock

Should the title of the competition be Math instead of Match? Is this a typo of some sort? [Moderator edit: fixed post]

A convex quadrilateral $ABCD$ inscribed in a circle $k$ has the property that the rays $DA$ and $CB$ meet at a point $E$ for which $CD^2=AD\cdot ED.$ The perpendicular to $ED$ at $A$ intersects $k$ again at point $F.$ Prove that the segments $AD$ and $CF$ are congruent if and only if the circumcenterof $\triangle ABE$ lies on $ED.$