49th International Mathematical Olympiad Spain

Assume that the disposition of points is as in the diagram. Since $\angle EBF=180^{\circ }-\angle CBF=180^{\circ }-\angle EAF$ by hypothesis, the quadrilateral $AEBF$ is cyclic. Hence $AJ\cdot JB=FJ\cdot JE$. In view of this equality, $I$ belongs to the circumcircle of $ABK$ if and only if $IJ\cdot JK=FJ\cdot JE$. Expressing $IJ=IF+FJ$, $JE=FE-FJ$, and $JK=\frac{1}{2}FE-FJ$, we find that $I$ belongs to the circumcircle of $ABK$ if and only if $$FJ=\frac{IF\cdot FE}{2IF+FE}$$ Since $AEBF$ is cyclic and $AB,CD$ are parallel, $\angle FEC=\angle FAB=180^{\circ }-\angle CDF$. Then $CDFE$ is also cyclic, yielding $ID\cdot IC=IF\cdot IE$. It follows that $K$ belongs to the circumcircle of $CDJ$ if and only if $IJ\cdot IK=IF\cdot IE$. Expressing $IJ=IF+FJ$, $IK=IF+\frac{1}{2}FE$, and $IE=IF+FE$, we find that $K$ is on the circumcircle of $CDJ$ if and only if $$FJ=\frac{IF\cdot FE}{2IF+FE}.$$ The conclusion follows.