I am given that, for $JG$ an exterior angle bisector of $\angle CGF$ parallel to the angle bisector of $\angle FHE$, prove that $CDEF$ is a cyclic quadrilateral. I can prove that if the quadrilateral is cyclic, then the angle bisectors are parallel, but the reverse direction is proving troublesome...
Let $\angle CGF=2\alpha$, $\angle FHE=2\beta$ and $\angle KCH=\gamma$, as shown in the diagram.
$\therefore\angle CKH=\pi-\gamma-\beta$ and $\angle JGK=\frac{\pi}{2}-\alpha$ .
$\angle CDE=\pi-\gamma-2\beta$ and $\angle CFG=\gamma-2\alpha$
Thus, $\angle CKH=\angle JGK$ if and only if $\angle CDE=\angle CFG$, whence, exterior angle bisector of $\angle CGF$ is parallel to the angle bisector of $\angle FHE$, if and only if $CDEF$ is a cyclic quadrilateral.
$\blacksquare$