$\textbf{\textcolor{red}{CLAIM:}}$ $AFZI$ and $AEYI$ are cyclic quadrilateral. $\textbf{proof:}$ At first observe that $\frac{AF}{FB}=\frac{AC}{CB}=\frac{sin\angle B}{sin 120^{\circ}}=\frac{sin\angle B}{sin 60^{\circ}}=\frac{AD}{BD}$. So, $DF$ is angle bisector of $\angle ADB$.So $Z$ is incentre of $\triangle BAD$.Hence, $\angle FAZ=30^{\circ}$. On the other hand,$ZIF=180^{\circ}-\angle BIC=30^{\circ}$.So, $AFZI$ is concyclic. Similarly, $AEYI$ is concyclic $\blacksquare$ Now we have, $ DZ\times DF=DI\times DA=DY\times DE$.Hence, $FZYE$ is a cyclic quadrilateral $\blacksquare$