Prove that if $E F$ is parallel to $B C$, then $A B=A C, A D$ is the axis of symmetry of $E F Z Y$, and thus the quadrilateral is a cyclic quadrilateral.

If $E F$ is not parallel to $B C$, assume that the extension of $B C$ intersects the extension of $E F$ at $P$. By Menelaus' theorem, we have
$$\frac{\overrightarrow{A F}}{\overrightarrow{F B}} \cdot \frac{\overrightarrow{B P}}{\overrightarrow{P C}} \cdot \frac{\overrightarrow{C E}}{\overline{E A}}=-1 \text {. }$$

Given $B Z=B D=B F, C Y=C D=C E$,
and $\frac{A F}{E A}=1=\frac{X Z}{Y X}$, we have
$$\frac{\overline{X Z}}{\overline{Z B}} \cdot \frac{\overrightarrow{B P}}{\overrightarrow{P C}} \cdot \frac{\overrightarrow{C Y}}{\overline{Y X}}=-1 \text {. }$$

By the converse of Menelaus' theorem, we know that $Z, Y, P$ are collinear, thus,
$$P E \cdot P F=P D^{2}=P Y \cdot P Z .$$

Therefore, $E F Z Y$ is a cyclic quadrilateral.