First Solution: Let circles $\omega_1$ and $\omega_2$ meet again at $R$ (other than $D$), and let $\omega_1$ and $\omega_2$ intersect again with segments $AB$ and $AC$, respectively, at $P$ and $Q$ (other than $B$ and $C$). By the Power of a Point Theorem, we have $AP \cdot AB = AR \cdot AD = AQ \cdot AC$; that is, points $A, R$, and $D$ lie on the radical axis of the circles $\omega_1$ and $\omega_2$. Therefore, points $B, C, Q$, and $P$ lie on a circle. Consequently, $\angle AQP = \angle ABC$. To prove that $XY \parallel BC$, it suffices to show that $\angle AXY = \angle ABC$, or $\angle AQP = \angle ABC = \angle AXY$; that is, to prove that $P, Q, Y$, and $X$ lie on a circle.

Because $BCQP$ is cyclic, $\angle AQP = \angle ABC$. Because $BDRP$ is cyclic, $\angle PDY = \angle PBE = \angle ABC$. Hence $\angle AQP = \angle PDY$, implying that $P, Q, Y$ and $D$ lie on a circle. Similarly, we can show that $P, Q, D$, and $X$ lie on a circle. We conclude that $P, Q, Y, D$, and $X$ all lie on the circumcircle of triangle $PQD$, completing the proof.

Second Solution: It suffices to show that
$$\frac{AX}{XB} = \frac{AY}{YC}. \qquad (*)$$
Let line $AD$ and segment $BC$ meet at $G$. Applying Menelaus's Theorem to triangle $ABG$ and line $XF$, triangle $ACG$ and line $YE$ gives
$$\frac{AX \cdot BF \cdot GD}{XB \cdot FG \cdot DA} = 1 \text{ and } \frac{AY \cdot CE \cdot GD}{YC \cdot EG \cdot DA} = 1.$$
Hence
$$\frac{AX \cdot BF}{XB \cdot FG} = \frac{AY \cdot CE}{YC \cdot EG}.$$
By the equation (*), it suffices to show that
$$\frac{BF}{FG} = \frac{CE}{EG},$$
or $EG \cdot BF = CE \cdot FG$. By the Power of a point Theorem, we have $EG \cdot BG = GD \cdot GR = GF \cdot GC$, implying that
$$\begin{aligned} EG \cdot BF &= EG(BG + GF) = EG \cdot BG + EG \cdot GF \\ &= GF \cdot GC + EG \cdot GF = GF(GC + GE) \\ &= CE \cdot FG, \end{aligned}$$
as desired.