Since $\angle CAZ = \angle ZCA = \beta$ (angle between the tangent and the chord), it follows that $\angle XZY = \angle AZC = 180^\circ - 2\beta$.

Quadrilateral $CAC_1A_1$ is cyclic, so we have $\angle C_1A_1B = \angle YA_1C = \alpha$ and $\angle BC_1A_1 = \angle AC_1X = \gamma$. Since $\angle XAC_1 = 180^\circ - \angle CAZ - \angle BAC = 180^\circ - \beta - \alpha = \gamma$, from the isosceles triangle $IXC_1$ we get $\angle C_1XA = 180^\circ - 2\gamma$. Analogously, from the isosceles triangle $A_1YC$ we get $\angle CYA_1 = 180^\circ - 2\alpha$.
Let us show that the quadrilateral $A_1AXD$ is cyclic. Since quadrilaterals $A_1C_1BD$ and $CABD$ are cyclic, we have $\angle C_1DB = \angle C_1A_1B = \alpha$ and $\angle ADB = \angle ACB = \gamma$, so $\angle ADC_1 = \angle ADB - \angle C_1DB = \gamma - \alpha$. Furthermore, $\angle A_1DC_1 = \angle A_1BC_1 = \beta$, so
$$\angle A_1DA = \angle A_1DC_1 - \angle ADC_1 = \beta - (\gamma - \alpha) = \alpha + \beta - \gamma = 180^\circ - 2\gamma = \angle A_1XA,$$
and the quadrilateral $A_1AXD$ is cyclic.
Similarly, $\angle CDC_1 = \angle CDA + \angle ADC_1 = \beta + (\gamma - \alpha) = 180^\circ - 2\alpha = \angle CYC_1$, so the quadrilateral $CC_1DY$ is cyclic.
Now we have
$$\begin{aligned} \angle YDX &= \angle YDC + \angle CDA + \angle ADX \\ &= \angle YC_1C + \angle CBA + \angle AA_1X \\ &= \angle A_1C_1C + \angle CBA + \angle AA_1C_1 \\ &= (90^\circ - \gamma) + \beta + (90^\circ - \alpha) = 2\beta, \end{aligned}$$
so $\angle XZY + \angle YDX = 180^\circ$, from what we conclude that the quadrilateral $XDYZ$ is cyclic.