a. From $AZ = ZC$, it follows immediately that $[ZT]$ is the angle bisector of $\angle AZC$. Notice that $\angle XAB = \angle ACB = \angle BC_1A_1 = \angle AC_1X$ and

b. We may assume that $AB < BC$; in this case, $D$ is on the minor arc $AB$. Let $O$ denote the circumcenter of triangle $ABC$ and let $L$ be the midpoint of the segment line $[BH]$. The radical axis of two circles being perpendicular to the line connecting the centers of the circles, it follows that $BD \perp LO$. As $BD \perp DH$, we get that $LO \parallel DH$. $OLHT$ is a parallelogram, therefore $OL \parallel HT$, and now it is clear that points $D, H, T$ are collinear.

$$\angle ADT = \angle BDA - \angle BDH = 180^\circ - \angle ACB - 90^\circ = 90^\circ - \angle XAB = \angle AXT.$$

It is easy to prove (in a similar manner to a)) that $TY$ is the perpendicular bisector of the segment line $[A_1C]$. From
$$\angle CDT = \angle HDB - \angle CDB = 90^\circ - \angle CAB = 90^\circ - \angle BCY = \angle CYT,$$
it follows that the quadrilateral $CTDY$ is cyclic. This leads to
$$\begin{align*} \angle XDY + \angle XZY &= \angle XDT + \angle TDY + \angle XZY = \\ &= 180^\circ - \angle XAT + 180^\circ - \angle TCY + \angle XZY = \\ &= \angle ZAT + \angle ZCT + \angle XZY = 180^\circ, \end{align*}$$

which means that the quadrilateral $DXZY$ is also cyclic.