Let $O$ be the center of circumcircle of triangle $XPZ$ and let the line $XY$ intersect this circle for a second time at point $Q$. Suppose that $AA'$, $CC'$ are diameters of the circumcircle of triangle $ABC$.

Note that points $C'$, $Q$ are the reflections of points $B$, $X$ with respect to the line passing through $O$ which is parallel to $BC$, hence $C'Q \parallel BC$ and $\angle C'QY = 90^\circ = \angle C'AY$. Therefore the quadrilateral $C'AYQ$ is cyclic.

Let $D$ be the second intersection of the $A$-altitude with the circumcircle of triangle $ABC$. Note that triangles $DBX$, $A'CZ$ are symmetrical with respect to the perpendicular bisector of the side $BC$, hence $\triangle DBX \cong \triangle A'CZ$. Similarly triangles $AC'Y$, $CA'T$ are symmetric with respect to perpendicular bisector of the side $AC$, therefore $\triangle AC'Y \cong \triangle CA'T$. Note that $D$ is the reflection of point $A$ with respect to the line passing through $O$ which is parallel to $BC$, hence $\triangle AC'Q \cong \triangle DBX$
$$\angle AQC' = \angle DXB = \angle A'ZC$$
$$\angle A'TC = \angle C'YA = \angle C'QA$$
Therefore $\angle A'ZC = \angle A'TC$ and the quadrilateral $A'ZTC$ is cyclic.

Proof. Let $AQ$ intersect the circumcircle of triangle $XZQ$ at $P'$.
$$\begin{aligned} \angle XZP' &= \angle XQP' = \angle YQA = 90^\circ - \angle AQC' \\ \angle AQC' &= \angle DXB = \angle A'ZC \end{aligned}$$
Note that $\angle A'CT = 90^\circ$, therefore $\angle TZC = 90^\circ - \angle CZA' = 90^\circ - \angle AQC'$ hence points $Z, P', T$ are collinear and $P \equiv P'$ $\Box$

Let $AQ$ intersect the circumcircle of triangle $ABC$ at $R$ for the second time. $M, N$ are the midpoints segments $XP, PQ$ respectively and $H$ is the orthocenter of triangle $ABC$.

Point $O$ is the center of circumcircle of triangle $ABC$, therefore $N$ is the midpoint of chord $AR$. If $k$ is the distance of $O$ to the side $BC$, it is well known that $AH = 2k$. Note that $2MN = XQ = 2k$ as well, hence by Thales's theorem $M$ is the midpoint of segment $RH$. The homothety centered at $H$ with ratio $1/2$ takes the circumcircle of triangle $ABC$ to its nine-point circle. This homothety takes $R$ to $M$, thus $M$ lies on the nine-point circle of triangle $ABC$.

Lemma. 1 Let points $Z, Q$ lie on the circle $\omega$ with center $O$ and $l$ be an arbitrary line. $M$ is the projection of $O$ onto $l$. Points $Y, T$ lie on the line $l$ such that $MY = MT$. $YQ, TZ$ intersect $\omega$ for the second time at points $X, P$. If $PQ, ZX$ intersect $l$ at points $A, C$ then $AM = CM$

Proof. Let $P'$ be a point on $\omega$ such that $PP' \parallel l$. Note that quadrilateral $TPP'Y$ is a cyclic trapezoid $(P', Y$ are the reflections of $P, T$ with respect to the line $OM)$.

By law of sines in triangles $ATP, TZC, XYC, YQA$:
$$\begin{aligned} \frac{TP}{TA} &= \frac{\sin \angle TAP}{\sin \angle TPA'}, \quad \frac{TZ}{TC} = \frac{\sin \angle TCZ}{\sin \angle TZC} \\ \frac{CY}{XY} &= \frac{\sin \angle CXY}{\sin \angle YCX'}, \quad \frac{AY}{YQ} = \frac{\sin \angle AQY}{\sin \angle YAQ} \end{aligned}$$
Quadrilateral $ZQXP$ is cyclic, therefore $\angle APT = \angle YXC, \angle AQY = \angle TZC$. Hence
$$\frac{TP \cdot TZ}{TA \cdot TC} = \frac{YX \cdot YQ}{YC \cdot YA}$$
Note that the ratios of power of points $T, Y$ with respect to circle $\omega$ and circumcircle of triangle $APC$ are equal, hence circles $\omega$ and the circumcircles of triangles $YPT, APC$ are coaxial which means quadrilateral $APP'C$ is cyclic. This quadrilateral is a cyclic trapezoid $(AP, P'C$ are symmetric with respect to line $OM)$ hence $MA = MC$ $\square$

Claim 4. $A$ lies on the line $PQ$

Proof. Letting $M$ be the midpoint of side $AC$, by Lemma 1. for the circle $\omega$, line $AC$ and points $Z$, $Q$ the line $PQ$ passes through the point $A$. $\Box$

Let $AQ$ intersect the circumcircle of triangle $ABC$ at $R$ for the second time. $R'$, $N$ are the midpoints segments $XP$, $PQ$ respectively and $H$ is the orthocenter of triangle $ABC$.

Point $O$ is the center of circumcircle of triangle $ABC$, therefore $N$ is the midpoint of chord $AR$. If $k$ is the distance of $O$ to the side $BC$, it is well known that $AH = 2k$.
Note that $2R'N = XQ = 2k$ as well, hence by Thales's theorem $R'$ is the midpoint of segment $RH$. The homothety centered at $H$ with ratio $1/2$ takes the circumcircle of triangle $ABC$ to its nine-point circle. This homothety takes $R$ to $R'$, thus $R'$ lies on the nine-point circle of triangle $ABC$.

We call the foot of the altitude from $A$ as $D$. Consider points $E$ and $K$ such that $AEXY$ and $AKZB$ are parallelograms. Now, since $EX = AY = CT$, it follows that $ETCX$ is also a parallelogram, which implies $ET = CX = BZ$. Thus, $ETZB$ is also a parallelogram. The extension of $BE$ intersects the circumcircle of $ABC$ at $F$.

First, note that since $AK = ET$ and $AK \parallel ET$, $AKTE$ is a rectangle. Since $O$ lies on the perpendicular bisectors of segments $FB$ and $PZ$, $PZBF$ is an isosceles trapezoid, and therefore $FPTE$ is an isosceles trapezoid. Moreover, since $\angle AFE = \angle ACB = \angle ATE$, $AFTE$ is cyclic. Given all the mentioned results, the points $A, F, K, P, T, E$ lie on same circle.

Now, suppose $M$ and $N$ are the midpoints of $BC$ and $PX$, respectively, and $X'$ is the reflection of $X$ with respect to $AD$. Since $\angle KET = \angle KAT = \angle ACB = \angle EXD = \angle EX'D$, the points $K, E, X'$ are collinear. Now, since $\angle KPT = 180^\circ - \angle KET = 180^\circ - \angle KX'Z$, $KPZX'$ is cyclic, and thus $\angle X'PZ = \angle X'KZ = \angle KZC - \angle KX'Z = \angle B - \angle C$. By applying a homothety with ratio $\frac{1}{2}$ centered at $X$, we conclude that $\angle MND = \angle B - \angle C$, which implies that $N$ lies on the nine-point circle of $ABC$.