, 2022

Firstly observe that $OAXB$ is cyclic, with diameter $OX$, and $Y$ also lies on this circle since $OY \perp XC$. Hence:
$$\angle AZC = \angle XAB = \angle ABX = \angle AYX$$
and so $CYAZ$ is cyclic.

Let $M$ be the intersection of $YZ$ and $AC$ and let $CY$ intersect $\omega$ again at $W$. Using the new cyclic relation we get $\angle CYZ = \angle CAZ$ and then using that $ZA$ is tangent to $\omega$ we get $\angle CAZ = \angle CWA$, so $\angle CYM = \angle CWA$. Therefore the triangles $CWA$ and $CYM$ are similar. But $CW$ is a chord of $\omega$, and $Y$ is the foot of the perpendicular from $O$, hence $Y$ is the midpoint of $CW$. It follows from the similarity relation that $M$ is the midpoint of $AC$, as required.

Let $M$ be the midpoint of $AC$. We have $\angle CAZ = \angle CBA$ and $\angle ZCA = \angle BAC$ so the triangles $CAZ$ and $ABC$ are similar. The line $CYX$ is the $C$-symmedian of triangle $ABC$, and $ZM$ is the corresponding median in triangle $CAZ$, hence by isogonality $\angle AZM = \angle ACY$. So
$$\angle ZMA = 180^\circ - \angle AZM - \angle MAZ = 180^\circ - \angle ACY - \angle CBA \quad (1)$$
Now observe $\angle OMC = \angle OYC = 90^\circ$, so $CMYO$ is cyclic. Thus:
$$\angle CYM = \angle COM = \frac{1}{2}\angle COA = \angle CBA.$$
This shows that
$$\angle YMC = 180^\circ - \angle MCY - \angle CYM = 180^\circ - \angle ACY - \angle CBA$$
Combining this with (1) we get that $\angle YMC = \angle ZMA$ and as $A, C, M$ are collinear, it follows that $Z, M, Y$ are collinear as required.

As in Solution 2 we have that $CX$ is the A-symmedian of triangle $ABC$ and that triangle $ABC$ is similar to triangle $CAZ$.
Let $f$ be the spiral similarity which maps $AC$ onto $AB$ and let $g$ be the reflection on the perpendicular bisector of $AB$. Note that $f$ is a rotation about $A$ by an angle of $\angle CAB$ (clockwise in our figure) followed by a homothety centered at $A$ by a factor of $AB/AC$. By the similarity of triangles $ABC$ and $CAZ$ we have that $g(f(Z)) = C$, so actually $f(Z)$ is the other point of intersection, say $C'$, of $CZ$ with $\omega$.
As in Solution 1 we have that $CYAZ$ is cyclic. Therefore, letting $W$ be the other point of intersection of $CY$ with $\omega$, we have $\angle WAB = \angle WCB = \angle CAY$. We also have $\angle ACY = \angle ABW$. It follows that $f(Y) = W$.
Let $W' = g(W)$. Then $W' \in \omega$ and since $CW$ is the A-symmedian, then $CW'$ passes through the midpoint $N$ of $AB$. Now $CW'$ and $C'W$ intersect on the perpendicular bisector of $AB$ and therefore they intersect on $N$. It follows that $N = AB \cap C'W = Af(C) \cap f(Z)f(Y)$ is the image of $M = AC \cap ZY$ under $f$. Since $N$ is the midpoint of $AB$, then $M$ is the midpoint of $AC$.