Solution:

Apply this lemma to see that if we wish to send $X$ to $A$ and $Y$ to $C$ then $M$ is the centre of dilation ($M,X,Y,B$ are concyclic since $\angle MYB=\angle MXB=90^{\circ}$). Stop this dilation halfway, so that we send the midpoint of $XY$ to the midpoint of $AC$. In other words $N\rightarrow K$ as $Y\rightarrow C$, but then triangles $MNK$ and $MYC$ are similar hence $\angle MNK=\angle MYC=90^{\circ}$.

Here is the diagram using Asymptote. [asy][asy] import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen qqwwww = rgb(0,0.4,0.4); pen qqcccc = rgb(0,0.8,0.8); pen xdxdff = rgb(0.49,0.49,1); pen ccccff = rgb(0.8,0.8,1); pen qqttcc = rgb(0,0.2,0.8); draw(circle((6.45,-1.39),4.84),qqwwww+linewidth(1.6pt)); draw((10.2,1.68)--(1.72,-2.44),qqcccc+linewidth(1.6pt)); draw((1.72,-2.44)--(10,-4.68),qqcccc+linewidth(1.6pt)); draw((10.2,1.68)--(10,-4.68),qqcccc+linewidth(1.6pt)); draw((3.79,2.67)--(1.72,-2.44),ccccff+linewidth(1.2pt)); draw((3.79,2.67)--(10.2,1.68),ccccff+linewidth(1.2pt)); draw((3.79,2.67)--(10,-4.68),ccccff+linewidth(1.2pt)); draw((3.79,2.67)--(5.24,-0.73),ccccff+linewidth(1.2pt)); draw((3.79,2.67)--(2.54,-2.66),ccccff+linewidth(1.2pt)); draw((2.54,-2.66)--(5.24,-0.73),ccccff+linewidth(1.2pt)); draw((3.79,2.67)--(10.1,-1.5),ccccff+linewidth(1.2pt)); draw((3.79,2.67)--(3.89,-1.7),qqttcc+linewidth(1.2pt)); draw((3.89,-1.7)--(10.1,-1.5),qqttcc+linewidth(1.2pt)); dot((10.2,1.68),ds); label("$A$", (10.27,1.95),NE*lsf); dot((1.72,-2.44),ds); label("$B$", (1.33,-2.74),NE*lsf); dot((10,-4.68),ds); label("$C$", (10.03,-5.04),NE*lsf); dot((3.79,2.67),ds); label("$M$", (3.55,2.87),NE*lsf); dot((10.1,-1.5),ds); label("$K$", (10.29,-1.64),NE*lsf); dot((5.24,-0.73),ds); label("$X$", (5.22,-1.1),NE*lsf); dot((2.54,-2.66),ds); label("$Y$", (2.48,-3),NE*lsf); dot((3.89,-1.7),ds); label("$N$", (3.91,-1.99),NE*lsf); clip((-3.22,-9.67)--(-3.22,5.35)--(14.63,5.35)--(14.63,-9.67)--cycle); [/asy][/asy]

$\angle MXB=\angle MYB$ implies $MXYB$ is cyclic. So $\angle XMY=\angle XBY=\angle ABC=\angle AMC$. Add $\angle CMX$ to both sides, we get $\angle CMY=\angle AMX$. So $\triangle CMY\sim\triangle AMX$, and we get $\frac{MY}{MX}=\frac{MC}{MA}$, so $\triangle MXY\sim\triangle MAC$. Since $MK$ and $MN$ are medians of similar triangles $\triangle MAC$ and $\triangle MXY$, then we have $\angle XMN=\angle AMK$, which gives $\angle KMN=\angle AMN$. We also have $\frac{MK}{MN}=\frac{MA}{MX}$, therefore $\triangle MAX\sim\triangle MKN$. So $\angle MNK=90^{\circ}$.

Let $Z$ be the projection of M onto the extension of line $AC$. Then it is well known fact that points $X$, $Y$ and $Z$ are collinear and the corresponding line is called as simson line. Since $MXAZ$ is cyclic, $\angle MAZ$ = $\angle MXZ$ which implies $\angle MAC$ = $\angle MXY ---> (1)$ Since $MYCZ$ is cyclic, $\angle MYZ$ = $\angle MCZ ---> (2)$ From $(1)$ and $(2)$ above, $\triangle MYX$ and $\triangle MCA$ are similar. Since $N$ and $K$ are the mid points of the corresponding sides of these two similar triangles, we have that $\angle MNX$ = $\angle MKA$ or $\angle MNZ$ = $\angle MKZ ---> (3)$ From $(3)$, we have that $MNKZ$ is a cyclic quadrilateral with $\angle MZK = 90^{\circ}$. Thus, opposite angle $\angle MNK = 90^{\circ}$.

WakeUp wrote: Let $M$ be a point on the arc $AB$ of the circumcircle of the triangle $ABC$ which does not contain $C$. Suppose that the projections of $M$ onto the lines $AB$ and $BC$ lie on the sides themselves, not on their extensions. Denote these projections by $X$ and $Y$, respectively. Let $K$ and $N$ be the midpoints of $AC$ and $XY$, respectively. Prove that $\angle MNK=90^{\circ}$ .