Let $M$ be the midpoint of the base $AC$ of an acute-angled isosceles triangle $ABC$. Let $N$ be the reflection of $M$ in $BC$. The line parallel to $AC$ and passing through $N$ meets $AB$ at point $K$. Determine the value of $\angle AKC$.

(A.Blinkov) 
1. Let $M$ be the midpoint of the base $AC$ of an acute-angled isosceles triangle $ABC$. Since $M$ is the midpoint, $AM = MC$.
2. Let $N$ be the reflection of $M$ in $BC$. This means that $BC$ is the perpendicular bisector of $MN$.
3. Since $\triangle ABC$ is isosceles with $AB = AC$, $\angle BMC = 90^\circ$ because $M$ is the midpoint of $AC$ and $BC$ is the altitude from $B$ to $AC$.
4. Let $\angle NMC = \alpha$. Since $N$ is the reflection of $M$ in $BC$, $\angle MNC = \alpha$.
5. Let $X = BC \cap MN$. Since $N$ is the reflection of $M$ in $BC$, $MN$ is perpendicular to $BC$. Therefore, $\angle MXC = 90^\circ$.
6. Since $\angle MXC = 90^\circ$, $\angle XCM = 90^\circ - \alpha$.
7. Since $\angle XCM = 90^\circ - \alpha$ and $M$ is the midpoint of $AC$, $\angle MBC = \alpha$.
8. By the properties of isosceles triangles, $\angle ABM = \alpha$.
9. Note that $\angle KBC = \angle KNC$ because $K$ lies on the line parallel to $AC$ passing through $N$.
10. Since $KBNC$ is cyclic, $\angle BNC = 90^\circ$ because $N$ is the reflection of $M$ in $BC$.
11. Therefore, $\angle BNK = 90^\circ - 2\alpha$.
12. By the properties of cyclic quadrilaterals, $\angle BCK = 90^\circ - 2\alpha$.
13. Therefore, $\angle BKC = 90^\circ$.
14. Since $\angle BKC = 90^\circ$, $\angle AKC = 90^\circ$.

The final answer is $\boxed{90^\circ}$