Let $L=NK\cap BC$. Since $LN\parallel MC$ and $LC$ bisects $MN$, $MLNC$ is a parallelogram, but since its diagonals are perpendicular, $MLNC$ is a rhombus. This means that $AM=MC=ML$ or points $A,L,C$ lie on a circle with diameter $AC$ meaning $AL\perp BC$. Due to symmetry with respect to $BM$, we have $CK\perp AB$ or $\angle AKC=90^{\circ}$

Attachments:

parmenides51 wrote: 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) Since $M$ is the midpoint of $AC$ and $\triangle ABC$ is isosceles $\Rightarrow \angle BMC=90^{\circ}$ Let $\angle NMC=\alpha$ Since $N$ is a reflection of $M$ in $BC$ $\Rightarrow \angle MNC=\alpha$ Let $X=BC \cap MN$ $\Rightarrow \angle MXC=90^{\circ}$ $\Rightarrow \angle XCM=90^{\circ}-\alpha$ $\Rightarrow \angle MBC=\alpha$ $\Rightarrow$ By isosceles: $\angle ABM=\alpha$ Note that $\angle KBC=\angle KNC$ $\Rightarrow KBNC$ is cyclic Since $N$ is the reflection of $M$ in $BC$ $\Rightarrow \angle BNC=90^{\circ}$ $\angle BNK=90^{\circ}-2 \alpha$ $\Rightarrow$ By cyclic $\angle BCK=90^{\circ}-2 \alpha$ $\Rightarrow \angle BKC=90^{\circ}$ $\Rightarrow \angle AKC=90^{ \circ }_\blacksquare$