Blinkov A. D.

Point $M$ is the midpoint of the base $A C$ of an acute isosceles triangle $A B C$. Point $N$ is symmetric to $M$ with respect to $B C$. The line parallel to $A C$ and passing through point $N$ intersects side $A B$ at point $K$.

Find the angle $A K C$.
Let $L$ be the intersection point of $NK$ and $BC$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_60e93adedf73503f106cg-36.jpg?height=640&width=763&top_left_y=1298&top_left_x=653)

It is clear that $AM = MC = CN$ and $\angle MCB = \angle NCB$. But also $\angle CLN = \angle LCM$ (since $LN \parallel AC$), which means that triangle $CNL$ is isosceles, and

$LN = CN = AM$. Therefore, segments $AM$ and $LN$ are parallel and equal, so $ALNM$ is a parallelogram, and $AL \parallel MN \perp LC$. Finally, from the symmetry of the triangle with respect to the altitude $BM$, we get $\angle AKC = \angle ALC = 90^{\circ}$.

## Answer

$90^{\circ}$.